linear-algebra

[wintry steppe]

i feel like its trivial but idk

[brittle juniper]

What's the context?

what's x? what's x0?

[wintry steppe]

Let x0 be a fixed real number. Prove that {p0(x) = 1, p1(x) = x−x0, p2(x) = (x−x0)^2, p3(x) = (x−x0)^3} is a basis for P3.

and im using this fact to prove linear independence

[brittle juniper]

are you working on polynomials or polynomial functions?

[wintry steppe]

do you mean P3? if so then the professor is talking about the set of all polynomial of degree 3 or less. if not im not really understanding the question

[brittle juniper]

you want to prove that something isn't another thing scaled

if you're working with polynomial functions, the proof shall use properties of functions

if you're working with polynomials, the proof shall use properties of polynomials

if P3 is a set of just polynomials and not polynomial functions, then you may use the definition of polynomials

[wintry steppe]

looking at my notes it seems to be just polynomials

so something like $$ c * 1 =/= x - x_0 $$ ?

[stoic python]

Mother Russia:

[brittle juniper]

1 is the real sequence which first term is 1 and all other terms are 0

x-x0 is the real sequence which first term is -x0, second term is 1 and all other terms are 0

by definition of the scalar multiplication, for all real c, c•1 is the sequence which first term is c and other terms are c×0=0, and in particular its second term is 0, which is not 1 so by it's not the same sequence as x-x0

[half glade]

let $A \in \mathbb{F}^{n\cross n}, A ≠ 0$

[stoic python]

AmitN:

[wintry steppe]

x-x0 is the real sequence which first term is -x0, second term is 1 and all other terms are 0

im not understanding this. how is the first term -x0 and second term 1?

[half glade]

Prove or disprove that there exists B so that $B \in \mathbb{F}^{n \cross n}, B ≠ 0$ and $N(BA)≠N(A)$

[stoic python]

AmitN:

[brittle juniper]

$-x_0+X=(-x_0)\cdot 1+1\cdot X=(-x_0,1,0,0,...)$

[stoic python]

Tuong:

[brittle juniper]

$\sum_{k=0}^na_kX^k=(a_0,...,a_n,0,0,...)$

[stoic python]

Tuong:

[brittle juniper]

do you know that polynomials are sequences with finite support?

and that the thing that makes them "polynomials" is a fancy ring structure?

[wintry steppe]

i dont. thats beyond the scope of the class i think

i know about the polynomial rings and that stuff but i dont think its supposed to be applied here

or are you talking to the other guy

whoops lol

[brittle juniper]

Am talking to you lol

I guess you can just say that two polynomials are equal when they've got the same coefficients

and that for all real c, the coefficient of X in c•1 is 0 while the coefficient of X in X-x0 is 1

[wintry steppe]

ah that makes sense

so i think i can just use the same case for the other ones because theyll divide each other, but say i wanted to prove (x-x0) = (x-x0)^2 wasnt a linear combination, I could show that the coefficient of the x^2 term is 0 on the left and 1 on the right?

[brittle juniper]

"to prove that (x-x0) = (x-x0) ^2 wasnt linear combination" doesn't make much sense

[wintry steppe]

ah sorry, to prove that p1(x) = (x-x0) and p2 = (x-x0)^2 arent a linear combination

which would be to prove that there doesnt exist a c (x-x0)^2 = (x-x0)

i could divide both sides by x-x0 and get the same case as the first

but i wanted to make sure i understood what you wrote

[brittle juniper]

you want to show p2 isn't a linear combination of p1 and p0

[wintry steppe]

he gave us a fact to prove p2 isnt a linear combination of p0

Fact 2: Let n ≥ 1 be an integer. A polynomial of degree n cannot be written as a linear combination of polynomials of degree at most n−1

so i would only need to prove p2 isnt a linear combination of p1 basically

after using fact 2 of course

[brittle juniper]

that Fact 2 gives immediately that p2 isn't a linear combination of p0 and p1

[wintry steppe]

oh

what

[brittle juniper]

p0 and p1 have degrees <2

[wintry steppe]

crap

fact 2 is what we were discussing though right?

[brittle juniper]

kind of

[wintry steppe]

well i feel like an idiot now

oh well

at least i understand this stuff a little better

linear algebra is not my strong suit

[brittle juniper]

linear algebra can get a lot worse

here, it was just discussing definitions and some properties

[wintry steppe]

yeah idk why but the basics for linear algebra are hard for me to understand

also been 2 years since i took the class

the linear numerical analysis im taking is already getting tough so im not looking forward to it

[brittle juniper]

Well, good luck!

[wintry steppe]

thanks 😅

i appreciate the help

[brittle juniper]

You're welcome 🍻

[lone quail]

Could anyone explain me the solution for this please?

The solution from the textbooks is:

The subspace required is the plane containing r and passing through the origin. The generic plane containing r has equation given by:

x+y+2z-1 + λ(x-y-1) = 0

Then it says that the origin is in this equation if and only if lambda=-1, then substituting we have y+z=0, which is the equation of the subspace

Where did it get the equation of the plane from? And from where did it get that lamda = -1?

[fallow jolt]

@wintry steppe Write it out in equations

w + 2z = 0
x + 2z = 0
y -z = 0

w = -2z
x = -2z
y = z

Z is your free variable, since your system is under defined it can be anything

Hence, the values of w, x, and y depend on z

Does that make sense?

[sonic osprey]

Help

If n + 1 unit length vectors that have pairwise dot product strictly less than 1/n

Then they must be linearly independent

[pallid swallow]

hmm

[steady fiber]

unit length as in length 1?

[pallid swallow]

yeah

Suppose a0, a1, a2, ... an, not linearly independent. then
b0a0+b1a1+...+bnan=0 has nontrivial solution
Let largest magnitude coefficient b0, dot by a0
b0 + b1(a1 dot a0) + ... = 0
strange

done

@sonic osprey

[sonic osprey]

I'm not sober so I'm going to ask some dumb questions

Where's the contradiction

Oh

[pallid swallow]

the other numbers are too small to make it 0

[sonic osprey]

Even if they're all negative

I get it

[pallid swallow]

wait the question is false

consider (1, 0) and (-1, 0)

question should be:
If n + 1 unit length vectors that have magnitude of pairwise dot product strictly less than 1/n

[sonic osprey]

That's probably what they said to me

[winter acorn]

I don’t get the reasoning behind a)

I guess if there is a non-zero vector x that multiplies the right side of A and gives 0, then there must be a linear dependance of columns of A?

