linear-algebra

[half ice]

The column space is the span of
[1] [ 2 ]
[5], [ 6 ], and the other two columns lel
[9] [10]

[charred stirrup]

is that not simply all of the elements of the matrix?

[slow scroll]

2(1, 5, 9) + 3(2, 6, 10) + -3(3, 7, 11) - (4,8,12) is the linear combination that corresponds to
P(2, 3, -3, -1)^T

[charred stirrup]

OHH

i get it

thank you so much

a scalar is applied to each column - C (column 1) + C2 (column 2) ...

e.t.c

[slow scroll]

sick np. Basically the column space is like the range of a linear transformation, We know P will map to R^3, but the column space tells us exactly which vectors get mapped to

[half ice]

The span of a set of vectors is exactly that, a scalar is applied to each vector

The column space is the span of the columns

[charred stirrup]

yeah I was just unsure how to do a specific combination for the column space

[dusky epoch]

@half ice [1, 5, 9]^T, [2, 6, 10]^T

[charred stirrup]

is it true that the rank of column space = rank of row space?

[slow scroll]

yea. I think i said Col(A) = Col(A^T) earlier which is not true, but dim(Col(A)) = dim(Col(A^T)) definitely is

[charred stirrup]

thank you

the rank is determined by the number of basic variables right?

[slow scroll]

when you put a matrix in row echelon form, the column pivots correspond to the column vectors that make up the basis for the column space. When you transpose that matrix, the same columns make up the basis for the row space.

nullity is determined by the number of free variables

[charred stirrup]

you're really good lol

thank you

helped me link column pivots to basic variables

[slow scroll]

np. rank + nullity = columns makes sense because every column either corresponds to a free variable or a column in the basis of the column space

[snow snow]

How do I find a in all of this?

I should be able to solve this but I'm a little rusty.

[half ice]

That's a picture not a question

[snow snow]

BA = BC

[half ice]

... And the picture doesn't really convey a question

Wats going on here?

[snow snow]

"The point (a,5) has the same distance from (4,2) as (2,3). What is a?"

[half ice]

Ahh I see

[snow snow]

Sorry if I made that unclear.

It's annoying me a bit because I solved a question like this on a fairly recent test.

[half ice]

Well, we have pythag for that
|AB|² = (a - 2)² + (5 - 3)²
|BC|² = (a - 4)² + (5 - 2)²

[snow snow]

But I don't know if it's a right angle

[half ice]

It's probably not!

I'm taking the length AB, and turning it into a hypotenuse of its own right triangle

(Length of AB)² = (Width of AB)² + (Height of AB)²

[snow snow]

Uugh..

I don't know how to handle that.

(a-2)^2

[dusky epoch]

$(x+y)^2 = x^2 + 2xy + y^2$

[stoic python]

Ann:

[dusky epoch]

sound familiar?

[snow snow]

Ah

The quadratic rule, right?

That's at least what we call it here.

I'm taking a break, I'll be back in about 20 minutes.

[dusky epoch]

here = where?

[snow snow]

Sweden.

I'm back, and I've been focusing on that math problem so hard that I legit forgot how to fold a burrito and it took me like 30 seconds to remember.

[rare barn]

no idea what to do

[rare barn]

<@&286206848099549185>

[broken hawk]

well first of all, can you diagonalize A, cause that's the first step

[rare barn]

I did

P is [ 1 1]

........[0 1]

and D

[2 0]

[0 3]

[dense holly]

its the determinant that shows how a matrix transforms the unit square thing right

im tryna refresh myself on what to think of it as

[rare barn]

axctually i got it (i think) i think Q would have to be P^-1 which i can find right here

[slow scroll]

@dense holly watch the 3b1b video on determinants

[oblique crag]

Can anyone explain 15c to me?

I had b1 basis and b2 b3 not basis

But the book had it flipped, i dont get it. Doesnt independent=basis?

[rigid cypress]

!15m

gei

[oblique crag]

<@&286206848099549185>

[slow scroll]

what is W?

sry i can't read

[oblique crag]

span of s

so the 4 vectors

[slow scroll]

According to my wikipedia sources, $$(\text{Col} A)^{\perp} = \text{Null} A^T$$

[stoic python]

kxrider:

[slow scroll]

@oblique crag

so find a basis for the left null space of A to get the basis for the column space of span(S)'s orthogonal complement

[sage mantle]

hey , so one of my friends sent me her notes but i don't understand how she got that equation , could someone help me understand it please

[dusky epoch]

presumably, she makes x2 and x5 her free variables

and expresses everything else in terms of that

[slim laurel]

Can someone help me with part C? I've found from part b that all 2nd partial derivatives are positive. However, I don't understand how H would be PSD since aj and ak can be negative.

[dusky epoch]

all entries positive may generally not imply PSD

[slim laurel]

yes I'm just having trouble with part c which states that L is convex

meaning H must be PSD

i'm having trouble showing H is PSD using their definition a^T * H * a >= 0

[dusky epoch]

i mean ok, what'd you get for your partials

[dusky epoch]

oof

[snow snow]

@dusky epoch Because of your hint yesterday I managed to solve the problem, thanks! I was on that thing for hours.

[pastel aspen]

i have this determinant

i'm supposed to get Dn= n+1 but i got no idea what i'm doing wrong

first to second line i expanded following the first column, all of the other terms are 0,
second one i expanded following the first line, and all the other terms are also 0, which gives me Dn=2Dn-1 + Dn-2
characteristic equation of this doesn't give n+1

thats a matrix with the main diagonal having 2's and two extra diagonals with -1's in them

nvm i was missing a - sign

[winter reef]

ok then

Hi, I have a problem thats asking whether there exists a basis B such that matrix of transformation in that basis is M (given matrix). What are necessary equirements for that to exist? Same eigenvalues, is that enough? I know the matrix of the transformation in standard basis

[dusky epoch]

same eigenvalues may not be enough. you need M and the standard basis matrix to be similar.

[winter reef]

So finding whether they have same Jordan form should be enough, right? If they do then they are similiar

[dusky epoch]

similar iff same jordan form

yes

[winter reef]

tahnks

and waht does it mean if rank of the matrix - aI (a is eigenvalue) is 4 and the rank of matrix squared is also 4 in terms of Jordan form?

then it isnt similiar to any Jordan matrix?

[dusky epoch]

do you mean rank(A - aI) = rank(A^2) = 4?

[winter reef]

rank(A - aI)=rank((A - aI)^2)=4

[dusky epoch]

oh

so that means rank(A - aI) - rank( (A - aI)^2 ) = 0, which means that there are no jordan blocks of size exactly 1 for eigenvalue a in the jordan form of A

[winter reef]

well but then I get rank( (A - aI)^3 ) = 4 and rank( (A - aI)^4 )=4 as well and this transformation is in R4 to R4

so that means theres no Jordan form of A?

[dusky epoch]

huh what? there's always a jordan form

wait

what even

can you post the problem you're doing

i'm p sure you can't have (4, 4, 4, 0, 0) be the rank sequence of a matrix

[winter reef]

SRYS RY, i mean its alwasy 4

[dusky epoch]

ok wait.

your matrix is 4x4, right?

