linear-algebra

[acoustic path]

and if we have no free variables

[wintry steppe]

determinant might not even make sense here

[copper stratus]

Hmmmm

if the matrix with these vectors as columns has a non-zero determinant. The set is of course dependent if the determinant is zero.
Google's answer

[acoustic path]

how about solving for the kernel

na idfk

[calm hamlet]

You can indeed use the argument of dimension

[wintry steppe]

?

too many cooks in the kitchen

[round coral]

@copper stratus just use the definition of linear independence and linear maps

[rose umbra]

i asked the wrong question before. Is group can be defined as sub space if the vectors in it basiclly can create any vector in space? (talking only about Real numbers)

[half ice]

Anyway yeah I can do this. And so can you! @copper stratus

Take your linearly independent set and map it. Is this new set linearly independent?

[copper stratus]

Can you explain by map it? Like plan it out on a graph?

[wintry steppe]

i like this attitude kaynex 😌

[half ice]

So they did give you a function called f

I mean "map" by use the function on it

[round coral]

@rose umbra could you please word your question properly? Not quite clear what you are asking.

[half ice]

So naturally you'll need to make sure you know what a linearly independent set is

[copper stratus]

I mean I know what that is

It's when the only solution is trivial

right? @half ice

[acoustic path]

these are linear maps

[half ice]

Yeah, but more importantly here, a linearly independent set is one that only has one way to span 0

[rose umbra]

@round coral what is the actual meaning of sub space? I know how to find out if a group is a sub space of particular space. but what does it actually means of the group?

[acoustic path]

its what it sounds

[rose umbra]

part of the space?

particular part?

or it means still the whole space definded by only specific group of vectors

[half storm]

Just a vector space that's a subset of a vector space.

That's all

[rose umbra]

@half storm so just defined area in space

[round coral]

@rose umbra Ok first important thing, vector spaces are abelian groups. To be a subspace of a vector space say V, you need to be subset of V and also inherit the vector space structure from V .Is it clear?

[rose umbra]

@round coral so its what i said right? i just tring to simply it to my word

its area that built of the sub space vectors

the group can define every point in specifc area (depends on the group)

[round coral]

I have never visualized it as an area. Just see it as a vector subspace. All the vectors in the vector spaces are defined in the vector space. The points you are referring to maybe are the one dimensional subspaces of vector space

Also it would be better to stop referring to groups in this, it is unecessary, we already know that vector spaces are groups, no more need to talk upon that

[copper stratus]

Can someone help me find d?

I can't seem to think of a matrix

[round coral]

@rose umbra let me know if there is anything still unclear

[rose umbra]

still tring to analyze your comment lol

[calm hamlet]

@copper stratus what does non-regular mean?

[copper stratus]

a stochastic matrix is regular if some matrix power contains no zeroes

a stochastic matrix is a matrix where the columns of the matrix add up to 1 (a probability matrix)

nvm got it

[1,1]
[0,0]

[half storm]

@rose umbra Not quite. A linear subspace is just a subset of a vector space that is also a vector space

that's quite litearlly it

so like $P^{2} \subset P^{3}$

The set of all polynomials of degree two is a subspace of the set of all polynomials of degree three or less.

[stoic python]

TheDon

[nocturne jewel]

Not entirely sure how to go about this

I tried to develop an inductive hypothesis, but having troubles with that

The inductive hypothesis I got is: $(\sum_{i \neq j} a_ia_j) + \prod_{i=1}^n a_i$

[stoic python]

moshill1

[nocturne jewel]

(3x3 was a_1a_2 + a_1a_3 + a_2a_3 + a_1a_2a_3)

[calm hamlet]

@nocturne jewel what are the properties of the determinant which could be useful here?

[nocturne jewel]

Ummmm no clue tbh

[calm hamlet]

Do you know the bilinearity of the determinant?

Linear with respect to columns, and linear with respect to rows

[nocturne jewel]

multilinear?

Ik det(A) is multilinear, alternating, det(I) = 1

that thing?

[calm hamlet]

No. I meant that if you add a linear combination of rows (resp. columns) to a row (resp. column), the determinant is not changed

[nocturne jewel]

oh yeah

R_i + kR_j operations dont change the det

[calm hamlet]

Yep

[nocturne jewel]

So do I add Rows/Columns or treat them as already added?

Like in the back of my head I understand why that property is important, but I dont fully see how lol

OH @calm hamlet i add a copy of all the other rows to each row?

R1 <-> R1 + R2 + .. + Rn
R2 <-> R1 + R2 + ... + Rn
...

[calm hamlet]

You want to make the most zeros appear

What I did is that I substracted R1 to each other row

[nocturne jewel]

just multiply the matrix by 0

OH

THEN ITS TRIANGULAR

[calm hamlet]

No

But almost

[nocturne jewel]

oh

[calm hamlet]

Then you use the development of the determinant with respect to a row (or a column), and try to find a nice property about the determinants you have

[nocturne jewel]

development?

[silver tree]

Hello everyone! I have a question that kinda makes me question my knowledge in Linear Algebra.

If I have a linear transformation / mapping f : R^3 -> R^3 and its kernel has the following parametric equation:

s * <3,-1,3> + t* <1,5,5> where t and s are real numbers.

The questions is which vector is in the image for f, in which I have 2 vectors, which is:

<11,2,14> and <8,8,16>

With Linearcombination, it is easily seen that <8,8,16> is in the kernel - but vector <11,2,14> isn't - that's why I would think that <11,2,14> is in the image, but it is <8,8,16> - can anyone tell me why?

[wintry steppe]

picture room

[silver tree]

That's what we call in our native language - I don't know the english word for it :/

I apologize.

[wintry steppe]

image is correct

which you wrote

[nocturne jewel]

Oh I thought you were translating their username

[calm hamlet]

$\begin{vmatrix} a & b & c \ d & e & f \ g & h & i \end{vmatrix} = a\begin {vmatrix} e & f \ h & i \end{vmatrix} - b \begin{vmatrix} d & f \ g & i \end{vmatrix} + c \begin {vmatrix} d & e \ g & h \end{vmatrix} $

[stoic python]

Svet L'octogone

[nocturne jewel]

yeah

[calm hamlet]

Things like that (development with respect to 1st row)

[nocturne jewel]

we call that expansion

[calm hamlet]

Oh

I'm not a native xD

[nocturne jewel]

lmao it's fine

everything will be 0 except column 1, row1, and the diagonal right?

[calm hamlet]

So try to expand the discriminant with respect to a nice row or column, you will find interesting properties about the "sub-determinants"

Yes

[nocturne jewel]

col 1 is -a1 (except entry 1,1), row 1 is the same, and the diagonal is just a_n ?

