linear-algebra

[heavy crown]

yes I understand the first and second but not the third

[dusky epoch]

if $E$, a $2 \times 2$ matrix, is neither invertible nor 0 then it can be written in the form $E = uv^T$ where $u, v$ are column vectors of size 2

[stoic python]

Ann

[heavy crown]

aha I see. I think it's because in that case |E| = 0, so E has one row or column with 0 so the dim would be less than 4

[dusky epoch]

"E has one row or column with 0"

i mean, it is possible for all entries of E to be nonzero and det(E)=0 nonetheless

[heavy crown]

yes like 0 0 as row ?

[dusky epoch]

the matrix $\bmqty{1&2\2&4}$ is singular

[stoic python]

Ann

[heavy crown]

oh, right

yea

so why it says the dim is 2?

[dusky epoch]

"it"?

i was the one who said it

twice

[heavy crown]

haha, true

[dusky epoch]

can i use this fact in my explanation?

[heavy crown]

yes

[dusky epoch]

okay let $E = uv^T$ and consider the defn of $U_E$

[stoic python]

Ann

[dusky epoch]

we will have $U_E = { (Xu)v^T \mid X \in M_2(\bR)}$

[stoic python]

Ann

[dusky epoch]

i claim this is equal to ${ wv^T \mid w \in \bR^2}$

[stoic python]

Ann

[heavy crown]

yes that's true

[dusky epoch]

yeah, and that in turn has dimension 2

it's spanned by $\bmqty{v_1 & v_2 \ 0 & 0}$ and $\bmqty{0 & 0 \ v_1 & v_2}$

[stoic python]

Ann

[heavy crown]

but those were vectors of size 2?

[dusky epoch]

?

oh sorry, i should've clarified, $v = \bmqty{v_1\v_2}$

[stoic python]

Ann

[dusky epoch]

U_E is a subspace of M_2(R), its elements are 2x2 matrices

[heavy crown]

oh okay I see

yes

[dusky epoch]

okay so bringing this all back

we now know that if E is invertible then U_E = M_2(R) and if E is singular then U_E is a proper subspace of M_2(R)

does this make sense?

[heavy crown]

yes

[dusky epoch]

okay

[heavy crown]

because of the span

[dusky epoch]

so you had your problem say that {AE, BE, CE, DE} is linearly independent

and AE, BE, CE, and DE are all members of U_E by construction - as i hope is obvious

fuck. messages not sending

there we go

so we have that U_E contains a linearly independent set of four points, meaning that dim(U_E) = 4 [as this is all happening in a vector space of dimension 4 !]

[heavy crown]

right, so because it is linerally independent it means dim = 4

then U_E = M_2(R)

[dusky epoch]

yes

[heavy crown]

and then its invertible

hmmm

[dusky epoch]

yes that's it

[heavy crown]

haha

[heavy crown]

thank you very much ❤️

[vital arch]

Cam anyone help me with a question

(x/y)+(y/z)+(z/x) = 1

Does this equation have any integer solution?

[dusky epoch]

wrong channel

[limber sierra]

the space of real 2x2 matrices under matrix addition is isomorphic to R^4, yes

vector spaces do not encode vector-vector multiplication

so the fact that you can multiply matrices is irrelevant

erm

[nocturne jewel]

eigenvectors are from linear transformations, not vector spaces

Eigenvectors are just the vectors in the domain space which just get scaled when the transformation is applied

[limber sierra]

the correspondence between linear transformations and matrices does not work well in a vector space of matrices.

Av = Iv is nonsense

[nocturne jewel]

$A:\mathbb{R}^{2\times 2} \to V$ for whatever V you want

[stoic python]

Mosh

[nocturne jewel]

if you mean A is the transformation

[limber sierra]

unless A is supposed to be a function here

rather than a matrix

[nocturne jewel]

You can have the matrix representation of A

you wouldnt happen to have a specific question would you?

$I^2=I$, yes

[stoic python]

Mosh

[nocturne jewel]

If A represents a linear transformation from a 2-dim space to another 2-dim space

then yes, that's the definition of eigenvalue/vector

what?

