linear-algebra

[winter harbor]

https://math.stackexchange.com/questions/1480125/moore-penrose-inverse-as-least-squares-solution

Yeah, he makes a Saul gives a very nice geometric approach to the least squares method. It might be helpful to visualize the Moore-Penrose inverse too.

[grave oasis]

Cool, thank you. Does an intro to linear algebra class typically cover psuedoinverses? I don’t think I saw it in the syllabi I was looking at for my city’s university. Is it more of an advanced topic?

[winter harbor]

I suppose it depends on the emphasis of the book

For instance

This isn't a particularly famous reference outside of Brazil

But Elon's book covers pseudo-inverses and the least squares method

And it is a pretty standard textbook reference here for Linear Algebra (It even won a damn book award, idk how he managed to do it with a Linear Algebra Book)

I suppose books/courses with emphasis on analysis, engineering and computer science might cover the moore penrose inverse.

It makes sense, since this is one of the most basic linear regression methods.

[grave oasis]

I see, yeah. And it makes a lot more sense now after watching the video.

[stable wasp]

Why don’t I get the same answers?

[limber sierra]

hows this

become this?

if youre multiplying R_1 by -1, surely you should apply it to the entire row

[stable wasp]

I switched the rows so the determinant should change signs

[limber sierra]

thats not what im talking about

howd the (-1 0 1) row become (1 0 1)?

[stable wasp]

I also factored out -1 from row 1

[limber sierra]

then it should be (1 0 -1)

[stable wasp]

Damn that’s true💀💀💀

Thank You😂

I still didn’t get the right answer

Oooh shit
Found my mistake. Thank you!!!

[vocal dock]

very ingenious

thank you so much :))

[hot swallow]

is a linear transformation invertible if its both onto and one to one?

[winter harbor]

yeah a linear transformation between vector spaces of same dimension is invertible iff it is both onto and one to one.

We can actually ask something weaker. A linear transformation between finite dimensional vector spaces of same dimension is invertible iff it is injective iff it is surjective (As a consequence of Rank-Nullity)

[hot swallow]

so if lets say the kernel is {f in F: f(-4) = 0 } that means that linear transformation is not invertible right casue its only surjective?

[winter harbor]

If the kernel is non trivial, then yes, the linear map is not injective, thus not invertible.

[hot swallow]

ah thank you

[vocal dock]

is there a general way of determining if a matrix is positive semidefinite ?

[winter harbor]

Yeah, there are some criterions.

For instance, a symmetric matrix is positive semi definite iff all of its eigenvalues are non negative, i.e bigger than or equal to 0.

Some texts take this as a definition even

For some classes of matrices, say hermitian matrices, there's the sylvester's criterion to test for POSITIVE definiteness.

Idk about something similar about positive semide-finiteness tho.

[vocal dock]

i see

for an example like this

[gleaming knot]

https://math.stackexchange.com/questions/510366/criterion-for-positive-semidefinite-matrices

[vocal dock]

would you advise i go about finding the eigenvalues of A to prove that it is semidefinite?

[winter harbor]

Ah, so it seems that a slightly different version of the Sylvester's Criterion that I know works for semi definite matrices too

[winter harbor]

No, in this case you can just use the definition.

[gleaming knot]

yep!

[winter harbor]

Or try a proof by a contradiction

Say ''suppose A has a negative eigenvalue, then...''

the nice thing of trying a proof like this

Is that via this equation, we can obtain information about the squares of the eigenvalue of A

via A^2 (the eigenvalues of A^2 are exactly the squares of the eigenvalues of A)

notice that A^tA is positive semidefinite

this may give you the information you need to proceed

[verbal ivy]

what book or youtube video helped helped you guys with linear algebra?

[winter harbor]

Is A necessarily positive semidefinite tho? I am not sure.

[vocal dock]

haha its not as obvious to me but ill definitely look over the hints you guys gave me

[vocal dock]

why not?

[winter harbor]

It's just that I am not sure at the moment if those conditions imply A is positive semidefinite.

Positive semidefiniteness is stronger than being symmetric, which is what we want to prove.

[vocal dock]

what do you mean by stronger?

[winter harbor]

We usually define a symmetric/hermitian matrix to be semi-positive /positive definite and don't talk about the non-symmetric case, (talking about non-hermitian positive definite matrices doesn't even make much sense, for instance)

so we include in the definition of semi-positive definitess and positive-definiteness the requirement for A to be symmetric (real case) and hermitian (complex case)

that's what I mean by being a stronger requirement

[prime fossil]

for a question like this

can i just find the solution set of the homogenous equation of A, and then translate it by [1, 2, 3]?

[violet pecan]

can the zero vector be a basis for a nullspace of a matrix? i got the zero vector as the basis and i don't think that's correct since isn't the zero vector linearly dependent?

[winter harbor]

The zero vector can't be a basis, yes.

Think about the {0} vector space

Intuitively, it should have dimension 0, right?

But the dimension of a vector space is defined as the cardinality of a linearly independent generating set

if the zero vector was a basis for {0}

we would have that {0} would have dimension 1

so we want a linearly independent set of cardinality 0 to span {0}

and this is precisely the empty set

the basis for the {0} vector space is the empty set (even tho this may sound unintuitively at first)

That's one way to get an intuition on why the basis should be the empty set instead of the 0 vector tho.

[violet pecan]

ooh okay thank you, i might have to digest this lol

[winter harbor]

https://math.stackexchange.com/questions/1909645/empty-set-is-the-only-basis-of-the-zero-vector-space-0

[viscid lily]

How do I do least square between two polynomials?

I have "solved" it by using the points given, but I dont think that is what is meant

[lavish jewel]

i think this is what they meant, @viscid lily

[quaint anchor]

Hey is this the right place to ask questions about DSP filtering?

[lavish jewel]

either here or #numerical-analysis , sure

[quaint anchor]

Cool. I'm doing some work with LinkwitzRiley filters. Audio Crossover filters.

When you split an audio signal into low and high bands using LR filters, If you sum those bands back together, it is equivalent to running your signal through an Allpass filter set to the same cutoff frequency.

LP -> ----\
HP ->     /----

When the AP filter output is multiplied by -1 and summed with the LP+HP output, silence is produced.

Here's where I'm running into trouble:
I've got this filter configuration for the 3 bands:

fc0    fc1        fc0  fc1
LP1 -> AP2 ->  ----\
HP1 -> LP2 ->      /----\
    -> HP2 ->           /----

and my inverted allpass filters are:

fc0          fc1
invAP1  -> invAP2 -> multiply by -1.f 

I'm expecting to be able to sum the output of invAP1 -> invAP2 with the output of the 3 bands and end up with 0

invAP1 is an Allpass filter.

if I had to write it as an equation, I'd start here:

LP + HP = AP

and then:

(LP+HP) + AP*-1 = 0

which leads to:

(LP1+AP2) + (HP1+LP2) + (HP2) = (invAP1 + invAP2)

[lavish jewel]

probably a better fit for the other channel but oh well 😛 so what's the question

[quaint anchor]

It seems that my assumption about summing the filtering is wrong.

This works as expected:

(LP+HP) + AP*-1 = 0

But when I split into 3 bands, and then use 2 inverted allpass filters set to the same cutoff frequencies and invert the output, I don't get silence in the output.

[lavish jewel]

are you downsampling at each filtering stage?

[quaint anchor]

no.

[lavish jewel]

what are you calling inv AP

[quaint anchor]

I'm doing this:

    using Filter = juce::dsp::LinkwitzRileyFilter<float>;
    
    //      Fc0     Fc1
    Filter  LP1,    AP2, //band 0
            HP1,    LP2, //band 1
                    HP2; //band 2

    //      Fc0     Fc1
    Filter invAP1, invAP2; 

[lavish jewel]

in frequency domain these look like you're just multiplying by appropriately shifted rectangle windows (unless you're doing something funky to the phase)

[quaint anchor]

The filtering is accomplished like this, and maybe this is the cause of the problem

1. I have an array of 3 filter buffers:

std::array<juce::AudioBuffer<float>, 3> filterBuffers;

2. I copy the input buffer into these 3 buffers

    for( auto& fb : filterBuffers )
    {
        fb = buffer;
    }

3. I process LP1, then AP1 with filterBuffers[0]
   I process HP1, then LP2 with filterBuffers[1]
   I process HP2 with filterBuffers[2]

    auto fb0Block = juce::dsp::AudioBlock<float>(filterBuffers[0]);
    auto fb1Block = juce::dsp::AudioBlock<float>(filterBuffers[1]);
    auto fb2Block = juce::dsp::AudioBlock<float>(filterBuffers[2]);
    
    auto fb0Ctx = juce::dsp::ProcessContextReplacing<float>(fb0Block);
    auto fb1Ctx = juce::dsp::ProcessContextReplacing<float>(fb1Block);
    auto fb2Ctx = juce::dsp::ProcessContextReplacing<float>(fb2Block);
    
    LP1.process(fb0Ctx);
    AP2.process(fb0Ctx);
    
    HP1.process(fb1Ctx);
    LP2.process(fb1Ctx);
    
    HP2.process(fb2Ctx);

I then clear the input buffer, and then add each of the 3 filterBuffers to the input.

