linear-algebra

[dusky epoch]

who

[lavish jewel]

ah, someone did some cleanup on isle 3

[lucid glacier]

We sorted it

All good now

[earnest ferry]

Could someone please help me to find some applications of vector spaces?

[limber sierra]

ℝ^n is pretty handy for everything.

beyond that, function spaces are very very important to harmonic/fourier analysis and therefore modern physics

a lot of computing science takes place in vector spaces over a finite field (or modules over a ring Z/nZ), but CS people are unlikely to use that sort of terminology

and of course, linearity is a very very very useful property for a function to have

most of numerical analysis is trying to linearly approximate things

or at least polynomial approximate it

and linear maps are just vector space isomorphisms, so you can think of linearity as just a statement about vector space relations

etcetc

those are just some of the common examples

[earnest ferry]

tysm

[ornate holly]

how to learn about Vector Spaces and Subspaces ?
i cant understand these topics , can someone recommend me some resource to study?

[subtle burrow]

Just a quick sanitiy check, given some matrix $A$ and some vector $x$ with $\sigma_{min,A}$ being the lowest nonzero singular value the following it true? given finit dimension and real etc

$$\sigma_{min, A} ||x|| \leq ||Ax|| ,\forall x$$

[stoic python]

Dusto

[lucid glacier]

Yes, assuming you are considering 0 as a singular value too

And I think you need to assume A is square

[lavish jewel]

doesn't need to be square

it works for complex vectors and matrices too, btw

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

[lucid glacier]

Right, I misremembered

[subtle burrow]

alright just checked numerically and this inequality doesn't seem to hold up at least in l2

what does work out is the following

$$||x|| \leq \sigma_{max, A} ||Ax||$$

[stoic python]

Dusto

[lavish jewel]

that's a weird one lol

it shouldn't work if A is a singular vec and the singular val is less than 1

the inequality you wrote before was true, but this one isn't

you can make a counter example as A = [0.9, 0; 0, 0.9], x = [1;0]

[lucid glacier]

I've derived the inequality you gave before, it definitely works

[lavish jewel]

yeah, they mixed something up

[lucid glacier]

Btw the inequality
$$\abs{\abs{Ax}}\leq \sigma_{\max}\abs{\abs{x}}$$
Does work as well

[stoic python]

ShiN

[lavish jewel]

yep, same rayleigh

[mild current]

If complex no is a vector space, then why cant we define multiplication and division of two vectors like complex nos

[nocturne jewel]

division of vectors isnt a thing

and multiplying 2 vectors is not a thing defined in the vector space, it's defined in the inner product

[mild current]

I see

[wintry steppe]

C is just special

(it's a field and any field happens to be a vector space over itself)

but of course, not all vector spaces are fields

[lavish jewel]

those other operations you mentioned just happen to look similar in this special case

but for vector spaces, you only need a handful of properties for + and * (with a scalar)

anything else is extra

[faint dune]

I know Gram-Schmidt, but I always fail to use it in the exams cus the task is weird.

Find orthogonalbasis {yi} of P3 with <p,q> scalar, so that yi has degree i

The other task to proof its actually an scalar product was easy. But now I dont get what they want me to do exactly.

Am I even supposed to use gram schmidt here somehow?

my exam is in 9 hours

[prisma cairn]

In ML they often say XOR gate is not linear. Are they just ignoring the field {0, 1}?

[lavish jewel]

they do the operations over the reals, so yes

[half ice]

@faint dune
Note that φ is not y (discord doesn't have support for the φ character sadly, φ is the closest we get)

It seems like it just wants any orthogonal basis where every degree poly is used. Except, that's every basis

[sleek sundial]

what does this question even mean? confused about the 2nd part after the comma

[wintry steppe]

if you reflect the vector x in the line y = x you get Qx

it's asking you to find the matrix representing the linear transformation that is reflection in the line y = x

probably

the wording is a little strange

[half ice]

What do you mean "create" haha

[wintry steppe]

Make?

[half ice]

The image spans a plane. It just kinda already exists

[wintry steppe]

oh

how would i get that plane?

[half ice]

How do you normally get planes?

Oh you want a plane equation

[wintry steppe]

Yes

[half ice]

Yeah you could do that cross product trick

You have two vectors that are on the plane already, they are the columns

[wintry steppe]

Yes

okay

thank you

i did that i was just really unsure

[half ice]

Yaya it works. Feel free to ask if you have any others!

[wintry steppe]

i have one more

using the rank-nullity theorem

is that valid?

@half ice ?

[half ice]

If the output is a 2D space, then the output is a 2D space. You don't need rank-nullity to say that haha

[wintry steppe]

Right

okay

thank you

[faint dune]

@half ice I thought about it at night and my idea was to just write down v1=1 as degree 0 startvector

And then I determine v2 si that the scalarproduct is 0.

And the same for v3 but scalar with both is zero. Mybe for v3 I can also just take any vector and hope its not collinear and use gram schmidt 🦢

[lavish jewel]

it asks you for the degree of each one to be increasing tho

so yeah, the procedure you said is correct, but it's a bit annoying

[warm yarrow]

Applied linear algebra vs optimization for my math elective as a stats/ML emphasis major

[lavish jewel]

that's a tough choice, but i would argue that optimization anyways includes a fair bit of linalg

[half ice]

Linear algebra 100% haha

Both more applicable, and more general

[faint dune]

@wintry steppe .

what is this notation

divided difference, but there is functions in brackets?

[wintry steppe]

applied to gf ?

like gf(x2)- gf(x1) / (x2-x1) but just a guess

check your notes toge

[lavish jewel]

normally i would agree, but they did say ML

they'll probably pick up some linalg along the way

but the optimization stuff is super important in ML

especially if you want to come up with new stuff

it's all optimization

you don't even have to do it with linalg

[wintry steppe]

Yeah there's definitely linear algebra in ML but I think you'd pick it up just as well on a course in neural networks or something

Most of the linear algebra required is not very advanced

[north hedge]

in this question, is n supposed to be the dimension of V with the fi linearly independent?

because otherwise i dont think its true

(tag plz)

[marble lance]

No, I don't think n needs to be the dimension of V

@north hedge

[north hedge]

suppose V is R^3 and f maps e1 to 0, and e2 and e3 to 1. If g maps e1 and e2 to 0 and e3 to 1, then clearly the kernel of f is in g, but g isnt a multiple of f

@marble lance

am i missing something in this example?

[marble lance]

f(e2-e3) = 0 but g(e2-e3) = -1

So I don't agree the kernel of f is in the kernel of g

[north hedge]

oh I see

lol

idk why i missed that

[marble lance]

It's okay, it took me a second, haha

My first instict was that the kernel of f was just {(x,0,0)} and of g was {(x,y,0)}

[north hedge]

yes thats what i thought too

[marble lance]

Oops

[random axle]

Not learning linear algebra, but matrix algebra as part of a high school course. My question is regarding using matrices to represent linear transformations. Why is it that when deriving a matrix to represent a linear transformation, we apply the transformation to unit vectors i and j and the images of i and j become our transformation matrix?

[frosty vapor]

each column of the matrix represents the action of the transformation on each of the vectors in the ordered basis

so yes

[marble lance]

Let's say our transformation is T and we want to transform the vector (x, y). Now (x, y) = x i + y j. And by the properties of a linear transformation this means T(x, y) = x T(i) + y T(j) and this we can write as [ T(i) T(j)] [x, y].

