linear-algebra

[dim venture]

then I'll think of uniqueness afterwards

[native rampart]

Write PA=B and P=BA^-1 and show P is orthogonal. (A and B are matrices with columns a1,a2.. and b1,b2..)

[dim venture]

I don't follow how the statements PA = B and P = BA^-1 show P is orthogonal

[native rampart]

You have to show that separately

Show (BA^-1) (BA^-1)t=I

Using B Bt=I and A At=I (which implies A^-1 (A^-1)t=I)

[dim venture]

this is plenty to work with, I think I can show orthogonality

[native rampart]

Yes

[dim venture]

B Bt = A^-1 * (At)^-1
B At = A^-1 * (Bt)^-1
B At Bt A = I
(B At) * (B At)t = I
P * Pt = I
Pt = P^-1
therefore P is orthogonal

not sure if I did extra steps since matrix multiplication is not commutable

but in this case I think it is commutable

[native rampart]

2nd step is not correct
You will get B=A^-1 (At)^-1 (Bt)^-1

[dim venture]

B Bt = I
B = (Bt)^-1
B A^-1 = (Bt)^-1 A^-1
B A^-1 = (Bt A)^-1
(B A^-1) (Bt Att) = I (since (At)t = A)
(B A^-1) (B At)t = I
(B A^-1) (B A^-1)t = I (since A At = I and hence At = A^-1)
P Pt = I
Pt = P^-1

[strange crystal]

Does anyone have a good resource to explain the multilinearity of the determinant?

My professors explanation isn't making too much sense to me

[native rampart]

You might like hoffman kunze

[strange crystal]

Dang this book looks a lot better laid out than ours

ill give it a look

[native rampart]

B Bt = I
B = (Bt)^-1
B A^-1 = (Bt)^-1 A^-1
B A^-1 = (Bt A)^-1
(B A^-1) (Bt Att) = I (since (At)t = A)
(B A^-1) (B At)t = I
(B A^-1) (B A^-1)t = I (since A At = I and hence At = A^-1)
P Pt = I
Pt = P^-1
@dim venture (ABt)^-1 in fourth step

[dim venture]

wait what

[native rampart]

But yea,(AB)t being Bt At compensates for that

[chilly solstice]

Sorry for jumping in.

Have I applied the binomial formular correctly here: (from down to up)?

[native rampart]

Yea,it's fine

[weak patio]

Can someone help me with a)

someone help me please

my english is not good im sorry

[native rampart]

For a) 1(v1)+(-1)(v2)+(-1)(v3)+1(v4)=0 ,i.e.,nonzero scalars exist such that c1v1+c2v2+c3v3+c4v4=0

[sleek veldt]

could someone let me know why this justification holds? was just copying down notes and im not sure why this proves the set is linearly independent

[native rampart]

You say c1 sin(x) + c2 (cos (x)) =0,where 0 here is the 0 function.(i.e,f(x)=0 for all x)

So rhs is 0 if you substitute x and lhs becomes c2

So,c2 has to be 0

Similarly c1 has to be 0

[sleek veldt]

right but what does proving c2 = 0 for x =0 and c1 = 0 for x = pi/2 show?

[native rampart]

Do you remember the definition of linear independence?

If c1 f1 +c2 f2=0 then li implies c1=c2=0

[sleek veldt]

so if i can force c1 and c2 to be 0 for two different values of x then ive proved the set is li?

[native rampart]

well,you are showing the function defined by c1 f1 +c2 f2 is the zero function

So if the resultant function gives 0 for all values of x,and if you get c1=c2=0 is the only case that can happen,you have shown li

[sleek veldt]

ah okay thank you so much:D

[strange dove]

can someone help me with this

[spiral star]

hint: use absolute homogeneity

@strange dove did that help? :p

it just means that for norms you have $ \norm{\lambda x} = \abs{\lambda} \cdot \norm{x} $

[stoic python]

Flow:

[strange dove]

uh im still thinking

so for example if b = (1,0,0) and lambda = 3, the above statement would be false right?

[spiral star]

yes

if you take any x such that |x| = 1, then x is in B but something like 2x is not in B because |2x| = 2 |x| = 2 * 1 = 2

[strange dove]

is it really that simple?

[spiral star]

yes

takeaway: closed balls are usually not subspaces

[strange dove]

damn, i just wrote the midterm for this course and it was one of the hardest tests i've ever written but they're giving me questions like these?

[spiral star]

maybe it wasnt difficult :p

[strange dove]

that's true, or maybe i'm just not confident enough

i guess i'll find out when i see my mark and the average

[spiral star]

;)

[strange dove]

thank you btw @spiral star

[warm niche]

I think I asked this question but I didn't get a full answer

but when testing for linear independence why do we test if there exists only a unique / trivial solution to the linear combination of the vectors equal to the zero vector

and not just some abritrary vector?

[native rampart]

Because that ultimately reduces to checking for the zero vector

[warm niche]

Cuz I've seen a lot of definitions, define lienar independce with the zero vector in the definition

[native rampart]

Let c1v1+c2v2=a and d1v1+d2v2=a be such that c1 is not d1 and c2 is not d2 simultaneously

This would imply c1v1+c2v2=0 has a non trivial solution

[warm niche]

Correct

the only way that its true is if c1 and c2 is 0

which implies the zero vector

[native rampart]

So, It's the same condition

The converse is easy to show

[warm niche]

but why can't we extend this definition to any vector though?

[native rampart]

You can

[warm niche]

or am I missing the point

[native rampart]

But that's too much work,rather than just doing it for zero

[warm niche]

Ye thats what I thought

[weak patio]

Is B, C are the only ones true

[warm niche]

So linear indepence implies that there is a unique solution for any vector in its vector space

[native rampart]

Yes

[warm niche]

ok

[native rampart]

*unique representation

[warm niche]

but we just use the zero vector for simplicity ok

that makes me feel better lol

cuz it always felt weird why we just use zero, (I had the idea that its just ez to prove using zero but thx for validating me intution)

for example (a) is the vector space R3?

@weak patio

[weak patio]

yes

[warm niche]

oh wait

A is false

thats what u said nvm

[weak patio]

is b and c the only one that is true?

[warm niche]

I think c might be false

because u dont know if z is independent or dependint on x or y

I think B and D are true

but u might want someone else to check that

[weak patio]

Ok 👍

[warm niche]

for D if u know that x and y are dependent

no matter how many additional vectors u add

it will always be dependent because of the x and y vectors

[lucid cedar]

@desert robin That should prolly be in #prealg-and-algebra move the question over there and I can answer it

[desert robin]

whoops mb

[dim venture]

let a and b be matrices. If ab = 0 and b =/= 0, does that imply a = 0?

where a and b are both nxn

[wintry steppe]

no

[broken belfry]

Can anyone confirm that this will result in 7^n?

[golden drum]

let a and b be matrices. If ab = 0 and b =/= 0, does that imply a = 0?
@dim venture

Imagine the matrix with $a_{1,1} = 1$, and $a_{i,j} = 0$ else. And the matrix with $a_{n,n} = 1$ and $0$ else

[stoic python]

Enigsis:

[royal ore]

Does anyone know of an example of a vector space with an infinite set of linearly independent vectors?

[vapid kettle]

can someone help me with geomatry

[half storm]

@royal ore the vector space of polynomials

Any vector space of the form C([a,b])

There are a few. You won't be able to think of any for Euclidean space; all those spaces are finite dimensional.

Gotta branch out into more abstract spaces.