For the inverted filters, they get their own buffer:

juce::AudioBuffer<float> invAPBuffer;

it gets it's own copy of the input buffer before it is cleared:

invAPBuffer = buffer;

then the AP filters that'll be multiplied by -1 are processed:

    auto invAPBlock = juce::dsp::AudioBlock<float>(invAPBuffer);
    auto invAPCtx = juce::dsp::ProcessContextReplacing<float>(invAPBlock);
    invAP1.process(invAPCtx);
    invAP2.process(invAPCtx);

Then, that invAPBuffer is flipped by multiplying each channels' audio samples by -1:

        for( auto ch = 0; ch < numChannels; ++ch )
        {
            juce::FloatVectorOperations::multiply(invAPBuffer.getWritePointer(ch), -1.f, numSamples);
        }

then the invAPBuffer is added to the cleared input buffer, after adding all of the buffers in filterBuffers back to the input buffer.

So, no Downsampling that I know of, @lavish jewel

[lavish jewel]

mhm

[quaint anchor]

Sorry to go so deep into the implementation.

My thought was that if LP+HP - AP = 0, then maybe LP1+AP2+HP1+LP2+HP2 - AP_1 - AP_2 = 0

where LP1+AP2 is the first band of audio
HP1+LP2 is the 2nd band of audio
and HP2 is the 3rd band of audio, and they sum to a flag magnitude response.

[lavish jewel]

are you sure though

because you're cascading the filters

it sounds more like hp1 = lp2 + hp2

[lavish jewel]

what i'd try first is to plot the spectrum of each filter first

yep

after cascading, you're doing something like

hp1 = hp1 lp2 + hp1 hp2

[lavish jewel]

lp2 + hp2 is all of the spectrum again, too

indeed

[quaint anchor]

LP2 + HP2 is not all of the spectrum

[lavish jewel]

so lp1 + hp1lp2 + hp1hp2 is the full specturm

oh yes it is

it isn't only AFTER you cascade it with hp1

which is exactly what you drew

[quaint anchor]

Right right. But it is applied AFTER HP1 is filtered.

[lavish jewel]

yep, so the inverse filter needs to consider tht

[quaint anchor]

    LP1.process(fb0Ctx);
    AP2.process(fb0Ctx);
    
    HP1.process(fb1Ctx);
    LP2.process(fb1Ctx);
    
    HP2.process(fb2Ctx);

[lavish jewel]

it makes no sense to think of just lp2 + hp2 as you did, because that's just the full spectrum

[quaint anchor]

What do you mean "As I did"?

I'm splitting into 3 bands, and then summing those bands back together.
Then I'm running a separate copy of the input through 2 other Allpass filters set to the same cutoff frequencies that I used to split the original 3 bands, inverting that output, and summing it with the 3 bands.

[lavish jewel]

here

you'll have to check what exactly ap2 is inverting

presumably h1(l2 + h2)

not just l2 + h2

and not h1 + l2 + h2

[quaint anchor]

ok, maybe that should have said

(LP1 -> AP2) + (HP1 -> LP2) + (HP1 -> HP2) - (AP_1 -> AP_2) = 0

this ^^ shows the order of processing more clearly

[lavish jewel]

these are in freq domain, so it's better if you simply use elementwise products

the filters are multiplied in frequency domain

[quaint anchor]

"elementwise products"?

[lavish jewel]

hadamard products

[quaint anchor]

lol what?

sorry

never heard of that

[lavish jewel]

you're implementing these as FIR filters presumably

[quaint anchor]

I'm using the juce::dsp::LinkwitzRiley implementation. I have no idea what it uses internally.

[lavish jewel]

ok, this is definitely not a good fit for this channel then, since you're not treating the filters as linear transformations

better use the applied comp channel

[quaint anchor]

is this FIR?

template <typename SampleType>
void LinkwitzRileyFilter<SampleType>::update()
{
    g  = (SampleType) std::tan (MathConstants<double>::pi * cutoffFrequency / sampleRate);
    R2 = (SampleType) std::sqrt (2.0);
    h  = (SampleType) (1.0 / (1.0 + R2 * g + g * g));
}

as far as I understand it, LR filters are just Butterworth filters squared.

literally a pair of 2nd order Butterworth filters in series for both HP and LP sides

so you get 24db/oct cutoff

@lavish jewel it turns out I needed to copy the output of HP1 into the buffer that HP2 will use before LP2 is applied. Once I did that, the output produced silence.

[lavish jewel]

sure, because the filters are concatenated 😛

you only apply hp2 and lp2 AFTER hp1, otherwise it's again an all pass

that's what i had said above

[north forge]

Does anyone know of a good video series to learn linear algebra?

[lavish jewel]

mit OCW gilbert strang's courses

[half notch]

Hi can someone help me with this?

the last one

I found the basis v1,v2 of the plane and to find all the perpendicular vectors to that, I tried doing (c1v1+c2v2)(x y z)'=0 and solve for x y z

which didn't work

because I had the constants c1, c2 in the result and it didnt make sense

[frosty eagle]

Does anyone know how to solve this

[narrow moth]

the matrix is real so the eigenvalues come in complex conjugate pairs

[frosty eagle]

Thanks and I got the answer using properties of eigenvalues using the the 6 complex and complex Conjugate pair😁

[frosty eagle]

Is the working correct

[narrow moth]

yep

[wintry steppe]

$Suppose U_1, U_2, ... , U_m are subspaces of a vector space V over the field F. The U_1 + ... + U_m is the smallest subspace of V containing U_1,...,U_m.$

I am trying to understand what this theorem is saying. What does smallest subspace of V containing U_1, ..., U_m even mean? I know sum of subspaces is a set.

Does containing mean every element of each subspace U_1, ..., U_m is in the set of sums?

[winter harbor]

We mean smaller in the following sense, suppose that $\mathcal{S} \subset V$ is a subspace of $\mathcal{S}$ for which $U_{1} \cup \cdots \cup U_{m}\subset \mathcal{S}$, then $U_{1} + \cdots + U_{m} \subset \mathcal{S}$.

[stoic python]

MisterSystem

[winter harbor]

I.e, if you have another subspace S of V that contains the union of these subspaces U_1, ..., U_m

then U_1 + ... + U_m is a subset of such S

Notice the following

[wintry steppe]

how does this show S is the smallest subspace of V? sorry I am still trying to understand what your saying. I understand what you have written but I still putting it together.

[winter harbor]

Let $G \subset V$ be any subset of $V$ and denote
$$
S_{G} := {W \subset V , \vert , G \subset W , \text{and W is a subspace of V} }
$$
Then $S_{G}$ is a partially ordered set by inclusion, and has a minimal element, which is called the span of $G$ (I.e, the span of a subset $G$ is the smallest subspace of $V$ that contains $G$). So saying that $U_{1} + \cdots + U_{m}$ is the smallest subspace that contains $U_{1} \cup \cdots \cup U_{m}$ is the same as saying:
$$
U_{1} + \cdots + U_{m} = \text{span}(U_{1} \cup \cdots \cup U_{m})
$$

[winter harbor]

No, S is not the smallest subspace that contains the union of all the U_i

What I am saying is that

If we hve any other S

that contains the union of such all such U_i

then it must be the case that U_1 + ... + U_m is a subset of S

I.e, if $G = U_{1} \cUp \cdots \cUp U_{m}$, then $U_{1} + \cdots + U_{m}$ is the minimal element of $S_{G}$ in the sense I have described above.

[vocal dock]

maybe you can prove this using contradiction?

[stoic python]

MisterSystem

[winter harbor]

I.e, if $G = U_{1} \cup \cdots \cup U_{m}$, then $U_{1} + \cdots + U_{m}$ is the minimal element of $S_{G}$ in the sense I have described above.

[stoic python]

MisterSystem

[winter harbor]

So by smallest, we mean this.

[vocal dock]

ive heard that the trace of an idempotent matrix is equal to its rank

but i haven't been able to come across a nice proof of it

does anyone know of any that I can take a peak at?

[wintry steppe]

trace is the sum of the eigenvalues. what are the eigenvalues of an idempotent matrix?

[vocal dock]

0 and 1?