(making sure I did it right cause i suck at the large amount of operations at once things lol)

[calm hamlet]

What I would do is expanding with respect to the first column since all coefficients is - a1 except (1,1) and your sub-determinants will have a row full of 1 which will allow you to simplify the 1 (with the linearity) and find a nice property

[gray dust]

det not discrim

[calm hamlet]

Oh

Sorry xD I always confuse the words

They are too close

[gray dust]

ye it happens. always writing just det helps me avoid that

[calm hamlet]

Ty i'll remember the tip

[hollow finch]

say i have a linear transforming T: V->W and a corresponding matrix A with respect to the standard basis. V and W have the same dimension and the transformation is full rank.
would someone mind helping me

1. determine a basis B (for W) such that the matrix relative to said basis is diagonal (or at least upper triangular)
2. find the transformation matrix A' with respect to a given basis B (for W)

$T:\bR^3\to P_2,\quad T(a,b,c)=a-b+\frac{c}{2}+(b-c)x+\frac{c}{2}x^2$

[stoic python]

nix

[hollow finch]

$[T]_{\epsilon,\epsilon'}=\begin{bmatrix}1&-1&\frac{1}{2}\ 0&1&-1\ 0&0&\frac{1}{2}\end{bmatrix}$

[stoic python]

nix

[hollow finch]

is that at least right so far?

[hollow finch]

<@&286206848099549185>

[spice storm]

You can probably reduce it more @hollow finch

[hollow finch]

wdym

[brittle orchid]

Hey, anyone here?

I'm writing here to request some clarity on my notes as I (most likely out of my own lack of intelligence) am struggling to understand what's going on

Kindly feel free to @ me if you choose to respond 🙂 Thanks in advance 🙂

[strange crystal]

Why would the dim V be 2?

[wintry steppe]

gimme a basis of V

[strange crystal]

{1, t, t^2} ?

[hollow finch]

are they all in V?

[wintry steppe]

alright, you sniped it before i could even follow up

[strange crystal]

not sure i understand your question

[wintry steppe]

i'll go finish my pizza

[hollow finch]

sorry ;-;

[wintry steppe]

it's ok lol

[strange crystal]

oh wait

{1} isn't

[hollow finch]

yeah

[strange crystal]

because p(0) = 0 so 1 can't be in it

[hollow finch]

exactly right

so you have {t,t^2}

those are definitely in V

are they a basis for V?

[strange crystal]

well they're linearly indepedent and in the span of V so they'd have to be correct

[hollow finch]

yeah for sure

there arent any quadratics (or less) that pass through (0,0) that cant be written as at+bt^2

so we got it

based on that being our basis, whats the dimension of V then?

[strange crystal]

It's be 2 then since they're 2 vectors in its basis

[hollow finch]

exactly right

[strange crystal]

Just didn't realize how exact I could get this question using the condition p(0) = 0

[floral trail]

ayy

i need help

understandiing

my work

[strange crystal]

I just used the theorem that the dimension has to be <= P_2 since its a subspace of P_2

but glad I cleared that up

[hollow finch]

yeah thats a great way to think about it

nice job 🙂

[strange crystal]

I also have C on this question which I just completely blanked on

Did fine on the rest of that question

[hollow finch]

alright so lets talk about the derivative of an at most 2 degree polynomial

whats the highest possible degree of the derivative of an at most 2nd degree polynomial

[strange crystal]

1

[hollow finch]

there you go

[floral trail]

whats it asking me to do

?

[hollow finch]

this is not linear algebra. check the pin

[floral trail]

ahh alr

[wintry steppe]

maybe you can use condition (i) to find a formula for the matrix

[dreamy iron]

https://cdn.discordapp.com/attachments/488104216678760469/783916090329399316/162017091996745728.png

Is this hard to do?

[shrewd mortar]

it's literally trivial

null(T) is already a part of V

@dreamy iron it's just the identity transformation on null(T) into V

[native rampart]

I think ninny wants to prove null(T) is a subspace

[shrewd mortar]

i mean i took the pic to mean that 1 through 4 were assumed

[dreamy iron]

I can do that. I can prove null T is a subspace of the domain

[shrewd mortar]

right

so then just take the identity map

that suffices

[dreamy iron]

lol. It’s that easy?

[shrewd mortar]

yes, null(T) is already in V

[dreamy iron]

Where’s the trick

[shrewd mortar]

there is no trick

[dreamy iron]

Why am i stumped by this?

[shrewd mortar]

the question confuses me to some extent

are you sure that's the question

[dreamy iron]

It’s just the identity, is what the lecturer said too.....

But i don’t understand it

okay. Imma play with this

[shrewd mortar]

okay

what don't you understand about it say

[dreamy iron]

Oh. I need to show its injective.

[shrewd mortar]

indeed

it's easy to verify that it is indeed a linear map as well

n(a+b) = a+b = n(a) + n(b)

[dreamy iron]

Is this a set theory proof?

[shrewd mortar]

(a, b vectors that is)

n(cv) = cv = cn(v)

uhhh

it's basically just an immediate proof

so if you have n(v) = 0

what is n(v)

also by the way do you know that injective <=> trivial nullspace

i can show you why really quick

[dreamy iron]

<Insert heavy question mark >

[shrewd mortar]

so if you have M(v) = M(w) implies v = w, then that's just M(v) - M(w) = 0

[dreamy iron]

wait. That i do know

[shrewd mortar]

which is just M(v-w) = 0

[dreamy iron]

I can prove that

[shrewd mortar]

okay

[dreamy iron]

I can draw the picture that show the nullspace of T inside V.

I guess Im having trouble showing that map is injective..... even though I see it.

(This forms what is called a short exact sequence.... and I’ve referenced videos on that from YouTube.)

[dreamy iron]

Okay. So after a bit of finger exercises, I can prove that :

1. If the identity map exists between the nullspace of T and the domain of T, then such a map is injective

2. If the identity map exists between the nullspace of T and the domain of T, then such a map is a linear map.

How do I know that the identity map exists between the nullspace of T and the domain of T?

This feels like a stupid question.

It’s like asking how do I know I can assign elements of a set to itself

OR like asking how do I know a subset of some ambient set maps to itself in the same subset

If I just declared “An identity map exists between the nullspace of T and the domain of T”, then would I have to prove that??

[hollow finch]

i dont see why you cant just define it. nullspace of T is a subspace of the domain of T so nothing crazy happens if you define T_1(n)=n since youve already proved its injective right?

this question is a bit confusing to me in the first place

[dreamy iron]

Okay i will define such an object exists

[uncut wave]

tex:
Let $A$ be an $n * n$ matrix with real entries such that $A^2 + I = 0$, then n is even.

Proof:
$det(A^2) = det(-I)$
We have $det(-I) = (-1)^n$ - (1)

If $\lambda$ is an eigenvalue of $A$ then it satisfies the characteristic equation $ x^2 + 1 = 0$

So we have $\lambda^2 + 1 = 0$

$$ \lambda = +i, -i$$

Since the determinant of a matrix is the product of its eigenvalues then we have

$$ det(A) = i(-i) = 1$$

$$ det(A^2) = det(A)^2 = 1$$ - (2)

From $1$ and $2$ we have $$ 1 = (-1)^n$$ so we have $n$ is even.

Is this proof correct?

Also is the result true when we have complex matrices ?

[stoic python]

Zenquiorra

[half forge]

can someone explain too me why this question is false

im not understanding it

[dusky epoch]

@uncut wave since when is det(A^2) = det(-I)?

@half forge T-cyclic subspace?

[half forge]

im not sure what they mean by that

im lil confuse on that

[dusky epoch]

that's a problem

clear yourself up on that maybe

i have a hunch tho: is the T-cyclic subspace generated by v just the span of {v, Tv, T^2v, T^3v, ...}?