[wintry steppe]

please help me I do not understand this at all

I don't know how to do this

[limber sierra]

this is not linear algebra

try #prealg-and-algebra , #precalculus , or one of the ten questions-_ channels (as long as theyre not occupied)

[wintry steppe]

my bad sorry

[rotund jetty]

How do I find the dual of an optimization problem that isn't in standard form?

Like I know the dual of the problem

Subject to Ax >= b ```
Is 
```Maximize b^{T}y 
Subject to A^{T}y = c, y >= 0```
I've also seen specific examples of different cases, but I'm confused about how to do this where the problem doesn't "look like" Ax >= b and is in a more general form (no specific numbers, just vectors and matrices)

For example, what about something like 

```Minimize c^{T}x subject to 
Ay + x >= b
z^{T}y >= d 

Or something like that, where you have extra variables - what do you do in that case?

[rotund jetty]

<@&286206848099549185>

[golden drum]

Does anyone know of an example of a vector space with an infinite set of linearly independent vectors?
@royal ore

$\mathbb{R}$ as a $\mathbb{Q}$ vector space

[stoic python]

Enigsis:

[wintry steppe]

I have a linear map T: Rn -> Rn. I need to show that if there's a nonempty open set S such that T(S) is also open, T is bijective. I get that you can get around the open set thing by just considering an open ball, but I don't know what to do past that.

[thorn robin]

@wintry steppe What do you know about the image of a linear map?

[wintry steppe]

Well, I know that T is continuous so if T(S) is open then S is also open

[thorn robin]

Not really what I'm looking for, I'm only asking about the linear structure

The image of a linear map is....

[wintry steppe]

I'm confused by what you're asking

[thorn robin]

It's a linear subspace

Right?

[wintry steppe]

Oh, yeah that's true

[thorn robin]

Okay

So you know that the image is a linear subspace, and also that it contains an open ball

Do you have a guess about which linear subspaces of R^n contain an open ball?

[wintry steppe]

well, in order to contain an open ball of dimension n, it would have to be itself, right?

[thorn robin]

It will have to be all of R^n, yes
If you want to see why, just take the center of the ball v_0 and n vectors v_1,...,v_n that are also in the ball but in different directions from the center, and then since the image is a linear subspace it contains the n vectors v_j - v_0 which are linearly independent

Does that make sense?

[wintry steppe]

Yeah, I get that

[thorn robin]

Okay good, so the image is all of R^n

Is that good enough to show that it's bijective?

[wintry steppe]

We need to show it's injective and surjective, right?

If the image is R^n, that would mean it's surjective?

[thorn robin]

Yes

Is it also injective?

[wintry steppe]

oh, yeah

ah

that's neat

wait hold on

is it injective

Wouldn't it have to be because it maps from R^n to R^n

[thorn robin]

Yes, there's a big famous theorem for this

[wintry steppe]

what would that be?

[thorn robin]

Rank-nullity theorem? I assume this is what you were referring to with Wouldn't it have to be because it maps from R^n to R^n

[wintry steppe]

oh fuck right

[thorn robin]

Right so you're done

[wintry steppe]

that's cool

[wintry steppe]

help im despo

[robust pond]

despo?

[frosty vapor]

despoceeto

[gray dust]

use #prealg-and-algebra

[midnight hedge]

Can someone explain what this notation means

is it just saying Ax is a linear transformation

[gray dust]

it's a function that sends x to Ax

[midnight hedge]

so the transformation is the function that transforms x to Ax

so it would mean x transforms to Ax via a function operation

[gray dust]

another example, $x\mapsto x^2$ is a function that squares its inputs

[stoic python]

RokettoJanpu:

[midnight hedge]

okay interesting. Is there a name for that symbol

[gray dust]

it's the same as saying "a function f defined by f(x)=x^2 for all x", i can just speak of the function w/o going through the trouble of naming it

[midnight hedge]

ok then x --> Ax
is saying. define a function f by f(x) = Ax

something like that

[gray dust]

don't focus on the name, this is a way to refer to a function without needing to name it, it's all about how the function maps inputs

[midnight hedge]

the function is just Ax then right and the variable is just x

[gray dust]

no Ax isn't the function

[midnight hedge]

but in your example x --> x^2; wasn't x^2 the function

[gray dust]

the whole thing "x |-> Ax" denotes a function that sends an arbitrary x to Ax

[midnight hedge]

okay so its not a function. its denoting a function without saying a function explicitly

[gray dust]

it denotes a function by giving a rule for how it acts on inputs, without giving it a name

[midnight hedge]

and the rule is that the vector x is multipliced by the matrix A?

[gray dust]

yes any arbitrary x is sent by the function to the product Ax

[midnight hedge]

what do you mean by sent by the function?

is it that the arbritary x becomes Ax through operations from the function

[gray dust]

the product Ax is the output of the function for any x you input into the function

[midnight hedge]

oh okay, yeah that makes more sense

[gray dust]

we phrase that as "the function maps x to Ax"

[midnight hedge]

okay thanks for the explanations

[gray dust]

you're welcome

[half forge]

can someone help me with latex

i wannna add a matrix here

We need to show that $S$ is a subring of $M(\R)$.First, we need to prove if $S$ is nonempty, such that
\begin{align*}
    A\cdot \textbf{0} &= 
\end{align*}```

[wintry steppe]

use

\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}

to get matrices in latex. separate the entries of each row by &, and go to a new column using \\

[half forge]

i wanna write this

[wintry steppe]

so put two matrices next to eachother

[half forge]

i want that whole matrix equation in the middle

okie

[wintry steppe]

\begin{pmatrix}
a & b \\
c & d
\end{pmatrix} \begin{pmatrix}
e & f \\
g & h
\end{pmatrix} 

[floral thistle]

$A \cdot \mathbf{0} = \begin{bmatrix}1 & 1\ 1 & 1 \end{bmatrix} \begin{bmatrix}0 & 0\ 0 & 0 \end{bmatrix}$

[wintry steppe]

they will show up next to eachother

or you can do it for them

jk

[stoic python]

Max Hetfield:

[half forge]

loll

[wintry steppe]

google and stackexchange tend to be very good for latex-writing questions

that's how i learned all of the latex i know and it's how i learn more

[half forge]

hmm my code came out weird

i tried both codes

[wintry steppe]

well, first, you have to put it in \[\] or $$$$ or \begin{equation}\end{equation} or whatever you prefer

[half forge]

let me try it

can you do me the format code loll

oo nvm i got it loll

anyone know subrings?

[native rampart]

How is S a subring? There is no multiplicative identity in S

[half forge]

not sure but it has to satisify these rules

[golden drum]

How is S a subring? There is no multiplicative identity in S
@native rampart A ring without Identity

@half forge , You just have to prove that S is close under sustraction and close under product

[golden drum]

And S not empty set

[half forge]

heres the original problem

[golden drum]

Yes, I saw it

To prove that is a subring, you have to prove those properties

[half forge]

i read my textbook that that it has to satisfy four things but my teacher said that we have to show if its a subset ?

[golden drum]

A subring of a ring is a subset of a ring that is a ring

But, S is inside the ring M(R), then, it already have the associativity and Distributivity Property

You need to prove that is closed under adittion and exists inverses (Or equivalent closed under sustraction) and prove that is closed under multiplication

See that if you prove that, if B, C belongs to S then B - C belongs to S. That's closed under sustraction, you csn get 0 taking C = B

[unkempt hawk]

Is it possible to achieve reduced row echelon form on a matrix that doesn't have equal dimensions (e.g. 2x3 Matrix)

Ah, in my example, there would just be a column of zeros?

[golden drum]

There you have one

$\begin{pmatrix}

1 & 0 & 2 \

0 & 1 & 3

\end{pmatrix}$

[stoic python]

Enigsis:

[unkempt hawk]

In this format, how would I know if the matrix is dependent/independent based on the values in reduced echelon form?