I’m not sure and I don’t get the picture

[fallow jolt]

You’re exactly on point

Linear independence means the only x that can make Ax = 0 is the 0 vector

[slow scroll]

What you said is true, and depending on what knowledge ur allowed to assume, that might be all the explanation they are looking for. It’s important to mention that this applies only to square matrices tho

[fallow jolt]

So if there’s an x that’s not the 0 vector, the columns are linearly dependent, and therefore noninvertible

Does that make sense @winter acorn?

[winter acorn]

Yes totally thanks! and therefore non-invertible since the elimination will bring a zero in a pivot position right?

[fallow jolt]

You’ll get linearly dependent columns

Think about it that way

Don’t get too specific with the mechanics

[winter acorn]

Ok good and I have another quick question

I'll send it

[fallow jolt]

There’s more than one way to have linearly dependence, not just a 0 pivot

For example, an under defined system

[slow scroll]

“under defined system?”

[winter acorn]

free variables ?

[fallow jolt]

@slow scroll Less equations than unknowns

[winter acorn]

okok yes

[fallow jolt]

That’s what I’ve heard it been called anyways

[slow scroll]

Yea okay but that’s not linear dependence. That’s non-spanning

[fallow jolt]

Under defined leads to a free variable tho generally

Unless you have a weird edge case

[slow scroll]

Misread oops

[winter acorn]

I can’t quite wrap my head around these types of questions..

Better: I don’t know where to start

[slow scroll]

Ur gonna have to define U, I, and R for us

[winter acorn]

U is the upper triangular matrix u get when eliminating A, R is rref(A) and I is identity

[slow scroll]

So a) is asking for a matrix with no zero entries that reduces to the identity via Gaussian elimination?

[fallow jolt]

No nul coefficients means it’s linearly independent I’m assuming?

Idk what a nul coefficient means lol

[slow scroll]

I think it means no zero entries. But I’m not sure I can think of an example like that. Hmm

[winter acorn]

yeap non zero entries

a) is asking you to build a 3x3 with non-zero entries that once reduced to Upper triangular gives the identity

exactly right @slow scroll

I'm translating from french haha there might be some errors

[slow scroll]

I’m pretty sure such a matrix doesn’t exist. I only hesitate to say that because I am not really sure how to prove it.

Moving along for now, for b), have you ever used row operations to find the inverse of a matrix?

Or maybe a better question: are you familiar with what row operations are?

[winter acorn]

Yes I am!

[fallow jolt]

@slow scroll Technically the identity matrix itself would satisfy a, right?

The identity matrix itself is in upper triangular form

[winter acorn]

That's what I thought

[fallow jolt]

Does the 3x3 matrix you're trying to find have to satisfy all of those?

Or can you find one for each one seperately?

[slow scroll]

I think they are separate questions.

Okay so think about what it means for rref(A) to be the identity. There are a series of row operations that you multiply to the left of A that give you the identity. So what is this series of row operations?

[winter acorn]

Find one for each one

I think I got the first one

[fallow jolt]

If rref(A) = A then you can use any linearly independent matrix

Because by definition its rref will be I

[winter acorn]

So yeah linear dependance

I think they are separate questions.

Okay so think about what it means for rref(A) to be the identity. There are a series of row operations that you multiply to the left of A that give you the identity. So what is this series of row operations? ...Hmmmmm

[slow scroll]

So you know that row operations are just special invertible matrices multiplied to the left of A?

[winter acorn]

yup the E’s

Trying to figure it out in my head logically

[slow scroll]

Yea so call the composition of all those E’s, E. So what is the equation that describes the relationship between A and its rref?

[winter acorn]

Would you just do the inverses of those row operations then

To come back to A

So E^-1

*A

*I sorry

To get A

[slow scroll]

I think you’re on the right track, what I am trying to get you to see is that since rref(A)=I and EA=rref(A)....

[winter acorn]

What I see is that through an elimination you get I so if you want to come back to what A was you apply the elimination inverse to I

I could also see (more intuitively) that (I - E) would be A

[slow scroll]

I minus E = A???

[winter acorn]

That was off the top of my head 10 minutes ago I regress

Doesn’t make any sense now

But yes my answer would be that A is the identity

[slow scroll]

A has entries =0 tho

[winter acorn]

True that

[slow scroll]

I think you’re on the right track, what I am trying to get you to see is that since rref(A)=I and EA=rref(A)....

take a look at this ^

[winter acorn]

Ok I’ll ponder on that more

[slow scroll]

finish the sentence

[winter acorn]

EA=I

So then A = I*E^-1

[slow scroll]

okay yea yea. Just look at EA = I. What does this say about A?

[winter acorn]

That’s how It goes in my mind at the moment haha

So E is the inverse of A?

[slow scroll]

yep, and A is invertible. Therefore, any matrix that reduces to the identity is necessarily invertible

[winter acorn]

That feeling when something is finally unblocked forever

Thanks a lot mate

Although I do remember that proof now from my classes

Also I was thinking of inverses and such... if we think in 3D, only matrices defining a region with a certain volume are invertible right?

what I’m trying to get at is that if day we had a plane in 3D space passing by [0,0], the matrix defining that plane would not be invertible right?

Does the determinant of a matrix represent the volume or the area ?

[slow scroll]

yea i was going to say, that topic is touching on determinants. There is an interpretation of determinants as the ratio of the volume enclosed by the vectors (a parallelotope) to the volume of the unit n-cube. Whenever you have a non-invertible transformation i.e. a linearly dependent column in your matrix, you lose at least one dimension in the image space, giving you zero volume.

For example, if you have a 3x3 matrix with a linearly dependent column, the span of the columns (the column space) would have to be a plane in 3 space, which has zero volume, and therefore 0 determinant. So for square matrices, when the determinant is 0, the matrix is not invertible, which implies that there is at least one linearly dependent column somewhere

[winter acorn]

Ok cool!! That’s what I thought. So in 3D space if you have parallelotope as output then you can be sure that all of the columns of the matrix defining that shape are independant since that shape fills a 3D portion of R^3

If one column was dependant then you would have a 2D plane in R^3 and so on with two columns dependant you would have a line

Which kinda depicts the column space too

And if column space is a line in R^3 then null space is a plane in R^3 ?

