linear-algebra

[wintry steppe]

I have found two eigenvectors corresponding to 2 eigenvalues but idk how to proceed further coz i need 2 more eigenvectors

[stuck tendon]

Post the question. If the eigenspaces each have dimension 1 then you cannot diagonalise the operator

[spare widget]

the rotation in 2d whose inverse is equal to it is either a 180 degree rotation or the identity (try multplying the angle by 2).

[torpid barn]

right

how do we get R^TQ = Q^TR tho?

[wintry steppe]

I got it thanks 134...!!

[outer goblet]

im trying to learn inner product from linear algebra done right

can someone explain to me what P(R) would mean in this ocntext

context

[dusky epoch]

probably the space of all polynomial functions with real coefficients

@outer goblet

[outer goblet]

ah okay thanks i guess that makes sence

[bright current]

Let V be a vector space, and let {VI' ... ,Vm} generate V.
Let WI' ... ,wn be elements of V and assume that n > m. Then WI'· .. 'Wn
are linearly dependent. can somebody explain what this theorem is trying to say

[dusky epoch]

have you tried to copy text from a pdf

[hazy cape]

What does the curly E looking cymbal mean?

[dusky epoch]

"is in"

or "is an element of"

[hazy cape]

Ah

That makes sense

[lavish jay]

After solving the equation (A-4I)x = 0 i got $$ \begin{bmatrix} 0 & 2 & 3 & 3 \ 0 & -2 & h & 3 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 1 \end{bmatrix} $$ in this matrix i'll only have one free variable, since x1 and x4 = 0. So the dimension will be 1

no matter what h

[stoic python]

Guilhotina

[wintry steppe]

for the first question, remember how determinants and volumes are related. for the second remember that det AB = det A det B and det I = 1.

(the volume of the parallelepiped spanned by vectors v_1, ..., v_n is det(v_1, ..., v_n), alternatively the determinant of the matrix whose columns are v_1, .., v_n. order matters, up to a sign)

[dusky epoch]

if you take the absolute value of the determinant then order doesnt matter

[wintry steppe]

just work in characteristic two

of a certain matrix, yes

the determinant tells you when a matrix is invertible

and you found the determinant of C in the first part

yes

the problem didn't say you had to compute C^{-1}

only its determinant

[wintry steppe]

use the last two facts

i do not know what that means

use the facts that det(AB) = (det A)(det B) and that det(identity matrix) = 1 to find det(C^{-1}) if you know det C

this is quicker than finding the inverse of C and computing its determinant manually

is this exam currently going on?

you tell me

asking for help on exams is against the rules

<@&268886789983436800>

[tranquil steeple]

idiot

[wintry steppe]

you are certainly not helping your case

[tranquil steeple]

you are

[loud halo]

Thanks

[wintry steppe]

good riddance

they already got warned by roketto a few days ago for being a prick

24 hour mute too

[loud halo]

Yeah that was so silly

"sarcasm"

[wintry steppe]

"your mom is too"

[loud halo]

what's sarcastic about saying you're taking an exam...

[stuck tendon]

Yes, dimension is 1 for all h, but actually x1 is free and x2=x3=x4 are 0. rref below

[amber osprey]

D is the answer right

bcs it depends on the system

[spare widget]

yes it depends on the system

[amber osprey]

what do they mean by $\mathbb{R}^{3}$ in $\mathbb{R}^{4}$ heelp

[stoic python]

hmmmm

[stuck tendon]

It is a misprint. The "in R^4" shouldn't be there

[amber osprey]

omg

[tribal willow]

why is the latter not a subspace?

[dusky epoch]

fails closure under addition

also fails closure under scaling

[tribal willow]

ah ok

[native rampart]

Well That would be an affine space though

[stoic python]

anamono

[tribal willow]

my guess would be no, right?

because no additive inverse

[faint lintel]

yup

you're right and that's exactly one of the reasons

also if 0 is not a natural number in your definition

then you also have no additive identity

[tribal willow]

yeah but there seems to be differing opinions on whether 0 in N in this server haha

idk if it is or isnt

[faint lintel]

it depends on who you ask

most mathematicians are wrong say that 0 is not a natural number

[tribal willow]

lmao

[faint lintel]

alot of CS people say it is

[dawn ocean]

@tribal willow

if you're interested in the more unique vector spaces you can construct one out of any field so you can use field extensions or F_p (integers mod p)

if you use the integers mod p you end up with like a lattice with the points as your vectors

also additionally there is the algebra stuff like modules (which use rings instead of fields and are much more flexible) and other things in algebra that have quite unique structures for vector spaces

[tribal willow]

i know what a module is but not a lattice :^)

but thank you i will keep them in mind

i know a lattice is something you put your plants on and they grow up on it

[dawn ocean]

i don't think the thing i described is officially a lattice (as an official name) it just has a lattice structure but i'm not sure

[tribal willow]

ah kk

[dawn ocean]

a lattice is an algebra structure tho

[tribal willow]

why is

the row space of

i get the first three rows

no idea where the fourth one came from

[dawn ocean]

-4a1 + a2=a4

[tribal willow]

oh true

didnt consider lin combinations

or i guess elementary row operations

or both

[pseudo maple]

Guys can someone please help me how to do both these qs. For c matrix multiplication how its done idk how the matrix should look like for me to multiply

[stuck tendon]

For (a) you just take 2 polynomials p and q and show that T(p+q) = T(p) + T(q), and for a scalar lambda, we have T(lambda p) = lambda T(p).

For (c), you find the image of the basis elements under T, expand them in the basis, and then write the coefficients as the columns.
For example, T(x^2) = xd/dx(x^2) -x d^2/dx^2(x^2) = x(2x) - x(2) = 2x^2 - 2x = 0 * 1 + (-2) * x + 2 (x^2), so the last column is (0,-2,2)^T.

[pseudo maple]

thank you :)))

[verbal lodge]

Hey guys can someone help me here to solve for R, C is just a constant: G=(515/512)(R-C)-(R-C)%4120

[ocean galleon]

I'm learning LA for machine learning. Is there a benefit to not using a python console along the way? I'm using Gilbert Strang's free MIT lectures

like how good should I get at handwritten solutions to matrix math

[hushed ocean]

in 7.1.6, you can only define P1,P2,...,PN as a vector subspace right?

not a vector space because it's not closed under addition

[subtle walrus]

hm?

p_1, ... are polynomials are they not

and vector subspaces are vectorspaces

they just live in some other vector space

[hushed ocean]

huh, i thought maybe it wasn't because i thought it wasn't allowing linear combinations of x^n

[fair wren]

why isn't e^x in the span of {p_0, p_1,...}

[subtle walrus]

because you would need an infinite linear combination of polynomials

see the taylor expansion of e^x

and thats not allowed

[dawn ocean]

you should understand how the computation is done like do one or two problems by hand. (you don't strictly need to but if you want to read papers on machine learning they put the math in there not the code so you need to understand how to actually do the computations a little bit)

[ocean galleon]

cool yeah im mostly doing it by hand but wanna introduce numpy at same time

instead of just doing numpy after which seems like it will take a lot longer

[dawn ocean]

the calculations for linearly algebra are fairly easy so yeah as long as you understand the process then you can do the numpy for the rest

although tensor/multilinear stuff is a bit trickier

[ocean galleon]

only on chapter 2 of Strang

like just beginning chapter 2, but have been programming (minus the vector and matrix stuff) for a while

[dawn ocean]

i think there are like linear algebra books out there suited for programming like they talk about the hand method and how the computer does it

[hushed ocean]

Could you have a vector space who's vectors indeed are the polynomials extended with infinite sums of polynomials?

also can you check my proof for the fact that e^x isn't spanned. (and cry because I'm not a mathematician.) Suppose by contrad that you can construct e^x from a finite lin comb of polynomials.

then there must be a least n st cnx^n ... =e^x

Take the derivative of both sides, get that the left hand= x. Now the only right hand satisfying that eqn is x

Who's antiderivative is .5x^2, which is trivially not equal to e^x

Qed by contradiction

[wintry steppe]

i don't know about that derivative computation

at all

[keen sierra]

maybe simpler, any linear combination of polynomials is a polynomial, and if you take enough derivatives of a polynomial you get the zero function. that ain't true for e^x

[hushed ocean]

oh lmao

duh

Ty

🙂

[wintry steppe]

no nonzero polynomial is its own derivative

[hushed ocean]

Sorry guys it's been years since i took a math course

I have a proofs based course in computer science that does a short deep dive in linalg which is what's the impetus for this

And ty. Forgetting basic properties 🥴

[keen sierra]

what other topics does the course cover?