[golden drum]

With dependent and independent do you mean the column vectors ?

Or row vectors

[unkempt hawk]

Column vectors

[golden drum]

Well, I think if you're working in R², they're dependent, because is a set of three vectors, and R² have dimension 2

R² or F² for any Field F

[unkempt hawk]

What about row vectors then?

[golden drum]

If you get the last row 0, I think they're dependent

[unkempt hawk]

Ah, great. That's helpful. Thank you 🙂

[half island]

'Shouldn't the highlighted operator be a + instead?

[unkempt hawk]

Returning to earlier, how could I tell by looking at the reduced row echelon form if the row vectors are dependent/independent if there are 4 3d vectors?

[dusky epoch]

four 3D vectors are always LD

[golden drum]

If the space have dimension n then, every set with k > n elements is linear dependent

In this case 4 > 3

Then, they're linear dependent

[unkempt hawk]

Thanks guys 🙂

[severe cedar]

If you have n eigenvectors corresponding to some eigenvalue, does this imply that the eigenvalue is of multiplicity n?

or can the multiplicity be lower/higher than the number of eigenvectors?

[native rampart]

Yea, if the n vectors are linearly independent

[whole solstice]

am scared to ask

but here it goes

i gag when i look at this

can someone help me

what even is P(R)

is that powerset of R?

[dusky epoch]

P(R) seems to be the space of all real polynomials

[severe cedar]

at my uni at least that is the notation for the polynomial space with real-valued coefficients

so just evaluate the transformation for linearity

[native rampart]

Yea, That's a weird proof

[serene tulip]

yeah I started with the hint

[dusky epoch]

suppose there exists a vector x such that (ST-I)x = 0, therefore STx = x therefore TSTx = Tx therefore Tx ∈ ker(TS-I)

likewise symmetrically in the other direction

[serene tulip]

I got to TSTx = Tx

from my own attempt

[dusky epoch]

ah wait there's a hole

i didn't prove Tx wasnt zero

[serene tulip]

yeah that was my issue

[dusky epoch]

it's not hard but it's a necessary gap to fill

assume x to be in ker(ST - I) \ {0}

so that x ≠ 0

then STx = x ≠ 0 will imply Tx ≠ 0 (as otherwise STx would be 0 implying x itself is also 0 which it isn't)

[serene tulip]

STx = x ≠ 0 will imply Tx ≠ 0 this is the step I am stuck on

specifically

[dusky epoch]

Sv ≠ 0 implies v ≠ 0

[serene tulip]

that's a property of linear transformations

[dusky epoch]

yes

and im applying here with Tx as my v

[serene tulip]

I understand that

I am going to look for that property of linear transformation in my profs notes

thanks for the help

[dusky epoch]

its the contrapositive of v = 0 => Sv = 0 ¯_(ツ)_/¯

[serene tulip]

It makes sense

I just like restraining myself with tools the prof give in class as good habit

basically so I am not fighting with the prof too much

[royal ore]

how would V be written out?

[dusky epoch]

wym written out

[drowsy yarrow]

Under what conditions is the linear map $F:\mathbb{R}^m \to \mathbb{R}^n$ given by $F(x) = Ax$, where $A$ is an $n\times m$ matrix, either injective, surjective, or both? I know that if it's both it has to be a square non-singular matrix, but I'm struggling with the individual cases.

[stoic python]

telegram sam:

[native rampart]

If m<n it can never be surjective

If m>n it can never be injective

[drowsy yarrow]

Ah yeah I established that too. But what are the restrictions on A assuming we've met the restrictions on n and m?

[native rampart]

Column rank=n would imply surjection

And if the column vectors are linearly independent,injection

Think of the columns of the matrix as a spanning set and x to be a set of scalars (Say,v1,v2..vm are the column vectors, take A(x1,x2 ..xm) to mean x1v1+x2v2...xmvm)

There are m column vectors, each belonging to R^n and you use them to span a subspace of R^n

[solid bough]

any householder transformation expert here?

[royal ore]

wait so for my above question, how can i prove it's a vector space?

[half storm]

Its a subset of the vector space of 3 x 3 matrices.

Because of that, you already know that alot of the axioms are automatically satisfied.

The four that you've got to verify is that it contains the zero matrix, and that it's closed under scalar multiplication and vector addition.

Then you're done.

I don't know what that fourth property is supposed to be.

[royal ore]

thanks

so for the zero matrix, how would it be verified?

i know A2,1=0, but how would it be shown in a proof

[warm niche]

Why does RREF actually mean in terms of vectors?

Becuase in linear algebra RREF makes sense in the context of system of equations

But what I'm learning right now is about column space and row space

[native rampart]

It's a convenient way to check the rank of rowspace

[warm niche]

And seems pretty disconnected from system of equations

so there is a way to find rowspace without RREF @native rampart ?

[native rampart]

You always know the rowspace

[warm niche]

Oh rite

I guess the basis of the rowspace

is what I mean

[native rampart]

Well you can write a bunch of linear equations and solve manually

[warm niche]

but thats just RREF

oh

RREF finds dependent vectors

I was trynna relate to RREF to system of equations

I think it, at least for me, to isntead think of it as a method to find dependent vectors

[spice ruin]

Can anyone tell me how I did this wrong?

The problem was just to solve the augmented matrix in vector form

[limber sierra]

what happened to the 1 in position 4, 2

howd it become a 0 between the second and third matrix

[spice ruin]

-_- sheeet

i just didnt write it down right

[wintry steppe]

Let $A = span{x, 1}$ and $B = span{1+x+x^2-x^3, 1+x+x^2}.$ Show that all real polynomials of degree 3 can be expressed as $A \oplus B$

[stoic python]

Tzucki:

[warm niche]

I think the way to do this

is to represent the sets of A and B as matricies

and then

add them?

and check if it rank is full

[wintry steppe]

i need to explicitly show that i can let $$ax + bx^2 + cx^3 + d = A \circleplus B$$

[warm niche]

but check if det is non zero?

[wintry steppe]

cant use det or rank theorems

[warm niche]

ye rite

[wintry steppe]

so would ijust solve the system

[warm niche]

oh

[wintry steppe]

AB as in A plus B

direct sum of a and b

$ax + bx^2 + cx^3 + d = A \oplus B$

[warm niche]

I think so

U can then use system of euqations

[stoic python]

Tzucki:

[warm niche]

because polynomials are linearly indepedent

[wintry steppe]

ok let me try soemthing

[warm niche]

d will only equal the constants

so ig as long as a b c d are defined

it can span everything?

idk I dont think I know enought linear algebra to help lmao

[wintry steppe]

damn

good try though , anyone else online to take a look?

<@&286206848099549185>

[half moss]

I think i know

Speaking generally we could take

$B = span{v_1, v_2}span{1+x+x^2-x^3, 1+x+x^2}$ and then $B = span{v_1 - v_2, v_2} = span{-x^3, 1+x+x^2}$

[stoic python]

darkninja175:

[wintry steppe]

i think i figured it

thank you though

[half moss]

then we basically have the building blocks of our polynomials

[spiral star]

showing that the standard basis vectors are in the span suffices

[half moss]

yeah

[spiral star]

and thats really obvious

1 = 1
x = x
x² = (1+x+x²) - 1 - x
x³ = (1+x+x²) - (1+x+x²-x³)

[wintry steppe]

i see that

but the answer was expressing it as a sum of two vectors mulitiplied by some coefficient to make it equal to the polynomial

i figured it out using some matrix madness

[half moss]

I have a question guys. I'm struggling to understand how Givens matrices work. I understand the concept that we can induce zeros into a matrix column vector through rotation, but my sources say we "rotate" 2 coordinates at a time, not the whole vector. This confuses me

maybe I'm getting lost in notation

Can someone give me some sort of intuitive sense as to what a single Givens matrix is doing? (I'm looking to string them together for QR factorization later.)