[winter reef]

yes

I will paste the matrix in standard basis and the one I need to determine whther the basis exists

[dusky epoch]

are you sure you calculated the rank sequence correctly

[winter reef]

det (Br- λ*I) is (λ-1)^4

same with the standard matrix

and matrix in standard basis has two cages 2x2 if I checked correctly

Let me (br-λ)^2 again, Ill paste it

ok I think I screwed up multiplication

[dusky epoch]

,w Jordan form [ [0, 1, 0, 0], [-1, 2, 0, 0], [1, -1, 2, -1

bah fuck

,w Jordan form [ [0, 1, 0, 0], [-1, 2, 0, 0], [1, -1, 2, -1], [1, -1, 1, 0] ]

[stoic python]

Display results online and refine query

Requested by Ann#8192

[winter reef]

yesh

so yeah actually I see where I fkd up

r(br - I)^2 = 1

and r(br - I)=3

so that means its jordan form will be like one block 1x1 and one 3x3, right?

[dusky epoch]

uh

that's a decrease of 2, so that would signify two blocks of size 1

(and accordingly, one block of size 2)

[winter reef]

but q1 (number of blocks of size 1) is 4-3=1

q2=3-1=2

[dusky epoch]

$q_i = r(B^i) - r(B^{i+1})$

[stoic python]

Ann:

[dusky epoch]

number of blocks of size 1 = rank(B^1) - rank(B^2)

...

wait.

hang on.

yes. bc rank(B^1) itself is the number of blocks of size at least 1

[winter reef]

by book says$q_i = r(B^{i-1}) - r(B^{i})$

[stoic python]

dog:

[dusky epoch]

[winter reef]

Can there exist a matrix A nxn such that {I,A,A^2,....,A^n} is linearly independent?

[broken hawk]

uh, I think the answer is no, but I don't quite remember why

[proper crescent]

Cayley-Hamilton

[broken hawk]

...right

though iirc we used this fact to prove CH

[proper crescent]

What

[broken hawk]

but whatever, CH has many proofs

[proper crescent]

Cayley-Hamilton says the characteristic polynomial specifically kills off the matrix

[broken hawk]

not continuum hypothesis you silly

yea, but we used cyclic subspaces

in the proof

[proper crescent]

Isn't that a different story though, you're using there that A^nv, A^{n-1}v, ... aren't linearly independent, and that's more of a dimension count

[broken hawk]

I don't remember the whole proof, but I'm pretty sure the powers of the matrices came up

[winter reef]

Ok so, characterisitc polynomial can be up to nth degree right?

[proper crescent]

It's degree n exactly

[winter reef]

yeah

[broken hawk]

and that automatically gives you A^n = linear combination of previous powers

[winter reef]

wait why?

[broken hawk]

cayley hamilton says char_A(A) = 0

so $A^n + a_{n-1}A^{n-1} + \dots + a_0 I = 0$

solve for A^n

[winter reef]

Ok, I see it now, thank you

[stoic python]

Sascha Baer:

[winter reef]

And how would I use it (since its probably the same deal) to determine whether there exists matrix A in M 2x2(Q) such that A,A^2,A^3,A^4 is a basis of matrices 2x2(Q). Like what does the Q do here?

[broken hawk]

Q as in rationals?

[winter reef]

yes

[broken hawk]

I don't remember, was CH only over C or general fields?

[wet finch]

any field

any ring, even

(commutative)

[rare barn]

Using this, where A is an n x n matrix, how do i find:

e^A is defined :

where Bm is

[vale arrow]

What do you want to find ?

[rare barn]

e^0n and e^In

[vale arrow]

Exp(0_n) ?

Oh ok

For exp(0_n)

Utilize the partial sum

You can show that for every m in N, B_m is I

[rare barn]

im confused though, how do you do it with a matrix...

[vale arrow]

Exp(A) and B_m are defined

Use their definition to find exp(0_n)

To continue
B_m is I for every m in N, because (0_n)^m is 0_n for every m in N*

So the sum of the series will be I

So exp(0_n)=I

So far so good ?

I_n

[rare barn]

yes i understand that

[vale arrow]

Ok, one second to put my phone on charger to tell you how to do the second one

Ok

As for I_n

Use the definition of the partial sum again

You have for every m in N, (I_n)^m = I_n

So for every m in N we have:
B_m=I_n+I_n/1!+I_n/2!+...+I_n/m!

[rare barn]

got it

[vale arrow]

Sorry my phone died

[rare barn]

nw

[vale arrow]

Ok let's continue

You can take the sum into the matrix I_n

So for a fixed m, you will have on the diagonal of I_n, the sum 1+1/1!+1/2!+...+1/m!

When you take n to infinty, the sum of the series will be exp(1)=e

So you will have e on the diagonal

Take e out, you will have e times I_n

So exp(I_n)=e.I_n

Ok ?

[rare barn]

makes sense

[rare barn]

@vale arrow i think i know what to do here, im just confused by what they mean by a^3=0_3 and b^3= 0_3

[slow scroll]

A(A(A))) = the 3x3 zero matrix

[rare barn]

how does that help here though

[slow scroll]

e^A = I + A + A^2/2!
because every term after A^2 is just zero

e^A = I + A + A^2/2!
e^B = I + B + B^2/2!

so does e^(A+B) = ( I + A + A^2/2!)(I + B + B^2/2!) ? Well, presumably not, but thats what you gotta prove

[rare barn]

ohhh

got it

ty sm!

[slow scroll]

np np

[rare barn]

wait, so now i can just add the matrices and show that theyre not equal right

[slow scroll]

yea u can do that

[rare barn]

@slow scroll srry 1 more question, for doing e^A+B, can I use B_m=I_n+(A+B)/1+(A+B)^2/2...

[slow scroll]

hmm the tricky part is that (A+B)^n isn't necessarily ever zero, so the series e^(A+B) doesn't necessarily ever terminate

[fierce bolt]

Trying to make sure I understand this. Is the dimension of the null space of (A - λI) the amount of free variables of rref(A - λI)?

[slow scroll]

yea. thats true for any matrix

[fierce bolt]

oooh cool, thanks for the clarification

[slow scroll]

np

[fierce bolt]

lol it feels nice to understand something in this class for once

[rare barn]

@slow scroll then how would you do it?

[slow scroll]

@rare barn im not quite sure tbh...

[rare barn]

hmm alright, could u help me w/ 1 more q

[slow scroll]

maybe :p

[rare barn]

ill ask my teacher abt the other one

[slender yarrow]

well since A and B are strict upper triangular matrices (ie upper trig + diagonal filled with zeroes), A+B is also one

so it will 'decay' like A and B

(not rigorous at all ik, just showing the idea)

[rare barn]

i see the ides

idea*

@slow scroll you still there?

[slow scroll]

yea im thinking

[slender yarrow]

well how do you find the power of a diagonal matrix ?

[slow scroll]

oh wait a second

[rare barn]

PD^nP^-1

[slender yarrow]

i'm not talking about diagonalisation lad

[slow scroll]

the diagonal entries of a diagonal matrix are just A^n_ii = d^n right?