[wintry steppe]

yeah

[vocal dock]

is the first part that trivial

maybe im just really dumb 😔

[winter harbor]

Notice that every idempotent matrix is diagonalizable

because its minimal polynomial is either x, x-1 or x(x-1)

all of which split into distinct linear factors

so it is similar to a diagonal matrix

with 1s and 0s

what is the rank of such a diagonal matrix?

[wintry steppe]

hot

[winter harbor]

notice that the rank of a matrix iand its rank are preserved under similarity/conjugacy classes

[vocal dock]

ahhhh i see

that makes things very clear

thank you so much :))

[winter harbor]

[wintry steppe]

[wintry steppe]

trace = sum of eigenvalues comes from writing out the char polynomial by definition and by comparing with a product of (t - eigenvalues)

[blissful vault]

I'm trying to find the determinant of each elementary matrix. This is the row swap matrix. Did I state it and show it correctly? Any improvements?

[teal grotto]

i really only think you need the last sentence

maybe flesh out the formula a little bit to make it more clear what you’re doing

det(S_ij) = det(e1,…,ej,…,ei,…,en)
= -det(e1,…,ei,…,ej,…,en)
= -det(I_n)
= -1

if i < j

but the pretext you have now doesn’t hurt. it’s just my style to try and make things as concise as possible

[blissful vault]

When writing a proof for this, would you show the other case?

[teal grotto]

when i > j you mean?

[blissful vault]

Oh, I meant i=1 and j=n.

[teal grotto]

ehh, it’s kinda implied that this covers it. any swap of two arguments of the determinant changes sign by anti symmetry. you could write it like this instead:
det(…, ei,…, ej,…)

since we really only care about i and jth entries

[blissful vault]

I see. so you chose the first way because most people would understand?

Sometimes I'm confused how to write stuff like this out haha

So I'm trying to figure out how others choose.

[teal grotto]

uhh, it’s just how i wrote it the first time. you could always add clarification at the end if you feel the need to show the case where i = 1 and j = n

[teal grotto]

what u wrote is fine. i could understand it clearly

[blissful vault]

ah, good. Thanks for looking it over and the comments.

[pure oyster]

Can I ask a question about multi-linear algebra here?

[winter harbor]

Yeah

Sure

[vocal dock]

does anyone know any good resources for learning matrix calculus?

i took linear algebra some years ago and would appreciate any resource for beginners

[slow scroll]

define matrix calculus

[noble swan]

I'm not sure how I'm supposed to go about doing this. I was given the hint to use Theorem 1

Am I supposed to go about showing c_1 & c_2 span y(t)?

[ionic laurel]

^^ i was about to ask that problem as well

[noble swan]

LMFAO

[ionic laurel]

how would you check for the 0 vector

would you substitute t for 0? or multiply a scalar 0

[noble swan]

Those rules are for showing it's a subspace, no?

[ionic laurel]

yes

a subspace is a vector space

if you can prove it is a subspace

it is automatically a vector space

[noble swan]

Ah

Well, in that case, you substitute 0 for the c's, yeah?

[ionic laurel]

that is what i would imagine

if c1 and c2 are arbitrary numbers then

if those are 0

then 0 + 0 is 0

i can prove the other conditions

well that solves my question

[noble swan]

Anything times 0 is 0, too

Ey, good teamwork LOL

[ionic laurel]

👍

i suppose when they say use theorem 1 which is the span

c1 {coswt} and c2 {sinwt} so that span{coswt,sinwt} is a vector space

which means that it is a vector space

[noble swan]

That seems a lot cleaner

[ionic laurel]

personally i think

checking the 3 conditions is easier

than using theorem 1

[noble swan]

It works out really nicely if you can

Since you can just pull out the variables to get your span

[ionic laurel]

yeah that is true

[vocal dock]

is there a way of determining if A is positive semi definite?

i started out with the eigenvalues of A

but i got no where with that

as i couldn't find a relationship between the eigenvalues of A and A^T A

[torn stag]

@vocal dock What's your definition of positive semi definite?

[vocal dock]

x^T A x > 0 for all x

[torn stag]

@vocal dock Does it also require that $A^T = A$?

[stoic python]

IlIIllIIIlllIIIIllll

[vocal dock]

nope

i think there are conditions on symmetric matrices being positive semidefinite

idk for sure tho

[torn stag]

Ok, well we can reduce to the symmetric case easily

Let $B = \frac{A + A^T}{2}$, the symmetric part of $A$.

[stoic python]

IlIIllIIIlllIIIIllll

[vocal dock]

oh i actually understand that proof

so I can tell that A is symmetric

[torn stag]

Then $A - B$ is antisymmetric, so $x^T A x = x^T B x$.

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

Since $B$ is symmetric, there is an orthonormal basis of $\mathbb{R}^n$ consisting of eigenvectors of $B$

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

and you can deduce that $x^T B x \geq 0$ for all $x \neq 0$ if and only if all eigenvalues of $B$ are $\geq 0$.

[stoic python]

IlIIllIIIlllIIIIllll

[vocal dock]

mhm

i have a hard time understanding why the eigenvalues are all greater than 0

well im not even sure the eigenvalues ARE greater than 0

[tight grail]

for this question

For 2 vectors X, Y, and 1 scalar a,

if X = a*Y,

then (x1-ay1) + (x2 - ay2) + ... (xn - ayn) = 0,

Meaning X and Y cannot be linearly independant, as it breaks the linear independency property of vectors

Therfore, two vectors can only be linearly independant if one is not a scalar multiple of the other.

Is this a good answer?

[torn stag]

@vocal dock $\mathbb{R}^n$ has an orthonormal basis ${v_1, \dots, v_n}$ of eigenvectors of $B$. So given $x \in \mathbb{R}^n$, $x = (x, v_1)v_1 + \dots + (x, v_n)v_n$, where $(\cdot, \cdot)$ denotes the dot product.

Thus $Bx = \lambda_1(x, v_1)v_1 + \dots + \lambda_n(x, v_n)v_n$.

[stoic python]

IlIIllIIIlllIIIIllll

IlIIllIIIlllIIIIllll

[torn stag]

so $(Bx, x) = \lambda_1(x, v_1)^2 + \dots + \lambda_n(x, v_n)^2$

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

@tight grail You can't (and don't need to) use coordinates $x_1, y_1$, etc. here.

[stoic python]

IlIIllIIIlllIIIIllll

[novel elk]

tag me when this question is finished

[tight grail]

what do you think I should do instead cause like

that was a similar proof I used for proving two vectors are not linearly independent if one is a scalar multiple of the other one

[torn stag]

@tight grail also you didn't explain why that implies $x, y$ are linearly dependent

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

@tight grail and you still have to prove the other direction too

@tight grail that if $x, y$ are linearly dependent, then one is a scalar multiple of the other

[stoic python]

IlIIllIIIlllIIIIllll

[tight grail]

thanks for the help

I'll have a reattempt at the question (and read theory)

@novel elk ayo I finished, if u wanna pop a question in

[wintry steppe]

So... can someone suggest a good book to learn linear algebra? I barely managed to pass last year (online class/covid was really annoying) and I know barely anything so now I need to study ;-;

[tight grail]

@wintry steppe I have "Introduction to Linear Algebra" by Gilbert Strang

which I issued out like 7 months ago from the library but never gave back cause of covid

I felt like it was a pretty good book, teaching everything from the start

in the perspective of someone who might know how to

multiply matrices together, but pretty much nothing else

[wintry steppe]

thanks so much I'll look for that book

[tight grail]

ye, other than that tho I'm kind of struggling myself lmao

so see if someone else has something else maybe too

[slow scroll]

if you want something more theoretical, hoffman kunze, friedberg insel spence, or axler

[torn stag]

The book here is good: https://mtaylor.web.unc.edu/notes/linear-algebra-notes/

[wintry steppe]

thanks too

[wintry steppe]

thanks

[tight grail]

thanks for the suggestions

so in terms of the definition of linearly independent vectors, I can understand that,
for some scalars, a, b, the two vectors X and Y are considered linearly independent of one another if
aX + bY = 0, as long as at least one of a or b is not 0

but then in the proofs, they just say that "X = (-b/a) * Y", such proving that a != 0

or

Y = (-a/b) * X

proving that b != 0

but doesn't the fact that Y = (-a/b) * X

mean that Y is a scalar product of the vector X

which goes against the original question of "prove two vectors are linearly independent if and only if one is not a scalar multiple of the other"

question and answer circled

they're using different symbols tho

[nocturne jewel]

Yeah the solution should be v,w are dependent

[sour cloud]

can anyone help me with this? I have an equation AB = 0 and i need to find B where B is not equal to 0. so i have matrix A : ```
[
-4 -1 -2
0 -4 -8
2 2 2
]

and the null space i found it to be [1, -2, 1] but im confused to what to do with the null space ?