[gray dust]

do you know what to do

[royal ore]

no clue on where to start

[wintry steppe]

are you allowed to use dimension theorems

kerT = {0} => dim(kerT) => 0 is injective

[royal ore]

would it be?

[gray dust]

do you know how to prove an if & only if statement

i asked him, don't answer for him

[wintry steppe]

both ways can hold with dim theorem you just need to explain it properly

[royal ore]

i never proved injective for a linear map yet

so i wouldn't know how to do this statement

[wintry steppe]

have you learned dimension/ rank null theorem

[royal ore]

i've done dimension but not for null

[wintry steppe]

if you havent you probably need to prove it some other way

dim(kerT) + dim(imT) = dim V for vector space V

where t is a linear transformation

[half moss]

well first off, and if & only if statement requires proving implications in both directions. So you need to do two proofs.

essentially

[royal ore]

so what would be the first thing we need to prove

[gray dust]

@wintry steppe you don't need to invoke rank nullity

[wintry steppe]

you could do it multiple ways, sure

[gray dust]

@royal ore i asked if you know how to prove an "if & only if" statement

[royal ore]

not really

[gray dust]

@half moss i got it

[half moss]

well we could start with the forward implication. so and injective linear map T, prove it's kernel must be {0}

The backward implication would be, given a linear map that has ker{0}, prove it's injective

it's two proofs

[gray dust]

first do you know what "if & only if" means

[royal ore]

so first prove kernel is {0} and then prove the linear map is injective

what does if and only if mean

[half moss]

it means a logical equivalence, or that the implication works in both directions.

[gray dust]

let's say we have 2 statements p and q. we say "p if and only if q" if we have that p implies q (we say "p only if q") AND q implies p (we say "if q then p")

[royal ore]

ok, so thats why we're proving both forward and backward implications

[brisk fractal]

the forward implication is a bit more straightforward, it might be wise to start with that

[royal ore]

so for example the linear map is injective if ker t = {0} and ker t = {0} if the linear map is injective

[half moss]

yes. if we had "If p then q", then that's a regular implication. but if we have "p if and only if q" it's basically saying they mean the same thing, they're interchangeable, aka the implication works both ways

yeah

[royal ore]

okie

[half moss]

you've got it

[gray dust]

@brisk fractal @half moss can you both leave this channel, i don't like helpers dogpiling

[brisk fractal]

sorry, my bad

[half moss]

sure thing

[gray dust]

let's say we have 2 statements p and q. we say "p if and only if q" if we have that p implies q (we say "p only if q") AND q implies p (we say "if q then p")
so if we're proving p if and only if q, we must prove 2 things, that p implies q AND q implies p

[royal ore]

ok so how would we start to prove that T is injective if kerT={0}

for the first proof

[gray dust]

so for example the linear map is injective if ker t = {0} and ker t = {0} if the linear map is injective
so that's the structure of what you're doing, prove ker(T)={0} implies T is injective AND T being injective implies ker(T)={0}, if you do those then you can conclude ker(T)={0} if and only if T is injective

so you're proving now that ker(T)={0} implies T is injective. what's your first step?

[royal ore]

we have to show vectors that are linearly independent?

[gray dust]

what vectors?

[royal ore]

those part of T?

[gray dust]

not sure what you mean

[royal ore]

so there's no vectors involving the proof?

[gray dust]

have you ever proven an implies statement?

[royal ore]

not yet, and haven't done injection

i've done linear maps but with vectors

one-to-one

[gray dust]

ok you have a definition above, you'll use it

let p and q be statements. to show p implies q, you assume p is true then show q is true

[royal ore]

yes for the first proof we're trying to show kerT = {0}

meaning we already know T is injective for the first

so show kerT = {0}

but then the second proof would be us assuming kerT = {0} is true but proving T is injective

[gray dust]

yes for the first proof we're trying to show kerT = {0}
meaning we already know T is injective for the first
assume T is injective then show ker(T)={0}. that'll show T being injective implies ker(T)={0}
but then the second proof would be us assuming kerT = {0} is true but proving T is injective
assume ker(T)={0} then show T is injective. that'll show ker(T)={0} implies T is injective

[royal ore]

okie

[gray dust]

assume T is injective then show ker(T)={0}. that'll show T being injective implies ker(T)={0}
1st step assume T is injective. do you know how to show ker(T)={0}?

[royal ore]

not sure how to show it for the kernel

do we have to write T is injective because each V can only go to one W

[gray dust]

you must be thorough. in proving this

T being injective implies ker(T)={0}
you say every step. so you include this too
1st step assume T is injective.

[royal ore]

so we assume t is injective for first step

then next is it being injective implies ker(t)={0}

[gray dust]

T not t

[royal ore]

yeah

[gray dust]

your next steps are toward showing ker(T)={0}

[royal ore]

would we have to show with dimensional theorem as they said before?

or null theorem

[gray dust]

no

have you ever proven set equality?

[royal ore]

yeah but that was last year's course with discrete math

[gray dust]

you'll do it again to show ker(T)={0}

[royal ore]

they have same elements but don't have to be in same order right

if we compare two sets that are equal

[gray dust]

there's no such thing as order in a set

@pallid rampart no

[royal ore]

yeah it just matters about the elements themselves

[gray dust]

what's the definition of set equality?

[pallid rampart]

I haven't even said anything yet

[royal ore]

Two sets are equal if and only if they have the same elements.

does the if and only if as well

[gray dust]

that's not a usable definition here. give me the textbook definition

[royal ore]

no textbook on me atm

but

[gray dust]

do you know how to read that

[royal ore]

For all x, x is a element of A and is equivalent to x being a element of B

[gray dust]

x being an element of A is equivalent to x being a element of B
not wrong wording but it may help to break it down into 2 statements. x is in A implies x in B, AND x is in B implies x is in A. you can also reword it as, A is a subset of B AND B is a subset of A

[royal ore]

so what does this have to do with ker{T}=0

[gray dust]

to show ker(T)={0} you must show ker(T) is a subset of {0} AND {0} is a subset of ker(T) (this bit is considerably easier)

[royal ore]

well isn't {0} a subset of every set

or that's only when it's null

[limber sierra]

{} is a subset of every set, but {0} is not.

[royal ore]

yeah sorry

[limber sierra]

to show {0} is a subset of ker(T), you'll want to think what 0 [i.e. the 0 vector] gets mapped to by a linear map

as a hint: 0 = x - x, for some (any) vector x

then use linearity

[nocturne jewel]

Could someone explain dependence/independence of vectors?

like.. vectors are dependent if you have a non-trivial sol'n to [A|0] for A=[v_1,v_2,...,v_n}

and independent if the sol'n is only the 0 vector?

[mild igloo]

would anyone mind answering a quick question?

[wintry steppe]

need it done in under 25 mins, its urgent

and idk the answer

[limber sierra]

@wintry steppe this is not linear algebra

see #prealg-and-algebra

[mild igloo]

if I have the equation L(t) = <1, 2, 1> + t<-3, 4, 1> and I give t a value and solve do I get a vector or a point on the line?

[limber sierra]

@nocturne jewel that is one way to define it, yes

[mild igloo]

I need to find two points on the line and I belive (1, 2, 1) is the initial point

[royal ore]

so in the proof i have to put 0 = x -x for some vector x?

sorry i'm just so confused lol

[wintry steppe]

@limber sierra can you still help?