So it only matters how you transform the basis vectors, since this then determines how all other vectors are transformed

[T(i) T(j)] is the matrix whose first column is T(i) and whose second column is T(j). And [x, y] is the vector we are transforming as a column

@random axle

[random axle]

I get it now, thanks!

[marble lance]

👍

[wintry steppe]

This from an old exam, I tried to take the eigenvalues and find the eigenvectors

For 1

But got a single span

So means its not linear

So confused

[hard drum]

'it's not linear'?

[frosty vapor]

?

what

ok walk us through what you did

[wintry steppe]

Linear independent

[hard drum]

what is not linearly independent?

[wintry steppe]

1,2,0,1 are the eigenvalues since its a upper triangle

Eigenvalue 1 only has 1 span but there are 2 off them

[hard drum]

the whole point of the question is you only get 1 eigenvector for 1

except for one value of x, for which you get 2, and so A is diagonalisable for that x

[frosty vapor]

you need to choose the right x which gives you two eigenvectors for the eigenvalue 1

so that you have 4 linearly independent eigenvectors

which gives u diagonalizable

[wintry steppe]

So your saying solve for eigenvalue 1 and find x which gives 2 eigenvectors

[hard drum]

ye, find x such that there are 2 linearly independent eigenvectors with eigenvalue 1

or equivalent find x such that

$\mathrm{null}\begin{pmatrix} 0 & 3 & 10 & x \ 0 & 1 & 4 & 3 \ 0 & 0 & -1 & 4 \ 0 & 0 & 0 & 0\end{pmatrix} = 2$

[stoic python]

nuclearpotat

[wintry steppe]

Ok i got it

I solve down to first line and x = 17 which was the answer

[hard drum]

yup nice

[steep harbor]

Does anyone have good resources for understanding the proof of spectral theorem?

[grim bridge]

the proof of what fact precisely ?

[steep harbor]

Any normal operator m on a vector space v is diagonal with respect to an orthonormal basis for v

I think I should said spectral theorem

[wintry steppe]

are you looking for a proof or for intuition on the theorem? you should be able to find a proof in any half-decent linear algebra book, maybe even on wikipedia

[grim bridge]

I think it would help specifying a step you can't understand

[steep harbor]

My book isn't a linear algebra book so it glosses over the proof as an aside

I think I'll look at the Wikipedia page for now then

[grim bridge]

first you should understand why a symmetric matrix is diagonalizable

(aka gauss decomposition of quadratic forms)

and then you just need to check the eigenspaces are orthogonal using the symmetry

that's just calculation

and then you can easily deduce the spectral theorem

[vital shard]

Hi, can anyone help me understand this proof? Why is <w1, vj> is equal to <u, vj>?

[dusky epoch]

<w_1 - u, v_j> = 0

@vital shard

[vital shard]

is it because w1 is orthogonal to u ?

what does "orthogonal project of u on W" mean?

[lavish jewel]

think of extending the orthogonal basis of W to one of V

[wheat jasper]

Any idea whats the answer to q1

I got u+ w = b1 + b2 +b3

[lucid glacier]

Is there a convenient way to visualize (in lower dimensions) sets of the form
$${v\in \mathbb R^n: \langle v,x_0\rangle = \alpha}$$
For some fixed $\alpha,x_0$?
With the standard dot product

[stoic python]

ShiN

[lucid glacier]

Ofc when alpha=0 it's just the originally complement and it's trivial

[lavish jewel]

idk, all vectors of the same length and making the same angle w.r.t x0?

aren't these hyperplanes?

this is of the form ax1 + bx2 + cx3 + ... + d = 0

you can build it backwards from the def of a plane

[lucid glacier]

A hyperplane can be affine right?

[lavish jewel]

say you know a point p0 is on a plane with normal x0. any vector w = u - p0 is on the plane if <u - p0, x0> = 0

then <u, x0> = <p0, x0>, where <p0,x0 > = alpha

and if alpha is nonzero, it's affine indeed

[lucid glacier]

So if you have some vector $p_0$ that satisfies the condition, then this is just
$p_0+{x_0}^{\perp}$?

[stoic python]

ShiN

[lavish jewel]

no

[lucid glacier]

But that's what you said basically

Oh wait

[lavish jewel]

u - p0 = {x0}_orthocomp

[lucid glacier]

Sorry I misread

Yeayea I read it wrong

I'm trying to understand geometrically why this forms a plane tho

Or a hyperplane

[lavish jewel]

set p_0 = 0 and try it out in 2D

a hyperplane in 2D is a line

then try changing p_0 to something else and see what happens

i constructed the set from the definition of a plane up there, too

[lucid glacier]

Yeye

Ty

I'll play around with it

[lucid glacier]

I think this construction works too tho, since for every orthogonal vector $u$ satisfies
$$<u+p_0,x_0>=<u,x_0>+<p_0,x_0>=\alpha$$

[stoic python]

ShiN

[lucid glacier]

It's just easier for mr to understand affine subspaces like this

[lavish jewel]

hmm in the previous one though, the alpha was directly related to the distance from the origin

here you have 2 constants

[lucid glacier]

You choose some arbitrary vector for this tho

[lavish jewel]

arbitrary yes, but anyhow projected onto x0

[lucid glacier]

This works since for every 2 vectors in our set you have p_0-q_0 is orthogonal to x_0

So they're in the same coset

So they generate the same affine subspace w.r.t the orthogonal complement

[lavish jewel]

it'll work, i'm just saying the interpretation of alpha is lost

[lucid glacier]

Ah

Well I guess

If you take a normalised vector it might be a better interpretation

[lavish jewel]

i should've said that, yes

i meant to have x0 be the normal of the hyperplane, with unit norm

[lucid glacier]

Ah ok

[short magnet]

Hi there, I'm trying to prove the following statement : "All hyperplanes of Mn(R) (the vector space of n by n matrices with coefficients in R) contain an inversible matrix". The approach I came up with kind of differs with the answer provided, which was a little bit more involved.
Let H be a hyperplane of Mn(R), thus there exists a linear functional f such that H = Ker(f) . All linear functions over Mn(R) are of the form Tr(A.) with A a matrix different from 0. To prove the result, it suffices to show that there exists i in [1,n] such that E_i,i is in H (i.e. f(E_i,i) = 0). (E_i,i is an element of the canonical basis) if such an E_i,i exists then we're done. If not then we consider E = E_i,i - f(E_i,i)/n * I_n (I_n being the identity matrix) . We have f(E) = 0 and E invertible.

(i = 1 for example, if for all i in [1,n] f(E_i,i) != 0)

Is my reasoning correct ?

[tight stag]

hi

I'm trying to write a program to convert a triplet of 3D points into Euler XYZ angle representation

I found this website that seemingly does this directly but I don't understand what it's doing so it's hard to write my code based off of its output when I hardly even understand what it's doing

If anyone can help me to understand the workflow I need to go through in order to convert a set of three points in 3D space into Euler angles, it would mean a lot to me

https://www.andre-gaschler.com/rotationconverter/

This is that website btw

It says that it normalizes all inputs to quaternions

tbh I don't have a solid grasp on what that means. It just converts a triplet of 3d points to a quaternion? Then it just converts that to Euler angles?