[slender yarrow]

yea, if the d_ii are the original entries in A

[slow scroll]

i was trying to figure that out by summing rows*columns, but ofc that isnt' necessary 🤦

$D^2_{jk} = \sum_{m=1}^n D_{jm}D_{mk}$ but when $j=k$ we get $D^2_{jk} = \sum_{m=1}^n D_{jm}D_{mk}$. Now, $D$ is a diagonal matrix so $D_{jk} = 0$ except when $j=k$. That means all of the terms of the sum are zero except when $j=m=k$. Therefore $D^2_{jk} = 0$ when $j\neq k$ or $D^2_{jk} = (D_{jk})^2$ when $j=k$

[stoic python]

kxrider:

[slow scroll]

@rare barn

$e^D_{jj} = \sum_{m=0}^{\infty} \frac{D^n_{jj}}{m!}$

[stoic python]

kxrider:

[rare barn]

hmm that makes sense

is that all we need to show it?

also dont you mean E^d_jk

dont see why its _jj

[slow scroll]

i think so. To prove that the thing i wrote above holds for all D^n, you may have to use induction or something, but idk if your teacher cares for you to get that pedantic about it.

e_jj^D corresponds to the diagonal entries of e^D. same thing as e_jk^D when j=k

[rare barn]

oh yep, got it, and i think my teacher is fine with this tbh

sorry, but i have one LAST question

[slow scroll]

okiee

[rare barn]

tysm 😄

Let A be a diagonalizable matrix, so that A = PDP^-1. Show that e^A= Pe^DP^-1

[quaint heart]

Well we know that (PDP^{-1})^n = PD^n P^{-1} right

I think it follows from that pretty easily

@rare barn can you see why?

[rare barn]

no...

[slow scroll]

plug stuff back in the series and substitute stuff

[wintry steppe]

helppppp

http://mathb.in/32664

[rare barn]

what do i plug back into the series though....

[slow scroll]

$$e^A = \sum_{m=0}^{\infty} \frac{A^m}{m!} = \sum_{m=0}^{\infty} \frac{??}{m!}$$ hint

[stoic python]

kxrider:

[rare barn]

since A = PDP^-1 can i use that?

[slow scroll]

ye

[rare barn]

ok, then how do you show e^D

[slow scroll]

well, A^m = (PDP^-1)^m = P D^m P^-1.....

u can take P and P^-1 out of the sum....

[rare barn]

i dont see where this is going...

[slow scroll]

$$e^A = \sum_{m=0}^{\infty} \frac{A^m}{m!} = \sum_{m=0}^{\infty} \frac{PD^m P^{-1}}{m!} = P\left(\sum_{m=0}^{\infty} \frac{D^m}{m!}\right)P^{-1} $$

[stoic python]

kxrider:

[rare barn]

i had this too'

oh

oh

dude im so stupid, i didnt realize that the sum there is e^D

my bad

[charred stirrup]

can you only identify pivots in reduced form? what about just echelon form?

[half ice]

I figure you generally identify pivots in rref anyway?

[slow scroll]

i usually identify pivots in row echelon

i think the only time i really use rref is when i am finding null space or something

[charred stirrup]

we identify pivots by the first non-zero entry of the row, but consequently, the column that pivot in is defined as a pivot column?

[slow scroll]

the column that the the row pivot entry is located in is a pivot column

[wintry steppe]

hey guys, does swapping the rows of a matrix change the determinant?

[slow scroll]

yea, it flips the sign of the determinant

[wintry steppe]

thank you!!!!

[slow scroll]

np

[wintry steppe]

what if theres a 3x3 matrix, determinant is given prior to change
one of the value of row 1 is doubled
how might that change the determinant?

[slow scroll]

doubles the determinant. Its a linearity thing

[wintry steppe]

how might something like this change the determinant?

sorry for multiple questions :/

[charred stirrup]

does your teacher/professor want you to come up with a general solution like what rider is saying?

cause you could just take the determinant, wouldn't that give you the answer?

[wintry steppe]

im currently prepping for finals and the prof gave up practice exams..... but there are no solutions to them

[slow scroll]

The idea is that for a matrix [v1 v2 v3], the determinant satisfies the properties
D(av1,v2, v3) = aD(v1, v2, v3)
D(v1 + v2, v2, v3) = D(v1, v2, v3) --invariant under column operations
D(v2, v1, v3) = -D(v1, v2, v3) --obtainable from other properties

because the determinant works out to be the product of the diagonal entries of a triangular matrix, det[v1 v2 v3] = det[v1 v2 v3]^T

[wintry steppe]

@charred stirrup it ends up as 5x -2xy -39 = 0

as i was trying to solve for determinant

[half ice]

You can't take the determinant directly, you need to use determinant properties, as kxrider said

[charred stirrup]

im just retarded dont worry about what I said

[wintry steppe]

@slow scroll @half ice thank you! Ill look more into determinant properties

thank you for taking your time to help me 😃

@charred stirrup you too ❤

[oblique crag]

So on #17 for the reflection

I got the exact same answer as the solutions. But without the 1/65 can anyone explain where the 1/65 is from

[plush mural]

I feel like it should be (3,-2)

[wintry steppe]

guys

if E is a complex space, why E is also a real space ?

[dusky epoch]

because R is a subset of C 👀

[slender yarrow]

tfw C is a subset of R @dusky epoch 👀

[dusky epoch]

[slender yarrow]

complex space => real space isn't true though

[dusky epoch]

why not

you can take a complex space and just inherit scaling from it but for real scalars only, and get a real space

[slender yarrow]

yea time to die ig

[solemn basalt]

Hey friends! I am currently learning proof in my linerar algebra course, and don't really know how to continue at this exercise im working on.

The question is
(n+2)^3 + n^3 (divisible by 3) => n (not divisible by 3) , n is a natural number

What I have done: We learned a lot of complete induction but that does not seem to be the way to go

wait

wrong channel

[brittle juniper]

That's not really linear algebra x')

indeed

[solemn basalt]

sorry ill be on my way!

our course is called linear algebra so I thought it would be that

welp

[brittle juniper]

#elementary-number-theory is maybe more appropriate

[solemn basalt]

nah I'm supposed to write in a help channel, just forgot the discord server I am on

but thanks!

[brittle juniper]

( ^_^)

[summer dagger]

is a basis for a subspace another way to think about it just the fewest amount of vectors to describe a span?

[dusky epoch]

that's exactly what a minimal spanning set is

[summer dagger]

so I'm wrong?

[dusky epoch]

no, you're right

you're exactly right

i'm just saying there's a word that captures your intuition exactly

[summer dagger]

ok ok ty

[wicked trellis]

Can anyone help me with (6)

[dusky epoch]

what was your answer for (5)?

as in, yes or no

[dusky epoch]

ok so you have your P and your D, great

so now

consider this:

consider this a hint

[wicked trellis]

So I will have to invert p first as well

[dusky epoch]

seems so.

[wicked trellis]

So I’m stuck because d^2019 is still pretty complicated to write down

[half ice]

Using diagonalization to get powers is black magic

[winter reef]

Or u can find Jordan form

[dusky epoch]

how is D^2019 complicated to write down?

D is a diagonal matrix

[winter reef]

Wait So thats True for any A thats diagonalizable?

[dusky epoch]

@wicked trellis you know how diagonal matrices behave under multiplication, right?

@winter reef what are you talking about

[winter reef]

A =PDP^1 where d is diagonal only if a is diagonalizable?

[dusky epoch]

a) ^-1, not ^1
b) that's literally the definition of diagonalizable

[wicked trellis]

It would be my values 4^2019, 5^2019, 5^2019 in a diagonal wouldn’t it

[dusky epoch]

yes!