[hushed ocean]

It's abstract interpretation! (A theory of sound static analysis). So it also covers a lot of math topics actually

order theory, induction and coinduction, generalizations of a limit, linalg, a little bit of number theory, a little bit of topology

[keen sierra]

interesting, i wouldn't have thought that topology had much role in cs, well maybe metric spaces if they're talking about convergence

[hushed ocean]

The prof is a strict french guy in his seventies

[wintry steppe]

based

[hushed ocean]

and expected we wouldn't forget our math 🙂

[hushed ocean]

Well it's a watered down computer sciency definition of topology. You can use topology to define safety and liveness properties (properties of prefixes of program executions and their limits)

The limits = the liveness properties

because you can't observe them in any finite execution (eg say you're trying to prove that a server always responds to a request forever)

[keen sierra]

ah well, it never hurts to know more math! 😁

[crude schooner]

If I have a vector space V where $dim(V) \geq 1$ consider $V^3$. I put the following relation on the vector space. $(x_1,y_1,z_1) ~ (x_2,y_2,z_2) \iff z_2 - z_1 = 0$. Given the following element $t = (t_1,t_2,t_3)$ in $V^3$. Can we construct $t_i$ elements in $V$ such that $\sum t_i$ is an element of the quotient where $t + \sum t_i$ is an element of the quotient?

[stoic python]

Runner

[pearl elm]

Not sure I understand what I’m supposed to do here

like i think there is a super clean way they want me to do this with clever algebraic manipulation that works with matrices, but I feel like I can do it the really long way

[fair wren]

Did u prove a?

[crude schooner]

If I have a vector space V where $dim(V) \geq 1$ consider $V^3$. I put the following relation on the vector space, $(x_1,y_1,z_1) \sim 0 \iff z_1 = 0$.

Given the following element $t = (t_1,t_2,t_3)$ in $V^3$. Suppose that the class of $[t = (t_1,t_2,t_3)]$ is non-zero in the quotient. It then follows that $t_3 \neq 0$. Can we construct $t_i$ elements in $V$ such that the class of $[-\sum t_i]$ is zero in the quotient, where we are given that the class of $[t + \sum t_i]$ is zero? If so, how many $t_i$ do we need to accomplish this?

[stoic python]

Runner

[pearl elm]

I am talking about approaching both parts of the problem here.

[wintry steppe]

there are relatively "clean" (i.e. not writing out explicit entries) ways to do the problem. have you done it the "really long way" first?

for the second one, you can do it just using non-degeneracy of the dot product. i'll let you figure out how

[wintry steppe]

it's really unclear what you're asking. are these "t_i elements in V" different from the coordinates of [t] you just introduced? what's the equivalence class of their sum? you defined the equivalence relation on V^3 but you're taking the equivalence class of something in V. and so on

can you give context for the question?

[subtle walrus]

no, vector spaces are too general (they can be defined over any field) and you cant make sense of infinite sums in that generality (you would need some notion of convergence, which general vector spaces do not have)
the proof works something like that, assume e^x = some polynomial of degree n and take the derivative n times on both sides; contradiction

[jade burrow]

if i want to find the number of solutions to an equation a+b+c = d where a,b,c can only come from a set like {0, 1, ... ,(p-2)} for some prime p, is this equivalent to finding the number of solutions mod(p-1)?

[subtle walrus]

is this how the question is asked?

then no, addition mod n is different than addition

[jade burrow]

i'm looking at the unique terms in a generic 3 variable homogeneous polynomial of a particular degree, n, over a finite field of characteristic p, and essentially the choice for the powers boils down to solving a + b + c = n(p-2), then n(p-2) - k(p-1) for k in natural numbers but the equation has to have solutions in the set {0, 1, ... , p-2} so i'm trying to wonder how to adapt the usual find solutions of an equation to finding solutions only in this set

[fair siren]

Hello! Could anyone direct me in to any resources that I could use to study these topics? Preferably problems with solutions

[outer goblet]

just basic matricies, not abstract linear algebra?

[sturdy turtle]

hi guys, i have a NxN matrix with elements in gf(2), and i ve raised to the power of E, how can i recover the original matrix?

[dusky epoch]

raised it to the power of E?

what's that supposed to mean?

[sturdy turtle]

like u have
A = matrix(GF(2),[[0,0], [1,1]])
and then
A = A^e

with e being a generic number integer positive

[dusky epoch]

generic?

[sturdy turtle]

yes like a number

[dusky epoch]

is it known

[sturdy turtle]

let s suppose
e = 31337

[sturdy turtle]

yes

[dusky epoch]

so you have an unknown GF(2)-matrix A and you know e ∈ N and B = A^e and you want to recover A...

hm.

[sturdy turtle]

no i have the matrix but already raised to e

[dusky epoch]

??

isn't that what i just said?

you know B and you know e and you want A

[sturdy turtle]

so i have
e = 31337
A = a^e

i need to recover 'a'

[sturdy turtle]

yeah right, sorry man

[dusky epoch]

ok so same idea different notation

[sturdy turtle]

yep

i ve looke up for matrixes root

i ve found this one on stackexchange https://math.stackexchange.com/questions/646919/nth-root-of-a-matrix

[dusky epoch]

well this seems impossible in the general case.

[sturdy turtle]

i can t understand it but maybe you can, but i think that u cant find the root in binary matrixes

[dusky epoch]

like if you have $A^2 = \bmqty{1 & 0 \ 0 & 1}$ then you cannot tell if $A = \bmqty{1 & 0 \ 0 & 1}$ or $A = \bmqty{0 & 1 \ 1 & 0}$ --- and in fact this counterexample works over any field at all, even $\mathbf{GF}(2)$.

[stoic python]

Ann

[sturdy turtle]

if u want i have the matrix B

i can send it to you so you can see it

[dusky epoch]

sure i guess

is it invertible

and what is its size

[sturdy turtle]

50x50

[dusky epoch]

yikes!

[sturdy turtle]

and e = 31337

[slender frost]

I've proven part a, not sure how to do b

Any help would be appreciated xxx

[native rampart]

@slender frost you know the 1+x+x^2...x^k=(1-x^k+1)/(1-x) thing?

It's exactly the same thing

[slender frost]

I don't get why $(I-G)^{(-1)} = (I + G + G^2 + ... + G^{k-1})$ though as someone irl recommended that to me as well

[stoic python]

df_trivial (Amy)

[stuck tendon]

To prove that, you can show $(I + G + ... + G^{k-1})(I - G)=I$. The result follows by uniqueness of inverses for square invertible matrices

[stoic python]

1345631

[slender frost]

I don't get it though expanding that out would just yield $I - G^k$ no?

[stoic python]

df_trivial (Amy)

[native rampart]

Yea That's the correct one

That's exactly what the question asks you to prove

[slender frost]

How is I-G^k =I?

[native rampart]

It's not

It's I-G^k

[slender frost]

Ohhhhh

I'm being dumb, I see lol

Thanks

[native rampart]

I mean I guess the next part is "G is nilpotent " so show that or something

[outer goblet]

how do i generally determine surjectivity and injectivity for transformations?

like for this as an example

[native rampart]

For injectivity,you check the null space

If null space has a nonzero vector it's not injectivr

[outer goblet]

but what would i put as abcd then

just e_i?