[slow scroll]

yea pretty much. And its clear that these define subspaces since they contain the origin, the sum of any point on plane with another point on the plane is a third point on the plane, etc, etc.

And if column space is a line in R^3 then null space is a plane in R^3 ?

yea. This follows from rank-nullity theorem.

i guess just keep in mind that we have been in the realm of square matrices this entire time. With non-square matrices a lot of the stuff we've been talking about doesn't necessarily hold anymore

[winter acorn]

Ok perfect @slow scroll 💯

[undone garnet]

rank(A) = 1

prove that ABA = trace(AB).A

any idea?

[dusky epoch]

if $A \in \bbR^{m \times n}$ is of rank 1 then it can be written as $uv^T$ where $u \in \bbR^m$ and $v \in \bbR^n$

[stoic python]

Ann:

[blissful vault]

if matrix A -> R2 + 2R3 = B
then I ->R2 + 2R3 = E
AE = B right?

it doesn't seem to work

[ember zenith]

number 4?

sorry i’m a hs sophomore so this ones a bit touch

[wintry steppe]

hi

[ember zenith]

i got x = 0, y = 4/3, and z = 4

don’t know where to go from there because i have to write a line, not a point

[ember zenith]

hm okay

i’ll look into it

[wintry steppe]

so i have a matrix ∈ M2

and X^2 = (4 0 / 4 4)

how can i find X

/ is a new line

4 0
4 4

[half ice]

Take a general matrix
a b
c d
Square it. You'll have a system of 4 equations

[wintry steppe]

ok that was simple

but i have another one

I did a

i'm getting into matrix

[tame mural]

For matrices, do people normally consider the rows or the columns as dimensions? Or both?

[dusky epoch]

what do you mean

[tame mural]

when translating word problems into matrices

I had the intuition that every row corresponded to an "entry" of information

and that piece of information had several dimensions, such as

3 apples, 2 bananas, 3 carrots

And so sometimes the word problem might say you shopped 3 separate times under the same prices

And I would put that as 3 separate rows in a matrix

And each column would in my mind correspond to a "dimension", such as apple banana carrot

or sometimes spatial dimensions

but then when I'm doing matrix multiplication

esp. with non-square matrices

I wonder what the heck is going on

and also when I'm reading about this stuff on the web

it sounds like people just say that both column and rows are dimensions

[north sandal]

Exam time baby.

[jagged saffron]

how would I even start for this?

[dusky epoch]

consider the characteristic polynomial of T

[jagged saffron]

so it would just be p(t) = $\prod (t-\lambda_i)^{a_i}$

[stoic python]

Victoria:

[dusky epoch]

not quite

[jagged saffron]

hmm is that over C?

and not R

[dusky epoch]

you don't need to write it out by itself

the charpoly of T has real coefficients

anyway

consider $\chi_T(t) \cdot \prod_i (t - λ_i)^{-a_i}$

[stoic python]

Ann:

[dusky epoch]

this is a polynomial

see what you can say about its roots

[jagged saffron]

its roots are the roots of $\chi_T(t)$?

[stoic python]

Victoria:

[jagged saffron]

is the negative exponent suppose to be there

[dusky epoch]

no

and yes

as in

yes the negative exponent is supposed to be there

no the roots of the poly i wrote are not all the roots of chi_T

[jagged saffron]

hmmm

all the roots of chi_T minus the eigenvalues?

[dusky epoch]

all the roots of chi_T except the real eigenvalues of T yes

[normal gust]

Hi, I need help with setting parameters for systems augment 0

they should all be zeroes*

[half ice]

What should all be zeroes?

The system reads:
x = 0
z = 0
0 = 0
There's no constraints on y

[normal gust]

oh yeah, the last row is all zeroes, I forgot about y

I'll name it t, lol

[blissful vault]

i don't understand what this wants me to do

what does ~ mean

[charred stirrup]

@blissful vault the tilde can mean the two augmented matrices are not equal

[blissful vault]

how would the statement make sense thoughj

also that's for programming

i'm in linear algebra

[charred stirrup]

@blissful vault yeah you're right, then tilde in this case means the two matrices are row equivalent

[blissful vault]

what does that mean @charred stirrup

?

[charred stirrup]

you can change one of the augmented matrices into the other using only the elementary row operations

[blissful vault]

ohhh

[charred stirrup]

You can also think of it stating the two matrices have the same row space

[blissful vault]

so E1E2E3(...)Ek = Q

and E1E2E3(...)Ek A=P

[charred stirrup]

Yeah you can use the unit vectors like that, but I'm not sure how to check whether that'd be right

feels right?

[blissful vault]

ayt thanks for the help

[charred stirrup]

np gotta give back here when i can

[rose grotto]

how 2 od this

nvm

[normal gust]

multiply it out

[rose grotto]

I should be fine

ty

kk

[normal gust]

lol yeah

set up a couple equations and then solve for it

[rose grotto]

do I do

a 1 b 0 * a 1 b 0

@normal gust

do I multiply the matrix by itself

[normal gust]

Yes, you multiply the matrix by itself, since it's squared

you don't even need to solve

[rose grotto]

@normal gust how do u get a^2+b

on the top

left

[normal gust]

1st row times 1st column

[rose grotto]

oh ya ty

[tough idol]

Any ideas on how to start the first one?

[rose grotto]

#help-8

[slow scroll]

@ionic ravine #prealg-and-algebra

[undone garnet]

A, B is M_2

AB=BA=A+B+I

prove that det(A+B)-det(AB) = 1+trace(AB)

hm...I found weird when I solve this problem

here's my proof

det(AB) = det(A+B+I)

and we've know that

det(A+B + xI) = x^2 - trace(A+B).x + det(A+B)

so det(A+B+I) = 1 - trace(A+B) + det(A+B)

det(A+B) - det(AB) = det(A+B) - (1 - trace(A+B) + det(A+B))

= -1 + trace(A+B) = -1 + trace(AB - I) = -1 +trace(AB) - trace(I) = -3 + trace(AB)

is problem or my proof wrong?

[late rampart]

If A is a matrix such that A^2=I, do we always have that A is symmetrical?