[wicked trellis]

Unless I just leave it as that form it would be a really large number

[dusky epoch]

leave it as is.

[wicked trellis]

Oki, thanks =)))

[winter reef]

And whats P? Idebtity Matrix from standard basis to basis of egein vectors?

If not then what basis should it be?

[wicked trellis]

Ya

[half ice]

Take a matrix A.
Let's say that there's a way to write A = PDP¯¹ such that D is diagonal
Then we say A is diagonalizable and it's easy to take the powers of it

[wicked trellis]

Does this make sense

Multiplying those matrixes looks pretty complicated

Oh nvm

It’s not

Feel like I messed up somewhere

[half ice]

Yes

If that's the correct D idunno

But that's the right way to take the power

[wicked trellis]

Okay

Thanks

Also that’s not RREF because you still have non zero values in v3 in the second and first row @charred stirrup

[charred stirrup]

Am I using the incorrect rules?

[wicked trellis]

Yea The matrix you put is in REF form but not RREF form

Because if you look at the third column there’s non zero values above the pivot

[charred stirrup]

AHH I SEE

leading ones are defined by the "1" in each ROW

and the entries of the column it's in must then all be zero (except for the 1)

thank you two

@wicked trellis @half ice

[wicked trellis]

Yes

[dense holly]

if i added another row of zeroes would this still be a basis for R3

[slow scroll]

u mean the columns? no, but it will be a basis for a 3d subspace of R^4

[dense holly]

sorry im not clear on what you just said

i did mean another row of zeroes

[slow scroll]

err i wasn't clear, im assuming u r asking about whether the columns of the matrix span a certain space?

[dense holly]

this is what im tryna answer

i get span but im unclear on bases

whats the difference

[slow scroll]

a basis for a vector space is a set of vectors that spans the space (you can represent every vector in the space as a linear combination of the basis), and representation of each vector in the space is unique.

if {v1, v2, v3} is your basis, representation of vectors in the space are unique if the only way to represent the zero as a linear combination is
0v1 + 0v2 + 0v3 = 0

(1, 0, 0) , (0, 1, 0), (0, 0, 1) form a basis for R^3 because i can represent every vector as a linear combination
(a,b,c) = a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1) for all a,b,c in R

and the only way to write the zero vector is when a,b,c = 0

[half ice]

I feel you're giving us the reduced matrix, and not the set itself. Note the very important distinction between the two.

[summer dagger]

If B is invertible, then the determinant of A equals the determinant of BAB^-1

kinda lost on how to show this if someone could give me a pointer assumming that its cause BAB^-1 = A

[slow scroll]

det(BAB^-1) = det(B)det(A)det(B^-1)

[summer dagger]

det(A) = det(B)det(A)det(B^-1)

[slow scroll]

don't want to give it away

[summer dagger]

ahh you get 1 = det(B)deat(B^-1)

gonna assume thats a property i should know

[slow scroll]

indeed 1 = det(I) = det(BB^-1)

[summer dagger]

ty

[slow scroll]

np

[quartz kraken]

can someone kindly explain to me the overarching idea to differentiate PMD and SVD? I know they are both applications of eigenvalues/eigenvectors, but I don't understand their applications very well.

[half ice]

@summer dagger
More generally, det(A¯¹) = 1/det(A)

Which is why zero determinant matricies have no inverse.

[oblique crag]

Can anyone help me understand 7

[slow scroll]

it looks like there are multiple answers to that to me...

[oblique crag]

Yeah theres multiple but theyre askin which can be possible

Its not multiple choice

I got it figured out though, thanks guys cx. I do need hep with a proof though

[slow scroll]

okiee

[oblique crag]

#50 if u were offerin to help 😛

[slow scroll]

think about what it would mean for the original statement if v_n was a member of span(v1,v2, ..., v_{n-1})

[sage mantle]

my teacher said in class that a polynomial of degree 0 = - infinity but can't 8 for example be a polynomial of degree 0 for example 8*x^0

[dusky epoch]

did your teacher actually say that the zero polynomial has degree -infinity?

[sage mantle]

well it's written in the notes of a friend

as an arbitrary choice we say that deg (0) = -infinity

[dusky epoch]

deg(0) = -infinity

0 here refers to the zero polynomial

[sage mantle]

oh okay i see

[broken hawk]

it's not really that arbitrary a choice imo

it's the only choice which preserves order of degrees in a sensible way and keeps them sensible wrt addition and multiplication

that is deg(f+g) ≤ max(deg(f), deg(g))

and deg(fg) = deg(f) + deg(g)

[plush jacinth]

stupid stupid question

so i have this vector space C[0,1]

of all continuous f: [0,1] --> \mathbb{R}

with the weighted inner product <f,g> = \int_0^1 f(x)g(x)x dx

ok so say i have three functions

i want see if they;re orthogonal

can i still just toss them in a matrix and see if det(matrix) = \pm 1

[dusky epoch]

no

[plush jacinth]

or should i do the inner products individually

aiight thx

[brittle juniper]

matrices in infinite dimension, that's news to me

[plush jacinth]

-_-

[wintry steppe]

how rg(A) = 1 ?

There is 0 column(0,0,0) ?

[half ice]

What's rg(A)?

Is that rank? Rank is 1 as you can write this as a single row of 1s

[wintry steppe]

rank

yes sry

@half ice isn't the rank the non zero column number ?

bcs here if I row reduce, it is the non zero row numer

[half ice]

If I fully row reduce, I can write:
[1 1 1]
[0 0 0]
[0 0 0]

See there's only one pivot, meaning rank is one.

[wintry steppe]

so if we row reduce, we check the number of non zero row ?

or we can check dim(vect(column)) ?

[half ice]

Either

The dimension of the column space is the same as the amount of pivots

[little cairn]

Does anyone know if there are any particular special properties of matrices that are symmetrical across the diagonal?

I've been working with adjacency matrices and noticed that every undirected graph is symmetrical across the diagonal and was wondering if this could somehow be used to imply anything else

[proper crescent]

Orthogonally diagonalizable

This is the spectral theorem

[little cairn]

Awesome @proper crescent thank you very much i'll check it out

[full palm]

hello

check if the subset W is a vetorial subspace of the vetorial space V.

V = M_2 , W = { (a b). a, b, c are real numbers}

M_2 = square matrix 2x2

below that (a b) there's a (-a c)

what i did is to prove if W is a subspace of V, but i'm not sure if that's right. Could someonbe help me, please?

[rare barn]

So I'm supposed to diagonalize A by writing A=PDP^-1

ive already found eigenvalues and eigenvectors

[lyric sphinx]

then if X_1 and X_2 are your eigen vectors you can let P:= (X_1 X_2)

[rare barn]

got it

ill do that rn,then i got 1 follow up q

is the diagonal matrix just the eigenvaluyes

[lyric sphinx]

y

[lyric sphinx]

should be it

[full palm]

i'll try to be more specific with my question (i think)... if i prove that the subset W, is a subspace.... is ok to say that W it's subspace of V?

[rare barn]

@lyric sphinx P is a singular matrix though...

[lyric sphinx]

@full palm W is indeed a subspace of V

@rare barn how so?

[rare barn]

wait no

my bad

i was using 0 and 0 instead of 1 and

1

[full palm]

by that matter, W must be a subset of V. Right?