[uncut wave]

Can I not do
A^2 = -I
And taking determinant on both sides?
@dusky epoch

[dusky epoch]

oh yeah, my bad.

pointed out the wrong thing

det(-I) = I^n ??

the right hand side is the identity matrix

[uncut wave]

Sorry, that was a typo, it has to be (-1)^n

[dusky epoch]

no, it's (-1)^n.

[uncut wave]

Yes yes, assuming that, is the rest of the proof correct?

[dusky epoch]

okay so this is a bit rounabout but yeah seems ok

[uncut wave]

Ok, can you please comment on what happens in the case of complex matrices?

[dusky epoch]

no statement can be made about the parity of n in this case.

take $A = iI_3$

[stoic python]

Ann

[uncut wave]

Ok, that makes sense. Thank you!

[brittle orchid]

I'm writing here to request some clarity on my notes as I (most likely out of my own lack of intelligence) am struggling to understand what's going on

https://cdn.discordapp.com/attachments/540211747613704221/783846054621610025/unknown.png

Particularly in the 3x3 matrix which is the formal notation for the real canonical form

Please @ me if responding, thanks in advance!

[brave pier]

Sorry for a very basic question, I do not understand change of basis very good atm. But how do you determine the "Change-of-coordinates" matrix from a Basis?

[tame mural]

wut is ℝ and wut is cake?

[uncut wave]

tex:
Find the eigenvalues of a differentation operator D,
$$ D : (P_n(R) - P_n(R) $$

So, suppose $p(x)$ is an eigenvector for the eigenvalue $c$ , so we have $$D(p(x)) = c p(x)$$

Let $p(x) = a_0 + a_1 x + a_2 x^2 ...... + a_n x^n$ , we have $D(p(x))$ has degree $n - 1$ for a polynomial $p(x)$ of degree $n$.

Now how do I find value of $c$ here?

[stoic python]

Zenquiorra

[native rampart]

There is no eigenvector

[uncut wave]

@native rampart So what about I + D?
$$ ( I + D ) p (x) = c p(x) $$
$$ p(x) + p_prime(x) = c p(x) $$
$$ p_prime(x) = ( c - 1) p(x) $$
Is there no eigenvalue for this $ I + D$ either?

[stoic python]

Zenquiorra

[uncut wave]

Sorry, here p_prime(x) = D(p(x)), I didn't write the latex right

[rose umbra]

what the arrow above the X means

is this just another sign for vector?

[toxic dome]

yep

an over arrow is an indicator of a vector, along with bolding or a tilde underneath :)

[half ice]

Mind you, being a member of R⁴ implies vector as well

[native rampart]

If (I+D) has to have an eigenvector,if has to be 1(if the eigen vector has degree n,look at the x^n term)

[half ice]

At least usually I suppose

[toxic dome]

^

[rose umbra]

ty

[native rampart]

That would imply D(p(x))=0,which means p(x) is a constant polynomial

[uncut wave]

@native rampartOk, so in the case of D(p(x)) = c p(x) , as you said there is no eigen vector because we are getting an eigenvalue of 0 ?

[brave pier]

how would I find the matrix for an image? It goes from R4 to R3 and I got some samples where the input into R4 gives R3 values. Do I create a matrix of R4=R3 and I add 0's at the bottom of R3 values?

Since I have the values, for example: T(1,-1,0,0)=(3,0,0). Do I create a matrix of the values that are given?

[uncut wave]

And for I + D case, we are getting (I + D)(p(x)) = p(x), means p(x) is a constant polynomial and eigen value here is 1

[native rampart]

You are getting (I+D)(p(x))=p(x)

[uncut wave]

Oh yes yes, that was typo.

[native rampart]

Yea, that's it

[dusky epoch]

p_prime
you know you can just say p' right

[uncut wave]

Oh alright!

[stoic python]

mirzathecutiepie

[native rampart]

Let's say the span of both sets are same,that means the vector you removed will be in new span

Yes

[desert rapids]

Something I can actually do XD

[dire thunder]

axler's proof of this is elegant btw

[dire thunder]

@wintry steppe well

Since the list is not linearly independent there is another way to represent 0 as linear combination, except for having all coefficients zero

brb

wot

somehow this is proof of linear dependence lemma

it says basically that if you find vector in span if (v_1, ... v_k) you can remove this vector from the list

wait

it is another theorem

[stoic python]

mirzathecutiepie

[dire thunder]

yes

i mean that is

like if no vector was in span of previous ones you actually would have linearly independent set

i mean {(0,1), (1,0)} is linearly independent

but {(0,1), (1,0), (a,b)} is not

as you can easily verify

is not this theorem proof of which you were screening

sorry occupied please move

(to flaming)

[topaz wadi]

Ok sorry

[dire thunder]

ok so let me see

ok yes

but yes flaming

[topaz wadi]

?

[dire thunder]

you basically project to the plane containing one of the vectors

iirc

[topaz wadi]

Oh

Thanks

[dire thunder]

@wintry steppe

[stoic python]

$\frac{f(b)-f(a)}{b-a}=f'(\xi)$

$\frac{f(b)-f(a)}{b-a}=f'(\xi)$

[dire thunder]

(i am lazy to relatex axler)

[stoic python]

$\frac{f(b)-f(a)}{b-a}=f'(\xi)$

[dire thunder]

probably this explanation would be clear tho it is ~same

WELL

well

not necessarily w_1

just some of w's

but yes

but uhh

u_1 is in the span of w_1 with the rest of the w's, and therefore w_1 is in the span of u_1

this may be worded better

as some w_j is in span of (u_1, w_1, ..., w_{j-1})

thus we can remove it

you will get used

[stoic python]

mirzathecutiepie

[dire thunder]

idk

you can try to prove it

or find countexample

but looks to be true

oh wait no

@wintry steppe

what if v = a_1v_1 + ... + 0v_s+a_nv_n?

[wintry steppe]

linear algebra

[dire thunder]

privyet ttera

[topaz wadi]

When we find the dot product of two dimensional vectors, we are basically projecting one vector onto another (in other projecting one vector onto a one dimensional space). I’m wondering if we take the dot product of a three dimensional vector A with two other three dimensional vectors C and B, are we projecting A onto a two dimensional space formed by B and C?

If so will A.B and A.C would be the two components of A in the plane that B and C are in?

A simple yes or no for both the questions will be sufficient 🙂

@dire thunder

[half ice]

@topaz wadi
Note the important relation
a•b = |a||b|cosθ
Which works in 2D and 3D.

The length of the projection of a onto b is given by right triangle trig:
|a|cosθ

So a•b/|b| gives the length, even in 3D

[topaz wadi]

Yeh I figured

Thanks

So basically instead of A.B it would be A.B/|B| and similarly for A.C?

@half ice

For the components

[half ice]

@topaz wadi
Oh I see, haha I misunderstood. No the process is different for a plane.

• Get the normal vector of the plane. I'll call it μ. This measures the direction that any other vector can "not be in" the plane.

• Project A onto μ. This is the height that the vector is off.

• You want A - ProjA(μ)

[honest imp]

would the first and last be the only ones that work?

[topaz wadi]

@half ice alright thanks

[dire thunder]

@wintry steppe because in statement of linear dependence lemma, it speaks about k > 1

somehow

but in ur case he does not restrict v_1 to be nonzero

and 0 is in span()

but like in theorem u assume u_1 being nonzero

no independent list can contain zero vector

[dire thunder]

and in theorem u assume that (u_j) is independent

for j step also notice

that u do not remove any of u's

since u's are independent

happens

i advice you not focus on proof

i mean get the idea

read further

sometimes return back

[honest imp]

would these be the correct dimensions? felt like i found them correctly

[tame mural]

Is it always possible to find the hamiltonian path through the cayley graph of a symmetric group, with 2 generators (such as a transposition and and n-cycle)?