[tight grail]

@nocturne jewel ok I actually fully understand now lmao

thanks for the help

I read "independent" as "dependent" as well so just got even more confused until I read ur comment and re-read notes

[nocturne jewel]

independent: only trivial span to 0
dependent: Nontrivial span to 0 exists

[sour cloud]

how do i make every column of B in null(A) ?

[wintry steppe]

AB = 0
A^(-1) * A * B = A^(-1) * 0
B = A^(-1) * 0
so what about you try to find

the inverse of A?

[sour cloud]

i need to find B tho

[wintry steppe]

if you find the inverse of A you can equal B

[sour cloud]

wait how ?

your multiplying

[wintry steppe]

AB = 0
A^(-1) * A * B = A^(-1) * 0
B = A^(-1) * 0

A^(-1) *A is the identity matrix

since B = A^(-1) * 0 ... B can A^(-1) or 0

since he's not 0 he's the inverse of A

I don't know if I explained right my english is very limited ;-;

[nocturne jewel]

(matrix)*(0matrix) is the 0 matrix

so you've said B=0

[sour cloud]

wait you cant get inverse matrix tho

b is not equal to 0

AB = 0 and B is not equal to 0

i have to make every column of B in null(A) but i dont know how to do that

i found the null space to be [1, -2, 1]

[nocturne jewel]

nullspace isnt a lone vector

it's a subspace

[sour cloud]

but i got the null space to be that

i need to find 3x3 matrix B

i need to find all matrices of B such that AB = 0

the nullspace is span( [1, -2, 1] )

[wintry steppe]

matrix A is 3x3 and B is also 3x3 so how 0 = 3x1 ?

[sour cloud]

its not equal to 0 tho

thats just the null space of A

idk how to use that to figure out how to find B

i couldnt find any resources online all of them just explained how to find the null space but not the matrix

[wintry steppe]

I don't know if is right but I found B if 0 = [ 0 0 0 ... ]

[sour cloud]

how ?

[wintry steppe]

if 0 is the null matrix

try to isolate B

on the equation AB = 0

[sour cloud]

nope

[nocturne jewel]

You suggested that already, and it's still flawed

[sour cloud]

i did that too got it wrong

yea

[wintry steppe]

okay ;-;

[sour cloud]

but the null space is [1, -2, 1] what do i do with this lol

[wintry steppe]

you can always do the system no?

but I don't see how B is going to be 3x3

[sour cloud]

i tried that but got it wrong as well

its a 3x3 matrix

[nocturne jewel]

Are you asked for a solution or all solutions B?

[wintry steppe]

if A is a matrix 3x3 and 0 is a matrix 3x1 and you have AB = 0 .... B can't be 3x3

[nocturne jewel]

???

where are you getting 0 is a 3x1?

[sour cloud]

0 isnt a 3x1 lol

[wintry steppe]

null space [1, -2, 1]?

[sour cloud]

i think ur confused with null space

[wintry steppe]

yeah I'm confused

[nocturne jewel]

null(A)=span([1,-2,1]^T)

[wintry steppe]

;-;

[sour cloud]

yea

exactly @nocturne jewel

[wintry steppe]

I'll use google to translate just a second

[nocturne jewel]

but yeah... .

answer this

[sour cloud]

but do you know what i do with null space ?

[nocturne jewel]

yes

found a math stack post about it

[sour cloud]

just matrix B

3x3 matrix thats it

[nocturne jewel]

yeah but all B or a B

[sour cloud]

all B

[nocturne jewel]

So let $v\in\text{span}([1,-2,1]^T)$ then define $B:=vv^T$

[stoic python]

Mosh

[nocturne jewel]

with of course v being a column vector

https://math.stackexchange.com/questions/243160/finding-matrix-b-is-not-zero-matrix-where-ab-0

[sour cloud]

its says that B is an inverse of A?

but an inverse doesnt exist for my matrix

yea this is really difficult im not understanding at all

[nocturne jewel]

If B=vv^T and v!=0, then AB=Avv^T=0v^T=0

For any nonzero v in null(A)

[torn stag]

null space of A is spanned by one vector

just make all columns of B that vector

[sour cloud]

what vector ?

[torn stag]

Take any nonzero vector in the null space of $A$

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

then let $B$ be the matrix whose columns are all that vector

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

The main point is $AB = [Ab_1 Ab_2 Ab_3]$, where $b_1, b_2, b_3$ are the columns of $B$.

[stoic python]

IlIIllIIIlllIIIIllll

[sour cloud]

so im multiplying each non zero vector to null space A

so if i have -4 i multiply that with 1 ?

[sour cloud]

for this matrix

and the null space is [1, -2, 1]

wait nvm im wrong :/

but we dont know the colums for b how are we using null space?

[sour cloud]

i dont get what you mean by this

still got it wrong 😢

well i give up i been working on this for a week lmao

[torn stag]

@sour cloud You want to find $B$ such that $AB = 0$ right

[stoic python]

IlIIllIIIlllIIIIllll

[sour cloud]

yes

[torn stag]

Do you agree that $AB = [Ab_1 Ab_2 Ab_3]$?

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

That is, the $j$th column of $AB$ is $A$ applied to the $j$th of column $B$.

[stoic python]

IlIIllIIIlllIIIIllll

[sour cloud]

yes

but we dont know the columns for B

its ok imma just give up i already got it wrong so i struggled too long for this

[olive beacon]

hi guys <@&286206848099549185>
is the converse of this theorem true?

[nocturne jewel]

dont ping helpers right away...

also this isnt LinAl

[long dome]

No

[hard drum]

Harmonic series

[tranquil steeple]

A =

   1.0000 + 2.0000i   1.0000 + 0.0000i   1.0000 + 0.0000i
   1.0000 + 0.0000i   1.0000 - 2.0000i   1.0000 + 0.0000i
   1.0000 + 0.0000i   1.0000 + 0.0000i   0.0000 + 1.0000i

>> det(A)

ans =

  -0.0000 + 4.0000i

[distant charm]

Can someone please help me out with this not sure How to write the answer

[granite kraken]

Need help with problem 1

Here's my attempt but I need to somehow show a elation between dim(R(T+S)) and dim(R(T)+R(S))

[winter harbor]

Given $w \in \text{Im}(T+S)$, then $\exists v \in V$ such that
$$
w = (T+S)(v) = T(v)+S(v)
$$
So $w \in \text{Im}(T) + \text{Im}(S)$, thus we have shown:
$$
\text{Im}(T+S) \subseteq \text{Im}(T) + \text{Im}(S)
$$
And so:
$$
\text{dim}(\text{Im}(T+S)) \leq \text{dim}(\text{Im}(T)+\text{Im}(S))
$$
Notice that we have, by finite dimensionality
\begin{align*}
\text{dim}(\text{Im}(T)+\text{Im}(S))
\

\
\text{dim}(\text{Im}(T) \cap \text{Im}(S)) + \text{dim}(\text{Im}(T)) + \text{dim}(\text{Im}(S))
\end{align*}
And since these are non negative integers, we thus conclude:
$$
\text{dim}(\text{Im}(T)+\text{Im}(S)) \leq \text{dim}(\text{Im}(T)) + \text{dim}(\text{Im}(S))
$$
Thus,
$$
\text{dim}(\text{Im}(T+S)) \leq \text{dim}(\text{Im}(T)) + \text{dim}(\text{Im}(S))
$$
As we wanted to show ; $\square$

[stoic python]

MisterSystem

[soft elm]

basic question but how would I go about doing part (ii) here

managed to determine a linear combination of the transformations and turn it into a single matrix for the overall transformation. is it enough to show that that matrix has linearly independent vectors?

[livid notch]

Can someone please help me with number 30

[forest quiver]

I kind of get how he got to line 3

But does anyone have a proper explanation?

Where P_theta(x) is a projection onto line L

and H_theta(x) is symettry about line L

[worldly bear]

i know i should row reduce this matrix, but i’m not sure what to do from there

like how to find the row and column spaces

[vale yoke]

Is the null space of A the same as null(A)?

[wintry steppe]

could you write down the definitions of "null space of A" and "null(A)"

i am 100 - epsilon percent sure they are

#groups-rings-fields message

[distant charm]

@wintry steppe thx

[remote hatch]

If $$(u,v)=u^{T}Bv$$ is an inner product then why must B be invertible?

[stoic python]

CottonBall

[winter harbor]

Because if that is a real inner product, then B must be positive definite. And positive definite matrices are invertible, since they have strictly positive eigenvalues, thus they have non zero determinant as well (since it is the product of all its eigenvalues).