[undone garnet]

no

0 0 1
2 1 -2
1 0 0

[wintry steppe]

Hello, how would I go about finding two R3 bases B and C, when I know of two matrix representations of the same Linear Transformation?

As you see one of the representations has the preimage in base B and the image in base C

while the other uses the canonical bases

[rose grotto]

How do I get 2a if they don’t give me the numbers for A

cuz u can’t just multiply -3 by 2

[dusky epoch]

det(2A) is indeed not 2det(A)

momo shush no spoilers

@rose grotto try expressing 2A as a matrix product

as in, MA where M is some matrix

rather than 2, a scalar

[fallow jolt]

Go for it lol

[dusky epoch]

@rose grotto?

[rose grotto]

Would everything in the matrix be 2

for M

Or do I do -3^3

[dusky epoch]

no, M is not the matrix consisting of all 2's

and no, det(2A) isn't (-3)^3 either, nor -3^3 for that matter

[rose grotto]

Ohh

I do 322*2

-322*2

ya

[dusky epoch]

\*

[rose grotto]

It’s cuz like discord does that thing

s

Italic

What does transposing a determinant do

It doesn’t change it right

[late rampart]

Thanks @undone garnet . Can I ask you how you found it? Was it pure guess, or was there a reasoning there?

[rose grotto]

@dusky epoch what am I supposed to do in D

[undone garnet]

@late rampart becuase I've done it before 😄

[dusky epoch]

@rose grotto once again try expressing the matrix whose det you are asked for as a matrix product with A

[rose grotto]

Would I do -3*5-3 @dusky epoch since it’s on one row rather then 3 id do it once

[dusky epoch]

did you do what i said to do

[strong bison]

i don't understand question b

could someone please explain it to me

[snow oasis]

Hey guys i went through like 90% of a problem but got stuck at the very end

[gray dust]

#real-complex-analysis

@strong bison what does it mean that a set of vectors S is a basis for R^3?

[strong bison]

@gray dust they form the basis of all the vectors in R^3, meaning you can make any vector in R^3 using through addition of the vectors in S?

[gray dust]

@strong bison sure. are you familiar with the R^3 standard basis (e_x, e_y, e_z)?

[strong bison]

@gray dust no sir

[gray dust]

perhaps you learned them as the i, j, k unit vectors

[strong bison]

@gray dust do ring a bell, but can't clearly remember how to apply

[gray dust]

e_x, e_y, e_z is just different notation for i, j, k. for example, how would you write the following vector in terms of the standard basis vectors? @strong bison

$\langle 1,2,3 \rangle$

[stoic python]

RokettoJanpu:

[rose grotto]

4a Asks me to get the determinant right

because it has the ||

ohh

its the

magnitude nvm

for B

would I do like BC/Magnitude and since its the opposite would I do - infront

like negative it since its other way

[gray dust]

👍🏽

[frosty vapor]

cirrect

[rose grotto]

ty

-4

sec

tys

what do I do

for 4e

I tried doing like

AB

AC

then A + sAB + tAC or something

???

[rose grotto]

@frosty vapor

[rose grotto]

<@&286206848099549185> anyone know what to do for C

I tried A + s(ab) + t(AC)

ty

[barren zealot]

@rose grotto A + s(BC)

What you’re thinking of, is the equation of the plane ABC

Which is e part

[rose grotto]

ya im asking for E part

I meant E not C

my bad

[barren zealot]

Oh, then it’s correct what you wrote

I believe

Is it not correct?

[rose grotto]

one sec

nvm

its right

idk why I asked for help my bad

I have another question tho

[barren zealot]

Hmm, go ahead

[rose grotto]

20. c

How do I tell what to multiply the determinant by

[barren zealot]

Hmm, let me see

I don’t think you multiply it by something, just take the determinant and compare it to when you have

You know ad - bc = 5

Calculate the det of the matrix in c part, and hopefully it’ll have some relation to ad - bc

$2ac + 4bc - 2ac - 4ad \ -4(ad - bc) = -20$

[stoic python]

hadeedji:

[rose grotto]

hi

kk

thats the

asnwer

[barren zealot]

Cool

[rose grotto]

ty

[north sierra]

im having trouble with this question if someone could help me thatd be great

[half ice]

Can you think of a matrix A such that
Ae1 = y1
Ae2 = y2?

Just expand what I wrote there, but include a general matrix for A

[rose grotto]

for e whats the quick way it was reffering to

I would do it like

id get the cofactors, adjoint it then inverse

whats the fast

[north sierra]

A = [2 0]
[5 0]

@half ice ?

[half ice]

That maps e1 to y1! But it doesn't map e2 to y2

[north sierra]

oh

i gotta find one matrix that maps to both?

[half ice]

Yep

[north sierra]

so do i just guess and check

[half ice]

If you'd like

Try, even if only in your head, to multiply some general matrix by e1, e2

[north sierra]

theres not easier way to do it?

[half ice]

The answer is pretty simple

Ultimately you can take what I wrote and let A be a general matrix
a b
c d
You'll get a system of equations

But this question doesn't need that treatment. The form of A is pretty simple

[north sierra]

@half ice hey so

[2 -1]
[5 6]

.would be A

[half ice]

@north sierra
Yes it would

[north sierra]

so i dont really get this question tbh

how is (5,-3) connected to any of e1,e1,y1,y2

like how can i find the range without even knowing the function

[half ice]

So you found the linear transformation.
T(e1) = y1
T(e2) = y2

The linear transformation, like all linear transformations, can be represented with a matrix

[north sierra]

yeah

so how could i possibly find T((5,-3))

[half ice]

T(x) = Ax
For any vector x

So you can find what T does to any vector now

[north sierra]

oh

so

[2 -1]
[5 6] multiplied by [5]
[3]

[half ice]

Yes lol

[north sierra]

lol that was weird typing it in discord

haha

[north sierra]

@half ice what if the roles were switched?

so like e1 and y1 would be the same

but e2 would be (5,-3) and y2 would be (13,7)

and we have to find the image of (0,1)

[half ice]

What's (13,7)?