[lyric sphinx]

yeah

but you have it wrong

you only try to prove that W is a subspace of V

if you know that W is a subset of V

[full palm]

got it!

[rare barn]

i got this

second part of this q is:

[lyric sphinx]

you have proven that A = PDP^-1

compute A^n

[rare barn]

so is it x_n= (what i just computed)^n [x1 x2]?

kinda confused

<@&286206848099549185>

[full palm]

Could someone please confirm if i've made the correct steps to prove that W is a subset of V, please?

[rare barn]

any1?

[full palm]

<@&286206848099549185>

[rich forge]

yoo can someone confirm the differentiation operator on the basis {1, x, x^2}

is equal to
[0 1 0]
[0 0 2]
[0 0 0]

[wintry steppe]

looks right

[rich forge]

and then the hermitian transpose is just
[0 0 0]
[1 0 0]
[0 2 0]

but this gives eigenvalues of all 0

[dusky epoch]

that is correct

[rich forge]

is hermitian transpose of a matrix the same as adjoint of a matrix

[rare barn]

generally do u need to show steps for matrix multiplcation on tests or can you just write the answer

[dusky epoch]

what does your prof say?

[rare barn]

its an online course so i have no idea

[slow scroll]

i don't think i have ever shown steps for matrix multiplication, but i have also never taken a test in LA 🤷

[half ice]

Matrix multiplication is very likely not what you'll be tested on for an LA final exam

So no don't worry about the steps

[rare barn]

yeah, what i was thought. just wanted to make sure

[rare barn]

Hi another formatting question

Show that if : $$ D = \begin{bmatrix} d_1 & 0 & ... & 0 \ 0 & d_2 & ... & 0 \ \vdots & \vdots & \ddots & \vdots \ 0 & 0 & 0 & d_n \end{bmatrix} $$ is a diagonal matrix, then $$e^D = \begin{bmatrix} e^{d_1} & 0 & ... & 0 \ 0 & e^{d_2} & ... & 0 \ \vdots & \vdots & \ddots & \vdots \ 0 & 0 & 0 & e^{d_n} \end{bmatrix} $$

[stoic python]

Zo:

[rare barn]

how do i just make this into one line

[slow scroll]

Show that if : $ D = \begin{bmatrix} d_1 & 0 & ... & 0 \ 0 & d_2 & ... & 0 \ \vdots & \vdots & \ddots & \vdots \ 0 & 0 & 0 & d_n \end{bmatrix} $ is a diagonal matrix, then $e^D = \begin{bmatrix} e^{d_1} & 0 & ... & 0 \ 0 & e^{d_2} & ... & 0 \ \vdots & \vdots & \ddots & \vdots \ 0 & 0 & 0 & e^{d_n} \end{bmatrix} $

[stoic python]

kxrider:

[slow scroll]

well, in general, $$ $$ makes a new centered equation, while $ $ is just inline math

[rare barn]

got it, didnt realize the difference between $$ and $

[bold python]

I have a matrix $$ \begin{pmatrix}0.4 & 0.3& 0.3\0.3 & 0.5 & 0.2 \ 0.3& 0.2 & 0.5\end{pmatrix}$$ and I want to find the eigenvector corresponding to the eigenvalue $\lambda_1=1$ so i set up that $v_1=\begin{pmatrix}x_1 \ y_1 \z_1\end{pmatrix}$ which gave me the expanded matrix $$ \begin{pmatrix}1.4 & 0.3& 0.3&0\0.3 & 1.5 & 0.2 &0\ 0.3& 0.2 & 1.5& 0\end{pmatrix}$$ row reducing this would give me the identity matrix does that mean that since $A=I$ giving us $I_3v_1=v_1$ so the eigenvector of $A$ is $$v_1=\begin{pmatrix}1 \1\1\end{pmatrix}$$?

[stoic python]

⨋ρ:

[lyric sphinx]

I don't get the last column of your expanded matrix why are there 0's?

But yeah your v1 is indeed an eigenvector, you can check that Av1 = v1

[bold python]

I mean I did $$\begin{pmatrix}0.4 & 0.3& 0.3\0.3 & 0.5 & 0.2\ \ 0.3& 0.2 & 0.5\end{pmatrix} \begin{pmatrix}x_1 \y_1\z_1\end{pmatrix}=\begin{pmatrix}x_1 \y_1\z_1\end{pmatrix}$$ giving us the system of eq's $$\begin{cases}0.4x_1 +0.3y_1+0.3z_1=x_1\0.3x_1+0.5y_1+0.2z_2 =y_1\ 0.3x_1+0.2y_1+0.5z_1=z_1 \end{cases}$$

[stoic python]

⨋ρ:

[lyric sphinx]

You micalculated your expanded matrix

[bold python]

whut?

how

oh yeah

true

should be negative

-0.6 -0.5 -0.5 i guess

[lyric sphinx]

Yeah

Reducing the expanded matrix gives you a basis of eigenvectors for the eigenvalue 1 then if that's what you asked

[bold python]

I'm genuinely confused

How did he get this

After taking the determinant of that

I didn't get = 1

got a bunch of $\cos{\phi}^2 + ....$

[stoic python]

⨋ρ:

[lyric sphinx]

He takes out the -(R + scos phi) from the 2nd column and the s from the 3rd because the determinant is multi linear

Then you can expand the resulting determinant against the 3rd row

[bold python]

I get that part, but that means that the determinant should be = 1 right?

[lyric sphinx]

Yeah

[bold python]

expand the determinant against 3rd row?

[lyric sphinx]

http://demonstrations.wolfram.com/33DeterminantsByExpansion/

[bold python]

ty

weird

didn't get 1

got $cos^2\phi cos^2\theta+cos^2\phi sin^2\theta+sin^2\phi sin^2\theta+sin^2\phi cos^2\theta$

[stoic python]

⨋ρ:

[lyric sphinx]

correct

thats also $\cos^2(\psi)(\cos^2(\theta) + \sin^2(\theta)) + \sin^2(\psi)(\sin^2(\theta) + \cos^2(\theta))$

[stoic python]

oumid:

[bold python]

ah

now I see

Thanks oumid

[lyric sphinx]

np

as a side note these matrix are called rotation matrix

you can compute A^2 you'll get the identity matrix

so their determinant are either 1 or -1

[prime rock]

is there a website where i can bury myself in linear algebra problems?