[remote hatch]

How could you show that without having B being positive definite? Because the follow up question to that is why must B be positive definite

So in my case i dont think i can say thats it positive definite first

[winter harbor]

Sure

Suppose that $\exists v \in \mathbb{R}^{n}$ for which $Bv = 0$. Then, $v^{t} B v = 0$. Since $B$ is an inner product, it must be the case that $v=0$. Thus, $B$ has trivial null space and as such is invertible.

[stoic python]

MisterSystem

[winter harbor]

Just suppose Bv = 0 and multiply by v^t on the left on both sides

Use the fact that uBv^t is an inner product, thus a positive definite bilinear form.

And this means that if v^tBv = 0, then v=0

So B has trivial null space/kernel

And is invertible

[worldly bear]

why is the row space of A = row space of the upper triangle reduced matrix

but the column spaces aren’t =

[remote hatch]

ahh ok thanks

[winter harbor]

Suppose that you have a matrix
$$
A = [ u_{1} , \cdots , u_{n}]
$$
That is given by rows $u_{1}, \cdots, u_{n}$. Then, the row space of $A$ is defined as
$$
\text{Row}(A) = \text{span}(u_{1}, \cdots, u_{n})
$$
Now, the reduced row form of $A$ is obtained by applying elementary row operations to $A$. I.e by either switching two rows or for $i \neq j$ substituiting the $i-th$ row of $A$ by $u_{i} + \lambda u_{j}$ for some scalar $\lambda \in \mathbb{K}$ in your field.
\
\
Now, it is clear that the span of vectors is not changed if we just permute them. What we have to verify tho, is that the following equality holds:
$$
\text{span}(u_{1}, \cdots, \underbrace{u_{i}}{\text{i-th position}}, \cdots, u{n}) = \text{span}(u_{1}, \cdots, \underbrace{u_{i}+ \lambda u_{j}}{\text{i-th position}}, \cdots, u{n})
$$
For any given $i,j$ with $i \neq j$.
\
\
This is indeed the case, and I would encourage you to verify this by proving the following:
\
\
Let $u,v \in V$ vectors in a vector space $V$ over a field $\mathbb{K}$ and $\lambda \in \mathbb{K}$ a given scalar then:
$$
\text{span}(u,v) = \text{span}(u,v+\lambda u)
$$

[forest quiver]

Parentheses 2nd equation

[nocturne jewel]

Giving Aldium a run for his money god damn

[forest quiver]

Well, is this channel free now?

You seem to really enjoy Linear Algebra

Is it your favorite topic?

[stoic python]

MisterSystem

[forest quiver]

or course

[winter harbor]

The same argument hold if you want to prove that the column space is preserved by applying column operations.

[winter harbor]

No lmao

It's just the only topic I am currently comfortable with in giving actual explanations

I mostly study topology actually

[zinc timber]

I do like yr preamble setup, it's nice

[winter harbor]

I am studying diff top, symp geo and a bit of riemann surfaces at the moment.

But I don't want to go on #point-set-topology rn

[winter harbor]

[forest quiver]

Differential topology is a thing?

Jesus

[winter harbor]

Yeah, it is. Why?

[forest quiver]

Have you taken differential geometry ?

[winter harbor]

It depends. For instance, I haven't formally studied classical differential geometry (say of curves and surfaces in R^3) in the usual way, say via Do Carmo or other introductory books.

But I have background on Smooth Manifolds

Via Lee and Tu's book

I know a bit of diff top from Hirsch, which is the reference I am using rn.

[forest quiver]

Manifolds are just curved spaces rifht ?

[winter harbor]

And Symplectic Geometry

[forest quiver]

Well have fun with that

[winter harbor]

Manifolds are generalization of curves and surfaces to higher dimensions, yes.

[forest quiver]

Oh nice

[zinc timber]

I took number theory instead of diff topo

[forest quiver]

I hope to learn as much as you

I start university next year

[winter harbor]

The idea is that a smooth manifold is locally modeled on euclidean space. Meaning that at each point it "looks like" a piece of n-dimensional euclidean space.

[forest quiver]

Yeah ok I get that

[winter harbor]

So it generalizes the idea that smooth surfaces "close" look like a plane.

[forest quiver]

Oh right

[winter harbor]

[forest quiver]

Thag must have some applications

Like everywhere

[zinc timber]

I dont' attend the classes anyway , kind of regret taking it tbh

[winter harbor]

Yeah, Riemannian Geometry (or to be more precise, Pseudo-Riemannian Geometry) is pretty much the basic mathematics behind General relativity. Symplectic Geometry was pretty much conceived at the time to study problems in Hamiltonian mechanics and dynamical systems and so on...

There's also computational diff geometry, which is a meme????

Jk

It is actually useful to model stuff in computer graphics

[forest quiver]

Will you continue math in grad school?

[winter harbor]

And to model certain biological phenomena

Have you heard of Stephen Smale?

[forest quiver]

Nope

[winter harbor]

He was one of the major Differential Topologists in the last century (He proved the h-cobordism theorem). But he also dedicated himself to apply these techniques in other fields. He made some research in the protein folding problem, for instance, using techniques of diff topology, also has done some stuff with algorithms which is very cool.

[forest quiver]

Interesting

Do you see yourself doing something great like that someday?

[winter harbor]

https://youtu.be/rOjPnRxn-6c

Smale at IMPA

[winter harbor]

I hope so

[winter harbor]

Idk lol

[forest quiver]

But anyway, I hope you don't mind if I send my qeustion here

[winter harbor]

I just want to do math research, teach a little bit and don't worry about these things so much

If I end up actually doing some major work

It would be cool

But not something that I am thinking on rn

[winter harbor]

Sure

[forest quiver]

How did my teacher get to line 3 here? Where P_theta(x) is a projection of vector x onto line L. And H_theta(x) is symmetry of vector x about line L.

[winter harbor]

God damnit, why am I ranting in the linear algebra channel of all places

[forest quiver]

I mean it makes sense what my teacher did kind of

Like I could see how LHS = RHS there

but that's just intuition, whats the actual logical reason?

[winter harbor]

By symmetry you mean reflection?

[forest quiver]

Oh yeah

Actually no

Wait idk actually

[winter harbor]

By line three you mean what exactly? Could you highlight it?

[forest quiver]

the equation with green underline

[winter harbor]

In a sense, your teacher is arguing by pictures here.

You could make all of this precise

[forest quiver]

I figured

OK so it's not something that's on there then

Oh wait it kind of makes sense

though

[winter harbor]

By actually constructing the projection matrix and the reflection matrix (which is called a householder matrix) and doing all of these computations explicitly.

[forest quiver]

Yeha I get it

[winter harbor]

https://en.wikipedia.org/wiki/Householder_transformation

[forest quiver]

Ah thanks

I don't really like wikipedia math articles, to be honest

they are always filled with jargon

and I get easily lost

but I will give it a try

ty

[winter harbor]

This is just if you really want to make this argument precise

Like, you could just argue by picture

And I think it would be fine

[winter harbor]

@languid scarab https://math.stackexchange.com/questions/4126189/proof-of-cayley-hamilton-using-krylov-subspaces?rq=1

You might like the proof given by levap in the comments.

The downside of this proof is that it heavily uses the fact that we are working over a field (we are using the fact that F[X] is a PID here)

But the Cayley-Hamilton theorem works for unital commutative rings, which Atiyah-McDonald proves in more generality.

[topaz linden]

Quick question: this means that the domain and codomain are the same finite dimensional vector space right?

No I meant when we say an injective linear map on a finite dimensional vector space, we assume that the domain and codomain are the same right?

[teal grotto]

on a finite dimensional vector space makes me think T : V —> V

[wintry steppe]

soryr im being dumb

i need sleep

[topaz linden]

I'm just thinking that a function on a set S means that its a function from S to S

[wintry steppe]

yes. it's not true if the spaces are different

this is what the word "operator" is for and im angry theyre not using it

[teal grotto]

*have different dimension

[wintry steppe]

yes

let's pretend i hit enter too early

[topaz linden]

Wait I'm confused

[topaz linden]

Is this true or false?

I'm grading hw and a bunch of people have said false which is making me question my sanity

If I read that, I would say that it is true

Because a linear map on something means that that something is both the domain and codomain

[teal flume]

I am not quite sure but I think that you can't say that a linear map alsways has the same domain as codomain

[wintry steppe]

under your interpretation it's correct. if the domain and codomain are different, it's still true if they're of the same dimension. otherwise, it's false

you should probably ask the prof you're grading for

[teal flume]

A linear map can map values from 2 to one dimension for example (correct me if I am wrong), then domain and codomain are different

[wintry steppe]

since your concern is more of a linguistic one than a mathematical one

[teal flume]

Yes ask the prof definitely

[dire thunder]

this is not linear algebra

this is #prealg-and-algebra or #precalculus

@daring ingot

[daring ingot]

aight

[coarse sandal]

How do i approach this problem

[winter harbor]

Just apply the definition of matrix multiplication/how a matrix acts on a vector.