And is it reasonably easy to find it?

[viscid kernel]

@honest imp Well I im blt good at working with polynomial spaces but I do know that the sum of the dim of ker and range should be 3 ( 1ste example ). Since dim(V) = dim(ker) + dim(range)

[edgy kraken]

how would I go about solving this?

[hollow finch]

how do you normally do an orthogonal diagonalization?

maybe i spoiled the surprise of what this problem is asking, but i assume your professor already taught you orthogonal diagonalization and expects you to make the connection based on what the problem asks of you

[edgy kraken]

the diagonalization we were taught is PDP^-1

[wintry sphinx]

so do you mean to say that changing the letters used to denote matrices somehow changes the meaning of the theorem / equation / formula?

You should note that for orthogonal matrices Q, Q^-1 = Q^T

[edgy kraken]

lmao forgot about that one

[desert rapids]

Cursed = yes

[quartz compass]

yeah it should be flipping the other way

[wintry sphinx]

I disagree with the use of the letter A

it has some reflectional symmetry

B is more appropriate

[mild igloo]

is a scalar multiple of the eigen vector of a matrix also considered an eigen vector of that matrix?

[hollow finch]

yes

[mild igloo]

okay

[hollow finch]

its a good exercise to prove it

[mild igloo]

so if were pretending we dont know the eigen vectors of the matrix and were asked to check and after multiplying the vector is the same as the eigen vector that would confirm it is an eigen vector?

cuz technically the vector is a scalar multiple of itself being multiplyed by 1?

[hollow finch]

Yes it would be an eigenvector with eigenvalue 1

[mild igloo]

so for example I have to check if the vector [3, 3, 3] is an eigen vector of the matrix and after multiplying I get [0, 0, 0] that means it IS an eigen vector?

sorry it might seem like im asking the same question over and over

[hollow finch]

Vectors in the null space are indeed eigenvectors with eigenvalue zero

[mild igloo]

alright

thank you!

[wintry steppe]

@wintry steppe I know it is kinda late, but I get it this way:
Suppose w1, ..., wn spans V, that is to say: every vector in V can be written as linear combination of w1, ..., wn. If we add the vector u1 of the linearly independent list {u1, u2, ..., um}, we have:
u1, w1, w2, ..., wn, but u1 is a vector in V, that means we can write it as linear combination (as said at beginning) and then it is linearly dependent, then it also allows us taking off a element that satisfies Linear Dependence Lemma (it works for u, but it is useless, since it returns into the list {w1, w2, w3, ..., wn} that we already know it spans V) and if we take off a w, the remaining list spans V. When you add another u in the list, you will have the same thing:
it is linearly dependent (since the list without this late u added spans V) and then we can take off another w. We repeat this process until we have {u1, u2, ..., um, w1, w2, ..., wk}. It shows us the spanning list needs to be bigger or even equal (if it were equal, then it will be only {u1, u2, ..., um}). I do not know if it is wrong somewhere, but as I have noticed: that is how I have thought about this theorem.

Some proofs of linear algebra are really bit intuitive 😔

[mild igloo]

if im asked to find a matrix p that diagonalizes a matrix A its the matrix of eigen vectors?

and then P^-1 AP = A

??

[gray dust]

use linearly independent eigenvectors for P's cols

[mild igloo]

Alright thats what I thought thank you

D would be equal to A?

[gray dust]

no, why would it?

we say P diagonalizes A if P^-1 AP=D where D is diagonal

[mild igloo]

oh

But to find d then I would have to do out p^-1

and the instructions specifically say I dont have to do that

oh

its identity matrix

which would make all non diagonal entries 0?

[1, 0, 0]
[0, 2, 0]
[0, 0, 1]

[gray dust]

A is symmetric which means smth special about w/e P you make

[mild igloo]

you said the p was the matrix of eigen vectors right?

sorry im not seeing how thats relivant

[gray dust]

P not p

[mild igloo]

yes sorry

[gray dust]

A is symmetric so its linearly independent eigenvectors are orthogonal

[mild igloo]

how does that affect my answer

P * P^1 = I right?

I just dont see how that means anything for the equation

if I dont need to calculate P^-1 that means the contents of P and P^-1 are irrelivant

A*I = D?

[gray dust]

do you know what's special of P if its cols are orthonormal?

[mild igloo]

no

it cannot be inverse?

[gray dust]

do you know what orthonormal means

[mild igloo]

no I thought you just meant orthogonal

[gray dust]

what does orthogonal mean

[mild igloo]

perpendicular

[gray dust]

not the geometric sense

[mild igloo]

then im not sure

[gray dust]

dot product 0

[mild igloo]

so PA would be 0?

zero vector?

sorry I was mistaking dot product for something else

[gray dust]

i'm talking about orthogonal vectors

[mild igloo]

yes

okay dot product = 0

[gray dust]

orthonormal vectors are orthogonal AND each have length 1

[mild igloo]

okay im missing the peice of knowledge that ties all this together then

still dont quite understand what this means in terms of the equation of D

I know what P vector consists of

I assume this all means that P^-1 will have ome special property?

[half forge]

can someone tell me why this question is false

[gray dust]

@mild igloo you should remember for the rest of your life that if a matrix has orthonormal columns or rows then its transpose is its inverse

[mild igloo]

okay so thats what I was missing then

HOWEVER(Sorry im being a dumbass) I still dont see how that effects P^-1AP

cuz P^-1* P is still equal to identity matrix right?

[gray dust]

A being symmetric means you can find orthogonal eigenvectors to make P. if you also normalize em, they're then orthonormal. then P^-1=P^T

[mild igloo]

right I have P tho

even with that ruel doing that would still be finding p^-1

which she said we didnt need to do

[earnest vessel]

@half forge think about the case v≠0, T(v)=0

[gray dust]

then P^-1=P^T
that's what she meant. just transpose P rather than doing longer computation

[mild igloo]

that means that I wouldnt be able to cancle P and P^-1 as identity matrix

which I thought I couldnt but wasnt sure

cuz matrix multiplication isnt associative?

but it is

sorry bro im confused out of my mind

I understand your rule and that I can use it to simply get the inverse of P

but if matrix multiplication is associative and I can do P*P^-1 which is identity matrix idk why I would need to know any of that to do this problem

okay I get it Im not allowed to do that and I HAVE to do the calculations out

and that method makes it much easier to do inverse

So without normalizing the vectors of matrix P I wouldnt be able to even take the inverse of P beacuse its det is 0 like you said

[gray dust]

i never said det=0

[mild igloo]

well when I throw the matrix in an inverse calculator thats what it tells me

oh you said dot product 0

how is it possible that the matrix can not have an inverse but once we normalize it it does

[gray dust]

i never said that

[mild igloo]

I didnt say you did

I put the vector P in a inverse calc online

it said no inverse det = 0

then I normalized it

and I could inverse it

and it matched what you said that it was the transpose

should that not have happened?