The scipy python library has an as_euler() function that I can allegedly pass quaternions into as input and it'd return em as Euler angles

But I don't know how to convert a triplet of points to a quaternion — all my googling led me to quite high-level technical discussion

I'm a quick learner but all the results I found about this came with the assumption that I already know my shit with linear algebra, which I don't.

I understand simple 3x3 matrix manipulation

Dot products and cross products

I'm a little iffy on what normalization is. I don't even know if I need normalization for the programming task at hand

Any insight into this would be mind-bogglingly relieving

[earnest ferry]

Any idea what's the answer to the b part of this question?

[lavish jewel]

for example, what can you say about the rank of A when t = 1?

how about if t = 0?

[icy orchid]

@tight stag quaternions are generally the standard for computing 3D rotations, so i'd recommend familiarizing yourself with how they are used to that end. There are a lot of good resources on them and they can be helpful for converting between representations

speaking of quaternions, i was wondering if anyone knows of any terminology to describe "planes spanned by a scalar and a vector" vs "planes spanned by two vectors"?

[lavish jewel]

wdym a plane spanned by a scalar and a vector

spans are defined for sets of vectors

[icy orchid]

the set of all quaternions of the form a+bu for some pure quaternion (vector) u

[lavish jewel]

how do you define the sum of a scalar and a vector?

[icy orchid]

what is your preferred notation for quaternions?

[lavish jewel]

ah i see the issue. for whatever reason the real component is also called scalar

that's pretty confusing

[tight stag]

I'll read more into them and get back to you here. But, just to clarify, are you suggesting that it may be necessary to convert from a triplet of points to a quaternion, to then convert the quaternion into a set of Euler angles?

[lavish jewel]

i would prefer denoting the quaternion as some 4d vector

then there is no confusion

[icy orchid]

that's actually my bad Edd. i have a bad habit of referring to the real part as such

[lavish jewel]

with that out of the way, there is no difference

the "scalar" is also a vector, e.g. some multiple of (1,0,0,0)

[icy orchid]

but, there is a difference! Span{1, u} forms a subalgebra isomorphic to the complex numbers, while Span{u, v} does not

[lavish jewel]

this is very application-specific because quaternions have additional structure

from the point of view of spans and vectors with no additional structure, they're the same

so you'd have to ask someone that works more with this stuff :x sorry

[icy orchid]

maybe not necessary, but it might be easier, and quaternions are helpful to know because they are so ubiquitous for 3D rotations

[wintry steppe]

How do i even go about this? translation: "determine the matrix that represents, in the canonical basis of R3, the projection onto the plan pi defined by the equation x+2y+3x=0 in parallel to the line D defined by the equation x/3=y/2=z

[lucid glacier]

Maybe i'm missing something but I don't see why Ei,i being in H suffices, since Ei,i itself is not invertible. Also not exactly sure I see why E satisfies f(E)=0, wouldn't it be

$$f(E)=f(E_{i,i}) - \frac{f(E_{i,i})}{n}f(I_n)= f(E_{i,i})(1 - \frac{tr(A)}{n})$$
?

[stoic python]

ShiN

[short magnet]

Oh my god yes damn that was stupid of me

thank you mate :""/ I'll see if I cam rework this method

[lucid glacier]

np

[wintry steppe]

can anyone help me with this?

[silver heath]

In matrices is ABCD = (AB)(CD)??

[limber sierra]

matrix multiplication is associative, so yes

but A((BC)D), A(B(CD)), ((AB)C)D, etc. also work

theyre all the same

[silver heath]

ok ty

[silver heath]

I know this is a tall order... but would someone be willing to proofread my homework? it's taken me like 4+ hours to type it up and don't have the energy to proofread

[wintry steppe]

you're doxxing yourself

[silver heath]

meh

i dont care anymore

but ill reload with no name

[earnest ferry]

can someone explain how to do this please

[lavish jewel]

think about the rank of A

[earnest ferry]

Rank of A will be n

[lavish jewel]

mhm

and what's the rank of A^T

[earnest ferry]

then rank of (A^T) should also be n

but how will I show that?

[lavish jewel]

actually i think that part isnt needed

for linear transf9rmations, if the arg is 0, the result is 0

[earnest ferry]

?

[silver heath]

I think he is trying to say that for every linear transformation T, T(0) = 0

[jade oxide]

this is an n-1 dimensional plane going through the origin. The basis will need n-1 vectors to span it and they will all be perpendicular to (1,1,1,1....1). I will leave it up to you to come up with these vectors

[drowsy flower]

Hey I have question on singular value decomposition

So I found the eigenvectors of V and normalized it

but the problem is it already does not have rational entries

I am not sure if I have made a mistake or if I am missing a step but how can I find a matrix with rational entries only??

[lavish jewel]

you can't just do an eigenvalue decomposition and expect the entries to be rational

[drowsy flower]

the question asks me to find matrices with rational entries so I am assuming it should be somehow possible for this case. I am not really sure how though

[lavish jewel]

i'm honestly not sure how this is done, but not with the regular method for sure

[wintry steppe]

im given this matrix and the questions says that i have to change the elements in the lower triangle area to make it symmetric. does this mean that i should add or subtract another matrix from this one?

[lavish jewel]

can you show the original question?

[wintry steppe]

Change the elements found in the matrix's lower triangle area so that it can become:
i) symmetric
ii) hermitian
iii) upper triangular

i just want to know if im supposed to subtract or add some other matrix from the original one given in the question

[glacial terrace]

Hey guys

this ok?

[lavish jewel]

if it doesn't say how... i'd honestly just replace the elements

[lavish jewel]

i think you're supposed to show the transpose distributes over the sum

[wintry steppe]

literally just change any number individually how i like?

doesnt sound very right, i thought id have to do matrix operations to get to those answers

[lavish jewel]

without any other context... i would say so, since they ask you to only change the elements in the lower triangular part

you can do it with matrices if you want, but you're anyway adding and subtracting lower traingular mats

[glacial terrace]

What I did is wrong?

[wintry steppe]

thx

[lavish jewel]

it's incomplete

i'd look at the elements of aA + bB, and those of (aA + bB)^t, and finally those of aA^t + bB^t

[glacial terrace]

wdym look at the elements?

[lavish jewel]

a matrix A has elements A_i,j

A^t has elements A_j,i

[glacial terrace]

I see

thanks

[wintry steppe]

is it possible to find the magnitude of a 1x2 matrix?

afaik in order to find the magnitude of a square matrix you need to find its determinant, but i dont think there is a way to find the determinant of a 1x2 matrix

[drowsy flower]

Sorry I missed your message. Yeah apparently in the hint given, we have since third eigenvalue is 0, then it belongs to Null(A^T A) = Span{v_3} and first eigenvalue = Null(A^T A - 9*Identity matrix) = Span{v_1, v_2} where v_1, v_2 are eigenvectors for 9 and v_3 is eigenvector for 0. So the suggestion was to find a basis {w_1, v_2} of first eigenvalue such that w_1 has rational entries and is linear combination of v_1 and v_2 and then apply gram schmidt on {w_1, v_2} and normalize it.
Quite hard to understand what my prof is saying tbh...

definitely not regular way yeah

[wintry steppe]

***The following vectors are given:
|f1>= ξ|e1 > +|e2>
|f2>= |e1 > +ξ|e2> +|e3>,
|f3>= |e2> +ξ|e3>

Where |e1>, |e2> and |e3> are orthonormal and ξ is a real number. For which values of ξ are |f1>, |f2> and |f3> linearly independent? For which values are they linearly dependent?***

Can someone help me out this this? I am wondering if I need to do this:

c1|f1> + c2|f12> + c3|f3> = 0

is this what i have to do? the problem is that i will end up with four scalars c1,c2,c3 and ξ

is that fine anyways?