You will see that the dot product will appear

[last girder]

How do I find the matrix A in the second part

[winter harbor]

We want to find the matrix of T with respect to the basis B_0, right?

So all we need to do is to find how to write e_1 in the basis (b_1,b_2,b_3). The coordinates of e_1 with respect to such a basis will give you the first column of the matrix of T wrt basis B_0.

Analogously, finding the coordinates of e_2 wrt B_0 will give you the second column of this matrix, and the coordinates of e_3 wrt B_0 will give you the third column.

[last girder]

would you mind explaining why basis (b_1,b_2,b_3) and not basis (e1, e2, e3) when we want matrix T with respect to B0

[lavish jewel]

idk if a slight change of working might make it clearer. you know the transformation T in the basis B, but want to find out what the transformation is for vectors in the basis B_0

so you can compose the transformation T in the basis B with a transformation that changes the a vector in the basis B_0 to the basis B

given a vector in the basis B_0, find its representation in the basis B, and then apply the transformation T to it

[hexed burrow]

If M is a matrix, $\lambda$ eigenvalues, and $\vec{v}$ is an eigenvector, is the following true?
$$(\bf{M}\vec{v}=\lambda\vec{v})^{}$$
$$\bf{M}^{}\vec{v}^{}=\lambda^{}\vec{v}^{*}$$

[stoic python]

june

[lavish jewel]

taking the complex conjugate distributes over products and sums, so it looks ok

[hexed burrow]

Ok perfect thank you!

[brave lintel]

I know to find the inverse of a 2 * 2 matrix A you take 1/det(a) * adj(A), but how this generalise to matrices of higher dimensions? Does it even generalise?

If not how would i find the inverse of some arbitrary n * n matrix?

[sharp shell]

if u take the matrix [ A I ] and turn the A into I with row operations the I will have turned into A^-1

assuming its invertable

[lavish jewel]

in the real world, one doesn't usually try to invert matrices directly

but yeah, gauss jordan works if you need it explicitly

or decompositions that yield easy-to-handle matrices

[brave lintel]

wow looks like i need to learn la pretty soon

my knowledge is not at the level yet to understand this

but ty all anyways

[dark brook]

If I'm given 3 vectors (1x3) and I need to find out if those vectors are an basis for R^3.
Would the best way first to setup homogenous equations and find the RREF - or would there be an easier, or more obvious way?

[lavish jewel]

that works, but you can also rref without augmenting the matrix

[dark brook]

I'll do that instead and see what it brings! Thanks!

[dark brook]

If I get something like

1 0 0
0 1 0
0 0 1

It would probably suggest that we have 3 linear independant vectors, where x1, x2 and x3 must be 0, right?

[teal grotto]

a matrix's rref is the identity if and only if the matrix is invertible. so the columns (and rows) you started with have to be linearly independent and spanning

[dark brook]

what

[teal grotto]

if u row reduce a square n by n matrix and you get the identity matrix, the columns (and separately the rows) of the original matrix form a basis for R^n

[dark brook]

Which it does, but how can I interpret the results? If the system needs to linear independent, x1, x2, x3 in v= v1x1 +v2x2 + v3x3 = 0, right? So x1,x2,x3 = 0 for them to be linear independent? Or am I missing something

[lavish jewel]

sure, this happened implicitly

remember there was an extra column of 0s

those are still "there"

this is telling you that each one of the x_i = 0 is the only solution

meaning the v_i are linearly independent

we ignored the column of 0s because any row operations will anyway yield a 0 column

[dark brook]

Just to make sure I understand what you are saying. Is this how you meant it?

[lavish jewel]

$\begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

[stoic python]

Edd

[dark brook]

I see

This seems like an easier solution (above), but what would happen if there was a 2 or 3 in one of the 0's from the right side? Would that cause them to be dependant and therefore not an basis for the R^3?

[winged prairie]

Any hint on how to do this

[dusty pond]

Could someone walk me through and explain this? I'm not sure what to do - thanks so much!

[slow scroll]

If lambda is an eigenvalue and v is it’s eigenvector, then T(v) = lambda v. Now apply the assumption of a)

Then for b), think about what the determinant of T can be

[torn stag]

@dusty pond (b) is not true

@dusty pond (b) is true if $V$ is finite dimensional though

[stoic python]

IlIIllIIIlllIIIIllll

[winged prairie]

any clues on this?

[torn stag]

@winged prairie Show $P \cap (M + N) \subset M + (P \cap N)$, then show $M + (P \cap N) \subset P \cap (M + N)$.

[stoic python]

IlIIllIIIlllIIIIllll

[winged prairie]

i did the first part but im struggling to do the second part

[torn stag]

why?

[winged prairie]

idk i just got stuck

how would you do it?

[torn stag]

Let $x \in M + (P \cap N)$ be arbitrary.

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

There exist $m \in M$, $v \in P \cap N$ such that $x = m + v$.

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

Now use the fact that vector spaces are closed under addition

[winged prairie]

ok ty

got up to here

is this even correct? and if so how do i continue

[torn stag]

Yes, $m \in P, v \in P \implies x = m + v \in P$

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

Now you have to show that $x \in M + N$.

[stoic python]

IlIIllIIIlllIIIIllll

[winged prairie]

ok ty

is it enough to say that since m belongs to M and v belongs to N, x belongs to M + N?

[warped trellis]

Anyone know any good sites to practice Solving Sytems of Equations with graphs?

[torpid socket]

Can someone help me with c

[tropic wadi]

This should be in #real-complex-analysis

[torpid socket]

I posted it here since I’m taking this in linear algebra 😅

[winter harbor]

Use the fact that $|z|^{2} = z \cdot \overline{z}$ and apply (b)

[stoic python]

MisterSystem

[dark brook]

When finding the null space of matrix, why is it that x_3 = 1?

I've found $\begin{pmatrix}
x_1 \
x_2 \
x_3 \
\end{pmatrix} =
t_3 \begin{pmatrix}
-1 \
-2 \
t_3 \
\end{pmatrix}$

[stoic python]

HrJonas

[dark brook]

But I just can't seem to get why t3 needs to be 1?

Why not 0 or 2?

[remote hatch]

Because your equations would be $$x_{1}+x_{3}=0 \text{ and } x_{2}+2x_{3}=0$$

[stoic python]

CottonBall

[remote hatch]

And so if you set $$t_{3}=x_{3}$$ then when you plug $$x_{1}, x_{2}, x_{3}$$ into the vector and factorise $$t_{3}$$ out of it you get 1

[stoic python]

CottonBall

[remote hatch]

Sorry for bad formatting lol

$\begin{pmatrix}
x_1
x_2
x_3
\end{pmatrix} =
\begin{pmatrix}
-1t_3
-2t_3
t_3
\end{pmatrix}$

[stoic python]

CottonBall

[shell kindle]

Can someone explain to me how they got [1, 1] on the bottom row??

[limber sierra]

subtract the first row from the second

it seems they did the first step of gaussian elimination early for some reason

[shell kindle]

Oh I see, thanks!

[limber sierra]

hint: bases must be linearly independent.

in particular, this means you cant have more basis elements than the dimension of your space

so take another look at (1)

[hidden heart]

Hello, can someone help me with this question. Here is the translation: Let S1, S2,... be infinite subspaces of a vectorial space V, such that: (I hope the notation is clear). Let S= U Si (again, hope the notation is clear). Prove that S is a subspace of V

[limber sierra]

have you heard of zorn's lemma?

[hidden heart]

no

[limber sierra]

thats... a problem since i think this statement relies on it

oh wait

never mind

im just being dumb

just run the standard subspace checks:

• show that S is nonempty
• pick any u, v in S. show u + v is also in S
• pick any u in S and any scalar k. show ku is also in S

are you having trouble with any of those checks?

[hidden heart]

to be honest, I have troubles with vectorial spaces as a whole

;-;

[limber sierra]

lets go step by step

first off, can you show S is nonempty?

[hidden heart]

mmm

S1, S2 and blablabla are all subspaces of V

V is a vectorial space, so it cant be empy, and thus the subspaces of V cannot be empty

The union of non empty sets must be not empty

Does that make sense?

Now, Idk how to put that (if correct) in a more formal language is my problem

[limber sierra]

even thats a bit roundabout

just say that S_1 is a subspace, so it must be nonempty

and so any union containing S_1 is nonempty

(since, as you said, the union of nonempty sets is nonempty)

thats enough

[hidden heart]

oh, makes sense

[limber sierra]

you had the right idea but you didnt need to talk about V at all

now lets try the next step

pick vectors u, v in S. can you prove u + v is in S?