[limber sierra]

well, you want to figure out whether 0 is in ker(T), so you want to figure out whether T(0) = 0 @royal ore

but we know 0 = x-x for any vector x

so instead write

T(x-x)

but what do you notice if you apply linearity?

[nocturne jewel]

@mild igloo you get vectors, but vectors "=" points if the tail is (0,0)

so when t=0, you get [1,2,1], which "is" (1,2,1)

[royal ore]

so in the proof what am I writing first

I have so far:

[mild igloo]

right but if t=1 I get <-2, 8, 3>?

[wintry steppe]

@limber sierra hey can you help

[nocturne jewel]

@royal ore probably define the vector x as [x_1,x_2,...,x_n] if you're in R^n space

[royal ore]

Assume T is injective
T being injective implies kerT = 0
We must show that kerT is a subset of {0} and {0} is a subset of kerT

[nocturne jewel]

oh

ignore me then lol

[royal ore]

do i need to send the question again

[limber sierra]

T being injective implies kerT = 0
This is what you want to prove, but yes

and it should be

kerT = {0}

since a kernel is a set

[mild igloo]

im sorry I just dont know what a tail is

[limber sierra]

not an individual vector

but yeah

[nocturne jewel]

@mild igloo vector arrow has the arrowhead and the not arrowhead part... the tail is the not arrowhead part

[royal ore]

so what would come next to write

i'm not sure how to put what you said in writing

[nocturne jewel]

so the arrow goes from origin to the point

[mild igloo]

oh okay but I have no way to know if the line passes through the origin?

[nocturne jewel]

it does with lines and planes

[limber sierra]

just flesh out my argument

"to show {0} is a subset of kerT, we want to show that 0 is in kerT"

"this means that T(0) = 0"

[nocturne jewel]

points = vectors in the context (not equal but lack of better symbol)

[limber sierra]

and then prove T(0) = 0

[wintry steppe]

smh bad server

[mild igloo]

so If I was asked to find two points on the line (1, 2, 1) and (-2, 8, 3) would be acceptable answers?

[nocturne jewel]

for t=1?

[mild igloo]

yes

[nocturne jewel]

double check the numbers, x component is right

[royal ore]

so we know T(0) = 0

[nocturne jewel]

but y and z are off

[royal ore]

but were you saying to prove it linearity before

[nocturne jewel]

(if the L(t) you gave is correct)

[mild igloo]

(-2, 6, 2)

[nocturne jewel]

ye

[mild igloo]

okay so if I was then asked to find a unit vector parallel to the line I would have to calculate a vector given two points on the line?

[nocturne jewel]

So in general, L(t) is [some point on the line] + t[Direction vector]

[mild igloo]

right

[nocturne jewel]

if a vector is parallel to the line, it's parallel to the direction vector, as the direction vector tells the line where to go from the initial point

so find the unit vector of the direction vector and you're done

[gray dust]

@royal ore no you DON'T know that. you set out to prove it. once you prove T(0)=0 then you can conclude {0} is a subset of ker(T)

[royal ore]

so

[mild igloo]

ah I see thank you so much

[royal ore]

T(0) = T(0+0) = T(0)+T(0) ?

[nocturne jewel]

np

[gray dust]

that's correct but justify it

[royal ore]

like say what proves it?

[gray dust]

why can you say this?

T(0) = T(0+0) = T(0)+T(0)

[royal ore]

a property of linear transformation

[gray dust]

namely what?

[royal ore]

Linear Transformation (or Map) on a Vector Space

[gray dust]

i mean what property of linear maps

[royal ore]

not sure which property its called

lemme look in my notes if i have it

[gray dust]

respects vector addition. T(x+y)=T(x)+T(y) for all x,y

[royal ore]

oh ok

so that proved t(0)=0 right

[gray dust]

T not t, and you lack 1 step

[royal ore]

what step is it

like where between

[gray dust]

T(0) = T(0+0) = T(0)+T(0)
ideally you must justify every step, every equation in between. justify T(0) = T(0+0), justify T(0+0) = T(0)+T(0), then justify how that leads to T(0)=0

you appear to have little experience in proof writing so i think you should do it, but if you enough of these then it becomes overkill, which is why i just asked you to justify T(0+0) = T(0)+T(0)

[mild igloo]

one more for you guys

is there some k that will make the matrix have no solution

or 1 solution

or infinite

[warm niche]

well

[mild igloo]

I know there is 1 that will give infinite

[warm niche]

u can use determinants

if u know one will give an infinite amount of solutions

[royal ore]

i'm not sure how to justify them specifically

[mild igloo]

well I know that -2 will make the bottom row all 0 which means infinite right?

[royal ore]

i just know they're vector addition like you said

[warm niche]

wait

nvm

[mild igloo]

if I evaluate to row echelon for and the bottom row is 0 0 3 | 0 that means one solution?

[desert vessel]

that means no solutions

[mild igloo]

okay so its impossible to make this matrix have 1 solution

[desert vessel]

wait sorry I read that wrong

that means one solution

x3=0

assuming your other variables are also basic

[mild igloo]

okay then its impossible for it to have no solutions?

[desert vessel]

put it into rref and that will answer your questions

[royal ore]

Linear map has a property where T(u1+u2) = T(u1) + T(u2)

[desert vessel]

it will have no solutions if it has a row of the form 0 0 0 | b for any b=/=0

[royal ore]

so for the T(0+0) = T(0)+T(0)

[mild igloo]

I thought that meant it has infinite

[royal ore]

what property is it called

[mild igloo]

if the bottom row is all zeros

[desert vessel]

I just said the bottom row is 0 0 0 b with b not equal to 0

[mild igloo]

ahhhh

okay

[royal ore]

actually in my notes there's a proposition of For any linear map,T,T(0) = 0.

wouldn't this be applied as well

[desert vessel]

you have to prove that T(0)=0

[gray dust]

i said

respects vector addition. T(x+y)=T(x)+T(y) for all x,y

at the very least you must do this

justify T(0+0) = T(0)+T(0), then justify how that leads to T(0)=0

[ocean sequoia]

I feel like I should know this but why does this require orthonormal matricies

b/c orthonormal matricies A^t * A = I?

does the | |Qx| | mean of length one?

[hard coral]

Can you specify your question a bit? What exactly surprised you?

Yes this is the norm of Qx

[ocean sequoia]

oh wait is thats because the norm of Qx is $\sqrt{(Qx)^t * (Qx)} $

wow i shouldve used latex

[wintry steppe]

tall thin matrices

[ocean sequoia]

hahaha its pretty applied

do i just not rememebr how to compute matrix norms

@hard coral I guess said a better way where does $(Qx)^t Qx$ come from

[stoic python]

brzig:

brzig:

[hard coral]

Well yes. In a eucledian vector space (?) The dot product is defined as $<x,y> = x^Ty$, the norm is defined as the sqrt of the dot product.