[quaint sage]

I am trying to figure out a 3-D geometry problem, and I am not sure if I can even describe it correctly... I think it might be some kind of projection problem?

If I have a point (P1) and and a plane (K), how do I figure out if another point (P2) is within the boundary formed by the point and the boundaries of K (see picture below sorry for ms paint).

[lavish jewel]

yep, projection and change of basis

you can make a local coordinate system on the plane and do a change of basis from the global 3D to the local 2D

once in the local 2D coords, you can easily measure distances

e.g. if you have a local orthonormal basis, it's as simple as projecting the point onto it and seeing if the absolute value of the projection is <= some quantity

[quaint sage]

Okay I am kind of following you up to here. What is that "some quantity" part? Presumably it changes in 3-D space

[lavish jewel]

no, it depends only on the size of your square K there

say you want it to be W x L "units"

you can arbitrarily make a local coord system with 0 in the middle of that square or rectange

then you need the projections to be |<point, local_x>| <= W/2

and similarly for L

[quaint sage]

Just so I understand here.

You are saying projecting the point onto the plane, not "onto the local othonormal basis", is that correct?

[lavish jewel]

there is no difference between those 2

[quaint sage]

And then I am comparing distances as defined in the local coordinate system?

[lavish jewel]

yeah, that works

the only thing that changes is at what point you subtract the reference origin

so i would do it like this

trace a ray from p1 to p2

find the intersection of that ray with an infinite plane that includes K

and then using a reference origin for K and an orthonormal basis, find out if it is within the boundaries

[quaint sage]

ahh that is a bit easier to visualize

[lavish jewel]

that last step requires a change of basis from the 3D world where the rays and the plane live to a cleverly chosen 2D basis

an orthonormal basis makes this simple

[quaint sage]

wait

nevermind

[versed topaz]

someone on twitter asked something interesting

take $\mathbb{F} = \C$ or an arbitrary field if you're feeling special

[stoic python]

a sham, a rock, a canal, panama

[versed topaz]

find four 4 by 4 matrices $A, B, C, D$ over this field such that the matrix $X = aA + bB + cC + dD$ has rank 2 for every $(a,b,c,d) \neq (0,0,0,0)$

[stoic python]

a sham, a rock, a canal, panama

[versed topaz]

i do not know why they think this exists, you can probably do dimension counting or something

I guess to put my thoughts on paper, we have $68 = 4^2 * 4 + 4$ variables and equations saying that all the 3 by 3 and 4 by 4 minors vanish, that at least one of the 2 by 2 minors does not vanish, and that at least one of the a,b,c,d does not vanish

[stoic python]

a sham, a rock, a canal, panama

[versed topaz]

there are a total of $\binom{4}{3}^2 \cdot \binom{4}{4}^2 = 16$ minors, so we're intersecting 16 hypersurfaces and then taking an open subset cut out by the two equations $abcd \neq 0$ and $\prod_{m \in 2\times 2\text{ minors of } X} m \neq 0$

so i think this line of thinking shows that there must be a solution, probably (for F algebraically closed so dimension stuff works out)

but I don't actually know how to find one, and if anyone could help that would be great

hm no this doesn't actually work to show there are solutions does it

because the open equations could just completely cut out the equations saying the minors vanish

[stoic python]

a sham, a rock, a canal, panama

[versed topaz]

but that probably isn't the case

anyways if anyone has thoughts about how you can find solutions please ping me

or thoughts on why solutions exist

[versed topaz]

Nvm: https://twitter.com/ExceptionalLiar/status/1423413241263312897?s=19

E8 (@ExceptionalLiar)

@hayden_wlog @SC_Griffith I think this is asking for four 2x4 matrices (A,B,C,D), where aA+bB+cC+dD=[v//w], v, w in F^4, and (v=tw for some t OR w=0) iff (a=b=c=d=0). Is that correct? If so, I have some bad news for you...

Twitter · Aug 5, 2021, 10:38 PM

Solutions won't exist in the algebraically closed case

Well, maybe. As the op points out, this assumes some restriction on the form of A, B, C, D

[wintry steppe]

Hey guys need some help checking this... I think I did it right? But I got slight different numbers for the solutions. I know the end result is the same because there are no solutions but... not sure if I messed up somewhere ?

My lighting is terrible hopefully thats better

[nocturne jewel]

yes there's no solution if you get 0=not 0

also there are RREF calculators online, which is quicker than asking someone here to do it for you

[wintry steppe]

@nocturne jewel I guess where I'm confused is more so if its okay to get different values for the solutions.. like if I did rref on this system again and got completely different numbers again

[nocturne jewel]

each matrix has a unique RREF iirc, so check your working or look for an RREF calc w/ steps

[wintry steppe]

I did but the rref calculators I found dont take AUG matrixes

they only show the rref of the matrix alone and thats exactly the same as what I have

[nocturne jewel]

https://matrixcalc.org/en/slu.html#solve-using-Gauss-Jordan-elimination({{1,2,1,1,8},{1,2,2,-1,12},{2,4,0,6,4}})

[wintry steppe]

Oh sweet thank you

Hmm wait I have a slight confusion

I thought every pivot row had to have 0

i mean column

shouldn't it be [ 1 2 0 1 ] for the top row

[nocturne jewel]

I mean you're just solving the system, so you can stop once you know there's no solutions

[wintry steppe]

Ah okay , so you can ignore the "standard rref" if you see its no solution

[nocturne jewel]

you stop solving once it's solved lol

if you get to x=2 for example... you wouldnt do anything else cause you've arrived at where you wanted to be, the answer

[wintry steppe]

Well im confused cause my teacher never said that but kahn academy said for it to be proper rref every pivot column should have nothing but 0's and 1's

[nocturne jewel]

yes that's true

[wintry steppe]

....

lmao

[nocturne jewel]

BUT the question isn't get the RREF of a matrix, it's to solve the augmented matrix

[wintry steppe]

So its another case of ignore this now because we dont care

Then wouldn't it just be REF

[nocturne jewel]

Not really, it's just pointless work to get the exact RREF

[wintry steppe]

So its a gray area between REF and proper RREF because we dont care and just want an answer

[nocturne jewel]

ill be honest I forget the exact difference but there is one

I just go w/ Gauss-Jordan until I don't need to anymore

[little crater]

If you want to fix you would have to flip some of the variables around

i assume

[wintry steppe]

Thank you for the help. Trust me ik that the rules get confusing and contradictory in higher mathematics. Sometimes you have to write a super rigorous exact answer or you get an F. Then other times its like.... eh... we know this so we can assume x,y,z lmao

[nocturne jewel]

pretty much REF is just getting the bottom of the pivots knocked out, RREF is getting both top and bottoms knocked out ( as much as you can, google the exact difference lol)

[wintry steppe]

huh

[wintry steppe]

Yeah that seems right. I'm not sure again why it really matters other than making it easier?