(as a hint: since u is in S, it must have come from some S_i, and similarly v must have come from some S_j. but note that if u is in S_i, then its also in S_(i+1), and S_(i+2), and so on. what does this mean?)

[hidden heart]

Si has strictly less elements than S_(i+1). So, the union of infinite subspaces of V, all being strictly bigger as i increases, must be V, and V is a vectorial space, so given any u, v in S, u+v is in S

I feel something doesnt add up, but its the best I could think of and I dont want to waste much of your time

oh, wait

I think i did other unnecesary roundabout xD

[nocturne jewel]

Is fourier space a thing?
Ie the subspace of functions which can be written as a linear combination of sines and cosines? Professor for modern physics mentioned it but didnt go into detail

I'd assume it's just a space with basis {1,sin(t),cos(t),sin(2t),cos(2t),...,sin(nt),cos(nt),...} or some other collection of n values

[winter harbor]

The theory of fourier transforms is deeply related to the study of L^2 functions on S^1 (which is a vector space), maybe that can give you a bit of insight.

[nocturne jewel]

in simple terms.

[winter harbor]

and what are you referring to is commonly called a hilbert basis, in contrast to the common definition of Hamel/algebraic basis we use in linear algebra.

[nocturne jewel]

again

simple words

No clue what L^2 functions are, nor S^1

[winter harbor]

$\mathbb{S}^{1}$ is just the unity circle, i.e the space:
$$
\mathbb{S}^{1} = { z \in \mathbb{C} , \vert , |z| = 1 }
$$
Now, given a subset $X \subseteq \mathbb{R}^{n}$, we can define what is called the space of square-integrable functions over $X$, which is commonly denoted by $L^{2}(X)$. It is defined as:
$$
L^{2}(X) = \left{f : X \subseteq \mathbb{R}^{n} \rightarrow \mathbb{R} , \vert , \int\limits_{X} |f(x)|^{2} dx < + \infty\right}
$$
This space is in fact a vector space over the reals and, when $X \subset \mathbb{R}^{2n}$ it can be endowed with the structure of a complex vector space. Now, it turns out that studying fourier series and fourier transforms of real periodic functions on the real line is most fruitful when viewing these as functions $f : \mathbb{S}^{1} \rightarrow \mathbb{R}$, i.e functions defined on the unity circle. And turns out that when restrict ourselves to studying square-integrable functions, the theory of fourier transforms is much more well behaved.

[stoic python]

MisterSystem

[winter harbor]

Just a really brief description lol

[nocturne jewel]

that isnt brief.

[winter harbor]

But anyways

Tl;dr

[nocturne jewel]

And the answer to my question was just "L^2 is the space of square integrable functions and S^1 is the unit circle in C"

everything else is fluff

[winter harbor]

Studying fourier series and fourier transforms is deeply related to studying square-integrable functions on the unity circle, and a lot of nice stuff that comes from fourier analysis comes directly from nice properties of L^2(S^1) as a vector space (like it being a reflexive space and so on)

I don't really know much analysis myself

[torpid socket]

Can someone help me with solving the mateiz

[nocturne jewel]

I also originally asked a yes or no but anyway

[winter harbor]

[nocturne jewel]

You have 2 questions on that ss, which one?

[torpid socket]

5

[winter harbor]

https://math.stackexchange.com/questions/184390/relationship-of-fourier-series-and-hilbert-spaces

Maybe this has better explanations than the ones I am trying to give

[nocturne jewel]

let x, y and z be the weight of the top, medium, and low quality cereals respectively

can you form the 3 equations that the preamble (the question text) gives in words interms of x y and z?

[torpid socket]

Yes

[winter harbor]

Bundles

[torpid socket]

:D

[nocturne jewel]

ok, what are they?

[torpid socket]

I don’t know how to use the bot :( but I will try

[nocturne jewel]

that's fine

just
write them
like this

(you also shouldn't need the bot for this btw)

[torpid socket]

3x1 + 2x2 + 1x3 = 39
2x1 + 3x2 + 1x3 = 34
1x1 + 2x2 + 3x3 = 26

Right ?

[nocturne jewel]

yes

[torpid socket]

Awesome!

[nocturne jewel]

so that's the system of linear equations they asked for, how do you turn that into an augmented matrix?

[torpid socket]

The matrix should be
3 2 1 39
2 3 1 34
1 2 3 29

[nocturne jewel]

Yep exactly

[torpid socket]

Now after that I can’t do

[nocturne jewel]

right so bit of notation, when I say [A|b] that means augmented matrix with coefficient matrix A and constant vector b

What you now want to do is perform row operations (EROs) to get [A|b] to something of the form [I|c] where the coefficient matrix is now the identity matrix (or as close as you can get to it) and c is the transformed b vector

[torpid socket]

I’m trying but its not working :(

[nocturne jewel]

can you post your working then?

[torpid socket]

Sure its a bit messy tho

Ignore the bi - bi 😅

[nocturne jewel]

so... not gonna lie

cant follow it really

[torpid socket]

I apologize:(

[nocturne jewel]

however, the tried and true algorithm is:
Make 1 in the (1,1)-entry then make all entries beneath it 0
Make 1 in the (2,2)-entry then make all entries beneath it 0
Same for (3,3)-entry.

You now have the "pivots" and an upper triangular coefficient matrix

Next part of the algorithm is:
Make all entries above the (3,3)-entry 0
and make all entries above the (2,2) entry 0

and this generalizes for nxn systems

[torpid socket]

Yeah its not working tho :(

I will try again

[nocturne jewel]

Then unfortunately it's your arithmetic

[torpid socket]

How can I prove them

[lavish jewel]

i would recommend to view this as a matrix-vector problem Ax = b, and then think about the null space and column space (range) of the matrix

[torpid socket]

I didn’t understand;-;

I don’t think I learned that yet

[lavish jewel]

maybe not by name

you've done RREF/gauss jordan, most likely

and seen cases where you get 1+ rows of all zeros

[torpid socket]

Yeah!

[lavish jewel]

right, it's the same thing

[torpid socket]

Wait +1 rows of all zero ?

[lavish jewel]

do you know what happens when you get a row of all zeros in gauss jordan?

[torpid socket]

No

[lavish jewel]

i doubt that's really the case cuz otherwise you don't know enough to solve the problem :x

[torpid socket]

Can you give me a small example

Maybe I have seen it but can’t remember since its 7am ;-;

[lavish jewel]

maybe you know them as "free variables"?

[torpid socket]

Let me google that

Hmm don’t think so

[lavish jewel]

i don't think you can solve the problem, then

[torpid socket]

Wait is it when you have 1

Like 1 00
010
001

[lavish jewel]

no

it's when you don't have that

[torpid socket]

Oh

[lavish jewel]

you need to have seen systems with no solution, 1 solution, and infinitely many solutions to solve this, and it seems you haven't

[torpid socket]

Oh

Thank you edd! ☺️

[gleaming knot]

Since you said subspace, I'm inclined to think you're looking for the space of functions which are a finite linear combination of those things. Am I right?

What have you tried? Have you used any definitions?

[quaint sphinx]

I tried to prove p(x) isn't empty but I'm confused

[gleaming knot]

Is proving p(x) isn't empty supposed to prove S isn't empty?

[quaint sphinx]

I had similar problem where p(0) is 0, but it didn't work out for this one

[gleaming knot]

Proving p(x) isn't empty doesn't make sense

The statement "p(x) isn't empty" doesn't make sense

[quaint sphinx]

yeah i meant S isn't empty, sorry

[gleaming knot]

Okay, showing S isn't empty is one step

did you successfully do that?

[quaint sphinx]

no actually im struggling with that part

[gleaming knot]

Hmm a common method for showing a set is nonempty is to show an example of an element in the set

[quaint sphinx]

if b and c are 0 wouldn't a = 1 ?

so there can't be the 0 vector?

[gleaming knot]

You're now proving something else?

what did you prove

[quaint sphinx]

no like to prove the set is nonempty I tried to show that the set contains 0 vector

[gleaming knot]

Did you actually prove the set doesn't contain the 0 vector instead?

[quaint sphinx]

so it's not a subspace?

[gleaming knot]

That sounds extremely unconfident

What's bugging you?