[native rampart]

a=1,b=-1,c=0,d=0 maps to (0,0,0)

So it's not injective

[stuck tendon]

Yes, meaning that the relation (I-G)^{(-1)} = (I + G + G^2 + ... + G^{k-1}) is true if G^k = 0, i.e if G is nilpotent of order k

[outer goblet]

so just find a random vector that disproves it? but what if it was injective, like how would i prove it completely

like this is how the solution showed injectivity but idk how does that make sence

[native rampart]

Well show there's no vector that does that

If something were to map to 0 here, (r1,r2,r3)=(0,0,0)

Which gives us the kernel has to be only zero

[outer goblet]

hm

[fair siren]

sorry for the late reply but we already have discussed matrices I was thining about just vectors and without the vector spaces

[outer goblet]

i sent u dm u can find pdf for that book somewhere online, prolly will have what ur looking for

[fair siren]

Thanks man!

[hushed ocean]

ok...
What would adding convergence look like then?

[subtle walrus]

you need some kind of metric

in R you can talk about "infinite polynomials" (power series) if they converge in the standard metric

more generally you will be doing some kind of functional analysis if you have an infinite dimensional vector space with a limit related structure

[outer goblet]

for a linear transformation T, is the dimension of range of T, the same as rank of T?

[stuck tendon]

yes, by definition

[outer goblet]

is range the same as the solution space of a transformation?

like the span of the output?

T:R^n -> R^m, v in R^n

range T = span Tv ?

am i understanding the definition correctly

[slender frost]

Is the spectral radius of a nilpotent matrix < 1?

[native rampart]

It's exactly 0

x^k being the minimal polynomial implies 0 is the only possible eigen value

[slender frost]

ahh

ty

[tribal willow]

why does any set with 0 vector count as linearly dependent?

because if i had a set with two vectors, say {0, 1}

[native rampart]

0=2(0)

[tribal willow]

then c_1 (0) + c_1 (1) for a non-zero c_1 does not equal 0

[native rampart]

Same vector,2 different representations

[tribal willow]

im a bit confused

wdym by same vector two different representations

[native rampart]

Ok,This is a point we don't usually discuss but when we choose a basis we want all vectors to be represented uniquely

If we exclude zero,this reduces to the usual \sum c_i v_i=0 \implies c_i=0

[tribal willow]

ah ok

thanks

[tribal willow]

given a set of vectors, if we express the vectors as rows of a matrix, if we cannot row-reduce it into the identity matrix, is it linearly dependent?

[native rampart]

Is the matrix square?

Well,Yes

If RREF is not I, the matrix is not invertible

[tribal willow]

or i guess i should say if all non-zero rows are not equivalent to the identity matrix, then it’s linearly dependent?

[native rampart]

What would that mean in the non square case

[tribal willow]

$\begin{bmatrix} 1&0&0\0&1&0\0&0&1\0&0&0\end{bmatrix}$

[stoic python]

anamono

[tribal willow]

like this is linearly indep

i think right

[native rampart]

The non zero rows form a linearly independent set yes

[tribal willow]

does the set of four vectors itself form a linearly independent set too?

[native rampart]

No

(0,0,0)

A LI set should not contain zero

[tribal willow]

ah ok

oh wait right because of what we talked about prior

set with zero vector is linearly dep

oops

[native rampart]

Yes

(3,0,0,0)=3(1,0,0,0)+0(0,0,0,0)=3(1,0,0,0)+10(0,0,0,0)

[hushed ocean]

Does anybody know of a proofs based linalg book with worked proofs so I can see when i get the proofs right and where I go wrong?

[slender frost]

I've proved that they are positive using the positive definite thing

not sure how to prove they are real

[hushed ocean]

Moreover this is only true because of the funny property of real vector spaces where the multiplicative identity for the field is contained in the vector space right?

[fleet sun]

1 isn't contained in R^n if n > 1

[young turtle]

Hi, idk if I am at the right server, I am struggling to find out how I am supposed to do this task anyone who can help ?

[young turtle]

<@&286206848099549185>

[teal grotto]

choose 5 points that the polynomial passes through and interpolate using, say, Lagrange polynomials

or a vandermonde matrix

[hushed ocean]

oh i mean, for example, the reason you can easily define the vector basis this way:

(1 0 0), (0 1 0), (0 0 1)

is because R^3 has the funny property of having the multiplicative identity and additive identity as possible elements for the first, second, and third element of its tuple

like in general, if you have a Vector space V over a field F, and the vector space is F^n, you'll be able to find a finite basis right

because you can just define it as e1 = (multiplicative id, add id, add id, ..., add id), e2 = (...), ...

where e1 is an n-tuple

[teal grotto]

what are your thoughts?

[signal dagger]

A vector space is in particular an abelian group under addition. Is this true in this case?

[hushed ocean]

can you pick any set of linearly independent vectors, then say their span = vector space
?

[gray dust]

the set {(1,0)} is indep but does it span R^2?

[hushed ocean]

it doesn't span R^2 yeah 🙂
can't we cheat and say it spans the vector space ({1, 0}) over R

ohh we need additive inverse though

[gray dust]

wdym ({1,0})

[hushed ocean]

({(1, 0), (-1, 0)}, +2, (0, 0)) as the vector space oops ({(1, 0), (-1, 0), (0, 0)}, +2, (0, 0))

😆

over R

[gray dust]

whats +2

[wintry steppe]

How do I find the matrix represention of this transformation?

With the basis of p2

figure out how L acts on the basis elements, express the results in terms of the elements of the basis, and put the coefficients in a matrix

[hushed ocean]

• over R^2 / extended pointwise to R^2

[gray dust]

exactly what vectors are in ur supposed space

[gray dust]

just these 3 vectors?

[hushed ocean]

yes

[gray dust]

but it must be closed under scaling

[hushed ocean]

oh yeah you're right

[wintry steppe]

So would it be dx/dp(1) + 1 for the first vector...

So 1, 0, 0?

[hushed ocean]

{(a, 0), (-b, 0), (0, 0)}, +2, (0, 0)

where $$a, b \in R$$

[stoic python]

jcob_the_student

[wintry steppe]

dp/dx, and yes, that'll be the first column

[gray dust]

that works but its a bit redundant

[wintry steppe]

do \mathbb{R} or \bR

[gray dust]

u can just say the set of all (a,0) with a in R

[hushed ocean]

yeah thats totally tru

[gray dust]

this is precisely span{(1,0)}

[wintry steppe]

Would the second one be 1, 1, 0?

Third then is 1, 1, 3?

[gray dust]

so if u wanna 'cheat' just recall spans are subspaces, so take the span of w/e vectors u want @hushed ocean

[wintry steppe]

the second column is correct but the third is not

That's the solution

I'm suck on the third vector

what's L(x^2)

Shit I'm dumb

Why the first element 0?

what's L(x^2)

2x

no

that's d(x^2) / dx

what's L(x^2)

[gray dust]

recall Lp=p'+p

[wintry steppe]

2x + p?

p being...

X^2

so L(x^2) = 2x + x^2

do you see how to get the third column now?

Ahhhhhhhhhhh

You know i understand this now

Thank you. Teachers here are awful.