[nocturne jewel]

[little crater]

If gives you x1 = the number doesn't it

[nocturne jewel]

Again I just run w/ GJ then stop when I'm done

[wintry steppe]

There's probably some fringe problems that require exact rref, but thats later

[little crater]

like you will divide it to make in 1 and all that

[wintry steppe]

I'm sorry but I am not understanding your English, no offense

[nocturne jewel]

yeah no clue what you're saying either

[little crater]

nvm i was talking about the REF thing but nvm

[wintry steppe]

Ah well thanks for trying to help lmao

[little crater]

like in the spots above and below the pivots you can just subtract off

[nocturne jewel]

yes that's gauss-jordan

[little crater]

yeah like this part

you could go further

and clear up all the numbers above i guess even if this is considered row reduced

[nocturne jewel]

that's not row reduced

that's REF

[little crater]

oh yeah

[wintry steppe]

In case you guys are curious. The main difference is that it is easy to read the null space off the RREF, but it takes more work for the REF.

I guess cause the upper and lower triangulars look nice its just pleasing to the eye , and slightly reduces the computation

Well... more computation during rref, but afterwards the system looks nicer

lmao

[viral patio]

Would anyone be able to link to me a video that explains linear transformations? I feel I need more information when it comes to calculating them

[little crater]

not sure what exactly you are looking for but there is this playlist https://www.youtube.com/watch?v=rHLEWRxRGiM&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab

[viral patio]

I have seen them. They are really good to understand what is happening. But it doesnt really go into solving the problems from what i remember. I am looking to solve problems like this.

[nocturne jewel]

apply linearity of T

[little crater]

tbh not that far kinda doing this on my own but would it not be something like this

[nocturne jewel]

$T[au+bv]=aT[u]+bT[v]$ for u v in the space and a b in the field

[stoic python]

Mosh

[little crater]

something like this maybe

idk if that is wrong or not

[nocturne jewel]

yes, that's called linearity like I said already...

[little crater]

idk never did any of this before

sooo

[nocturne jewel]

though dont post full solutions...

[little crater]

so is that basically linear transformation just a change of basis?

kinda idk

[nocturne jewel]

???

transpose is an operation, change of basis matrix is a matrix

[little crater]

oh

[little crater]

i meant linear transformation. isn't that what this is doing? Is that called a linear transformation right?

oh it says it in the title

[wintry steppe]

only invertible linear transformations are changes of basis

a non-invertible linear map doesn't take a basis to a basis so it can't be a change of basis

[wintry steppe]

[lime zinc]

in Sylvester rank inequality , when we have two matrices like A and B of order $m \times n$ and $ n \times m$ repectively then $rank A+rank B - ? \leq rank(AB) \leq min {rank A, rank B}$ what is ? here is it m or n?

[stoic python]

honey99

[dusky epoch]

n

[lime zinc]

so common term?

[dusky epoch]

sylvester's ineq says that if A is m by n and B is n by k then rk(AB) >= rk(A) + rk(B) - n

[wintry steppe]

${(x,0,0)|x\in\mathbb{R}}$ is a subspace of $\mathbb{R}^3$ right?

[stoic python]

jswatj

[wintry steppe]

im not crazy

0 is in there, x+y is in there and most definitely ax is in there

[teal grotto]

yea it’s a subspace. it’s the x-axis

[random axle]

Could someone explain what shearing is please? First time I've seen that term.

[limber sierra]

exactly what the images show

"bending space" along one direction

a shear is applied to transform the blue grid into the green one

which "bends" the shapes as well

[random axle]

ok, i see thankx

[lavish jewel]

maybe a visualization of this type can also help

[random axle]

So it seems like a rotation, but the points that are on the x-axis stay fixed (if you are shearing in the x-axis direction)?

[lavish jewel]

not a rotation, lengths aren't preserved

[random axle]

Oh i see

[wintry steppe]

how has nobody linked this yet

[dusky epoch]

what do you call a baby eigensheep?

[gray dust]

a lamb, duh

[random axle]

is that eigen stuff hard? it's a topic that crops up later in my chapter. I've noticed a lot of exam questions based on it

[dusky epoch]

eigenstuff is not hard in and of itself but it can be tricky to understand if you don't have a good grasp of the more basic stuff that comes before it.

[little crater]

How do we know what b equals in the case of yTb = 1

I just dont understand the right-hand side verfies that yTb = 1 statement

couldn't b = [1, 1, -1], [1, 0, 0], etdc

[nocturne jewel]

it's the same vector in Ax=b

[little crater]

ohh i guess it is saying if we are given b from the start?

oh yeah i see

it is from the x1 -x2 = 1 , x2 -x3 = 1, x1-x3 =1

[1,1,1]?

riht

[nocturne jewel]

of course you're given b

[little crater]

right*

[nocturne jewel]

you're given the system of m equations off the bat

[little crater]

yeah i see that now although not exactly sure what this question was getting at or showing

just says it is under the topic

[onyx cypress]

If det(U) > 1, can we prove that det(2*I-U^n) will become negative for big enough n? Or is this false. Any help would be appreciated

U is a real nxn matrix here

[wintry steppe]

oops

ignore that

[lucid glacier]

Is the n in U's exponent and its dimension the same n, or is U fixed?

I'm assuming the latter

Consider the case where the size of U is even, and all the eigenvalues of U are negative, then if n is even the eigenvalues of 2I-U^n might be negative, but for odd n the eigenvalues will all be positive, so det(2I-U^n) will be positive for all odd n in this case

Supposing U has all real eigenvalues in this case

[wintry steppe]

I know what i did wrong

i think

nvm

im confused

how would i turn a single vector into a line?

[lavish jewel]

wachu mean

like a line parallel to that vector?

[wintry steppe]

No

just a single vector

for example

$\begin{bmatrix}1\-1\3\end{bmatrix}$

[stoic python]

jswatj

[lavish jewel]

mhm

and what kind of line do you want

parallel to this?

passing through some specific point?

[wintry steppe]

oh

i guess parallel to this

this question makes no sense

[lavish jewel]

what's the original question

[wintry steppe]

so

[lavish jewel]

that's simple

what's the span of a single vector?

[wintry steppe]

null is a span?

spanning set*?

[lavish jewel]

null space is a vector space

it has a basis

[wintry steppe]

oh damn

[lavish jewel]

you found the null space to be of dim 1, yeah?

[wintry steppe]

yeah

[lavish jewel]

so it has a basis containing 1 vector

[wintry steppe]

yeah

wait

i think what i have is the basis

[lavish jewel]

"a" basis

not "the"

[wintry steppe]

oh right

a basis*

[lavish jewel]

it's easy to check if you have a basis for null(A)

Ax = 0

[urban moon]

hey guys i need help with this question

[wintry steppe]

Yeah i did that

and got a unique solution

sorry

not uniqe solution

only one basic vector*

[lavish jewel]

mhm

but you agree that any scaled version of that vector is still in the null space yeah?