[teal grotto]

try adding two elements in S together. note that the constant coefficient has to be 1 if a polynomial is in S.

alternatively, any subspace has to contain 0. 0 just….isn’t in that set. by definition

[gleaming knot]

That's something else you could do if you approached the problem by testing the linearity condition first

[granite kraken]

Can someone explain these 2. I'm very stuck. Thanks

[quaint sphinx]

It isn't subspace then

got it, thanks to both of you

[teal grotto]

so for two, if AP = PD for some diagonal matrix D, try showing that the columns of P form an eigen basis consisting of eigen vectors of A.

then you can do some nice computations to show that the eigen values of A are exactly the diagonal entries of D

for three, just show that those sets are closed under scalar multiplication and vector addition

@granite kraken

this is actually an equivalent condition for when a matrix is diagonalizable, that is, if and only if A admits an eigen basis

[vocal isle]

Hey folks I want to find an equation that governs the position of a point of light bouncing around a closed 2D reflective surface. The surface is defined in cylindrical coordinates.

f(r,theta) = 1

I know I can use a mirror matrix to calculate the ray of light reflected from a point of a surface.

Mirror matrix shown in red where (n1,n2) is the unit normal to the surface

Say the ray of light starts at some position inside the surface P0 and has the direction V0. How can I find an analytical expresit for Pn and Vn (after n reflections)

[fast beacon]

Hey can someone help me with a solution to this pls, I have no clue where to start
Find a parametric equation for the line L through the points A(1, 2, 1) and B(2, 1, 2). Compute the distance between line L and the point C(2, 3, 4).

[dusky epoch]

have you found parametric equations of lines before?

[spare widget]

Pamaretric equation of a line going through 2 distinct points $A$ and $B$ is given as: $c(t) = A + t (B-A)$. The distance between a point $P$ and a line $c(t)$ is the distance of the orthogonal vector from $P$ to $c$. Project $P-A$ onto $c$: $c\left(\frac{(P-A) \cdot (B-A)}{ |B-A|^2}\right) = A + \frac{(P-A) \cdot (B-A)}{|B-A|^2} (B-A)$, then the length of the orthogonal vector is: $\left|P-A - \frac{(P-A) \cdot (B-A)}{|B-A|^2} (B-A)\right|$.

[stoic python]

criver

[fast beacon]

Thank you

[winged prairie]

any clue on how to do part (i)

[lavish jewel]

show that even and odd functions in V are orthogonal, and that any generic function in V can be found as a linear combination of an even and an odd function

and i suppose that, to begin with, you'd have to show even and odd functions form subspaces, which you can do by checking closure under addition and scalar multiplication

[winged prairie]

ok ty

[hexed burrow]

For a symmetric matrix M, the passage matrix P is orthogonal, because the eigenvectors form an orthogonal basis right? So then is it true that the inverse of P is equal to the transpose of P, where P is comprised of the normalised eigenvectors in the columns?

[winter harbor]

Yup, that is correct. Indeed, that's precisely how we define an orthogonal matrix to begin with.

[hexed burrow]

Perfect thank you!

[winter harbor]

$O(n) = {A \in \mathcal{M}{n}(\mathbb{R}) , \vert , AA^{t} = A^{t} A = I{n} }$

[stoic python]

MisterSystem

[winter harbor]

This is the definition of the set of all orthogonal matrices.

Which is also a subgroup of the space of invertible matrices, and so is commonly called the orthogonal group.

[wintry steppe]

Can someone please explain to me the following sentence

What does it mean "the component of x in v"?

Like say the ONB is {u1,u2...} and it contains v

and if we want to express x then that would be x = au1 + bu2+ ... + kv+ ..

how does the inner product come into play?

[lavish jewel]

it's telling you that, using your notation, a = <x, u1>

an easy way to see this is as follows

say x is given in the canonical basis I

and we want the coordinates of x in a new basis that is orthonormal, let's call it B

then the coordinates of x in the basis B are some other vector w so that Bw = x

and if you want to find w, you'd need to do w = B^-1 x

since B is orthonormal, B^-1 = B^T

so w = B^T x

now notice that each component w_i of w is one row of B multiplied by x

i.e. w_i = b_i^T x

and b_i^T x is precisely <x, b_i>

[wintry steppe]

Ohhh

That makes total sense

thanks

[lavish jewel]

and so substituting this back into w in the expression Bw = x, you can see that each coordinate is given by the dot prod

[wintry steppe]

BB^T = 1 right ? so B^-1 = B^T

[lavish jewel]

if by 1 you mean the identity matrix, yeah

BB^T = B^TB = I for orthonormal matrices, as mistersystem explained just a few messages above

[wintry steppe]

Bx = w right?

[lavish jewel]

no

B^T x = w

[wintry steppe]

Bw = x makes sense

w is the coefficients

[lavish jewel]

yes

and the columns of B are the vectors that form the orthonormal basis

now multiply both sides of the eq by B^T

B^T B w = B^T x

but B^T B = I because B is orthonormal

so w = B^T x

[wintry steppe]

Thanks Edd!

[mystic dagger]

Given the set $V=[(x,y); x, y \in \mathbb{R}]$
and $(x_1,y_1)$ and $(x_2,y_2) \in V$,

$\lambda \in \mathbb{R}$,
$(x_1,y_1)\oplus(x_2,y_2)=(x_1+x_2, 0)$,
$\lambda\odot(x_1,y_1)=(\lambda x_1, \lambda y_1)$
why isn't V a vector space?

[median ocean]

Does anyone know how to do part c?

[mystic dagger]

i mean we could have y1+y2=0 why not?

i'm really stuck on that

how can i prove it's not a vector space???

[stoic python]

leonardomoura

[winter harbor]

Basically because $(V, \oplus)$ is not an abelian group, for the solely reason that it does not have an additive Identity.

[stoic python]

MisterSystem

[winter harbor]

More precisely, prove that $\not\exists u = (x_{2}, y_{2}) \in V$ such that $\forall v = (x_{1},y_{1}) \in V$ we have:
$$
v + u = u + v = v
$$

[stoic python]

MisterSystem

[winter harbor]

Just think about any element in V for which its second coordinate is non zero

Like (1,1)

[mystic dagger]

oh

ohh

but we couldn't have $y_1=1$ and $y_2=-1$?

and the sum would be zero

[stoic python]

leonardomoura

[winter harbor]

It doesn't matter, because (0,0) is not the identity of $(V, \oplus)$ in the first place (it doesn't have one, it's a commutative semigroup without an identity). To talk about inverse elements in a semigroup we have to have the notion of an identity element in the first place, and you are using the naive intuition that (0,0) should be the identity of this operation, when it isn't.

[mystic dagger]

oh

ok i'll think about it

thanks

[stoic python]

MisterSystem

[winter harbor]

Nada pô

[median ocean]

C please

What do i do after I combine the rig vectors to matrix

Eigenvector*

[winter harbor]

You consider the matrix whose columns are given by the linearly independent eigenvectors you have found

P will surely be invertible

And to verify A = PDP^-1

It is enough to verify AP = PD

Which is considerably easier

[dark brook]

When finding basis for column and row vectors, do you do it from the matrix of A or the RREF of A?

Also called C(A) and C(A^T)

[glad acorn]

RREF is needed for sure but the basis are from the original matrix

[marble lance]

If you are doing elementary row operations and finding the basis for the row space, then use the RREF form. You can use the original matrix, but if you swap rows at any point, it makes it hard to figure out which to use. Elementary row operations preserves the row space, so you can just use the rows from RREF

When you are finding the basis for the column space, then you need to use the original matrix

Because row operations do not preserve the column space

[dark brook]

Ahh okay, I'll keep that in mind thanks!

One last question. I am unsure why the final vector for N(A) is (-1, -2, 1)^T, when I've got this RREF and where I get that last 1 from

[glad acorn]

parametric form

[dark brook]

Whats that?

[glad acorn]

due to the fact that x_3 is the free variable

[dark brook]

But if x_3 = b_3 and b_3 = 0, x_3 needs to be 0 as well for it to work?

[lavish jewel]

x3 is not = b3 though

you don't enough enough about x3 to even find it, actually. the best you can do is say x3 = t, some parameter

[dark brook]

Yea we usually denote our free variables as t_3 or something like that but I just don't see why it needs to be 1

[lavish jewel]

it doesn't, that's the point

find all of the x_i in terms of t

and then any scaled version of that vector is still in the null space, by linearity

including the same vector scaled by 1/t

[lavish jewel]

exactly as phymath has done it

[glad acorn]

there are 2 pivot columns and 1 free variable column in A so you can use 1 parameter and one vector to represent the null space.

free variable column in RREF has the equivalence meaning to non-pivot column in RREF

[earnest dock]

if i have two bases B, C, does [I]^B_C = [I]^E_B x [I]^E_C ?

idk how to use the texit bot eh

$$[I]^B_C = [I]^B_E * [I]^E_C $$

is this correct?

or uhm sorry

[stoic python]

ischelle

[earnest dock]

the texit bot has it right

b->c = b->e * e->c

idk if that's correct

oh 😛

[lavish jewel]

kinda weird that you compose from the left

[earnest dock]

wdym?