Don't teach problems solving, only theorem. Tyty

Is that good for prøving this a linear transformation?

what have you tried

[wintry steppe]

if you've shown this to be true for every single c, p, and q, then yes. taking c = 1 this shows additivity, and taking q = 0 this shows it respects scaling

that said you should probably just show L(cp + dq) = cL(p) + dL(q) for all scalars c, d and vectors p,q, rather than working with the same scalars

[wintry steppe]

Well a set consisting only 1 vector which is non-zero is always linearly independent
So (a) would be the answer, I believe

you're right, (a) is one of them

but you have three other parts to look at

[wintry steppe]

a, c and d?

you are right

[hushed ocean]

A linear algebra question, bordering on criminally stupid (apologies), im not a mathematician, need help checking my comprehension

Suppose we have the vector space $$\langle [0, \pi ] \to \mathbb{R}, \vec{+}, \vec{0} \rangle$$ where $$\vec{+}(\vec{v_1}, \vec{v_2})(x) \triangleq \vec{v_1}(x) + \vec{v_2}(x)$$, $$\vec{0}(x) \triangleq 0$$ over the real numbers, with scalar multiplication defined $$\times(\alpha, \vec{v_1})(x) = \alpha \times \vec{v_1}(x)$$

double checking -- we can do this right? i was trying to make an abelian group plus a field (reals) plus a satisfying scalar multiplication. I believe the scalar multiplication satisfies all the axioms

[stoic python]

jcob_the_student

[wintry steppe]

the set of functions [0, pi] -> R with pointwise addition and scalar multiplication? this is fine.

[hushed ocean]

Choose $$x \mapsto x$$, $$x \mapsto sin(x)$$, $$x \mapsto cos(x)$$ as three linearly independent vectors (an exercise I proved last night.)

[wintry steppe]

\sin, \cos work

[hushed ocean]

You can't span with just $$x mapsto x$$ right? Becuase then you cant span the function $$x \mapsto c_1$$ where $$c_1 \not= 0$$

[stoic python]

jcob_the_student

[wintry steppe]

no, this space isn't spanned by the function x \mapsto x

[stoic python]

jcob_the_student

[hushed ocean]

you'd need sin and cos or x and cos

[wintry steppe]

neither is it spanned by {x, cos x, sin x}

this space is infinite-dimensional

[hushed ocean]

wait really? how'd it become infinite dimensional 😓

[wintry steppe]

it has no finite spanning set

[hushed ocean]

right, because you allow any function from $$[0, \pi] \to R$$

🤦‍♂️

[stoic python]

jcob_the_student

[wintry steppe]

for each $c \in [0, \pi]$, consider the function $$1_c(x) = \begin{cases} 1, & x = c, \ 0, &\text{otherwise}.\end{cases}$$ then ${1_c : c \in [0, \pi]}$ is an uncountably infinite linearly independent set

[stoic python]

TTerra

[hushed ocean]

right because there are inf. #'s in [0, \pi]

[wintry steppe]

if the space were finite-dimensional, it could not have an infinite linearly independent set

[wintry steppe]

Bro, tyty.

[wintry steppe]

someone explained like this.. "If you know the definition of linear independence, two vectors being the same automatically means linear dependent
Since you can apply a coefficient to one vector to get the other, in b, you multiply the first by 0.75, in c, you multiply the first by 1
So only two answers- a and d"

the sets in (a) and (c) are the same set

they are both {(0, 1)}

you would be right if the question was instead asking "are (0, 1) and (0, 1) linearly independent?"

you're correct for (b), though

[cedar patrol]

can someone help me with this. i already did it but im not sure about my answer.

[supple moss]

idk but u can put it in an online calc @cedar patrol

soz

[wintry steppe]

if you're not sure, plug your answer right into the system and check

[supple moss]

i dont get this can someone explain

[wintry steppe]

a + lambda b is a unit vector if it's norm is 1, and that's the same as its norm squared being 1

[supple moss]

norm?

do u mean normal

[wintry steppe]

|vector|

no

[supple moss]

oh

okay

[wintry steppe]

i mean the norm. |v| = sqrt(v dot v)

for v to be a unit vector means that this equals 1

for you, v = a + lambda b, and you need to find lambda making this true, so it boils down to solving the quadratic equation written down there

[supple moss]

ohhh

[hushed ocean]

another misproof, if you're a mathematician, please cry:

Why can't you just say for all vector spaces V, for all fields F, V trivially has a basis:
V spans V

so V is a basis

is it something like a basis must be the least set of vectors that spans V?

[molten ivy]

Yea, a basis has to be linearly independent, none of the vectors in the basis can be a linear combination of any of the others

Which is obviously not the case for the whole vector space

[hushed ocean]

oh ty! 🥴

[vernal cliff]

Prove that if $V$ is a finite-dimensional vector space, then the space of all linear transformation on $V$ is finite-dimensional, and find its dimension. I would like some criticism/proofreading of my approach. It's still incomplete, but it should be somewhat in the right track.

[stoic python]

informationgain

[neon jolt]

is the 0 vector considered to be an element of the span of any list (or empty list) of vectors span(v1, ..., vm)?

[vernal cliff]

Yep. If it's non-empty, just set all coef to zero. If it's empty, the span is just the zero vector (by definition, if I'm not mistaken).

[obtuse pelican]

@dire scaffold this looks good lol

[supple moss]

its not my solution lmao

[wintry steppe]

Let A be vector in K^n. Let W be the set of all elements B in K^n such that B is othogonal to A. Then is W K ^ (n-1)?

[wintry steppe]

take A = 0 and you get K^n. if A is not zero, however, then the resulting space is isomorphic to K^{n-1}

[wild hollow]

bit confused wether this peicewise function is a linear map or not. The second function defined for x<0, preserves addition and scalar multiplication over its domain, but doesnt map the zero vector in R^2 to R^3. So is the function not a linear map then ?

[wintry steppe]

no, f(0, 0) = (0, 0)

so your reasoning is false

[wild hollow]

nvm lol

[wild hollow]

if f(x,y) was just defined for x<0 would it still be considered a linear map ?

or are linear maps usually taken from the entire R^n to the entire R^m

[wintry steppe]

if it was only defined for x < 0, its domain wouldn't be a vector space, so it wouldn't be linear

[wild hollow]

thanks man appreciate it

[tacit pelican]

bit confused on wtf happened on the second line

how did they move the $P^{-1}$ there?

[stoic python]

sean

[tacit pelican]

even if you distribute it $(P^{-1}x I - P^{-1}A)P \implies P^{-1}xP-P^{-1}AP$

[stoic python]

sean

[tacit pelican]

am i just dumb

or am i missing smth

oh wait..

never mind im just dumb

thought x was a vector

[zinc timber]

x is not a vector

it's scalar

[tacit pelican]

yeah figured

[fringe burrow]

Could someone clarify please why the algebraic multiplicity of eigenvalue gives the size of Jordan block of the matrix with that eigenvalue on diagonal?

[native rampart]

T((x,y))=(x+y,x+y)

[split hull]

T(x,y)=(x,0)

[amber osprey]

what do they mean by proportional?

[stuck tendon]

Multiples of those vectors

[stuck tendon]

Take any basis {e1,e2} of R^2, and define T by T(e1)=e1 and T(e2)=e1.

[hushed ocean]

Hi guys, can I get some help with affine spaces:

I'm struggling with understanding WHY an affine space is closed under translation

the wikipedia example, of a plane that passes through the origin (I assume sth like $${(x, y, 0) \mid x, y \in R}$$) which is a vector subspace, then $${(x, y, 10) \mid x, y \in R}$$ makes sense (that it must not be a vector subspace, not closed under group addition)

but why the second would be closed under translation makes no sense! Unless you define the translation to somehow take only parts of the vector basis?

[stoic python]

jcob_the_student

[hushed ocean]

in other words, does the translation project onto $${(x, y, 10) \mid x, y \in R}$$

[stoic python]

jcob_the_student

[hushed ocean]

because i'd assume your vector is still $$R^3$$ (the one you're translating with)!

[stoic python]

jcob_the_student

[stoic python]

Jester

[wintry steppe]

so is the set of all vectors $(x, y) \in R^2$ such that $x + y + 1 = 0$ doesn't form a subspace

[stoic python]

Jester

[wintry steppe]

because (0, 0) is not in it?

[native rampart]

Ok so the logic behind affine spaces is "Ok I take a subset A of R^n and and element x of R^n ,So would {a-x|a in A} form a vector space?"