[wintry steppe]

Yes

wait

[lavish jewel]

aight

[wintry steppe]

how do you know that

[lavish jewel]

well

linearity

Ax = 0

A (c * x) = c A x = c* 0 = 0

[wintry steppe]

Ohhhhh

okay

so it can be lots of vectors

[lavish jewel]

not "lots"

infinitely many

[wintry steppe]

Yeah

[lavish jewel]

any scaled version of x

now thing for a second what this is, geometrically

[wintry steppe]

A line

[lavish jewel]

mhm

simple enough. a line can be parametrically described given a point on it and a direction that the line is parallel to

the line contains zero, since if we set c = 0, then c x = 0

so a line of the form p + c x, with p = 0, is simply c x

a line parallel to the vector x and passing through the origin

this is what you'd expect, since the null space is a subspace and needs to contain the origin

[wintry steppe]

right

[lucid glacier]

Solve 4 equations with 3 unknowns

[blissful vault]

Can someone critique my proof writing skills + correctness?

[wintry steppe]

wouldn't it just be that $p=tv$ where $v$ is the vector we have?

[stoic python]

jswatj

[wintry steppe]

since we can use the origin as p_0

[lavish jewel]

ur using a different notation from the one i picked, but i guess so

i used p as p_0 and c x as t v

[wintry steppe]

Oh sorry

[urban moon]

lambda 1,2,3 adds up to 0
so i dont think its as simple as solving 4 equations with 3 unknowns

[lavish jewel]

where do you get that the lambdas add up to 0?

[lucid glacier]

Why would they need to add to 0?

[wintry steppe]

is the im(A) a vector space?

[lavish jewel]

yes

[lucid glacier]

Yes

[wintry steppe]

where A is a matrix

[lavish jewel]

you can write A x as a linear combination of the columns of A, so it is the subspace spanned by the columns of A

[lavish jewel]

why do you say the lambdas add up to 0?

[steep harbor]

Need some help with the proof of spectral theorem

I'm trying to understand how PMP = lambda P

[lavish jewel]

that doesn't look right

[steep harbor]

Where M is a normal operator on V dim 1, lambda is and eigenvalue of M and P is the projector onto the egienspace

[urban moon]

i mean the lambdas couldn't be 0s

[lavish jewel]

all right. at any rate, you'd really just try to solve the system

you have a 4 x 3 matrix, so you'll end up having to parameterize something

[urban moon]

how do i do that?

[forest quiver]

IM a bit confused by this

I understood this at the time, but looking back I don't quite understand

If A has a pivot point in every row, doesn't that mean any augmented matrix is always consistent?

[last holly]

but this is saying if [A b] has a pivot

in the warning

[forest quiver]

wht do you mean?

OHHH

I think I get what you mean

[last holly]

ya

[forest quiver]

Wait

do you mean this:

[stoic python]

Tim O'Brien

[forest quiver]

b= [ 2 3 7 ]

but wait IM confused again

idk man fuck

[last holly]

lol

draw out some cases

ugh im bad at latex i should figure out how to do this

one sec

[forest quiver]

could you draw on paper ?

[stoic python]

pitabread

[last holly]

third column is b

what do you think of this

does the augmented column have a pivot?

and is it consistent? (considering col 3 is b)

[forest quiver]

There is not a pivot in every row

of A

sorry for lat ereply

reply

[last holly]

of A, but does the matrix [A b] have a pivot?

[forest quiver]

No it doesnt

[last holly]

you don't see/consider the 3rd row as a pivot?

[forest quiver]

Oh I mean it's a pivot it's just not row reduced echelon form

I guess I don't get what pivot means then

[last holly]

i mean the text is just talking about pivots when it talks about pivots

[forest quiver]

right

what is a pivot exactly?

Like whatws the definition

Just a term in a matrix with 0's below it?

[last holly]

informally it's the first non-zero entry in a row

i think it has to be unique too (column wise)

[forest quiver]

if its unique in a column that means row reduced echelon form

[last holly]

not sure about my last statement

[forest quiver]

I think first non zero entry in a row is good

[last holly]

yea

[forest quiver]

So let me try to understaned the theorem once more

[forest quiver]

this is inconsistent

[last holly]

correct

[forest quiver]

Despite having pivot points in each row

[last holly]

correct

[forest quiver]

so if Ax=b has a solution for every b, A has a pivot point in all rows

[last holly]

and if you find some example of a consistent one, there you have your two cases for [A b]

[forest quiver]

Ok wait so

[last holly]

it's not always inconsistent

when the pivot in the augmented column it is always inconsistent

in your example it would be consistent

(probably)

[forest quiver]

For Ax=b to be consistent, there has to be a pivot point in every row of A. But the fact that
[A b] has a pivot point in every row doesn't guarantee consistency.

Is this right?

[last holly]

hm it's not strong enough, your text is good in this

and also it's important to note it's talking about mxn

non square matrices

[forest quiver]

ugh

[last holly]

it's close

the reason i say it's not strong enough is that that's only one condition for consistency

[forest quiver]

Yeah I get that

[last holly]

actually

hm

it says a b c d are equivalent...

in which case you'd be right full stop

but then you'd have to be able to prove a <-> b <-> c <-> d

[forest quiver]

not looking for 100% proving everything at the moment

[last holly]

lol

[forest quiver]

Just want to understand

[last holly]

yea i get that

[forest quiver]

I think I do

[last holly]

i think you do

lol linear alg deep tho

so i don't

[forest quiver]

Yeah ikr

[last holly]

i always hear, "you can never know enough linear algebra"

[forest quiver]

Yeah

I am doing some Group Theory too and the crossovers are nice

[last holly]

ah yea

artin?

[forest quiver]

let me look

no not artin

But yeah I want to be fluent in Matrix Algebra to start Group Theory

but thanks for the chat I think I understand it?

Like 50% sure

[last holly]

lol that's better than 0

np

any time

[wintry steppe]

"you can never know enough linear algebra" only applies to useful linear algebra, i.e., not matrix reduction stuff :^)

[forest quiver]

Unfounded claim

[wintry steppe]

what do you mean "unfounded?" it came from me so it's fact

[forest quiver]

God has entered the chat

[tawdry sage]

Hi guyes
I have somehow good linear algebra back ground and i want to put it in practice via solving problems
So please some one introduce me to a good source for it (it is better to be a website or something)
:)

[last holly]

yea let the computers deal with matrices

[forest quiver]

Check out "Linear Algebra and its Applications" by David C. Lay

It has some cool problems

at least to a math n00b like me

[wraith patio]

all row swapping elementary matrices are symmetric right

[celest slate]

idk if this is where this goes

I’m learning about tensors and I see this

All of them have the transformations on the left-hand side of the tensor, except the (1,1) tensors which have one transformation on the left and one on the right

then this rule is given

and the issue here is that for the (0,1) and (0,2) tensors in the top chart, the forward transforms are on the left instead of the right like this rule suggests

so my question is, how do you know whether to apply transforms to the left or right side of a tensor given its type?

If I have a (1,2) tensor, for example, I know it’s got two forward transforms and one backwards transform when I change coordinates, but how do I know where to put them?

[celest slate]

I’m still looking for an answer btw

[wintry steppe]

***You are given the linear system:
kx-y=1
4x-ky=n

For which values of k and n does the system have a single solution (x,y)? For which does it have infinite solutions? For which is it impossible to solve?***

As you can see I haven't done any attempts on finding the values for infinite solutions as I'm rather unsure. Is my method correct for the rest?

[last holly]

in terms of infinite solutions, try graphing something with a single solution, and then try to imagine what an infinite solution answer between two lines would imply

[wintry steppe]

Which two lines?