[bold python]

Uhm, @lyric sphinx how does this work? I want to find an expression for the amount of cars in the city after n days

do I just simply multiply them in?

meaning that the amount of cars after n days is given by $$\begin{pmatrix}40^{n+1}(-1)^{n+1}\40^{n+1}+4.5^{n+1} (-0.5)^{n+1}\ 40^{n+1}-(4.5)^{n+1}+0.5^{n+1}\end{pmatrix}$$

[stoic python]

⨋ρ:

[lyric sphinx]

I dont get quite get your working

but if you have $X_{n +1 } = A X_n$ where $X_n = (x_n , y_n , z_n)^{T}$

[stoic python]

oumid:

[lyric sphinx]

then $X_n = A^n X_0$

[stoic python]

oumid:

[lyric sphinx]

if you have $A = PDP^{-1}$ you can easily compute $A^n$

[stoic python]

oumid:

[bold python]

I'm supposed to write $$\begin{pmatrix}30\60\30\end{pmatrix}$$ as a linear combination of the eigenvectors and then find an expression for the amount of cars in the city after n days

[stoic python]

⨋ρ:

[bold python]

I set that the in the beggining $$\begin{pmatrix}x_0\y_0\z_0\end{pmatrix}=c_1v_1+c_2v_2+c_3v_3$$

[stoic python]

⨋ρ:

[bold python]

Which I later found out to be $$\begin{pmatrix}x_0\y_0\z_0\end{pmatrix}=40v_1+15v_2-5v_3$$

[stoic python]

⨋ρ:

[bold python]

and since $$\begin{pmatrix}x_{n+1}\y_{n+1}\z_{n+1}\end{pmatrix}=A\cdot\begin{pmatrix}x_n\y_n\z_n\end{pmatrix}$$

[stoic python]

⨋ρ:

[bold python]

then I did what I first sent you

[lyric sphinx]

think its good then just check your calculations because $2 \times (0.5)^{n+1}$ isnt $(-1)^{n +1}$^

[stoic python]

oumid:

think its good then just check your calculations because $2 \times (0.5)^{n+1}$ isnt $(-1)^{n +1}$^
```Compile error! Output:

! Missing $ inserted.
<inserted text>
$
l.11 ...$2 \times (0.5)^{n+1}$ isnt $(-1)^{n +1}$^

I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.

[bold python]

nvm

what would I write there then?

I can't find any solutions for it

[lyric sphinx]

wym

[bold python]

nvm was thinking wrong

would 2(-0.5)^n+1 suffice?

[lyric sphinx]

you can simplify further

0.5 = 1/2

[bold python]

thanks again

[charred stirrup]

what does "row of zeros" mean?

matrix R will have atleast 1 row of all 0's? in any row?

[half ice]

Yes

[dusky epoch]

row of zeroes means row of zeroes

[charred stirrup]

matrix R of some square order, must be an identity matrix or just have a row of zeros in any row ?

i think i get what it's saying but i dont know why that's significant

[half ice]

Alone, it isn't lel. One might say it's obvious

[charred stirrup]

GOD bless you're good

thanks for the help like always

u and kx and ann

[half ice]

But following it, you'll have theorems based on how many zero rows there are

Those are important

And np! Feel free to ask if you have anything else

[earnest vessel]

You can give the null space without giving a basis

For example, the null space of the zero matrix is the whole space

Regardless of dimensions and bases

[sage mantle]

i have a midterm of linear algebra 2 tomorrow

damn good luck to you

the closer i get to the exam timewise , the more depressed i am

[half ice]

I would argue that you will always present the null space as a basis. Finding that basis is the same process

@sage mantle
Heyo, did you have a question?

[sage mantle]

yes but they it's actually multiple ones which is why i thought that it would be better to ask you per private message whether it'd be okay with you if i ask them to you privately and if you have time/you don't mind. I decided to change for the best and become more active in my studies but since it's quite a recent change , i have a lot of troubles ^^

[half ice]

Well, feel free to ask them here, if you'd like. That way if I'm not active, you might still get answers

[sage mantle]

i've been trying to understand how to compute the kernel a matrix and the basis using the kernel of a matrix but i don't really get it x.x

[half ice]

Have an example by chance?

[sage mantle]

yes one second

exercice 2 question 2

[summer dagger]

whats a good way to visualize null space?

[half ice]

@sage mantle
You want to solve
[1 1 1 | 0]
[1 1 1 | 0]
[1 1 1 | 0]

[sage mantle]

isnt there a diagonal of 0's?

[half ice]

As this will give you the vectors, that when passed through M + I, gives you the zero vector

[sage mantle]

but how would you solve that?

[half ice]

It's not consistent, so you'll need to solve with row reduction

[sage mantle]

until i have row reduced echelon form?

like with leading ones and all that?

[half ice]

[1 1 1 | 0]
[0 0 0| 0]
[0 0 0| 0]

Reduces to that, pretty clearly lel

[sage mantle]

my matrix is :
[0 1 1 | 0]
[1 0 1 | 0]
[1 1 0 | 0]

[half ice]

M + I

[sage mantle]

oh

okay i see

and when i get that result

what do i do with that?

[half ice]

It's worth taking a step back and seeing what's actually going on here. We're solving this system:
[1 1 1] [a] [0]
[0 0 0] [b] = [0]
[0 0 0] [c] [0]

We want to know what vectors we can pass through our matrix to get the zero vector. That's what the null space is!

If we carry out that matrix multiplication, we get one important line:
a + b + c = 0

So that's the rule. If each of the entries of our vector sum to zero, then M + I will map it to the zero vector

[sage mantle]

so that's for the kernel right ? not the basis

or maybe i misunderstood

[half ice]

This is just getting the idea of the rule that puts a vector in the nullspace

Now, if I hand you a vector, you can easily tell me if it belongs in the nullspace

So we're half-way there so to speak.

[sage mantle]

isn't there infinite answers for a + b + c = 0

[half ice]

We can describe the nullspace itself with its own basis. Take our equation:
a + b + c = 0
Only one of these two variables are fixed.
a = -b - c
Which means we can write this vector space as:
[-b - c]
[ b ]
[ c ]

We can write that as:
[-1] [-1]
[1] b + [0] c
[0] [1]

And thus those two vectors are a basis for the nullspace

[sage mantle]

what do you mean by fixed variables?

[half ice]

You're right that a + b + c = 0 has infinite solutions. Since it's one equation with three variables, (3 - 1 = 2) two of them are free, and the fixed one will be written in terms of the other two.

[sage mantle]

so we can choose which ones are fixed?

[half ice]

Yes. No matter which, you'll get an equivalent basis

[sage mantle]

okay so if my teacher asks me what is the kernel of this matrix

then i give those two vectors right?

[half ice]

Yes yes

To summarize:
We developed a rule that a vector belongs to the null space if its entries sum to zero

[-b - c]
[ b ]
[ c ]
Describes all of those vectors

That set of vectors is spanned by
[-1] [-1]
[ 1 ] [ 0]
[ 0] [ 1 ]

Since the basis has two vectors, the nullspace is 2 dimensional

[sage mantle]

i see , so kernel = the vectors and basis = the dimension?

[half ice]

Kernel is a vector space made of the vectors that your matrix will map to zero.

[half ice]

Every vector space can be described with a basis which is a set of vectors that span the entire space

[sage mantle]

i see , so the teacher won't really ask me for precisely for the kernel but rather for the basis?

[half ice]

They'll ask you for the kernel by having you state the basis of that kernel

[sage mantle]

i see

thanks ^^

[half ice]

Np. Feel free to ask if you have anything else!

[sage mantle]

dw there is a lot

[half ice]

Jk I'm sure it's fine

[sage mantle]

[frosty pewter]

help pls

[sharp lodge]

If it's in the nullspace of B is in in the nullspace of AB?

for a

[sage mantle]

@half ice when i do the row reducted echelon form , is my goal to get a row full of zeros? because i feel like i could get an identity matrix in the third question

[earnest vessel]

@frosty pewter for part (a) you have to notice that if Bv = 0 then (AB)v = A(Bv) = A0 = 0

And part (b) is similar

[charred stirrup]

do inverses of a matrix only exist if the matrix is a square order?

[sage mantle]

how would you answer exercice 5 question 1 ? i would compute the eigenvalue and then see if the eigenvector of one of the eigenvalues is a multiple of the u1 but since they asked about the eigenvector first , it looks like i fucked up the order

[half ice]

@charred stirrup
Yes.

And, if that matrix has a non-zero determinant

[vale arrow]

@sage mantle , calculate Su_1

It should be like this:
$S \cdot u_1 = \lambda \cdot u_1$

[stoic python]

Patrick Salhany:

[vale arrow]

By calculating it, you find λ and answer the second part of the question

@sage mantle

[sage mantle]

i see thanks

[vale arrow]

Welcome

[frosty pewter]

howwwwww

[sharp lodge]

Can you prove (3=>4)?

[frosty pewter]

no .-.

[sage mantle]

What is the sign after prove that in exercise 3 question 1?

[dusky epoch]

$\chi_C$

[stoic python]

Ann:

[dusky epoch]

this?

i think this is the characteristic polynomial of C

[sage mantle]

I see thanks

What do they mean by define ? Should I compute it or not

[dusky epoch]

well since you're going to use it in part 2

i guess it makes sense for you to compute it

[sage mantle]

Okay i See

[sage mantle]

For the second question exercise 2 , I found that the rank is equal to 1 but for the dimension , it depends on which n*n we are working on so how am I supposed to answer? I wanted to use the rank nullity theorem. Maybe I misunderstood since I don’t know what to do with the second part of the question

[lyric sphinx]

@sage mantle you can use rank nuity theorem the dimension depends of n

[sage mantle]

Well should I just show them for n=2 and n=3 and explain that the dimension depends on n

[dusky epoch]

give the dimension in terms of n

[sage mantle]

It would be R^n-therank

Like if it’s a 2*2 matrix , the dimension of the eigenspace is 1

3*3 = 2

And so on

Is the trace the eigenvalue for symmetric matrices?

[sharp lodge]

Think of the diagonal matrix yatori

Questione

Now contraction in V∗⊗V,is defined by the rule v∗⊗v↦v∗(v)

How rigorosuly spoken was this?

[dusky epoch]

uhhhhhh hh h h hhh

bad paste?

no clue what either mess of symbols is meant to be

[sharp lodge]

Not all elements of  V∗⊗V have the form u*⊗v so that's why it doesn't seem entirely rigorous. It does definr a function exactly but that's slightly a posteriori right?

Sorry Annn, I'll try fixing it

[dusky epoch]

i mean like

you know that to define a linear transformation it suffices to say what it does to a basis on its domain

right

[sharp lodge]

Yrah that worka

[dusky epoch]

this isn't exactly that but it's similar

[sharp lodge]

Yeah

I just realized my confusion was pretty stupid

I was thinking which elements of V*⊗V cant be represented as just u*⊗v. And can that even happen?
Of course. I just have to think in terms of matrices and ranks and everythongs pretty clear i think

Thanks, nice to see you back btw!!! Even though it's been a while

[sharp lodge]

[wintry steppe]

help? (my bad if its the wrong channel bc the problem is apart of my linear unit )

[summer dagger]

if a question askes to find the dimensions of null, col and row space is it assummed to find the bases of them first?

[broken hawk]

not necessarily, they exist independent of their bases

[summer dagger]

ty

[broken hawk]

but they would probably be easiest to describe in terms of some basis

[summer dagger]

If asked not to find the basis wouldnt it just be the rows for row and the span of the matrix

meant to ask the non bases case

[summer dagger]

https://gyazo.com/9d8b5d3539768c2ed45ba77405011f77

Fin an element Y in V that is not in W
confused as to how something could be in V and not W if they are equal

[dusky epoch]

are they?

what's the dimension of V?

[summer dagger]

2

[dusky epoch]

how'd you arrive at that

[summer dagger]

it has 2 matrices stored in it

[dusky epoch]

it?

what's "it"

[broken hawk]

can you find a basis of V?

that would give you the dimension for sure

(mind that V itself is a subspace of a 4-dimensional vector space)

(namely that of all 2x2 matrices)

[summer dagger]

oh vV is a vector space os f 2x2

[dusky epoch]

so let me make sure

you're claiming the dimension of V is 2