If there's such an x,A is affine space

[hushed ocean]

so to rephrase informally (sorry) you're saying that an affine space must be a subspace translated from the origin

[native rampart]

You care about the differences not the vectors itself

Yea

[hushed ocean]

so does the translation translate the vector space about the origin?

the translation operation

[native rampart]

There is no origin tho

[hushed ocean]

translation operation translates all points in the space?

[native rampart]

Translation just means instead of caring about A you care about the set {p+a | a in A}

Where p is the vector you are translating everything by

[hushed ocean]

when you do the translation operation, are you at least "usually" translating p

[native rampart]

Now clearly if A originally satsified the affine property
There's an x such that {a-x| a in A} is a vector space

Now this new set is also a affine space since {(a+p)-(p+x) | a in A} is the same set

Your "reference point" is now p+x instead of x as it was before

[native rampart]

Exactly p everywhere

[hushed ocean]

and every translation creates a new affine space that satisfies the affine property?

[native rampart]

Yea

We wanted a vector space like structure which is preserved under translation

Hence affine spaces came into existence

[vivid field]

How do i show that a set is a non-empty subset of P2 ?

[wintry steppe]

give an example of an element in it

[viral olive]

Guys maybe i am stupid but if you complex conjugate a scalarproduct then it just means that you complex conjugate both arguments

[empty hemlock]

Try it on a small example and see what happens.

[viral olive]

Sorry showed the wrong thing

i meant this one either way

i tried it twice and it worked out so far

[wintry steppe]

examples are not proofs

[viral olive]

which is why i asked

i can understand not explaining proofs and i am pretty sure it doesnt work but i dont find any source who argues against it is my problem and this is a small part of a bigger proof

[empty hemlock]

FYI: Inner products satisfy <u,v>* = <v, u>.

[wintry steppe]

Isn't row a perpendicular the nullspace?

So the answer is 5?

I'm pretty confident of tgat

Watch me be wrong lol

[empty hemlock]

Yes and yes.

[wintry steppe]

Noice

[wintry steppe]

its a

[wintry steppe]

How do I do this?

I can't reverse engineer the answer

[empty hemlock]

For a), you already have it. For b), there is a change-of-basis matrix that you have to compute.

[wintry steppe]

Do you take inverse of t or b?

[wintry steppe]

Above

[empty hemlock]

Huh? I haven't done anything.

[wintry steppe]

I was just pinging you

For what i said Above

[empty hemlock]

Ah.

Do you understand the change-of-basis idea?

[stuck tendon]

For (b), you can compute change of basis matrix, or do it directly.
T(1,1) = (1,-3+4) = (1,1) = 1(1,1) + 0(1,3)
T(1,3) = (3,-3+12) = (3,9) = 0(1,1) + 3(1,3)

[unique jasper]

Can anybody help me understand how to get the answer to this. Let A be an nxn matrix. Find the image and kernel of the linear transformation L(A) = 1/2 (A+AT ) :
Rn×n → Rn×n

[empty hemlock]

AT is the transpose of A?

[unique jasper]

Yeah

I know for the kernel I can set it equal to 0 and it'll give me something like A=-AT

[unique jasper]

But I'm thoroughly confused on the image

[empty hemlock]

Yeah, the image is the image. Not much you can say about it unless you know some facts beforehand about that kind of sum.

[unique jasper]

So how would I get the image off of an equation

[empty hemlock]

I suspect L(A) is always Hermitian.

[unique jasper]

Would it just be A=AT?

[hard drum]

There is a nice description of the image

Have a think

[hard drum]

No, what about e.g. a matrix with just i on the diagonal

[empty hemlock]

Yeah... only if A is already Hermitian is L(A) also Hermitian.

[hushed ocean]

pls don't answer this before veroegard's question:
my prof has a stray proof in his book that you can represent, for finite matrices

Tx = b

as the affine subspace that is x_0 + Ker(T)

where x_0 is a solution to Ax+b

i don't really understand it 🥴 i think he proves it with gauss Jordan, has anybody presented it nicely somewhere

[unique jasper]

Ok since the equation is 1/2(A+AT) I know the kernel would be L(A)={A:A=-AT}

But the image of L(A) would be given by what exactly.

Would it be A=AT, but then how would I explain how I got that?

[empty hemlock]

The image is just the set { L(A) | A is n x n real matrix }.

They probabaly just want some salient description of it.

[unique jasper]

Oh ok that makes more sense now thanks for the help.

If it's a salient description then I don't really need to explain myself

[empty hemlock]

Have a look at: https://en.wikipedia.org/wiki/Symmetric_matrix#Decomposition_into_symmetric_and_skew-symmetric

So the image consists of symmetric matrices.

This is why I mentioned Hermitian, since there's a similar result that I happened to know.

[unique jasper]

Ok I get it now thanks, I really appreciate the help.

[hushed ocean]

Ok:
my prof has a stray proof in his book that you can represent, for finite matrices

Tx = b

as the affine subspace that is x_0 + Ker(T)

where x_0 is a solution to Ax+b

i don't really understand it 🥴 i think he proves it with gauss Jordan, has anybody presented it nicely somewhere

[fringe fjord]

Well x0 is certainly in the solution.

Anything else in the subspace is x0 + y for some y in Ker(T), which means Ty=0.

But then T(x0+y) = T(x0) + T(y) = T(x0) = b as required. So the entire affine subspace are solutions.

[hushed ocean]

oh 🙂 that's totally true thank u

I feel like that's obvious and i should have been able to get it 🥴 but ty

[fringe fjord]

On the other hand if you have an x1 that is also a solution, then T(x0)=b and T(x1)=b, so T(x1-x0) = b-b = 0, so x1-x0 is in Ker(T). So every solution is in the subspace.

[bright current]

What is the dimension of the subspace of Rn consisting of those vectors
A = (a 1, ••• ,an) such that a 1 + ... + an = O?

pls help

[native rampart]

n-1

[dusky epoch]

did you mean 0 and not O?

[bright current]

ya i mean 0

[bright current]

i know the ans but how do i solve it

[dusky epoch]

well you can construct a basis for it pretty explicitly

${e_1 - e_n, e_2 - e_n, \dots, e_{n-1} - e_n}$

[stoic python]

Ann

[hidden bridge]

For this question, can I not find the linear transformation matrix that maps the basis B to E (standard basis) -- $At^{B->E}$ and wouldn't that be equal to the identity matrix? As the coordinate of S(v1) is (1, 0, 0) with respect to the standard basis, and the same goes for S(v2) and S(v3)

[stoic python]

ya boi

[hidden bridge]

And then, wouldn't the null space just be the zero vector and the range is just R^3?

[empty hemlock]

You can find the change-of-basis matrix from B to E. That's not the identity matrix though since it would mean v1 gets mapped to v1, not e1.

[hidden bridge]

Wait, how isn't it? Isn't T(v1) = e1, so T(v1) with respect to the standard basis is also just (1 0 0)?

Like, what would the change of basis matrix, then be?

[empty hemlock]

The identity matrix would do I(v1) = v1.

That's not what you want.

[hidden bridge]

oh true

[hidden bridge]

but i thought, we get At^B-->E from finding each image vector's coordinates with respect to the basis that we are mapping the vectors onto (in this case, E)

[empty hemlock]

I'm not sure what you mean. You can find the B to E matrix as the inverse of the E to B matrix.

And the E to B matrix is just the matrix whose columns are v1, v2, v3.

[hidden bridge]

ohhhh yes, mb haha, thank youu

[young turtle]

Hi guys, anyone who can give some ideas or help me around a bit with those two tasks...

[young turtle]

<@&286206848099549185>

[wintry steppe]

Since Q is subfield of C, then C is a vector space over Q. So how many dimensions does vector space C have over Q?

[dusky epoch]

continuum

[wintry steppe]

continuum?

[dusky epoch]

uncountably many

in fact even to prove that a basis exists you have to invoke choice

[wintry steppe]

what is C over R?