[quartz compass]

as an example, A=[1,0;0,2] has eigenvector [1;0] with eigevalue 1, and A is not the identity

[nocturne jewel]

if A is identity, then you have the identity map, in which any vector is an eigenvector with value 1

[limber sierra]

Av=lv for v being a matrix just means A is identity
what if v isnt invertible?

for example, Av = Iv is certainly true for all (linear) A if v is the 0 matrix

okay, so pick a slightly less degenerate example then

let v = (1, 0; 0, 0)

note that there are multiple linear A such that Av = v

for example, (w, x; y, z) mapsto (w, 0; 0, 0) works, but so does (w, x; y, z) mapsto (w, w; 0, 0) or (w, x; y,z) mapsto (w, x; x+y, 2x+z)

the problem isnt the "single element" part

its that v is noninvertible

perhaps you should be looking at GL(R, 2)

[quartz compass]

you could write it as a block matrix like [A,0;0,A] [u;v] = [Au; Av]

a basis for the eigenvectors are then just the eigenvectors of A concatenated with 0 on top and bottom

so the matrix v eigenvectors for a specific eigenspace are multiples of a single eigenvector of A in both columns.

v = [3e_1, -7e_1] for instance, given e_1 is an eigenvector of A

as matrices these clearly are never invertible since they're linearly dependent

try vector space of functions

like continuous functions on [0,1]

or finite fields maybe, just any random field you can get your hands on

[nocturne jewel]

Or just polynomial spaces

[quartz compass]

there's a game you can play where you turn on and off lights in a pattern by pressing a square

[nocturne jewel]

I hate that game

I can never win it

[quartz compass]

you can solve it by thinking of it as a vector space over F_2

(finite field with 2 elements)

I think it's easy to get the misconception that group theory is just mindless, abstract jerking off

[nocturne jewel]

I havent done Group so I cant answer this

[quartz compass]

but there's a purpose to the stuff it does and it's for stuff

like just abstracting for the sake of abstraction isn't really group theory or abstract algebra

I'm not the end all be all either lol, I don't know what you're actually doing past just trying to find more examples of vector spaces to play with to get a better understanding

so I don't know what automorphisms and maps you're talking about

[viral flint]

You need that matrix multiplication is linear of course.
And once you have shown it's an isomorphism of M_2(R), you know there is an inverse on M_2(R) but you don't a priori know that the inverse is also given by right-multiplication by a matrix. For that part you need to use the property of matrices that if a square matrix has no null space, it is invertible. Depending on how you prove that property, it could involve determinants at least.

Oh I thought you were asking about the question about Z = {AE, BE, CE, DE}. Ignore what I just typed then.

[night gazelle]

What are, in your opinion, the necessary prerequisites to start studying linear algebra by oneself?

[gray dust]

@night gazelle experience in proof writing/reading & matrix computation

[stable kindle]

patience

[next vapor]

if i have a set of commuting operators on a complex vector space V(dim n) and one element T with n distinct eigenvalues and a basis consisting of the eigenvectors of T, how do i show that every other element of the set can represented with a diagonal matrix

[gray dust]

@next vapor it's not true as is

take operators $\m{0&1\0&0}$ and $I$ on $\bC^2$. they commute but the former isn't diagonalizable

[stoic python]

RokabeJintaro

[gray dust]

but it DOES hold if we also require the operators be hermitian, in fact in that case they're simultaneously diagonalizable iff they pairwise commute

[next vapor]

but they both have just 1 eigenvalue dont they?

in the counterexample i mean

or am i missing something

[dusky epoch]

yeah i think youre missing that @gray dust

[gray dust]

my bad i missed 'n distinct eigenvalues'

@next vapor then it's actually true, try proving it for a pair of commuting operators and extend it to any set of pairwise-commuting operators

[next vapor]

i think i need to show that 2 commuting operators share eigenvectors right

so if A and B commute, and v is an eigenvector of A with eigenvalue λ, then A(Bv)=B(Av)=B(λv)=λBv

so Bv is also an eigenvector of A

hm not sure where to go from here

[gray dust]

@next vapor since A has n distinct eigenvalues, each eigenspace of A is 1-dim

you showed Bv is in the λ-eigenspace so it's a scalar multiple of v, thus v is an eigenvector of B

[odd kite]

I believe this holds whenever there is a complete set of orthogonal eigenvectors

even if eigenvalues aren't district or are complex

[next vapor]

oh yes thank you

that should be it right, since i can do the same thing with any 2 operators in the set

[next vapor]

thank you, ill have a shot at proving it

[gray dust]

you can't do this, no inner product is given

[next vapor]

yea i was just gonna take the standard hermitian

[gray dust]

yes extend to any set of pairwise-commuting operators

[next vapor]

alright, thanks alot

[crude ridge]

Does A(Bx) = (AB)x ?

[wintry steppe]

yes

associativity of matrix multiplication

[crude ridge]

Oh forgot about that. Also, if nul(A)=nul(B) then can you say nul(A-C)=nul(B-C)?