[gray dust]

P should be invertible even before normalizing its cols. but i won't check what you did

[mild igloo]

huh

the eigen vectors I used were [1, 1, 1], [-1, 0, 1], [1, -2, 1]

in that order

[gray dust]

ok?

[mild igloo]

w/e nvm

I did the determant its not 0

its invertible

[gray dust]

mhm

[mild igloo]

any chance u would wanna check my answer when im done?

[gray dust]

just use a matrix calc

[mild igloo]

well theyre giving me different answers depending on which one I use

which was the source of the whole why isnt the original matrix invertible

the inverse of the matrix before normalization != to the inverse of the matrix after normalization

sorry for being difficult @gray dust I appreciate the help

[gray dust]

just follow my instructions. you can play with the matrix calc later

you can find orthogonal eigenvectors to make P. if you also normalize em, they're then orthonormal. then P^-1=P^T
after that you're done

[mild igloo]

alright

is this even considered a diagonal matrix?

thats what I got after all the math

as far as im concerned thats not a diagonal matrix

meaning I did something wrong

[round coral]

well you are right this isn't diagonal matrix

better remove this message

[mild igloo]

dang nabbit

[round coral]

what's the question?

[mild igloo]

https://cdn.discordapp.com/attachments/540211747613704221/784161976011980800/dbed929462eb2dc2789150ab31327eb2.png

I think I just did my order of operations wrong

I did AP

then multiplied that matrix by P^-1

when I should have done P^-1*A

then * P?

[round coral]

well old days are gone, if you want to check your calculations, there are so many calculators online

I always check my matrix diagonalization either through verifying it or using symbolab

[mild igloo]

yea Im gonna check

just a little harder cuz I normalized by hand

[limber mantle]

Alright im not very good with maths so im not sure if this particular question belongs here

In part b

[round coral]

well not here

[limber mantle]

I tried equating the original equation to the inverse equation

where can i post this question?

iv ebeen struggling with it for the past 30 min

[round coral]

you can go to the question section , or precalculus

[limber mantle]

ah that makes sense my bad

[mild igloo]

better

@round coral is this considered diagonal ?

[round coral]

yeah it is

but I don't know if it is correct answer

in that question you don't need to make a diagonal matrix, you need the change of basis matrix , that you will get from the eigenvectors

[mild igloo]

I just followed the steps that RokabeJinatrou said

normalize vectors

create matrix p

translate to get p^-1

then do the math

P^-1AP = D

[round coral]

well I haven't studied yet what normalization means, but I can calculate eigenvalues, eigenvectors, diagonalization and change of basis fairly well just plainly

provided if it is a diagonlizable or then jordan form

[gray dust]

normalize means divide by own norm

[idle nimbus]

Is this geometry

?

[wintry steppe]

??

A normal vector means a vector which module equals one

if you want to normalise some vector, you need to make its module to equals one

intuitively:

we can compose this big vector by multiplying some times the red vector

then, we have:
||black|| = red . some times (| means modulo)

why not?

that is what linear dependence lemma states

hmm

you are right, my mistake

It is really unintuitive. I always struggle trying to understand how to use it in some exercises, because most theorem seem to be out of nowhere

I would prefer to work on real analysis while proving something than on linear algebra

me too

when talking about matrices and traces.. uff 🤔

my favourite branch of l.a is operators and stuffs like linear transformations

[brisk hemlock]

heyo. quick question: imagine you're tracking a moving object in a 3d space, in discreet steps. from a given frame of reference you have the pitch, yaw, and speed of the object, and you need to calculate the deltaX, deltaY, and deltaZ of the object. (Z = up/down, X=left/right, Y = forward/backward)

my equations look like this:

delta z: sin(yaw) = delta z / speed
delta y: tan(pitch) = delta y / delta z
delta x: tan(pitch) = delta x / delta y

is this right?

[brittle orchid]

Hi, I've been struggling to get a particular matrix to real canonical form, I've tried this quite a few times now, below is my working.

[wintry sphinx]

@brisk hemlock I wouldn't recommend calculating it that way; you'd also have to define your coordinate system

in particular, you need to define what axis the angles are being measured from

[brisk hemlock]

@wintry sphinx Ok, I get that. I'm not really sure how else to represent it, though, and I'm not really strong enough at math to understand most of this, anyway.

[dreamy iron]

So if i have an arbitrary vector space V and another arbitrary vector space W, then I can always find a linear map between them, namely the zero map, which is linear.....

Are there theorems that’s gonna guarantee that other maps exists between the two vector spaces, that’s not just the zero map?

[wintry sphinx]

if W is just a 0 vector

then you're kinda screwed

[gray dust]

@brittle orchid i'd redo it, try to make the off diagonal entries 0 instead

[brittle orchid]

is there any error in my working though?

[gray dust]

no but it didn't get what we want

[brittle orchid]

yup

and I'm struggling to manipulate the matrix beyond that to bring it to real canonical form

I mean, surely there must be some operation to bring it to the desired form, no?

[tame mural]

yes

it is called wolfrom alpha

it is the way of many students

[brittle orchid]

would it actually give me the step by step solution?

[tame mural]

well only if you pay lol

but yes

only the premium tier shows step by step

😄

[brittle orchid]

😂

[tame mural]

just enter the matrix directly

also that matrix is already reduced

the one you just typed in

[brittle orchid]

how is it already reduced

[tame mural]

you can't go any further

if you fully reduce a matrix

and you don't find the nice identity matrix

it means it's not invertible

[brittle orchid]

you mean it can't be reduced to real canonical form, rather?

since it should be a diagonal matrix

[tame mural]

Then one of your numbers went screwy somewhere

because that matrix you typed in has a 0 row

there's nothing you can do with that row anymore

[brittle orchid]

yes, so the matrix I'm typing in was the final step of my working, I couldn't get the matrix to real canonical form though... any idea where I went wrong? I've been stuck on this problem for the entirety of today

btw I've used two types of notation to represent the same row/column operation just to get used to both since I'm kinda new to this

I'm not performing each row operation twice and the column operation twice, in case it isn't obvious

[tame mural]

The final row reduced form should be...

[1 0 1, 0 1 -2, 0 0 0]

[brittle orchid]

This isn't reduced row echelon form / gaussian elimination

[tame mural]

That's according to wolfram

oh

you want the jordan normal form

sorry

am right?

[brittle orchid]

Nope, not the jordan normal form either, though that was last chapter

this is the "real canonical form"

From my lecture notes, for reference

[gray dust]

berry's reducing a quadratic form to a sum of squares

A is symmetric. it's a matrix representation of a quadratic form on R^3

[brittle orchid]

I did not know that, the question just told us to determine the real canonical form of the matrix but thank you, more information is always appreciated 😄

[gray dust]

eg $\m{1&3\3&1}$ represents a quadratic form on $\bR^2$
$$1x^2+3xy+3yx+1y^2=x^2+6xy+y^2$$

[brittle orchid]

ah yes I see what you mean

I was looking at the 2nd matrix (I've done the first one)

[gray dust]

i'd redo the reduction. this time i'd focus on making the entries below the diagonal 0

[brittle orchid]

okay

[stoic python]

RokabeJintarou

[brittle orchid]

does that mean there's certain "steps" which don't eventually lead you to the solution?

if you know what I mean?

because when it comes to solving a system of linear equations, for instance, you can just brute force muscle memory without thinking, you know what I mean?

you don't always have to make the most "efficient" decision

[gray dust]

it's like repeatedly adding 2 to an equation hoping you solve it. it's not wrong but you don't get anything

there's no right set of steps in row reducing but yes there are efficient ones

[brittle orchid]

Also, LockettoJampuu, hopefully you might be able to answer this question in a way I understand because I failed to understand others' explanations

Could you kindly explain the formal definition, in terms of the notation and all, of the real canonical form?