[dusky epoch]

2

{1,i} is the obvious basis

[wintry steppe]

then is C over Q same as R over Q?

since 2*inf = inf?

[dusky epoch]

yes they have the same dimension as Q vector spaces

[golden wasp]

does $\frac{1}{\vec{a}}$ hold any meaning?

[stoic python]

Noro

[dusky epoch]

no

[gleaming knot]

1/a where a is a vector has meaning only in an appropriate mathematical structure such as a Clifford algebra

Or if the vector space happens to be a division ring or even a field such as C

[bright current]

Let U, W be subspaces of a
vector space V. Show by the indicated method that
dim U + dim W = dim(U + W) + dim(U n W)

how do i do this??

[teal grotto]

what’s the indicated method?

[bright current]

idk ther is no method mentiones

mentioned*

[stuck tendon]

Note that since U,W are subspaces, U n W is a subspace. So a basis exists for it. Extend it to bases of U and W, say B_U and B_W. Show that B_U U B_W is a basis for U + W, and then it's a counting argument.

[grand hare]

Let M: R3 -> R3 a linear transformation represented by matrix A with respect to ordered basis B = {v1,v2,v3}.
Let S: R3 -> R3 a linear transformation represented by matrix C with respect to the ordered basis B

if v = v1-v2+2v3, find [(S ◦ M)(v)]B

<@&286206848099549185>

[hidden bridge]

what part of this, is confusing

[grand hare]

what

[hidden bridge]

i.e. what part of this are you struggling to get

\understand

[grand hare]

what would be the process here

specifically S(M)

[neon jolt]

Does anyone have any tips/advice for how to show/prove that a given list of vectors in V does or does not span V?

If I’m clever enough I know I could just try to intuitively come up with a counter example. But is there a way to do it algebraically?

[wintry steppe]

if you have dim V vectors you can put them as the columns of a matrix and calculate its determinant

[neon jolt]

Dunno what that means yet 😅. Dimensions are next chapter and matrices after that

[wintry steppe]

well it's an algebraic way of doing it

the determinant is non-zero if and only if those vectors are linearly independent

and since you have dim V of them, this is the case if and only if they span

[wintry steppe]

"algebraic" as in there's an easy, readily-available formula for computing the determinant

[neon jolt]

Is there another way to do it without these concepts? The answer key suggests there is a way to do it that can be tedious, but I’m ok with that for now before I eventually learn about determinants and dimensions

[wintry steppe]

all you can really do is check whether or not every vector in V can be written as a linear combination of the ones you have, then

[neon jolt]

Yikes. Okay maybe I’ll just try it once then move on to the next chapter

[wintry steppe]

i guess you can stick all of the vectors into a matrix and find its rank by e.g. row and column operations

i always forget that's a thing

[devout hazel]

Does anyone have a link or something to a decent 'cheat sheet' for Linear Algebra, just summarizing certain topics concisely?

[spare widget]

https://www.mathematik.hu-berlin.de/~wendl/pub/connections_appendixA_2.pdf

[sage gale]

I'm trying to solve for an eigenvector associated with an eigenvalue of 2 but when I do I get a row with all zeros so the corresponding matrix yields formulas for k1 and k2 in terms of k3 but there is no way to solve for k3, what should I do in this scenario

The part I'm stuck on is underlined in red

[dusky epoch]

well this is exactly what should happen

you would only find the eigenvector up to a constant factor

after all, if v is an eigenvector of A, then so is cv for any nonzero c...

so set k3 to whatever

[sage gale]

got it thank you

[unique jasper]

Does anybody know what the equivalent to a row space is?

For example the column space is the image. Kernel is the nullspace

[fringe fjord]

It's the orthogonal complement of the kernel.

[unique jasper]

ok thanks, I appreciate the help

[unique jasper]

Wait from what you said does that mean that the row space of a A is equivalent to the ImA^T?

Bruh I wrote kernel instead of image

[teal grotto]

y = A^Tx iff y is a linear comb of the rows of A, so yes

[molten ivy]

You say dimension is next chapter, but once you have that you can just see if the Dimension of V is n, that you have n independent vectors in your list, then they span all of V

[bright shale]

Hey, is there any free to use online tool that solves equations given a certain Ring ?

e.g. 358/150 * x +298 = 235 Ring of natural numbers modulo 419

[subtle gust]

how would we prove those 2 statements

i could only do half of the first one

managed to prove that N(A) is a subset of N(A^TA)

how do i prove that N(A^TA) is a subset of N(A)

[wintry steppe]

try using the fact that <x, Ay> = <A^Tx, y>

[subtle gust]

if that's the inner product

then we didn't cover that 🙂

[wintry steppe]

dot product, then

[subtle gust]

same thing

didn't cover that

i could only use the definition

AX=0

and A^TAX=0

[wintry steppe]

solve it using the dot product stuff and then try to eliminate any mention to it, then

you can probably do that

[subtle gust]

idk anything abt the dot product 💀

i just know the name

[subtle gust]

i proved the first half by multiplying both sides of this by A^T

[wintry steppe]

that is correct

[subtle gust]

and that proves that every element x in N(A) is also in N(A^TA)

looking for a similar way

to prove the second half

[wintry steppe]

try multiplying A^TAx = 0 by x^T

this hides any mention to inner products

[subtle gust]

The product wouldn't be defined tho?

nvm it would be lol

so A^TAX=0

X^TA^TAX=0

(AX)^TAX=0

i don't see how we would go from this to AX=0

[wintry steppe]

it's true for any vector y that if y^T y = 0, then y = 0. try proving it by writing out what y^Ty is

assuming everything has real entries

[subtle gust]

oh

i see

tysmmmm

wb the 2nd one?

[subtle gust]

.

rank(A)=rank(A^TA)

oh wait i think i have an idea

we just proved that the null space of those matrices is the same

so it follows that nullity(A)=nullity(A^TA)

rank(A)+nullity(A)=n

rank(A^TA)+nullity(A^TA)=n

[stuck tendon]

you may want to mention that the domain is finite dimensional, so you can apply rank-nullity

[subtle gust]

if we subtract them, and since the nullity is the same, the rank would be the same

right?

[wintry steppe]

right

[subtle gust]

yeah we only deal with finite dimensional spaces

[unique jasper]

What's the difference between an eigenspace and an eigenbasis

Isn't the eigenbasis just the basis of the eigenspace?

Also isnt the span of the eigenvectors of the matrix the basis of the eigenspace?

I'm sorry these question are probably very simple but I have dyslexia so reading these very long definitions with just variables is very difficult for me.

[wintry steppe]

"eigenbasis" usually refers to a basis of the domain of your matrix consisting purely of eigenvectors, not necessarily a basis of a single eigenspace

[wintry steppe]

a span cannot be a basis

linear independence will fail

[unique jasper]

Ok thanks man I appreciate the help. I get it now so the eigen vectors can be the basis of the eigenspace, but not the span of eigenvectors because that would mean it is not linearly independent. And as for eigenbasis it's a basis consisting solely of eigenvectors and so it may cover more than just a singular eigenspace. That's what I'm getting from it. If I'm wrong please let me know.

[hard drum]

So you seem to have understood an eigenbasis correctly, though your sentence before that seems slightly confused to me

For completeness: given a linear map T:V -> V (or you can think of a matrix if you wish I suppose), an eigenspace is a (non-trivial) subspace of V of the form ker(T - λI); equivalently, it's the set of all λ-eigenvectors where λ is an eigenvector of T

An eigenbasis is then a basis of V consisting of eigenvectors of T

Now one link between the two is that we can take a basis of each eigenspace. Taking the union of these bases over the eigenspaces, we'll get some linearly independent subset of V. One hopes that this'll form a basis of V, though this needn't be the case (for example, T may not even have an eigenvalue to begin with!) But if it does form a basis, we've got an eigenbasis

[unique jasper]

Thanks for that. It took me little while to digest but I think I got it.