[north sierra]

the image of (5,-3)

[half ice]

I think this question is significantly different from the original

[north sierra]

oh

[half ice]

What you're eventually getting into is a change of basis

If you find a basis of R², then you can define any linear transformation by just defining where the basis vectors go

[north sierra]

true yeah

[half ice]

So let's say your basis is (1, 0) (5,-3)

Which is a basis of R²

[north sierra]

ya

[half ice]

There's a linear transformation that takes it to any other basis of R²

You can find it by solving a system of equations

[north sierra]

i could turn (1,0) (5,-3) into a matrix and do marix multiplication with a vector right?

[half ice]

Nope. That just happened to work last time

[north sierra]

o

[half ice]

Since you're not mixing where anything goes, it's still the same matrix

You're just changing the basis we're paying attention to

[north sierra]

Since you're not mixing where anything goes, it's still the same matrix
You're just changing the basis we're paying attention to

for the question that we were doing earlier?

[half ice]

The question we did earlier, and the new one you just posted, both have the same matrix A

[north sierra]

o

the basis is just (1,0) and (0,1) right?

[half ice]

In the first one yeah

And if you say where those go, you've said where everything goes

[north sierra]

true

wait why is the basis not the same for the second one (for my second question (the one i made up))

.

[rose grotto]

I have a question plz

[normal gust]

don't ask to ask, just ask

[rose grotto]

How do I calculate C

Like Someone told me to get the DET of it

But how

[fallow jolt]

Go through the process, maybe it will simplify to something easy?

[rose grotto]

Would I do like
The 2x2 matrix déterminant method where I a d- b c all over 1 * the swapped version

@half ice can u help plz

[slow scroll]

@rose grotto how can you manipulate a matrix in such a way that the determinant does not change?

then think about how different operations change the determinant

[rose grotto]

adjoint idk

transpose

@slow scroll

if det(b) = 5

whats det(adj(b))

<@&286206848099549185>

if Det A = 1 and Det B = 2

Whats Det(AB)

and whats Det(ADJ(b))

[undone garnet]

any idea for b) and c)?

for b)

I assume that det(B) != 0

so

det(A).det(B)^k = det(B)^k.det(A+kI)

<=> det(A) = det(A+kI) for all k in N

I don't know how to conclude that this is impossible

or I would say

P(k) = det(A+kI) - det(A)

deg P = n

so

P has no more than n solutions

but

P(k) = det(A+kI) - det(A) = 0 for all k in N

so P(k) = 0

<=> det(A+kI) = det(A)

=0

so that

det(A+kI) implies A has -k as an eigenvalue

for all k in N

but A is n matrix, A cannot have more than k eigenvalues, contradict

so det(B) must be 0

[uneven bloom]

Yeah, you can’t have more than n solutions to the resulting polynomial relation.

Nice idea to see!

[undone garnet]

Thank you :D

[north sierra]

@half ice hey sorry for keep @ing you.

But I just thought of something and wanted to ask you:

earlier you told me to find a matrix A such that:
Ae1 = y1
Ae2 = y2

is that because we want to get a function (matrix) that will satisfy both Ae1 = y1 and Ae2 = y2
and once we find that matrix A we basically found the linear transformation and once we found the linear transformation we can know the image of (5,-3) because our matrix will work on any vector x?

i really hope you see this cause im so curious now lol

[undone garnet]

Im thinkingg for problem c

Haven't had an any idea yet

[uneven bloom]

I’m looking at eigenvectors

[undone garnet]

Hm..

You mean

B^k.v = 0 for all v

[uneven bloom]

Yeah, eigenvectors are maintained in a sense

We can go by contradiction and assume there is a nonzero eigenvalue

[undone garnet]

Hmm...ok, i think about it

[half ice]

@north sierra
Yes

[north sierra]

nice!

okay thanks

[uneven bloom]

@undone garnet Using tr(AB)=tr(BA) and the additivity of the trace, one gets tr(A^k)=0, implying all eigenvalues of the matrix are zero.

[uncut forge]

If $T^3=0$ but $T^2\neq 0$ is it true that the dimension of ker $T^3, T^2, T$ are strictly decreasing?

[stoic python]

AoiKunie:

[past badge]

Is there a way to prove two arbitrary vectors in $R^2$ are parallel if they are both orthogonal to a third vector in $R^2$? I can clearly see this intuitively but I'm required to give a somewhat rigorous proof.

[stoic python]

josh:

[uncut forge]

Let u be the third vector. It is sufficient to show that the orthogonal complement of u is one-dimensional

[undone garnet]

@uneven bloom

I get your idea

you mean

$\lambda_1^k + \lambda_2^k + ... + \lambda_n^k = 0, \forall k \in \mathbb{N}$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

but I don't know how we can imply $\lambda_1=\lambda_2=...=\lambda_n=0$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

oh

😄

let k = 2

$\lambda_i^2 \ge 0$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

so

$\sum\lambda_i^2 =0 \Leftrightarrow \lambda_i^2 = 0 \Leftrightarrow \lambda_i = 0, \forall i$

[stoic python]

Nguyễn Thành Trung:

[atomic flint]

http://prntscr.com/pmovql

So im struggling with this one, i think that it has something to do with one of the fundamental properties which must hold in order for a metric space to be satified

but im not sure how i would show this

[subtle walrus]

by finding a counterexample

[atomic flint]

what do you mean

[subtle walrus]

you know the properties of a metric?

[atomic flint]

yes sir

[subtle walrus]

so one of them must not hold

you can give a concrete example for that

[atomic flint]

yea, but like the thing which is confusing me is how would i do the one where d(x,x) = 0

[subtle walrus]

what do you mean "do"?

[atomic flint]

like prove for that d(x,x) = d(y,y) = iff y=x

as this one has |x + y|

[subtle walrus]

you are supposed to show that this is not a metric

[atomic flint]

yea

[subtle walrus]

also "d(x,x) = d(y,y) = iff y=x" is not a requirement of a metric

[atomic flint]

are you sure

[subtle walrus]

i am sure

you have "d(x,y) = 0 iff x=y"

and "d(x,y) = d(y, x)"

[atomic flint]

i mean i dont have that

idk

[subtle walrus]

why do you not have that?

can you give an example?