I understand informally: it is a diagonal matrix in which each diagonal entry may be equal to 1, −1 or 0, and where any entries of 1 that appear on the diagonal appear before any entries of −1 and any entries of 0 that may appear on the diagonal, and where any entries of −1 that appear on the diagonal appear before any entries of 0 that may appear on the diagonal

but I don't quite understand with reference to the 3x3 matrix over in the screenshot

[gray dust]

this is on bilinear forms. recall a quadratic form is a bilinear form evaluated on a vector paired with itself. that's what symmetric matrices can represent

A is symmetric. it's a matrix representation of a quadratic form on R^3
reducing a quadratic form to a sum of squares
by reducing to a sum of squares, we pick a basis of R^n, which means changing the usual coordinate system to smth else, in which the quadratic form that A represents now just looks like a sum/difference of squares

https://cdn.discordapp.com/attachments/540211747613704221/784255226113687585/150414892225134592.png

recall this. if we pick a certain basis of R^n then in that basis the below quadratic form then has ONLY square terms eg terms of x^2, y^2, z^2 etc no mixed terms eg xy, yz, xz. this corresponds to the real canonical form of A (having changed basis to the new basis) being diagonal with 1s, -1s, and 0s on the diagonal

[magic timber]

Hi, I have a quick question. If for example, you have the equations, x + y = 4 and 2x + 2y = 8 is it possible to add a third equation and get no solution when you add that third equation? Thanks in advance.

No, not at the top of my head.

would it be something like x + y = 20

kk

Thanks sm

[summer wagon]

Could someone help me do this question?

[dusky epoch]

what do you know about similar matrices?

[summer wagon]

Both have the same invertible so that the linear map is same for both

[dusky epoch]

what?

[summer wagon]

I mean I can simply find the determinants of A and B

Since they are not same they are not similiar right?

[dusky epoch]

you're right. det(A) ≠ det(B) implies A and B cannot be similar.

[wintry steppe]

Similar matrices: A and B are similar if $$\exists P\in M_n(\mathbb{R}) s.t$$ $$A = PBP^{-1}$$

[dusky epoch]

if there exists P such that A = PBP^-1.

[summer wagon]

I see

How about the second one?

[stoic python]

bacon prince

[dusky epoch]

do you know how to find the least-squares solution to Ax = b?

[summer wagon]

Not really

Could you please explain?

[dusky epoch]

there is a formula somewhere in your notes, isn't there

[summer wagon]

AtAx = AtB

?

[dusky epoch]

that sounds promising.

[summer wagon]

I simply find AtA and AtB, then the substitute them into the formula and thats it?

I simply find AtA and AtB, then the substitute them into the formula and thats it?

[blissful pagoda]

I think I know how to do this just confused on what the parameters are

Does this mean x_1 = -x_2 and y_1 = -y_2

And that every time you add vectors it must add to 0

Because if that's the case there's no way it's a vector space unless all vectors are 0 lol it's not closed under scalar multiplication or vector addition..?

(however I'm pretty sure I'm just interpreting it incorrectly)

[subtle walrus]

no

it means that you just define the sum of any two vectors to be 0

i say vectors, but

i mean elements of R^2

[blissful pagoda]

But then if I have two vectors (x_1, y_1) and (x_2, y_2) and they must add to 0 wouldn't that imply x_1 = -x_2 and y_1 = -y_2?

[subtle walrus]

no

we define our addition that way

like

[blissful pagoda]

I thought the addition of two vectors was like: (x_1, y_1) and (x_2, y_2) = (x_1+x_2, y_1+y_2)

[subtle walrus]

it doesn't have to be like that

that's what you would maybe call "standard addition"

but nothing stops you from defining e.g. (x_1, y_1) + (x_2, y_2) = (x_1 * x_2, 42*x_1)

[round coral]

beware this R ^2 may not be the \mathbb R^2 of reals, take it how its defined

[subtle walrus]

or in this case (x_1, y_1) + (x_2, y_2) = (0, 0)

yeah, in this case it not even sure if e.g. x_1 + x_2 even makes sense

it depends on the set R

[round coral]

you don't have to make any sense

[blissful pagoda]

Hmmm I think I have yet to learn this because every time I've added vectors so far it was like this: (x_1, y_1) and (x_2, y_2) = (x_1+x_2, y_1+y_2)

lol

But thank you that's interesting

I will have to change my way of thinking about the question

[subtle walrus]

maybe its easier to exchange the symbol +

and say, write $(x_1, y_1) \oplus (x_2, y_2) = (0, 0)$

[stoic python]

Lochverstärker

[round coral]

because you were dealing in R^2 vector space over reals, that is just one of the many vector spaces,

[subtle walrus]

and use that as your addition of vectors

[blissful pagoda]

Would this new interpretation even be closed under addition and multiplication then?

[round coral]

check that out

[subtle walrus]

under addition, sure

as long as 0 is in R

because every sum of two elements just gets sent to (0, 0)

[blissful pagoda]

Well addition i think yes because (0,0) must be in the subspace otherwise it is not a subspace

Yea

[subtle walrus]

under multiplication, it depends on the set R

and the underlying field

[wintry steppe]

yee it is vector space

[blissful pagoda]

There's nothing special with multiplication though it just states the normal properties of scalar multiplication on a vector so I think it's closed

[subtle walrus]

if R is the real numbers, sure

[blissful pagoda]

Wait I have it in my notes that "Zero vector 0 must be in V, such that for every u in V, u + 0 = u" which is not true here

It satisfies every axiom except this I think, damn

u + 0 = 0

[wintry steppe]

0 is alwys in R

and any vec space

[blissful pagoda]

I mean the second part. u + 0 = u is not satisfied because if I take u + 0 using the definition I have of vector addition, it must be 0

[subtle walrus]

you are right that there is no identity with respect to this addition

[blissful pagoda]

And this axiom is telling me that it must be u

Unless u is the zero vector then it works

[subtle walrus]

indeed

[wintry steppe]

yes it works only if it's zero or 1

so the set is {0,1}

[subtle walrus]

what

bacon, you make no sense

[blissful pagoda]

I think it satisfies 9 of the 10 axioms

Just not this one

[wintry steppe]

wait a minute, I am thinking of a simlar question

[blissful pagoda]

This was the question

Everything except 4 holds

😦

[subtle walrus]

i mean, 5 fails too

(if 4 fails)

[dusky epoch]

in the absence of axiom 4, axiom 5 is N/A

[blissful pagoda]

How so because you can still have a vector (0, 0). When I say 4 fails I mean the u + 0 = u fails but there is still a 0 vector

[subtle walrus]

if 4 fails, there is no zero vector

then you cannot talk about 5

because for it to make sense, you need a zero vector

[blissful pagoda]

But in this case wouldn't there be a zero vector (since the question states that two vectors add up to the zero vector in this vector space). It's just not satisfied that a vector + the zero vector = the vector added to the zero vector

[dusky epoch]

(0, 0) is not the zero vector until the truth of axiom 4 confirms it as such.

and here it does not.

there is no zero vector.