[last holly]

i'm just putting in two equations here to highlight what it means graphically for a system of 2 equations to share a unique solution:

(you can verify that (1,2) is indeed the solution to the system)

[wintry steppe]

Mhm

[last holly]

it's also useful to think about the relationship the lines would need to have to have no solution

(no point of intersection)

[wintry steppe]

Parallel

I don't remember if there is a way to express parallel vectors through an equation though

[lavish jewel]

you can look at the dot product of the normalized vectors

if it is 1 or -1, they're parallel

[celest slate]

can someone help me with my question

[small glen]

Can I get advice on how to solve this question? No answer needed.

[lavish jewel]

i'm not super familiar with that, lorem, sorry

[celest slate]

what channel should I ask in?

[winter flume]

wait what actually is linear algebra?

is it just a more graphic interpretation of concepts?

[hollow finch]

check the pins. namington has a good explanation.

[lavish jewel]

this channel is the right one, lorem, but people that might know are not around at the moment

[wintry steppe]

And they are orthogonal when it's 0 right?

[lavish jewel]

mhm

[silver heath]

Umm.... for this problem

Aren't they all true????

I mean d is definitely true.....

a,b,c can be true???

Am I missing something obvious?

<@&286206848099549185> ?

[stable kindle]

my linalg is meh but they all seem true??

idek

[wintry steppe]

why is a true?

rather

can you explain your answer to each of them?

i'll just give an example

{(1, 0), (0, 1)} is a spanning set for R^2 which does not contain a basis of the subspace spanned by {(1, 1)}

[silver heath]

well

wait wut

OHhhhhhhhhh

[little crater]

should it have said for all V and H maybe?

or like any V and H or something?

[steel river]

a is false I think

[little crater]

because i mean if you said (1,0) (1,1) is the set for R^2 but h happens to be a line in the direction (1,0)?

[wintry steppe]

?

i don't know what you mean

a basis of span{(1, 1)} has to consist of a single vector which is a multiple of (1, 1), and neither of the vectors in {(1, 0), (0, 1)} are of that form

[steel river]

Yeah this counter example seems right

[wintry steppe]

it is right

[little crater]

yeah i get that

so i guess it is implied for any H in V?

[steel river]

Yep

[little crater]

okay yeah then i see that

[steel river]

Yeah if the question went like "some spanning set" for option a then it will be true @little crater

[wintry steppe]

well

perhaps the wording in a is unclear (hell, in a, b, and c really)

experience tells me it means "for all" but it can definitely be interpreted as "there exists" regarding spanning sets

[steel river]

Yep

[little crater]

can someone explain the last part with "I − P projects onto the....."

like i wasn't sure how to know what space (I-P) was projecting onto

add information if you want to explain I think I might just need to sit on it and look over this post https://math.stackexchange.com/questions/2507116/i-p-projection-matrix?rq=1

[torn stag]

Well $P$ is the orthogonal projection onto $R(A)$, the range of $A$.

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

So $I - P$ is the orthogonal projection onto $R(A)^\perp = N(A^*)$

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

I guess $N(A^*)$ is what Strang calls the left null space

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

The general fact is if $P$ is the orthogonal projection onto a subspace $W$, then $I - P$ is the orthogonal projection onto $W^\perp$.

[stoic python]

IlIIllIIIlllIIIIllll

[little crater]

what is range?

this is how the spaces were defined

[torn stag]

Yes $C(A) = R(A)$

[stoic python]

IlIIllIIIlllIIIIllll

[little crater]

oh

[torn stag]

range is defined for any function, so I was applying it to the function $A \colon \mathbb{R}^n \to \mathbb{R}^m$

[stoic python]

IlIIllIIIlllIIIIllll

[little crater]

is that big p or little p?

[torn stag]

$P$ is capital P I think.

[stoic python]

IlIIllIIIlllIIIIllll

[little crater]

so when you make these projection matrix they are always square?

[torn stag]

Yes a projection maps a vector space $V$ to itself. So it is (represented by) a square matrix.

[stoic python]

IlIIllIIIlllIIIIllll

[silver heath]

well it says can... so it technically is still true....

[wintry steppe]

which message specifically are you replying to

[wintry steppe]

^this message later may clarify things

you should take the former interpretation for this problem

it's poorly worded

[silver heath]

hmm...

[silver heath]

i mean im pretty sure they can all be true in the correct circumstances....

[torn stag]

a is false

b c d are all true

[wintry steppe]

a is false when you interpret it as:
for all f.d. vector spaces V, and all subspaces H of V, *every* spanning set of V can be reduced to a basis of H.
it is true, however, when you interpret it as:
for all f.d. vector spaces V, and all subspaces H of V, *some* spanning set of V can be reduced to a basis of H.
the problem is that the exercise uses the article "a" to mean every when it is also correct to interpret it as "some." the wording is unclear. nevertheless, you should pick a to be the false one, since the others are true no matter the interpretation of "a"

if they wanted to explicitly say that some spanning set could be reduced to a basis of H then they probably would have said that explicitly ("there is a spanning set of V such that...")

of course, a is true in some circumstances. what if V = {0}? but you need to interpret the problem as quantifying over all vector spaces

and i gave an example to show that there's a vector space + subspace + basis for which it fails

i've typed too much

[stoic stag]

Can an orthogonal column matrix have A'.A= D, where D is a diagonal matrix? Or is it always normalized to an identity matrix?

[lavish jewel]

no need to normalize it

easy example, take A = 2*I, where I is any identity matrix

A'.A = 4*I

[stoic stag]

And how would this relation hold if A' is replaced by (A')*

I'm trying to prove something and I have never taken linear algebra before so sorry if these questions sound repetitive or dumb 😅

[lavish jewel]

it would be faster if you say what you're working on

you can conjugate the matrix i used as an example and nothing changes 😛

[stoic stag]

Could you please take a look at this and let me know if it's correct or even comprehensive at all?

[stoic stag]

Ah okay

[lavish jewel]

you didn't prove it

[stoic stag]

Ah crap

[lavish jewel]

i would suggest looking at the elements of A^H A

(H as in complex conjugate transpose or hermitian transpose)

[stoic stag]

I do know that the elements of A^H have real diagonals right?

Or am I wrong?

[lavish jewel]

sure but that doesn't matter here

wait

no

we still don't need it, but it's also not true

[stoic stag]

That's for hermitian matrices

[lavish jewel]

you can make up any matrix you like out of real or complex numbers, and then take the hermitian transponse or whatever you wanna call it

[stoic stag]

Ah I think yeah understood not for the conjugate transpose

[lavish jewel]

my hint is to write the matrix in terms of its columns for the first case, and in terms of its rows for the second

and recall that the dot product for vectors in C^n is v^H v

[stoic stag]

It would be I right? Since the column elements are orthogonal?

And that would follow that rows are orthogonal as well?

[lavish jewel]

what

you're mixing everything up

the two things are completely separate

and it also wouldn't be I

for the same reason i gave you earlier

you can have orthogonal vectors that don't have unit norm

they're even telling you it's diagonal, not an identity

[stoic stag]

Okay....I think I lack major understanding of this subject sorry 😞

[lavish jewel]

yep, i would recommend reviewing first

matrix multiplication, representing matrices with column and row vectors, dot products, orthogonality

that sort of stuff

[stoic stag]

Okay thank you so much for your help though 😄

[glossy sage]

Hello, so, I’m supposed to solve these linear systems, yet I can’t figure out why number 36, for which complete and coefficient matrices SEEM to have different ranks, still has solutions.