[hard drum]

Yeah i think it can take a while to get conceptually but I hope it helps

[umbral spindle]

Is the Nullity(A Transpose) equal to Nullity(A)?

[slow scroll]

no, not necessarily

[small vigil]

I am reading the paper "What is local optimality in nonconvex-nonconcave minimax optimization?", and I am confusing to check the fact why the characteristic polynomial shown in the figure can be expanded like p0(\lambda) + ... form?
Could you tell me the detail?

[young sentinel]

@tawny sinew about this, since the channel closed

isn't it false, since at least one of B, C, D has to have rank 1?

[dawn ocean]

yep

[plush dust]

Sorry for dumb question, but how he done this?

[dusky epoch]

factor λ^2 - 5λ + 6

[subtle gust]

Is it true that det(adj(A^-1))

= det(A)^(1-n)

[subtle gust]

Umm anyone?

[dusky epoch]

A^-1 adj(A^-1) = det(A^-1) I so adj(A^-1) = 1/det(A) * A

okay yeah checks out

[subtle gust]

Alr ty

[plain robin]

what could be the basis of this vector space?

[silver heath]

solve the system of linear eqs?

[teal grotto]

find a basis for the kernel of the matrix
1 2 0 3
2 -1 -1 0

[wintry steppe]

Hello, does anyone want to help me with a worksheet on vc?

[weak wharf]

How do I determine whether they are perpendicular, parallel or don't have any relationship?

[haughty berry]

Parallel: is one a scalar multiple of the other?
Perpendicular: is their dot product 0?

[weak wharf]

Oh it's just this

I tried reading on wikipedia and was getting a stroke

[haughty berry]

[young sentinel]

trying to learn math on wikipedia is kinda scary, i usually just end up getting lost in a web of endless definitions

[slow scroll]

You have to figure out what space you are in first

did you mean x_0e^t + x_1e^{-t}?

because x_0e^t + x_1te^t = (x_0 + x_1)e^t

okay but still this

You have to figure out what space you are in first

these are functions

like when you did this with polynomials, you were in the vector space of polynomials of degree <= n

my guess would be something like span{e^t} but idk i prob wouldn't assume

[proven lotus]

Hi, I'm in my first year of engineering and we just started with linear algebra. I was wondering if anyone knows any great recourses for an introduction to linear transformations?

I don't really understand the notes of the professor

[wintry steppe]

I have a good textbook

If you're interested in reading

I can send you the pdf

[proven lotus]

That would be great

[wintry steppe]

I got the eigenvalues, and I got the min

But 26*3 is 78..... right?

Or is this saying the absolute value of lamba can't be greater than 3?

So it's 3 x 3 x 3

[iron harbor]

So say I have two vectors (1,1,1,1) and (-1,-1,1,1) and they're orthogonal... But what does this mean for four dimensional vectors? Do they still span a two dimensional plane between them? How does one even picture a two dimensional plane set in some fourth dimensional space?

[subtle walrus]

orthogonal implies linearly independent, so they span a two dimensional subspace

you dont picture it

[proven lotus]

I don't understand the essence of a "basis" that is made up of linear independent vectors

What is it used for?

[subtle walrus]

it gives you unique representatives of every element in the vector space

thus allowing you to represent them with coordinates and represent linear transformations with matrices

[proven lotus]

Okay, thank you

[pearl elm]

@wintry steppe sorry I didn't get back to you sooner. Sorry I am still struggling with formality on that same question

[pearl elm]

So I think the problem is trying to formalize how to show I am doing multiplication from one matrix pair then to another

Honestly... this is how far I got. The chapter kind of poorly shown an example of how to expand the terms out and do clever algebra IMO. Not sure if you can give me a little more than a hint for this problem at this point.

give me a second to show a screen shot of what I got

so like I think I can visualize the dimensions but I am really struggling here how to show the algebra

[wintry steppe]

show that the left hand side equals the right hand side

x.(Ay) = (A^Tx).y

[pearl elm]

how am I showing how the terms expand out I guess is what I'm saying

cuz this non-commutative shit fucks with my intuition

[wintry steppe]

so write x = (x_1, ..., x_m), y = (y_1, ..., y_n), A = (a_{ij}), etc.

and just...

write out the left and right sides lmao

[pearl elm]

write them out as row and column vectors first?

Oh I messed up there a little

Hold on let me repost

[novel needle]

Could someone explain how they are able to write these dot products as partial derivatives? I know what projection means and all that but just dont see how it is a partial derivative.

[pearl elm]

Ok there sorry for reposting annoyingly fixed it now I think?

But there is a way to shorten this?

Or is this fine and I’m overthinking it

A^(T) is just nxm matrix

[visual hedge]

Can anyone explain why dim E = dim F means that the linear application is surjective?

[wintry steppe]

context? what's the linear mapping?

something from E to F satisfying some conditions?

[faint lintel]

yea you need some conditions for that to happen

[pearl elm]

Ok nvm I figured that problem out

Last one for the section. I’m a little confused with wording here

[wintry steppe]

why are you confused

if you don't want to do block multiplication shenanigans, you can just write out the entries of the left and right hand sides lol

[pearl elm]

I thought it would be that redundant, I just wasn't sure

[bright current]

Let A 1' ... ,Ar be generators of a subspace V of Rn. Let W be the set of all
elements of Rn which are perpendicular to A 1, ... ,Ar. Show that the vectors of
Ware perpendicular to every element of V. what exactly do i have to prove here, i mean its already stated in the que that W contains those vectors which are perpendicular to vectors of V

[dusky epoch]

bad copypaste again?

anyway, no, you are told that $W = { w \in \bR^n \mid \forall i \in 1:n, w \perp A_i}$.

[stoic python]

Ann

[dusky epoch]

you are asked to show that $(\forall w \in W)(\forall v \in V)(w \perp v)$.

[stoic python]

Ann

[dusky epoch]

after all, V does not consist of ONLY the A_i and nothing else.

[vocal isle]

Hey folks, does anyone know how to solve a system of equations which are complex using the pseudo matrix?

I have a vector Y (which is a complex n x 1) and a matrix M which is a n x 3 and I want to find the vector P which is a 3 x 1

Y = [M] P

I thought i could simply do a pseudo inverse
pinv(M)*Y = P

but i get some weird results, any ideas?

[wet stratus]

well do you know that such a solution P exists? otherwise the pseudoinverse calculates a vector P which minimizes the error ||MP-Y||_2

[tribal willow]

what makes rank-nullity thm notably significant?

[hard drum]

It basically tells you how linear maps interact with dimensions, which is incredibly important

For example it can often force the kernel or image to be a specific subspace

[lavish jay]

I have to do a assigment on monte carlo method with linear algebra

but I'm completely lost

[tribal willow]

i see, thank you

[lavish jay]

I read my book, but still didn't understand very well

Do you guys have some recommendation ?

[tribal willow]

where does the sum ba - sum ca = 0 part come from

second to last eq

[hard drum]

the sum of ca term is alpha from earlier

basically we've got two expressions for alpha and then are saying the difference between them is 0

[tribal willow]

for some reason i can’t find this on there lol

[tribal willow]

aw shit lmao thanks

[hard drum]

np lol

[tribal willow]

i was looking at the big equations

[hard drum]

ye dw

[tribal willow]

for a linear transform $T : V \to W$ and $\dim V \ne \dim W$, then $T$ cannot be isomorphic, correct?

[stoic python]

anamono 🍔

[tribal willow]

because T is not bijective

[dusky epoch]

T cannot be an isomorphism*

but yes

[tribal willow]

right yeah

ty

[wintry steppe]

hi anyone can help me figure out why this function is not in bilinear form?

but this is bilinear?