[blissful pagoda]

Oh

I did not know that

Thank you

[dusky epoch]

this isn't the vector space you're used to.

[blissful pagoda]

Okay makes sense ty to both of you

[wintry steppe]

zero vector is defined by axiom 4, not the other way round]

[blissful pagoda]

I kind of handicapped myself because I read the question incorrectly I thought it said confirm that it was a vector space not determine

So for the most part I was trying to prove it was lol

[tame mural]

I prefer the definition that involves a commutative group

it organizes a lot of those definitions into the behavior of something very familiar, the integers under addition

[acoustic path]

Agreed

[hushed dock]

How do you calculate the determinant of this?

why cant i just do 1*(-1)*1?

if we use laplace on a_13

[native rampart]

Why should you be able to do that?

[hushed dock]

then it's 1*det(0,-1;1,1)

then lapplace again on the first row

-1*det(1)

-1

[native rampart]

Yea, You can kind of do that

[hushed dock]

but it gives the wrong result

it should be 1

[native rampart]

It should be 1

[hushed dock]

yea but laplace would give me -1

[native rampart]

No, Laplace gives you 1

[nocturne jewel]

Triangular is when the diagonal is top left -> bottom right iirc

if you're thinking of the fact the det is the product of the diagonal entries

[shrewd mortar]

you get --1

[native rampart]

By laplace, you get 1*( 0- (-1) ) +0 * other terms

[shrewd mortar]

it's just

1*det(0, -1 \\ 1, 1) = 1*--1

[nocturne jewel]

Yeah you expand along the 1st row, which means only the last term is relevant

[shrewd mortar]

taking the det of the lower left 2x2

you are perhaps forgetting that for a 2x2 you take -(product of up diagonal)

that is why you get back to --1 = 1

[hushed dock]

but det(0, -1 \ 1, 1), wouldnt it be -1*det(1) if we expand across the first row?

[shrewd mortar]

the signs alternate when you do this method

[native rampart]

It's negative of that

[hushed dock]

wait why?

[native rampart]

0 * 1- (-1) * (1)

[hushed dock]

but im doing laplace expansion on -1 in det(0, -1 \ 1, 1)

[shrewd mortar]

@hushed dock when you are taking det via this method and your top row is like

a b c d ...

then it's actually a*det(stuff) - bdet(stuff) + cdet(stuff) - ddet(stuff)

etc

so you get a negative once again as you're using the 2nd term in teh top

giving back -1*-1*1 = 1

[native rampart]

In that you have to multiply the result with -1

[hushed dock]

ohhh

[nocturne jewel]

det = 0(thing) - 0(other thing) + 1(0*1- (-1*1))

[hushed dock]

ohhh shit

[shrewd mortar]

in any case you wouldn't reduce until det of a 1x1 matrix lmao

[hushed dock]

ohh shit man

[shrewd mortar]

just take the det of the 2x2

it's basically equivalent

[hushed dock]

its because of the (-1)

[shrewd mortar]

(i mean, it is, but you would probably still be hampered to perhaps a small but still nonzero extent by having extra steps)

[hushed dock]

i could also just sub R3 and R2 and multiply the diagonal entries

[nocturne jewel]

Yeah if you make the matrix triangular

[hushed dock]

alright thanks

[nocturne jewel]

Swap R3 and R1 mean you have -det(original matrix) = 1*-1*1 -> det = 1

[old flame]

one quick question, so $<u,u>= |u|^{2}$, then does $<u,v>=|u|^{2}|v|^{2}$ ?

[stoic python]

Otoro

[native rampart]

No

You are defining |u|^2 that way

[old flame]

what do you mean

[native rampart]

What does |u| mean?

Except for being sqrt(<u,u>)

[old flame]

|u| means its length right ? the magnitude

[native rampart]

That kind of doesn't make sense in most vector spaces

[old flame]

oh...

[native rampart]

Like if you have a space of polynomials and define a norm,How would you intepret length?

[old flame]

doesnt that depend on the inner product ?

[native rampart]

It would

[old flame]

so its always the square root of the inner product

I don't think for that space, length could be easily measured

[native rampart]

Yes

[old flame]

so for example $<u,v>$ then it just stays that way ?

[stoic python]

Otoro

[old flame]

so right now, I want to prove that $<v-Pv,Pv>=0$, I separated into $<v,Pv>-<Pv,Pv>$, then Im not sure what to do

[stoic python]

Otoro

[native rampart]

What else do you know?

[old flame]

I know that $P^{2}=P$ and $|Pv| \leq |v|$, I showed that $Pv \in range P$, $v-Pv \in null P$

[stoic python]

Otoro

[old flame]

Im trying to show the inner product be zero, in order to show that $range P = U$ and $null P = U^{\perp}$

[stoic python]

Otoro

[old flame]

@native rampart

[native rampart]

Is the space over R or C?

[old flame]

eh its just $F$

[stoic python]

Otoro

[old flame]

so could be either

[quartz compass]

or neither

[native rampart]

How would you define a inner product on an arbitary field?

https://math.stackexchange.com/questions/706193/how-can-i-show-that-if-p-in-lv-is-such-that-p2-p-and-pv-leq-v

[old flame]

Well all I know is that an inner product satisfies the 65 conditions

5*

[quartz compass]

I don't think there's a way to do it in general

in particular, trying to take the naive approach to defining a dot product over a p-adic field breaks because you can have a dot product equal 0 with the vector itself being nonzero

and unlike the complex number case where we can take the complex conjugate to fix it, there is only the identity field automorphism, so you'd have to do something more drastic

[bold python]

if I fix a vector x_0 in R^3 and have a sequence x_k in R^3 given by x_{k+1}= Ax_k, k>=0, how to I justify that | |x_k 1|| = C for all k>=1 where is a constant

[quartz compass]

are you asking what conditions you need on A for this to be true?

[bold python]

like I have a matrix A = (1 -2 2; 1 -2 1; 0 0 -1)

I have to explain why | | x|| =C where C is a cosnant >= 0

isn't it rather obvious since the norm is always either positive or 0 and cannot be negative

as negative length does not make sense

@quartz compass

i forgot to add that

(for all k >= 1 || xk | |=C)

[quartz compass]

something doesn't seem right, that matrix has determinant 0

[bold python]

it wants me to justify that norm of x_k = C>=0

where the vector x = (x1 x2 x3) in R3

[quartz compass]

yeah I'm saying that's just not true for that matrix

[bold python]

huh

[quartz compass]

maybe I'm wrong idk I didn't really work it out, I guess an easy way to check is try to prove by induction

$|x_{k+1}|^2 = x_{k+1}^T x_{k+1} = x_k A^TA x_k = x_k^T x_k = |x_k|^2$

[stoic python]

merowo (◡‿◡✿)

[quartz compass]

trying to relate the first to the last, the second to last equals sign is what we are really needing to prove

so we could try rewriting that as

$x_k^T (A^TA-I) x_k = 0$

[stoic python]

merowo (◡‿◡✿)

[bold python]

how would I go on about doing this with induction?

[quartz compass]

well in order to relate |x_k|=|x_{k+1}| it forces this condition on us

[dusky epoch]

what's A?