[stoic python]

Tobii:

[ocean sequoia]

ok

thanks

[idle echo]

How was the definition of matrix multiplication arrived at? it seems a bit arbitrary

@warm briar apparently not

[limber sierra]

are you aware of how to use matrices to solve linear systems

[idle echo]

@limber sierra yes

[limber sierra]

if so, then the general form of Ax = y should give you an idea, at least, of how to multiply a matrix by a column vector

the question, then, is how to extend that to general matrix * matrix

rather than just matrix * column vector

one answer is that we can consider Ax to be representative of a linear transformation T(x)

and then it turns out that, given another linear transformation S

composing S and T is the same thing as the definition of multiplication we give for their representative matrices, B and A

but what do i mean by "it turns out"? it still seems fairly random, surely?

well, let's consider the effects of AB on a column vector

hold on, extensive latexing incoming

thisll take a bit

say $A = \begin{pmatrix}1&2\1&0\end{pmatrix}, B = \begin{pmatrix}2&1\-1&1\end{pmatrix}$; then the corresponding linear transformations are $T_{A}(x) = T_{A}\left(\begin{pmatrix}x_1\x_2\end{pmatrix}\right) = \begin{pmatrix}x_1 + 2x_2\x_1\end{pmatrix}$ and $T_{B}(x) = \begin{pmatrix}2x_1+x_2\-x_1+x_2\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

but also $T_{A}(T_{B}(x)) = T_{(A)}\begin{pmatrix}\text{dot product of the first row of B and x}\\text{dot product of the second row of B and x}\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

so $T_{A}(T_{B}(x)) = \begin{pmatrix}\text{dot product of first row of A with (first entry first row of B and x, first entry of second row of B and x)}\\text{dot product of second row of A with (second entry first row of B and x, second entry second column of B and x)}\end{pmatrix}$

this might be a bit hard to visualize

but it agrees with how we want matrix multiplication to behave

and if we look at that resultant matrix

[stoic python]

Namington:

[limber sierra]

oops typo fixed

but yeah if we look at this

it seems like we've taken the dot product of the first row with A and

the first first column of B

for the first entry

and similarly, the second entry of the resultant vector is the dot product of the second row of A with the second column of B

I encourage you to try some examples by hand to "convince yourself" of this

to go back to the numeric example I was using before:

$T_{A}(T_{B}(x)) = T_A\begin{pmatrix}2x_1+x_2\-x_1+x_2\end{pmatrix} = \begin{pmatrix}(2x_1+x_2) + 2(-x_1+x_2)\2x_1+x_2\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

note that we basically plugged the rows of B into each row of A

but we did this "moving by columns" so to speak

(we started in the first entry of the columns, then went to the second entry of the columns)

and indeed if we interpret this as matrix multiplication:

$\begin{pmatrix}1&2\1&0\end{pmatrix}\begin{pmatrix}2&1\-1&1\end{pmatrix}\begin{pmatrix}x_1\x_2\end{pmatrix} = \begin{pmatrix}1&2\1&0\end{pmatrix}\begin{pmatrix}2x_1+x_2\-x_1+x_2\end{pmatrix} = \begin{pmatrix}(2x_1+x_2)+2(-x_1+x_2)\2x_1+x_2\end{pmatrix} = \begin{pmatrix}(2-2)x_1+(1+1)x_2\2x_1+x_2\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

the 2-2 is, of course, the dot product of the row {1,2} from A and the column {2, -1} from B

bleh i made an arithmetic mistake somewhere

too lazy to go spot it

but you should get the idea

basically, if we view it from the prespective of "applying transformations to somethign that's already been transformed"

as soon as we've "transformed" a vector

the information of the transformation (the matrix) is now organized not in the way it was originally organized

i.e. by rows

but instead by what variables it is paired with

i.e. by columns

and when we apply another transformation, this column-wise organization is what gets affected

hence why matrix multiplication takes the dot product of rows of the first matrix and columns of the second

[mild igloo]

what would be the inverse of a for this equation?

I is identity matrix

[hard coral]

Well the inverse of A is defined as a Matrix Z with A* Z = I
And matrix multiplication is associative

[stray nacelle]

are two zero vectors orthogonal?

[wintry steppe]

yes, their inner product is zero

[mild igloo]

(B^-1 * A^-1)C = I is inverse?

[hard coral]

but you asked for the inverse of A?

[mild igloo]

well I just skimmed some lectures and saw that AB = B^-1A^-1

only reason I said that

is it literally just A^-1

[hard coral]

so A^-1 should equal something and not be tangled up

[mild igloo]

ah

so solve for a?

[hard coral]

Well you could just see the solution in this case. As the definition is almost in front of you :')

If (AB)C = I and matrix mult is associative then (AB)C=A(BC) = I

How was an inverse defined again 🤔

[mild igloo]

im not entirely sure tbh

I could have just seen something out of context and gotten confused

Guess I need to rewatch those lectures I assumed that inverse meant to raise it to -1

well is that not the notation for an inverse of a vector

[hard coral]

that's just notation in most cases.

But the definition of an inverse for A would be the element which combined with A leads to the neutral element

[mild igloo]

yea I have no clue what that means definitely gonna need to rewatch the lecture

oh you wanna just put it in the form A*Z = I

so A * Z. Z=BC

[hard coral]

so BC is the inverse of A

(Well the right-inverse but that leads down another rabbit hole)

[mild igloo]

yeaaaaa I dont even wanna think of that and different type of vector norms

wants me to multiply the original equation by BC?

[hard coral]

Hard to say out of context

[mild igloo]

Didnt even realize my bad

nvm I got it I think

[golden drum]

what would be the inverse of a for this equation?
@mild igloo

BC, since Matrix multiplication is associative

[mild igloo]

I actually already calculated the inverse so I just multipled them and got a reduced echelon form matrix which is correct

if Im correct

😄

[dry echo]

Trying to understand singular value decomposition in comparison to eigenvalue decompsition. i can see in this image how the 3 parts are DERIVED from the matrix. but i dont understand how SVD derives its 3 parts.

[fallen karma]

Give an example of vector space $V$ and subspaces $U_1,U_2$ of $V$ such that $U_1 \cross U_2$ is isomorphic to $U_1+U_2$ but $U_1+U_2$ is not a direct sum. Hint: The vector space $V$ must be infinite dimensional.

[stoic python]

seth.delacroix:

[fallen karma]

If you can think of one, slide me a hint. I'm struggling coming up with an injective map that wouldn't force the spaces to be a direct sum.

[fallen karma]

Oh I thought of one I think

[serene tulip]

How would you show uniqueness whens supposing the Direct Sum of the M subspaces

I know it's going to use the fact that M_1∩...∩M_k = {0}

[elfin mist]

Uniqueness of x_i right?

[serene tulip]

yeah

[elfin mist]

Try a proof by contradiction? Assume there exists some v such that x_i in M_i is not unique, i.e. v = x_1+ ... + x_n = x'_1 + .... x'_n

now see that sum from 1 to n of (x_i - x'_i) = 0

and use the fact you mentioned earlier

[native rampart]

Isn't the condition M_i ∩M_j ={0} for all i not j?

[elfin mist]

now see that sum from 1 to n of (x_i - x'_i) = 0
@elfin mist
claim that x_i = x'_i for all i, and find a contradiction otherwise

Isn't the condition M_i
∩M_j ={0} for all i not j?
@native rampart
They're both equivalent conditions I think

[native rampart]

Because let M1=span{e1,e2} M2=span{e1,e3} and M3=span{e4}

There is no unique representation
(Let x=ae1 which can be written as b e1 (in M1)+ (a-b) e1(in M2) taking suitable b)

[serene tulip]

now see that sum from 1 to n of (x_i - x'_i) = 0
@elfin mist I can state (x_i-x'_i)∈M_i then idk what to do after that

I know the proof of the 2 vector space case (x_1-x'_1)=(x_2-x'_2) and (x_1-x'_1)∈M_1 , (x_2-x'_2)∈M_2

and then (x_1-x'_1)=(x_2-x'_2)∈(M_1∩M_2)={0}

but I don't know how to generalize that step for M_1, M_2, .... , M_k

can we state (x_1 - x'_1) = (x_2-x'_2)=....=(x_k-x'_k)=0?