[wintry steppe]

how?

basic properties of the determinant

it's even baked in to some definitions of the determinant that it's multilinear

what is the full question (i.e. not just part (b))?

to check if its bilinear and if so if its symmetric and positive definite

okay, so you need to check symmetry and positive definiteness

as in, check whether they hold or fail

[wintry steppe]

yes symmetric holds but positive definite fails

are you sure?

hmm maybe symmetric does not bcs u change the rows and columns so the determinant may not be the same.. sorry its just my first time studying bilinear forms

symmetry fails, right

the determinant is antisymmetric in the columns

oke makes sense and I guess its not positive definite bcs we cant know if the determinant is >0 ?

[winter harbor]

positive definite means that b(x,x) > 0 for x ≠ 0.

The determinant is not positive definite because it is alternating.

I.e

if a matrix has two columns that are equal

then the determinant is 0

In this particular case

we could take x = (1,0)

and we readily see that det((1,0),(1,0)) = 0

Even if (1,0) ≠ (0,0)

So the determinant is not positive definite.

[wintry steppe]

Thanks @winter harbor

[tribal willow]

is this what we commonly call Change of Basis?

[winter harbor]

Yup

[tribal willow]

dope ty

lin func is specifically a lin trans that goes from v space to field?

[wintry steppe]

from a vector space to its field of scalars, yes

[tribal willow]

ah kk

[rigid fractal]

hello

[alpine gale]

been tryna figure this out

someone mind helping me

[chilly hazel]

what is the correct mathematical notation for squaring the elements in a vector?

[dawn ocean]

<v,v>
vv^T or v^T v

or v.v

[chilly hazel]

that would give you a scalar tho

or a nxn matrix

[dawn ocean]

oh whoops

[chilly hazel]

I figured it out tho

[dawn ocean]

XA+B = C

XA = C -B

XAA^-1= (C-B)A^-1
X= (C-B)A^-1

[bright current]

there is a vector space V a subspace of vector space W then how can i show that union of W and Orthogonal W will make V

pls help

[chilly hazel]

here we go!

[wintry steppe]

sum. you need to write every vector in V as a sum of something in W and something in W^perp

[tacit pelican]

can someone explain

how

they

deduced this

[wintry steppe]

make the substitution i -> n - i in the sum

[tacit pelican]

alright lemme try that

[tacit pelican]

i see how that works, but i'm a bit confused on how sum indices work. as in when i make that substitution, do i only change stuff within the sum or do i change every instance of "i" i see? for example do i change
$$\sum_{i=0}^ng_if_{n-i} \to \sum_{n-i=0}^nf_ig_{n-i}$$ or $$\sum_{i=0}^ng_if_{n-i} \to \sum_{i=0}^nf_ig_{n-i}$$?

[stoic python]

sean

[tacit pelican]

like ofc in this question it makes sense that its the latter

but in general im always confused on how that works

like when i make a substitution what goes on inside my mind is "change everything"

but im not sure if thats what i do about the indices

[wintry steppe]

hmm

at least in this case, it's just doing the sum backwards

you don't have to think of it in terms of changing sum indices

[tacit pelican]

hmm

i see

alright well thank you

[bright current]

can someone give the solution to this que

[wintry steppe]

no

[wintry steppe]

No

[wintry steppe]

Do I take the inverse of t??

Then multiply it by b

Or do I take the inverse of b?

[wintry steppe]

Find the pivots first

That should give you the rank

Getting it to rref should give you a basis I think.

[stuck tendon]

Matrix wrt E is already given.
For matrix wrt B, find image of basis elements under T, and expand the result in the basis B and the write coefficients as columns.
Or find a transition matrix and calculate C^-1MC

[wintry steppe]

hi I have this (x − 3)(x − 1)^2 minimal polynomial, and want to find out the possible canonical forms (I found one). I would like to advice if:
[1 0
0 1] is a jordan block?

[zinc timber]

that is not a jordan block

[wintry steppe]

so the only jordan blocks are:
[ 1 1
0 1]
and [3] ?

[zinc timber]

your answer is wrong in many ways, first of all that's not a jordan block because you should have 1's in the first off diagonal (don't know the exact term)

second, you can't just find the jordan form without even knowing the dimesnion of the vector space

like you need to know the char poly

to know the geometric multiplicity

[wintry steppe]

sorry its 3x3 matrix

[zinc timber]

ok

then yes

[zinc timber]

this is the one but I hope you didn't just guess it out of blue just because I said the prev one is wrong

I hope you understood

[wintry steppe]

i found it weird thats only 2 blocks bcs the question asks find how many jordan canonical forms

[zinc timber]

then you gotta find number of jcf

btw they have said (x-3)(x-1)^2 is the mini poly right? or is it the char poly??

[wintry steppe]

mini poly

[zinc timber]

ok then you already have your answer

[zinc timber]

2 block?

2 blocks*?

oh

yeah

[wintry steppe]

wait [1] alone can be a block too right?

[zinc timber]

in your case, no

in general, yes

[wintry steppe]

but even if its bcs its 3x3 matrix u only got 1 jcf

out of curiosity why is not?

[zinc timber]

correct

[wintry steppe]

thanks @zinc timber

[zinc timber]

jcf should be a block diagonal, your
[1 1
0 1]
do not form a block diagonal if you consider [1] to be a block

[wintry steppe]

thanks again

[zinc timber]

[cunning violet]

why does a 2-dim vector (1 1) * (1 1)^t equal (1 1 , 1 1 ) matrix and not just 2?

shouldnt the correct answer be the lower one?

[wintry steppe]

the second one is correct. a 1 by 2 multiplied by a 2 by 1 is a 1 by 1, so just a scalar

[cunning violet]

but this cant be because the center matrix is defined by "..... (1...1) * (1...1)^t ", and with google the solution is the matrix

[wintry steppe]

idk you're probably doing it wrong then

do you mean $$\begin{pmatrix} 1 \ 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \end{pmatrix} = \begin {pmatrix} 1 & 1 \ 1 & 1 \end{pmatrix}?$$

[stoic python]

TTerra

[wintry steppe]

because that gives the matrix you're pointing at

[cunning violet]

yes

now im confused

[cunning violet]

above you said the answer is 2

[wintry steppe]

no, i didn't

[cunning violet]

here

[wintry steppe]

you are mixing up $$\begin{pmatrix} 1 \ 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \end{pmatrix}$$ and $$\begin{pmatrix} 1 & 1 \end{pmatrix}\begin{pmatrix} 1 \ 1 \end{pmatrix}$$

[stoic python]

TTerra

[cunning violet]

ofc theres a difference

fuck math lmao

[wintry steppe]

I would suggest online calculators with detailed steps for beginning bcs they explain steps really well!

Is that an acceptable proof????

Noticed I dropped the square

[tribal willow]

trying to make sense of why it’s called “annihilator”

we “annihilate” vectors in the sense that we take vectors from S subset V then send them to 0 in F?

[wintry steppe]

every element of S^0 annihilates S

[tribal willow]

yeah im just trying to make sense of why it’s called “annihilate”

[dusky epoch]

annihilate = send to zero

[tribal willow]

kk ty

[zinc timber]

looks ok

[dusky epoch]

also the annihilator of S is sometimes denoted Ann(S)

[wintry steppe]

you!

[dusky epoch]

me!

[zinc timber]

[tribal willow]

you!

[wintry steppe]

Thank you

[winter harbor]

"Annihilate" is a better expression than "Kill" at least

[tribal willow]

eradicate

[tribal willow]

this is a mapping L_alpha : V* to V?

or is it V to V*

[wet stratus]

V* to the field F. so basically an element of V**

[tribal willow]

oh right linear functional

[lilac tendon]

anyone know what this could be?

[noble swan]

I'm supposed to use the two basis vectors I defined as e to rewrite the four vectors listed in 2b

Is there an easier way for me to do this other than the row reduction method?

[supple moss]

i just dont understand

[fair wren]

B^n = the null matrix? right

[wintry steppe]

for what n

you should specify

[fair wren]

naturals

[wintry steppe]

which ones

[fair wren]

N

0

'>' 0