[atomic flint]

I have d(x,y) = |x+y|

and if x = y then wont it be d(x,y) = |2x|

[subtle walrus]

correct

[atomic flint]

but then d(x,y) doesnt equal 0

[subtle walrus]

well, for x not 0 at least, yes

but also correct

so, you can't have a metric

and you are done

[atomic flint]

oh damm

that was easier than i thought

thanks

[subtle walrus]

yw

[compact light]

If i do a linear aplication from R5 to R3 from the canonic base of R5 to R3, do i have to calculate the rank of the result to prove its linearly independent or is it by definition?

[atomic flint]

http://prntscr.com/pmpzck

so for part a, i got for $n=1, (1-\frac{1}{n},\frac{1}{n^2})$ goes to (1,1), then for $n=2 (\frac{1}{2},\frac{1}{4})$, n=3, $(\frac{2}{3},\frac{1}{9})$

[stoic python]

B1GW0LF:

[atomic flint]

so like would the limits for this one be as n tends to infinity, (1-1/n,1/n^2) tends to (1,0)

and then for part c, i get a circle, with centre (0,0) and radius 1

So like what would it tends towards

cause there is no end, or is that just a Euclidean space which doesnt have a limit

[dusky epoch]

and then for part c, i get a circle, with centre (0,0) and radius 1
what

ok like

your terminology is way off

in many places

[atomic flint]

yea dw about it i figured it out, and yea i was off

[limpid gale]

Hi, i have a question to you guys, problem says: find all matrices that

question might be trivial but i had my first matrices lecture today and tommorow i will have exercise classes and i dont have any idea how to do this kind of problems

[uneven bloom]

For 2 by 2 matrices, it’s necessary and sufficient that both the determinant and trace are zero. Can you go from there?

@undone garnet you can’t assume all eigenvalues are real. However, the result follows from Newton sums. After we deduce that all eigenvalues are zero, we can use Cayley Hamilton to conclude

[limpid gale]

oh on today lecture there was nothing on traces

maybe i should wait until tommorow and listen

i hope he will explain that to the students xD

[uneven bloom]

For a 2 by 2 matrix [a b] [c d], the trace is a+d and the determinant is ad-bc.

[limpid gale]

huge thanks <3

[half ice]

To keep more general, write a matrix
a b
c d
Then square it. You'll get a system of equations to solve

There are better methods. I can't imagine knowing them on day 1 though

[limpid gale]

ok, i will try

thank you too

[north sierra]

having trouble showing this

and even understanding lol

[half ice]

x = p + tv
Is an infinite line though p that has the direction of v

Show that a linear transformation either moves a line onto another, or maps it all to a point

[north sierra]

what does it mean that a line goes through p?

[uneven bloom]

P is a vector (corresponding to a point in R^n)

[coral elk]

Anyone know a good book on linear algebra? Wanted to take it next semester but there’s a conflict with my physics 2 class.

I wanted to at least get acquainted with it before I hit advanced material like quantum. I’m already picking some linear algebra from differential equations rn

[wispy kayak]

@coral elk

[jagged saffron]

given a min poly x^3+x^2-1 how can i construct two non similar real 3x3 matrices?

we have 1 real root of degree 1, so im kinda stuck

I don't think complexifying works either because the other roots have real parts

[unkempt robin]

Even with the hint they provided, how do I even begin to approach part A of this problem

Why should I + BA be invertible just because I + AB is invertible?

[dense holly]

i know you have linear dependence when your determinant is nonzero but what if you are working with functions and the determinant is 0 only at a certain value

[jagged saffron]

that does not make sense

the determinant of an linear operator T is defined as the determinant of the matrix which represents such an operator over any basis you choose

since similar matrices have the same determinant

therefore the determinant cannot just be 0 at a certain value

[dense holly]

@jagged saffron a lot of my determinants come out as polynomials like 6x^4 which is 0 when x=0

not sure how that contradicts what youre saying

[jagged saffron]

those are

minimal polynomials

or characteristic polynomials

you're doing M - $\lambda I$ and taking the determinant of this aren't you

[stoic python]

Victoria:

[jagged saffron]

the roots of this tell you the eigenvalues

[dense holly]

im using wronskian to show if functions are linearly independent or not

not looking for any eigenvalues

[jagged saffron]

oh you're doing fg' - gf'?

[dense holly]

yea

but for a lot of fxns

so back to the example of 6x^4 as my det would i say the fxns are LI or LD

[jagged saffron]

wait im completely wrong

if its nonzero it implies independence

thats it

they are independent on x not equal to 0

[dense holly]

ok do i need to state an interval you think or is it cool to just say the fxns are LI when x is not equal to 2,3,4, or whatever

[jagged saffron]

the interval is x not equal to 0

[dense holly]

ok thanks

[undone garnet]

@uneven bloom I have a question

A is nilpotent

then

trace(A^k) is always = 0

but

trace(A^k) is always = 0

can we imply A is nipotent?

[uncut forge]

Trace is the sum of the eigenvalues

A is nilpotent iff the only eigenvalue is 0

[undone garnet]

iff?

A is nilpotent, trace(A) = 0

but

trace(A) = 0

ahhhhh

woops

I'm wrong

[uncut forge]

Yeah a counterexample is diag(1,-1)

[quartz compass]

it's not a counter example because when k is even the trace is 2

[uncut forge]

Right, that also shows a counterexample can't be diagonalizeable

Turns out there are no counterexamples because it's true

https://math.stackexchange.com/questions/1229233/operatornametra-operatornametra2-ldots-operatornametran

[undone garnet]

hey guys

I have a question

$a \in \mathbb{R}$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

prove that A is diagonalizable

so I compute charac poly of A

$P(\lambda) = -\lambda^3 + (2a+1)\lambda^2 + \lambda - 2a-1$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

$P(\lambda) = (\lambda - 2a-1)(-\lambda^2 +1)$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

for a not equal -1 or 1

we have P(lambda) has 3 distinct solutions

so A is diagonalizble

but for a=-1, a=1, how can we prove?