<@&286206848099549185>

[elfin mist]

so maybe you could do this:
(x_1 - x'_1) = (x'_2 - x_2) + .... (x'_n - x_n)

[serene tulip]

can I state (x_i - x'_i) = (x_2-x'_2)+...+(x_i-1 - x'_i-1)+(x_i+1 - x'_i+1)+...+(x_k-x'_k)

[elfin mist]

LHS is in M_1, while RHS is not

[serene tulip]

LHS?

[elfin mist]

left hand side*

can I state (x_i - x'_i) = (x_2-x'_2)+...+(x_i-1 - x'_i-1)+(x_i+1 - x'_i+1)+...+(x_k-x'_k)
@serene tulip
You missed x_1 - x'_1

but that's okay

so maybe you could do this:
(x_1 - x'_1) = (x'_2 - x_2) + .... (x'_n - x_n)
@elfin mist
LHS here is in M1, RHS is not

also can we type in LaTeX on Discord? It's a little annoying with all the symbols

[serene tulip]

yeah you can

with $f(x)=x$

[stoic python]

Mango:

[serene tulip]

just use $$$$$

[elfin mist]

that's awesome! thank you

does my proof convince you?

[serene tulip]

I am having issue seeing what the RHS is tho

like is it the intersection of all the M excluding M1

or is the contradiction that RHS isn't M_1

[elfin mist]

Hmm I don't think it's the intersection, but we can argue that the RHS contains the sum of vectors from M_2, to M_n, and M1 intersect M_i is empty for all i = 2,3...n so M_1 is not contained in the RHS

or is the contradiction that RHS isn't M_1
@serene tulip
I think the contradiction is that M_1 is not contained in the RHS

What do you think

[serene tulip]

It makes sense

[elfin mist]

How should I go about this? I don't know where to start (part 1(a) to begin with)

[serene tulip]

for part 1(a) i Think you need to show vector space V for arbitrary vector space X, W such that X maps to V and V has a unique map to W

So you have show that if a map exist between to vector spaces then you can make a map that goes through a detour vector @elfin mist

[elfin mist]

Yes, agreed. Can that detour vector space be anything? I'm unable to "construct" it

[serene tulip]

It can be anything cause all you have to do is show existence and not define it

I don't know if this is correct but I would probably start X and V to be isomorphic so I would not lose any "information" through the transformation

@elfin mist

[elfin mist]

Ohh, that is a smart choice.

[clear arrow]

what

no construction is okay

its actually the only way

to show its existence

[if I'm not wrong]

@elfin mist you can consider for a) the vector space with basis the elements in X

its a free module basically, where the ring is a field

[solid bough]

how does one change the position of elements in a diagonal matrix using orthogonal matrix?

[fringe lantern]

The question is to check what the matrix T_c with respect to these two basises C and E are. I cant find similar exercises anywhere 😦

[rugged basalt]

Why did they let x = 3

Why can they do that?

[shrewd mortar]

Because it doesn't matter

(3,-1, - 2) is just one solution but you can multiply by any scalar to get another

@rugged basalt

[fleet echo]

Hey ya'll, I compleatly forget how to do these... someone mind explaining them rq?
https://cdn.smc.wtf/MxZ.png

[wintry sphinx]

the right column is f(-1), f(0), f(1), f(2), and so on

then you just graph (-1, f(-1)), (0, f(0)), (1, f(1)) and so on

[fleet echo]

Ah okay

thanks!

@wintry sphinx Sorry if you don't mind, so I just essentially solve the polynomial?

Or..?

Bit confused how to get f(#)

[wintry sphinx]

If I say that f(x) = x^4 + x^3 + 5x^2 + 8(9-x), then what I'm really saying is that f(<something>) = <something>^4 + <something>^3 + ...

[fleet echo]

AHHH

Okay yeah I'm slow haha

Thanks a ton!

[bold ivy]

is lines on a plane ?

[warm harbor]

Can somone explain me what REF (reduced echolon form) is for? I know for RREF, it's useful and important because it tells you exactly what the solutions are for a system of equations, but i don't see any usefullness or importance of a matrix of the form REF... 😦

like it's a matrix form that's half solved, so what is it for lol

[dusky epoch]

REF is the name for the intermediate stage between raw and RREF

[warm harbor]

yeah i understand that part, so it's just the stage between unsolved equation and RREF?

[dusky epoch]

i guess so?

[warm harbor]

like why didn't we come up wither another name for a stage between raw and REF then lol

[dusky epoch]

no

[warm harbor]

i just dont understand why someone came up with that notion

[dusky epoch]

RREF has two R's

it's reduced row echelon form

[warm harbor]

since it seems not useful to me

yeah i know

[dusky epoch]

cause every matrix in RREF is also in REF

[warm harbor]

yeah

[dusky epoch]

gaussian elimination can be separated into two stages:

getting into REF

and getting into RREF from REF

[warm harbor]

so does knowing REF make a difference?

[dusky epoch]

wym "knowing REF"

[warm harbor]

i meant the ultimate goal of getting a REF matrix is to get a RREF right

[dusky epoch]

if you absolutely insist

[bold ivy]

if a line is on a plane in R^2 then that means in R^3, lines is in a plane?

[vocal prairie]

R^2 is itself a plane

And yeah, every line lies in a unique plane.

[dusky epoch]

no ted

for a given line in R^3 there are infinitely many planes which pass through it

[bold ivy]

IM SO CONFUSED

what do you mean by which pass through it

[dusky epoch]

i mean exactly what i said

[bold ivy]

so a line passes through one plane

that doesn't make any sense

[calm hamlet]

If the line is included in the planes it makes sense

A line in R^3 belongs to infinitely many planes (you rotate around the line)

[jagged forum]

Suppose A is a symmetric square matrix, and \theta is an eigenvalue of A. Let U_\theta be a matrix whose columns form an orthonormal basis for the eigenspace space corresponding to \theta, and let E_\theta=U_\thetaU_\theta^T.

I am trying to show E_\theta is idempotent.

Does anyone have any hints?

[calm hamlet]

Parentheses lmao

[stoic python]

Svet L'octogone:

[jagged forum]

Oof sorry

[stoic python]

Svet L'octogone:

[calm hamlet]

What is (U_θ)^T equal to? @jagged forum

[jagged forum]

It's the transpose of U_\theta, so its rows form an orthonormal basis for the eigenspace corresponding to \theta

I am imagining we can show somehow that $(U_\theta)^T U_theta=I$

[stoic python]

Kähler:

[jagged forum]

I am very rusty on matrix multiplication (this isn't a course on LA), but we could probably multiply through and have the diagonal be the only entries so that the same basis vector meets (and evaluate to 1).

[calm hamlet]

What does that mean for U_θ that its columns form an orthonormal basis for the eigenspace corresponding to θ?

[jagged forum]

That I am not exactly certain on

[calm hamlet]

OK. Take a vector u of the eigenspace. What is its image by A?

[jagged forum]

Au=0

[calm hamlet]

No

F is an eigenspace of A corresponding to θ <=> for all u in F, Au=θ*u

[jagged forum]

ohh yes sorry

[calm hamlet]

In the eigenspace, the vector is just scaled

OK. Let's go in R^3. Write an orthonormal basis of R^3

[jagged forum]

Canonical is one example, and to get more for any 3 linearly independent vectors grahm schmidt constructs an orthonormal basis

[calm hamlet]

Yea

So what happens for the "orthonormal basis" of your exercice is that it is scaled of θ by the matrix

So for example if ((1,0,0,0),(0,1,0,0),(0,0,1,0)) is the basis of the eigenspace, then its image by the matrix is ((θ,0,0,0),(0,θ,0,0),(0,0,θ,0))

[jagged forum]

Is U_\theta orthogonal?