[dusky epoch]

are you sure you calculated those ranks correctly?

[glossy sage]

That's the first thing I'm unsure about in fact

[dusky epoch]

yeah well arithmetic mistakes happen to the best of us

#36 is an overdetermined system. if you disregard the third equation and solve the system formed by the other two you will get a unique solution, and said solution happens to also satisfy the third equation and so is a solution for the whole system

[glossy sage]

The thing is, the full matrix (so the matrix that includes the second member) has a non-zero determinant.

So, 3 is its rank.

If I remove the last equation, that would make the rank 2.

Unless I can calculate ranks for non-square matrices, but that is pseudomath.

According to the Rouché-Capelli theorem, if the rank for both matrices isn't the same, the system has no solutions.

[dusky epoch]

lmao what

rank is defined for any matrix, square or not. it's not pseudomath

[dusky epoch]

are you sure of that?

[glossy sage]

Maybe I'm mistranslating ranks. I'm studying them on an Italian textbook.

[dusky epoch]

,w determinant [[1,2,2],[2,1,3],[4,5,7]]

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Ann#0413

[glossy sage]

That was the one I was sure I had calculated right. Oh well.

Thanks.

[little crater]

2 of the first row + 1 of the 2nd = row 3?

[glossy sage]

It depends XD. Do you understand the concept of Gaussian elimination?

[little crater]

i was just talking about the example Ann has

looks like you really only have in that case 2 linearly independent vectors and the other 1 is just a combination of the others

[glossy sage]

This is a complete matrix. The first two columns are the coefficients, the last column is the second member.

[glossy sage]

It summarises exercise 36 here.

By the way, both my math books absolutely ignored square matrices when explaining how to determine them.

No wonder I am left like this.

[little crater]

yeah like for every unknown variable you should have at least 1 equation

which is i guess where the whole square matrix things come into play

like for x,y,z you would want 3 equations

but in 36

you have 3 equations and only after 2 unknowns so you only needed the 2 equations i believe

[glossy sage]

Yeah, that's a good rule of thumb

👍

[little crater]

Are you familiar with projections of vectors onto sub spaces?

[dusky epoch]

if you wanted to ping me out of the blue, you could've at least pinged me directly instead of replying to an irrelevant message.

[little crater]

Oops

[dusky epoch]

you shouldn't ping people out of the blue anyway.

if you have a question, post it here.

[little crater]

Okay...

So so we have like a vector [2,1] would we always with regards to projections and subspaces say that the sub space would be like [ {2,1}, {0,0}]?

actually i think i might need to sit on it to word it better

[nocturne jewel]

it's a linearly dependent set, not "2 independent vectors and 1 other one"

[little crater]

isn't that the same thing?

2 indepent vectors with the 3rd being dependent on the other 2

[nocturne jewel]

it's a dependent set

cause all 3 are lin combs of the other 2

[little crater]

yeah

isn't that what i said?

i mean i guess that is the correct term to call that

[wintry steppe]

[lavish jewel]

how do you choose which one the 3rd is?

[little crater]

yeah i mean you don't do you

[lone stirrup]

It's just a naming convention. Dependent set of vectors <=> at least one is linearly dependent on the others.

[nocturne jewel]

Yeah i was pointing out that we just call it a dependent set

you can reduce the set to an indep set by removing one of the vectors of course

[little crater]

So im trying to work out that whole cross product thing and so say if you have 3 vectors U, V, A which are in r^3 how do I relate the find the the part where vectors U, V share the same left null space

Like basically A is that vector you get when you do the cross product of U, V

but I was trying to work off the idea

V transpose A = 0 , and U tranpose A = 0

but how do i find the intercept of said left null space <@&286206848099549185>

graphically it looks like it is the interception of the 2 planes from the resulting left nulls space of each vector

[still lodge]

a system of equations can be though of as a matrix by just taking the coefficients and sticking them in a matrix then multiplying by the column matrix of variables

[little crater]

oh so i need another system dont i

[still lodge]

oh i was about to ask my own question

[little crater]

oh

[still lodge]

what is was gonna ask is just when thinking about a matrix as a representation of system of equations, then what is the column space

[little crater]

well I believe it is thought of as the span of the vectors in that matrix

[still lodge]

bc i understand that it's just the span of the columns, but what's the point of treating all the x components of a system as a single vector

[little crater]

basically what is even reachable given the vectors

[still lodge]

i should probably start typing messages as one big thing rather than a few small ones

[little crater]

do you have an example system

say maybe of 3 equations

[still lodge]

im watching the strang lectures

fuck i forgot the latex for matrices one sec

[little crater]

here is an example i geuss

guess

[still lodge]

yeah that sure

[little crater]

basically looking at it is just the span of the columns

i believe

basically what all those 3 vectors could possibly reach

in this case

[still lodge]

columnspace is R^3 since theyre all linearly independent i know that

[little crater]

well in our example they are not actually linearly independent

hard to show but from the image

they all sit on a plane

so they really only span a plane in R^3

but basically the column space is just what all possibly places through linear combinatons of adding the vectors together that you can possibly reach

well and scaling and substracting

[still lodge]

ight imma let it marinate for a bit then, merci

is that desmos btw?

[little crater]

https://www.geogebra.org/

they have the 3d calculator

[little crater]

this might also help https://www.youtube.com/watch?v=k7RM-ot2NWY&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=2

[still lodge]

yeah i havent gotten to his colspace video yet so im just gonna stick it out till then

merci

[little crater]

there is also this

https://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/

https://www.youtube.com/watch?v=8o5Cmfpeo6g&list=PLCpgggEg1JgqgRR7KztwkbYrwIi8AIla4&index=6

[still lodge]

exactly why im asking

[still lodge]

.

[little crater]

oh what are you studying it on your own

[still lodge]

yessir

[little crater]

same

;/

lecture 15 so far...

[stable kindle]

k = an + b
g^an exists in U because it's just (g^n)^a, closedness
then g^-an exists in U, inverses
g^(k mod n) = g^k * g^-an, so by closedness it's in U

wait this is #linear-algebra ?

[faint dune]

Ok I got it thank you! Also sorry for wrong channel, will delete in 5 min.

[silver heath]

Uh... I think this is true... but when i tried to write it out with the definition of similarity it didn't work out am I doing something wrong???

[stoic python]

Elonmosqito96

Elonmosqito96

[silver heath]

I tried expressing P as some change of coordinate matrix

but I am getting confused with the D

[lucid glacier]

This is not true, every invertible matrix is the change of basis matrix for some bases (In fact, it's can be the change of basis from some basis into the standard basis in F^n, can you see why?), but not every invertible matrix is diagonalisable

[silver heath]

I see

ty

[wintry steppe]

is this correct?

[wintry steppe]

<@&286206848099549185>

[teal grotto]

rank is the dimension of the image of A and the image of A is the span of its columns

so the image is a subspace of R^7

[wintry steppe]

oh shit

[teal grotto]

the null space is going to be a subspace of R^4 because of the dimensions of the matrix

and you can use rank nullity to figure out the dimension

the row space should be the span of the rows, which has the same dimension as the span of the columns of A, but is a subspace of R^4

[wintry steppe]

so im(A) is a subspace of R^m and null(A) is a subspace of R^n?

[teal grotto]

for an m by n matrix yea