I mean problem is prove A is diagonalizble for all a in R

[pallid swallow]

well, that's only 2 cases left

I guess you can substitute and try it out

[undone garnet]

you mean

diagona it?

but this problem have a) and b)

problem a) is what I

am asking

and problem b) is diagona it

[silver ore]

heya

I got a question

its about projections

[pallid swallow]

@undone garnet yeah just diagonalise in the case of a=-1 and a=1

@silver ore #❓how-to-get-help maybe because this channel may be crowded

[silver ore]

kk

[undone garnet]

oke

[sonic osprey]

How do I show

Given a finite dimensional vector space over C and two dot products on the space

That there exists a basis such that the vectors are pairwise orthogonal to both dot products

[sleek helm]

I think you’ll probably have to construct it

[sonic osprey]

Hm

Not sure how

Taking an orthogonal basis for the first dot product and trying to gram Schmidt for the second doesn't work I think

@pallid swallow help me again

[sonic osprey]

<@&286206848099549185>

[undone garnet]

I don't understand this paragraph a lot

can anyone make me get it?

https://math.stackexchange.com/questions/1229233/operatornametra-operatornametra2-ldots-operatornametran

[silver ore]

heya

I have no idea what to do

like at all

I know the first isomorphism rule, but I don't know how to actually use it

[dusky epoch]

(a) or (b)?

[silver ore]

a

[dusky epoch]

ok well

[silver ore]

what is the difference between (U+W)/W and just U/W?

[dusky epoch]

U/W doesn't make sense because W isn't known to be contained in U

and your exercise doesn't mention U/W at any point either.

[silver ore]

I see

ah, the question makes more sense now

what would be the kernel for this function?

will I get the answer if I manage to equate the kernel to UnW?

or is it something else that I need to do?

[brittle juniper]

"this function" you mean phi?

φ

[silver ore]

yep

[brittle juniper]

Well yeah, but you'll need to know Im(φ) as well

[silver ore]

what is lm?

[sonic osprey]

image

[silver ore]

oh

I though that was a L all of a sudden

soz

UnW is not the kernel is it

I don't get it

I don't know how to do this at all

[urban bough]

I have a question as to whether or not a proof is rigorous enough/correct

[silver ore]

wait

it is?

yeap

I think I got it

probably

@urban bough I think you can post here now

I don't have any more questions

[urban bough]

I posted in alpha tho

[spice ruin]

im just starting to learn linear, is there a way to use linear to decompose fractions easier than partial fraction decomp?

[silver ore]

@urban bough but I don't think its gonna be seen

probably

and @spice ruin what do you mean decomposing fractions

[spice ruin]

like in calculus when the denominator is factorable, you can set the fraction equal to multiple fractions being added together using a linear system of equations

im more a less just asking if there is a technique in linear that is not so clunky as to have to manually factor and decompose each problem

[wintry steppe]

Hey! So I'm still in high school but since I am going to study mathematics I was attempting some starter university problems and I'm really unsure how to approach them.
The problem I am looking at right now is as follows:

Show that for all v, w, x ,y ∈ R² \ {0}
(v⊥w and w⊥x and x⊥y) => v⊥y

How would I go about solving this? Do you know of any good resources that help a pre-university student to be better prepared for university proof based math?

[half glade]

vw = 0, wx = 0, x*y = 0

now you need to prove that vy = 0

[wintry steppe]

Yes, I did get to that point

[dusky epoch]

\*

that way discord won't eat your asterisks

even though they aren't really appropriate for dot product

[half glade]

yeah ik

I should use ${v}\cdot{w}$

[stoic python]

AmitN:

[scarlet hamlet]

hey quick question

to disprove this

would this be valid

hit me with a ping if ur here to help

ill go try to do b

[half ice]

@wintry steppe
I suggest trying instead to think of three vectors in R² that are each orthogonal to eachother

[wintry steppe]

Hmm, thank you I will attempt it

[half ice]

@scarlet hamlet
It's hard to read your writing, but a single counter example would be enough here

[arctic fox]

I have a quick question, are two vectors distinct even if one's a scalar multiple of the other? Are they distinct as long as the components are different values?

[gray dust]

@arctic fox if distinct means not equal, that's all there is to it. (2, 4) is distinct from (4, 8)

[arctic fox]

aight got it, thanks!

[gray dust]

for any two elements in a set, ie $x,y \in S$ then x and y are distinct if $x \neq y$

[stoic python]

RokettoJanpu:

[pallid swallow]

@sonic osprey So far what I have is WLOG the first dot product is standard, because we can take an orthonormal basis for the first dot product, so that the components of the vectors we construct are considered over the orthonormal basis. We are going to construct orthogonal vectors (with standard dot product, as well as the other dot product)

[urban bough]

Hello everyone

[pallid swallow]

hi

[urban bough]

I have a question about a proof that I have been trying to get help with for the past few days

[pallid swallow]

@sonic osprey okay so, now we start with the standard basis and we take all the vectors as column vectors in a complex matrix. We are allowed to do these operations: Right multiply by a diagonal matrix with nonzero entries, right multiply by any orthonormal matrix (which would preserve the standard dot product) I'm thinking we use orthonormal matrices which are just rotations of 2 vectors and slowly shift the vectors so that they form orthogonal vectors for the other dot product.

[urban bough]

May I please get help with it?

[sonic osprey]

Yeah I thought about that

Given any two vectors, it'd be true that you can rotate them so that they become orthogonal in the other dot product

But it seems that then changing the orientation of either vector by rotating it along some other plane would change its valid in the other dot product

[paper egret]

what is the definition [T]_B

the only thing i got is

[T(v)]_B = [T]_B * [v]_B

[paper egret]

<@&286206848099549185>

[left wolf]

i think it's the matrix of T in base B

[paper egret]

i kinda have an idea what's going on

but i just need a bit of intuition, i think there are still some gaps on what i know and dont know

[north sandal]

${∀}c\exists${realnumber}

[stoic python]

czloc:

[north sandal]

I don't know how to use TeXit

[strange vault]

Perhaps your real number could be x, and your set R is denoted with \mathbb{R}

so,

$\forall{c},\exists x\in\mathbb{R}$

[stoic python]

NighttimeFlux:

[strange vault]

Forall c, there exists an x in R (a real number x)

there you go matey

A bit sloppy, but yeah lol

[gilded veldt]

is outer product the same thing with cross product ?

same goes with inner product is same with dot product?