I know a matrix whose columns and rows are orthonormal are orthogonal, but not sure of just columns

It follows easily if so

[vocal prairie]

for a given line in R^3 there are infinitely many planes which pass through it
Sorry, conflated some things back then.

[bold ivy]

@calm hamlet I don't understand

@calm hamlet can you perhaps draw this on a pieace of paper

because im genuinly confused

[calm hamlet]

I can't here

But imagine an infinite plane in space

Imagine a line included inside the plane

You see that you can make the plane turn around the line, and the line is still included inside it

[bleak ginkgo]

Find a vector v in R○^3 whose projection on each standard basis vector is the same standard basis vector.

Does that mean the Projection from v to (1, 0, 0) has to be (1, 0, 0)
v to (0, 1, 0) has to be (0, 1, 0)
v to (0, 0, 1) has to be (0, 0, 1) ?

[soft burrow]

yes.

[bleak ginkgo]

So just (1, 1, 1) as v works?

the norm to each standard basis is 1

[marble lance]

Yes

[dawn remnant]

yes, and nothing else

[bold ivy]

@calm hamlet Whaaat, Damn imagine how to do this in R^4

if I have a hard understanding doing this in r^3 then rip me

[robust pond]

i have i think a dumb question

if i have the standard matrix of a transform, how do i know the dimension of the domain and codomain?

[brisk fractal]

Ax = y right

[robust pond]

it should be just R^m -> R^n right

[brisk fractal]

say A is a n x m matrix

[robust pond]

for an m x n standard matrix?

[brisk fractal]

yeah

[robust pond]

okay 😄

[brisk fractal]

n x m * m x 1 results in n x 1

[robust pond]

i was curious bc its not really clear where the input vector should live

like it could go on either side

[brisk fractal]

I think typically linear transformations are written T(x) = [T]x for the matrix [T] in the basis, and I believe the definitions of multiplication of matrices follows from composition written in that way

so like A(B(x)) = ABx and that's how matrix multiplication for AB is defined, by the composition

[robust pond]

hmm

could someone give me a hand in finding the bases of the null space of a matrix

im trying to follow along with this guide and my book but the dimensions dont seem to line up so i dont really understand how its supposed to be a solution

I'm looking at $\begin{pmatrix} 1 & 0 & 0 & 4 & 5 \ 0 & 1 & 0 & -2 & 2 \ 0 & 0 & 1 & 1 & -1 \end{pmatrix}$

[stoic python]

jan Niku:

[robust pond]

so following my book and this page https://yutsumura.com/how-to-find-a-basis-for-the-nullspace-row-space-and-range-of-a-matrix/

i get to $\begin{pmatrix} -4x_4 -5x_5 \ 2x_4 -2 x_5 \ -x_4 -x_5 \ x_4 \ x_5\end{pmatrix}$

[stoic python]

jan Niku:

[robust pond]

i dont really get what this is supposed to be representing, I get that x_4 and x_5 are free variables but how does this corrolate to the bases of the null space other than just its the output of the augmented matrix solving for Ax=0

[dawn remnant]

You start by asking "for what x vectors is Ax=0?"

where x is completely arbitrary

you then end up with the solution that involves two arbitrary coefficients still

This means that the null-space is two-dimensional.

Now you just need to find some basis. For that, just take two different pairs of (x4,x5) values and get vectors for them. For example:
$$x_4=0,x_5=1$$
$$\begin{pmatrix} -5 \ -2 \ -1 \ 0 \ 1\ \end{pmatrix}$$
$$x_4=1,x_5=0$$
$$\begin{pmatrix} -4 \ 2 \ -1 \ 1 \ 0\ \end{pmatrix}$$

[dusky epoch]

x_4

x_5

[robust pond]

so we only need two basis vectors since the nullspace is 2 dimensional within R^5, but they'll have 5 elements each?

oh i guess that makes sense

[dawn remnant]

Yup, the nullspace of the matrix is a 2-dimensional subspace in a 5-dimensional space of vectors.

[robust pond]

because that would draw out a 2d plane

gotcha 🙇‍♂️ thanks

[stoic python]

ConfusedReptile:

[robust pond]

fuck online homework

nvm got it

[bleak ginkgo]

~ w has a positive z coordinate.
All three vectors are unitary.
The angles between them are all equal to π/3
.
Then find the volume of the parallelepiped they determine.

Any starting tips for that? I'm really lost

[wintry sphinx]

well

I don't think this is uniquely determined, so you can pick w = [0, 0, 1] and u = [sqrt(3)/2, 0, 0.5]

then you can solve a system of linear equations to get a value of v whose dot product with both of these is 0.5

as for finding the area of the parallelepiped, use https://en.wikipedia.org/wiki/Triple_product

a v that works is [2/sqrt(3), sqrt(2)/3, 0.5]

[bleak ginkgo]

thanks, I'm trying this

Just I don't think that v works

[elfin mist]

@elfin mist you can consider for a) the vector space with basis the elements in X
@clear arrow
Hmm, I'm not able to proceed from here :(

So you're suggesting V = linear-span(X)?

I'm not sure what the map is here, exactly. Let's say X is finite and has k elements x_1 through x_k. Where is x_i mapped to?

[robust pond]

hey im not sure what the heck im doing wrong here

[stoic python]

jan Niku:

[robust pond]

trying to find bases for the null space

i get uhh

[stoic python]

jan Niku:

[robust pond]

are my dimensions wrong or something?

the vectors seem right just from $x_1 = -x_3-5x_4$ and $x_2 = 4x_3 - 4x_4$

[stoic python]

jan Niku:

[robust pond]

anybody

[marble lance]

x1 = -t - 5s, x2 = 4t- 4s, x3 = t, x4 = s. So (x1, x2, x3, x4) = t(-1, 4, 1, 0) + s(-5,-4,0,1)

So the first one seems wrong

@robust pond

[robust pond]

t and s?

oh

[marble lance]

Got it?

[robust pond]

oh was it just a sign error

sorry one moment i want to see if its correct

dang that was a dumb mistake to get caught up on

thanks @marble lance 🙇‍♂️

[marble lance]

Np

[low prawn]

can anyone help me with a problem?

[robust pond]

[low prawn]

are linear *

[golden drum]

What do you have?

Just use the definition

[low prawn]

I have very wrong ansers

[robust pond]

this is what youre trying to check btw

[low prawn]

isn't t(0) = 0 another one?

[gray dust]

T(0)=0 follows from the above

[low prawn]

I see I see

imma redo it and make it legible and post

[marble lance]

Remember if it's false, you only need to show that one of those conditions don't hold

For specific points

[low prawn]

what am i doing lmaf

i succ

[wintry steppe]

nice cat

[robust pond]

cat

[wintry steppe]

[robust pond]

i dont really understand what you're doing

[low prawn]

Evidently, nor do i

Can someone do one of the problems please so that I can follow through

[golden drum]

Do you understand the definition of linear trandformation? @low prawn

[low prawn]

not anymore

[low prawn]

I got it

the first one is not a linear transformation because it fails the t(0)=0

[golden drum]

Yes

👍

[low prawn]

Ima

[dusky epoch]

,rccw

[peak perch]

can someone explain

like

[dusky epoch]

explain what

[peak perch]

how to get to the last gauss elimination

[dusky epoch]

add twice row 2 to row 3

[peak perch]

do you guys mind if you explain step by step because i really dont know what is going on

[dusky epoch]

have you done gaussian elimination before at all?

like, are you familiar with the concept of row operations?

[peak perch]

no and not familiar