[bold python]

A is the matrix with columns (1 1 0) (-2 -2 0) and (2 1 -1)

@dusky epoch

if I take the eigenvectors and check the norm of it and find that its >=0

which is a basis of R3

and if those are >= 0 it should hold that every vector is >=0?

@gloomy girder

[dusky epoch]

...

[bold python]

eigenvectors of A*

[dusky epoch]

the norm of any vector is >= 0.

[bold python]

hmm true lol

[dusky epoch]

can you post the entire problem again

i wasn't present when the convo started

[bold python]

okey

Fixed a vector x0 in R^3 and consider the sequence xk in R^3 given by, x_(k+1) = Axk for k>=0, explain why the norm of xk = C for all k >=1 where C >=0

where norm xk = (x1^2+x2^2+x3^2)^1/2 when x = (x1,x2,x3)

and A is the matrix I told you about

[dusky epoch]

uhh

what is x0?

this is clearly not true for x0 = [1; 0; 0].

x1 would be [1; 1; 0], which has a different norm

[wintry steppe]

Can I help?

[dusky epoch]

you?

be my guest...

[wintry steppe]

Which question?

Just did my Linear algebra test today!

[dusky epoch]

Fix an arbitrary(?) vector x_0 in R^3 and define a sequence of vectors x_k given by x_k = Ax_{k-1}, where A is the matrix [1 -2 2; 1 -2 1; 0 0 -1]. Show that the norms of the x_k are all the same.

[wintry steppe]

Ok

[dusky epoch]

good luck proving a false statement, i guess.

[bold python]

Show that norm of x_k = C for all k>=1 where C=>0

[wintry steppe]

$$ x_k =A^k x_0$$

[bold python]

@wintry steppe i was thinking of checking the norm of the eigenvectors of A, if they are >=0 shouldn't all vectors also be >= as the eigenvectors is a basis of R^3?

[stoic python]

bacon prince

[dusky epoch]

@bold python are you sure nothing else is known about x_0?

[bold python]

nope, it just said fix an vector x_0 in R^3

[dusky epoch]

also, the norm of the eigenvectors will ALWAYS be >=0. this information tells you nothing.

[bold python]

ah i see damn

[dusky epoch]

i'm saying your statement is FALSE ffs.

why do you still insist on proving it.

[bold python]

hmm, it says "explain" that || x_k| | = C

[wintry steppe]

$$norm(x_k)^2 = x_k ^T x_k$$

[stoic python]

bacon prince

[bold python]

its the exam from last year

[hollow finch]

Is it an orthogonal matrix?

[dusky epoch]

do you have the exact statement of the problem @bold python yes or no

[dusky epoch]

it is not

[wintry steppe]

A is not invertible lol

[bold python]

Yes I gave the exact statement of the problem

[dusky epoch]

i refuse to believe that you didn't omit anything

[bold python]

lemme

give you it agian

[dusky epoch]

as stated the problem is FALSE

[bold python]

its in norwegian

[hollow finch]

Clearly something is lost in translation

[dusky epoch]

taking x0 = [1;0;0] as counterexample
|x0| = 1 but |x1| = sqrt(2)

[bold python]

I showed that A is dioagnolizble prior

Fix a vector x0 in r3 and consider the sequence xk in r3 given by xk+1 = Axk k>=0

explain that the norm of xk = C for all k>=1 for a constant C>=0

[dusky epoch]

are you sure you have the right A???

[wintry steppe]

Yeah, the determinant is zero

[bold python]

yeah

A is diagonalizable cause dim of eigenspace = the multiplicity of eigenvalue

he eigenvalue -1 has multiplicity 2

[dusky epoch]

uh yeah

no

this is just

not true

take $\bd{x}_0 = \bmqty{1\0\0}$, then $\bd{x}_1 = \bmqty{1 \ 1 \ 0}$

[stoic python]

Ann

[wintry steppe]

It doesn't matter, the statement seems false,
$$ x^T_{k+1} x_{k+1} \neq x^T_{k} x_{k}$$

[stoic python]

bacon prince

[dusky epoch]

$\nrm{\bd{x}_0} = 1$ but $\nrm{\bd{x}_1} = \sqrt{2}$

[stoic python]

Ann

[dusky epoch]

these are not the same

end of

[wintry steppe]

$A^TA \neq I$

[stoic python]

bacon prince

[dusky epoch]

bacon prince congrats you've come to the same conclusion i've been trying to point at here

[wintry steppe]

Yeah

[dusky epoch]

@bold python do you still insist on proving 2+2=5

[bold python]

hmmm wth

[dusky epoch]

am i not getting through to you?

your statement is false, it cannot be proved! you're asking for the impossible!

[bold python]

you are but

[quartz compass]

hey you came the same conclusion I came to before you too haha

[dusky epoch]

but...?

[bold python]

when they say norm xk = C, do they mean they have the same length?

[wintry steppe]

Ask your prof to correct it I guess

[bold python]

all of the vectors have the same lengt h

[wintry steppe]

Yee

A is not an isometry

[bold python]

ah

[wintry steppe]

For isometry,

$$A^{-1} = A^\dagger $$

[stoic python]

bacon prince

[wintry steppe]

Good

[quiet heron]

can someone help me with this

[hushed dock]

@quiet heron reduce the corresponding matrix

for each of the linear systems

[hollow finch]

if you haven't learned how to convert systems of equations into matrix equations than just solve them like you would in beginning algebra

[quiet heron]

so do u mean rudce the equation untill i find the x and y?

[hollow finch]

how do you determine if an equation has infinitely many solutions?

[quiet heron]

if the X and the B are the same right?

[hollow finch]

no

do you mean X and B as in AX=B?

[quiet heron]

no

i meant y=Ax+B

[hollow finch]

yeah not quite

when you reduce the equations to solve for x and y there are 3 possible cases

do you know what they are?

[quiet heron]

no i do not

[hollow finch]

just to make 100% sure, this is linear algebra right and not elementary/intermediate?

[quiet heron]

it is intermediate

[hollow finch]

then you are in the wrong channel. check the pin

you want #prealg-and-algebra

[quiet heron]

ohhhh ok sorry

[acoustic path]

Yo is $oplus$ and $ominus$ just used to show addition and subtraction between vectors/matrices

[stoic python]

Coolie Whipperino

[acoustic path]

Crap

This

[quartz compass]

no, people use regular + and - for that

[acoustic path]

Ight

[gray dust]

oplus can denote a direct sum in different contexts, that of matrices & that of vector subspaces

[wintry steppe]

is it true that in every diagonalizable linear transformation 𝑇 there is a line 𝐿 in ℝ𝑛 through the origin such that 𝑇(𝐿)=𝐿?

[quartz compass]

do you know what an eigenvector is

[gray dust]

T diagonalizable is sufficient but not necessary

[wintry sphinx]

T is almost always diagonalizable

[wintry steppe]

@quartz compass isn't it a vector that just changes by a scalar when we apply a linear transformation to it?

[quartz compass]

yeah good

now think about how that might apply to your question

[wintry steppe]

ahh so it must be true

[quartz compass]

ah, and why is that?

[tame mural]

I believe the diagonalization part is a distraction, IMO?

Don't all non-trivial transformations leave some vector on the same span?

[gray dust]

what's a nontrivial map

[tame mural]

Something other than the 0-matrix

[wintry steppe]

ohhh nooo