let M1 = [ 1 0, 0 0 ], M2 = [ 0 1, 1 0 ], M3 = [ 0 0, 0 1 ]. these are all matrices in V.

find me a nontrivial linear combination of these which sums to zero.

[summer dagger]

is V an arbitrary Matrix in this case?

[dusky epoch]

no, V isn't a matrix

[broken hawk]

V is the vector space defined in your question

that is, the space of all symmetric 2x2 matrices

[thin bloom]

So if we talk about the vector space of a matrix do we have to prove the properties of a vector space using the matrix as a vector?

[broken hawk]

yes, if I understood the question right

you can think of matrices as vectors rearranged into a square for all intents and purposes

(but unlike other kinds of vectors there’s a multiplication on the space of matrices, so they’re special in that regard)

[thin bloom]

?

[broken hawk]

(in abstrac algebra terms, matrices form both a vector space and a ring)

[thin bloom]

ring?

[summer dagger]

ok i think i get it more now basically W is a section of V so creating a symetric matrix that would be outside one of the given span matrix?

[broken hawk]

a ring is basically a generalization of the structure of ℤ: a set on which you can add and multiply, but not necessarily divide

[dusky epoch]

wdym by "section"

[summer dagger]

space

would prob be a better word

[broken hawk]

nico, if you want me to detail this further, let’s move to #groups-rings-fields

so as to not interrupt ann and avocado

[dusky epoch]

W is a subspace of V indeed.

[thin bloom]

I'm fine I have a test in lin soon so I'm trying to understand concepts of lin here

[dusky epoch]

but the important thing is that you understand that dim(V) is not 2 @summer dagger

[broken hawk]

I kinda feel like it’s not the worst idea to explain what it means for the matrices to form a vector space, the ring bit is just interesting extra info

[summer dagger]

um i honestly don't know what the dim(V) would be

[dusky epoch]

sigh

okay, i've failed to explain this properly

[summer dagger]

nah i'm just slow af

[broken hawk]

here’s a basis for V

which also disproves dim(V) = 2

[earnest vessel]

That's the weirdest basis ever xd

[broken hawk]

okay it’s a bit overly complicated I guess

but it works

there’s an easier one :p

[dusky epoch]

it's not the simplest one you could think of yeah

[broken hawk]

it’s the second simplest tho :P

[dusky epoch]

[broken hawk]

here’s a more straightforward one

[summer dagger]

to get that basis did you think of it in terms of the values being diagnoal of eachother so when you take the transpose

[earnest vessel]

To think of that basis you think of how a symmetric 2x2 matrix looks like

a b
b c

[broken hawk]

it’s actually pretty easy to prove that this is a basis:
the space of 2x2-matrices has dimension 4, this set is linearly independent, not all 2x2 matrices are symmetric. therefore this must be spanning

[earnest vessel]

And then you write a matrix for each of the entries a, b, c so that you may express this as a linear combination

(how do you write your matrices)

?

[broken hawk]

see #bots

where I generated the image

[earnest vessel]

Fancy

[broken hawk]

made it show the source code too, now

so you can copy stuff out

[summer dagger]

So ig the ansewr to my question would be
5 1
1 1

[dusky epoch]

an answer

there are many matrices here which will do

[summer dagger]

ty

[thin bloom]

Is this correct?

dim(Col(A)) is the number of pivots columns since Col(A) is the span of the column vectors of A

[broken hawk]

sounds right

[thin bloom]

The linear combination of column vectors of A will reduce dependent column vectors

?

I'm still trying to grasp this concept of dim

[broken hawk]

oh yea what I wanted to elaborate on:
you can add matrices, and you can multiply them with scalars; those operations behave how you’d want them to (with things like distributivity); you can check that quickly by noting that e.g. you can simply view a matrix as a rearranged column vector; the way you write it doesn’t affect how the operations work
This means the set of matrixes (of a given shape) are a vector space

matrix multiplication doesn’t play into the vector space structure, it’s an additinoal thing you can do with matrices

[earnest vessel]

In most contexts you can read dimension = degrees of freedom

[broken hawk]

not sure what you mean with “reduce dependent column vectors”

[earnest vessel]

That is why the space of symmetric matrices had dimension 3

You have 3 entries with which you can play around

The fourth one is fixed by the matrix being symmetric

[broken hawk]

also mind that a set doesn’t have a dimension yet, the column space is the subspace generated by (spanned by) the vectors in the columns

[thin bloom]

since the number of basis vectors in the column space of A will be the dim(Col(A)) any redundant column vectors will be "discarded"

[broken hawk]

discarded in order to turn the set into a basis, yes

you can write each of those redundant vectors as a combination of the pivot ones

which should be easy to convince yourself of

just start with the last one, multiply it to get the last coordinate right. then add the penultimate one in the right amount of times to get the second-to-last coordinate right, etc

[earnest vessel]

In the "degrees of freedom" point of view that means that whatever you could do with the vectors you remove, you can do with the remaining vectors anyway

[thin bloom]

nice and the dim(Col(A)) = dim(row(A)) since the number of pivot columns = number of pivot rows

[wintry steppe]

if f is a linear applicaion : E -> F , rank(f) <= dim(E)

I think it's true, is that right ?

[earnest vessel]

Yes

By definition

rank(f) = dim(Im(f))

[wintry steppe]

yes

[earnest vessel]

Ohh

You wrote E

It is still true

[wintry steppe]

I just wanted a confirmation

[earnest vessel]

But for a different reason

[wintry steppe]

bcs it's linear the base of im(f) is the im of the E base right ?

[earnest vessel]

Exactly

Well

It's not a basis

It's a spanning set

That's why the dim is lower or equal and not just equal

[stiff surge]

I think this is something super basic that I've missed somewhere but

[-1-i -2
1 1 - i ]

apparently gives

x = [ 2
- 1 - i ]

I don't understand what the x is

[slow scroll]

@stiff surge what do you mean by "gives?"

it looks like a matrix and a vector to me

[stiff surge]

uhh so apparently

okay so the context is

?

why is my text getting deleted

A = []

[slow scroll]

@dreamy coral #prealg-and-algebra

[stiff surge]

apologies, lemme get paint

i'm trying to get A's eigenvectors

and one of the eigenvalues is (2+i)

and when I do A - (2+i)I

I get

and from that, apparently I can know that the eigenvector is

but I don't know

where that vector came from

I am assuming (-1-i)a + (-2)b = 0 and (1)a + (1-i)b = 0 will give me a=2 and b=-1-i

but how

is there a way to cross eliminate that I'm not seeing

[slender yarrow]

(you can multiply by complex numbers yk)

[stiff surge]

yk?

[slender yarrow]

yk : you know

$$\begin{cases}(-1-i)a - 2b = 0 \ (1+i)a + (1-i)(1+i) b = 0\end{cases}$$

[stoic python]

emeric75:

[broken hawk]

oh no im dumb

[slender yarrow]

that's a way of cross-eliminating dope

[stiff surge]

still not sure

oh

since the first can be interpreted as

(1+i)a + 2b = 0

i can substitute (1+i)a with -2b

[slender yarrow]

if you want ye

[stiff surge]

is that the only way?

do i need to have it as two equations to simplify?

[slender yarrow]

or you can just do gaussian elimination

$$\begin{cases}(-1-i)a - 2b &= 0 \ (1-i)(1+i) b -2b &= 0 : L_2 +L_1 \to L_2\end{cases}$$

(but for 2 by 2 system whatever)

[stoic python]

emeric75:

[slender yarrow]

(and (1+i)(1-i) is just 2)

$$\begin{cases}(-1-i)a - 2b &= 0 \ 0 &= 0\end{cases}$$

[stoic python]

emeric75:

[stiff surge]

ahhh

thank you 🙏 🙌

[slender yarrow]

💤 good night (or whatever)

[summer dagger]

is there a trick for creating a idempotent matrix?

[lyric sphinx]

you can take the matrix of any projection

[stiff surge]

[ -4i -4
1 -4i ]

how do you find a non-zero eigenvalue for it? (It's already in A-λI form)

[lyric sphinx]

in dimension 2 any matrix has exactly 2 complex eigenvalues

and the trace of the matrix equals the sum of these eigenvalues

[stiff surge]

uh

OH

sorry eigenvector

eigenvector

😭

not eigenvalue, eigenvalue is 4i, hence why it's -4i in the given matrix

[lyric sphinx]

solve for X the equation AX = 4iX

alternatively if you have an eigenvector for 0, you can take any other vector such that these two form a basic of C^2

[stiff surge]

C^2?

[lyric sphinx]

and it will be an eigenvector for 4i because the matrix is diagonalizable

[stiff surge]

diagonalizable?

[lyric sphinx]

$\mathbb{C}^2$

[stoic python]

oumid:

[stiff surge]

shoot wait i don't think i know those terms

[lyric sphinx]

diagonalizable means that the direct sum of the eigenspaces spans the whole space

[stiff surge]

so... that means...

eigenvector exists...?

[lyric sphinx]

wym

[stiff surge]

i don't get it

[lyric sphinx]

whats ur definition of eigenvectors

[stiff surge]

wait

the coordinates in space that will multiply by a scalar no matter how many times the space is stretched by a given transformatio matrix

i think

[lyric sphinx]

uhhh

too vague

[stiff surge]

oh

[lyric sphinx]

an eigenvector X for an eigenvalue lambda of the matrix A is a non zero vector such that AX = lambda X

[stiff surge]

okay

i am getting something strange

okay so the original matrix is

0 -4
1 0

and the eigenvalue turn out to be (if i am correct)

+4i and -4i

but both are giving me

0
0

[lyric sphinx]

the eigen values are $\pm 2i$

[stoic python]

oumid:

[stiff surge]

OH

right cuz it's square root of 4

oops

[summer dagger]

If w is not in Span{v1, v2, v3} and {v1, v2, v3} are linearly independent, then
{v1, v2, v3, w} is also linearly independent

Is this true b/c there is no combination of v1,v2,v3 that makes w

[wintry steppe]

it's true

[broken girder]

i know how to diagonalize the symmetric matrix corresponding to the given quadratic form, but how do i find the angle alpha?

[broken girder]

i feel like i can do it in reverse

so i diagonlize first, get my new bases, then do change-of-coordinates from the new bases to the original x,y bases of the point farthest away along the major axis, then do arctan(y-coordinate/x-coordinate)

[summer dagger]

does row reducing change what my determinant is

[dusky epoch]

in general, yes. but you can keep track of exactly how it is changed by each row operation