[cobalt tartan]

For part b of this question, I'm having some trouble

I'm able to show that $U'' \cap W = {\vec{0}}$

[stoic python]

Liria ^(;,;)^:

[cobalt tartan]

But I'm not sure how to show that $U'' \oplus W = V$

[stoic python]

Liria ^(;,;)^:

[cobalt tartan]

My idea is that if we assume there exists some $\vec{v} \not\in W$, then we have that, by the givens we some unique $\vec{u} \in U, \vec{w} \in W} such that $\vec{v} = \vec{u} + \vec{w}$

But I'm not sure what to do from here

[cobalt tartan]

Like

Another idea was that because L:U->W, by part a, we must have that L is an isomorphism

But that's not true

Because L:U->W is any linear mapping from U to W, not necessarily an isomorphism

Is that rigHT?

[dusky epoch]

yes

[cobalt tartan]

Yea so like it doesnt' matter that L(0) = 0 either right?

Like I know that L(0) = 0 because if that were'nt true then we wouldnt' have that $U'' \cap W = 0$

[stoic python]

Liria ^(;,;)^:

[cobalt tartan]

Like the only theorems that I know that relate the intersection and the direct sum are

And this I guess?

But not really

Ann, do you have any suggestions?

Wait

L is any linear mapping, not necessarily injective and surjective

As long as it's linear

Like it's easy if L is an isomorphism

[undone garnet]

A, B, C is M_n

prove that if r(A) = r(BA) then r(AC) = r(BAC)

any idea?

[dusky epoch]

consider A, B and C as linear transformations from R^n to R^n

[undone garnet]

😮

got it

I solved it

[uneven bloom]

I just looked at null spaces

[undone garnet]

nullspaces?

can you show me?

here how i did

my idea is to prove dim(ker(AC)) = dim(ker(BAC))

indeed

let x in ker(AC), then

[uneven bloom]

The kers are the null spaces

[undone garnet]

ACx = 0 => BACx = 0 <=> x in ker(BAC) or dim(ker(AC)) <= dim(ker(AC))

for y in ker(BAC), then

BACy = 0 => Cy in ker(BA) = ker(A)

so A.Cy = 0 <=> y in ker(AC) or dim(ker(BAC)) <= dim(ker(AC))

[dusky epoch]

kernel = nullspace

[undone garnet]

so dim(ker(AC)) = dim(ker(BAC)) => rank(AC) = rank(BAC)

[uneven bloom]

Exactly the same argument

[undone garnet]

😄

oh

[dusky epoch]

the two terms mean the same thing in the context of linalg

[undone garnet]

I got it

thank you

[undone garnet]

hm...

A is nxn matrix

a_ii = 0 or = 2000

a_ij = 1 for i != j

prove that rank(A) = n or rank(A) = n-1

😐

I have no idea to solve this problem

any idea?

[quartz compass]

is this two separate problems

what does a_ii =0 or =2000 mean?

[undone garnet]

element in diagonal is 0 or 2000

[quartz compass]

...

are you flipping a coin to determine the diagonal elements?

nevermind, good luck

[dusky epoch]

let S = {v_1, v_2, ..., v_n} be linearly independent and let S' = {v_1 + u, ..., v_n + u} for some fixed vector u. i think dim span(S) should then be greater than or equal to n-1.

oh and u != 0 i guess for convenience

try proving this @undone garnet

no guarantees but rn it seems intuitively clear to me more or less

[undone garnet]

oke

I'll try

I'm thinking about

let A = [v_1, v_2, ..., v_n]

such that

v_ii = 0, v_ij = 1

that makes det(A) != 0

so

A = [v1 + u1, v2 + u2, ..., vn + un]

uii = 0 or 2000

😐

[dusky epoch]

i... was doing that specifically to avoid the messy numerical details

[wintry steppe]

Would anyone be so kind as to verify my proof to

Show that for all v, w, x ,y ∈ R² \ {0}
(v⊥w and w⊥x and x⊥y) => v⊥y

Proof:

**
w.x = 0 and w.v = 0 => v and x are collinear and it follows, that y.v = 0 => y⊥v
**

[empty copper]

Seems good

[wintry steppe]

Thanks!

[undone garnet]

@dusky epoch

I have an idea 😄

use mod 2

A mod 2

a_ii = 0

a_ij = 1

and det != 0

so

A also != 0

so rank(A) = n >= n-1

seem ok?

[dusky epoch]

hang on

is the determinant of A mod 2 actually always going to be 1

that doesn't sound right to me

[undone garnet]

hm...

[dusky epoch]

gonna check that rq

[undone garnet]

det(A mod 2) = (-1)^(n-1).(n-1)

so when n is even

det(A mod 2) always odd

but for n is odd hm...

[dusky epoch]

yeah so that only works when n is even

i don't think this is a good avenue at all

if you had some other real number instead of 2000 you wouldn't be able to go modular at all

[undone garnet]

hm....

no

when n is odd

we take

determinant (n-1)x(n-1)

of A

and n is odd then n-1 is even

using same method

we get det(A_(n-1)x(n-1) mod 2) != 0

so

rank(A) >= n - 1

I mean

A_11

drop first row and first column

[dusky epoch]

still feels kinda cheaty that you're using modular arithmetic in the first place

[undone garnet]

hm...

I feel it ok

cause I

don't know another method yet

[dusky epoch]

i mean the thing is this seems like a special case of something much more general

and what if instead of 2000 it was 2001

or 2000.5

[undone garnet]

I think

problem say 2000

for this reason

using modular

[dusky epoch]

there's a certain rule of thumb

if a problem has numerical data that looks like a recent year number

there is VERY LIKELY nothing special about that number

like if your problem's talking about 2013×2013 matrices... the chances are so remote to be nearly nonexistent that there's something significant that holds for those matrices but not ones of any lower size

[undone garnet]

hm...

hope to see another idea

[undone garnet]

exist B such that B^2=A?

[brittle orchid]

Hey, could someone help me understand this?

I'm a little confused with the notation

[gloomy arrow]

Can you have more than one eigenvector to one eigenvalue

[gray dust]

a scalar multiple of an eigenvector is also an eigenvector

[normal gust]

Notice sometimes you have to set up parameters when solving the augmented matrix of A-lambda*Identity

after you plug back in the eigenvalue

the column of those coefficient of parameters are your eigenvectors, and sometimes you'll have more than just 1

[gloomy arrow]

Gotchu

And eigenvectors are all parallel right

Wait

No I mean they are orthogonal?

[half ice]

Not necessarily either

[gloomy arrow]

I see

Right okay that would make sense

[half ice]

They do form a subspace