[wintry steppe]

let A and B be your favorite distinct invertible operators and let C = A.

eg Ax = x and Bx = 2x as operators R -> R. both have null space {0}. if we take C = A, then A - C has null space R, but B - C has null space {0}

[crude ridge]

Oh that's interesting, thank you

[next vapor]

can i get a hint on this problem

i have a finite dimensional inner product space V with basis v1,...v_n and orthogonal basis w_1,...w_n obtained through the original basis

how do i show that |v_1|^2+...+|v_n|^2>=|w_1|^2+....+|w_n|^2

[next vapor]

any hint? <@&286206848099549185>

[spare crystal]

I want to prove that if $T : \mathbb{R}^2 \to \mathbb{R}^2$ is a linear transformation that preserves the magnitude of angles, i.e.,
$$
\frac{\langle x, y \rangle}{|x||y|} = \frac{\langle Tx, Ty \rangle}{|Tx||Ty|} \quad \text{for all $x, y \in \mathbb{R}^2$},
$$
then $|Tx| = \lambda|x|$ for all $x \in \mathbb{R}^2$ and a constant $\lambda$.

I haven't done linear algebra in a while, so I'm not super sure where to start. Any hints are appreciated 🙂

[stoic python]

Frank

[spare crystal]

Intuitively, T is a rotation + reflection + scaling, so it's clear, but I'm not sure how to prove that

[thorn robin]

@spare crystal Still stuck on this?

[spare crystal]

yep

[thorn robin]

Ok

I can guide you through what I did

It's not the cleanest proof

I'm just gonna work with |<x,y>|^2 |Tx|^2 |Ty|^2 = |<Tx,Ty>|^2 instead for |x| = |y| = 1

Since this identity holds for unit vectors iff your original identity holds for all vectors

And makes it so we don't have to worry about |Tx| = 0

Anyway step 1: if you evaluate T at x = (1,0) and y = (0,1) you should find that ab + cd = 0

This will help simplify the next calculation (and it also says that the columns of T are orthogonal)

Step 2: more generally, evaluate T at x = (cos t, sin t) and y = (1, 0)

You should find that the identity reduces to (a^2 + c^2)*((a^2+c^2)cos^2(t) + (b^2+d^2)sin^2(t))*cos^2(t) = (a^2 + c^2)^2 * cos^2(t)

To get to this you need to use the fact that ab + cd = 0 to deal with some cross terms

Anyway just choose a t so that cos(t) \neq 0 and divide by cos(t)

And also divide by a^2 + c^2, which you can do as long as T \neq 0 (since you can check that a = c = 0 implies b = d = 0 by evaluating at x = (1/sqrt(2), 1/sqrt(2)) and y = (0, 1))

This gives you (a^2+c^2) * cos^2(t) + (b^2+d^2) * sin^2(t) = a^2 + c^2

[sweet vine]

why is this true? i dont see it

helo seion

[thorn robin]

Which you can manipulate to get b^2 + d^2 = a^2 + c^2 by choosing a t with sin(t) \neq 0

So to summarize you have ab + cd = 0 and b^2 + d^2 = a^2 + c^2

i.e. the columns of T are orthogonal and have the same length

So you can write T = lambda * S where S is an orthogonal matrix

This is what you wanted

[spare crystal]

oh hmm, i will try to work that out, thanks so much!!

[thorn robin]

No problem lol

Like I said it's not the cleanest proof

I would hate to do that for n > 2

But I think it all works

[spare crystal]

yeah lol

[dreamy iron]

Hi #linear-algebra

SCENARIO 1:
Given a vector space V, with basis a,b,c,d; I can prove that

a+b, b+c, c+d, d is also a basis for V.

I proved the above claim.

I have not proved the following::
SCENARIO 2:
Consider a basis,
a,b,...., z.

a+b, b, ..... , z is also a basis.

Is the second scenario like a special case of the first scenario.

[dusky epoch]

it's not

in scenario 2 you presumably have 26 basis vectors for a 26-dimensional space

which is not at all a special case of scenario 1, which deals only with 4-dimensional spaces

[dreamy iron]

lol, just say n, not 26

[dusky epoch]

if you want n then use proper notation

like $v_1, v_2, \dots, v_n$ and $v_1+v_2, v_2, \dots, v_n$

[stoic python]

Ann

[dusky epoch]

those are the lists you are considering

[dreamy iron]

thanks!

[dusky epoch]

of which the former you are requiring to be a basis

[dreamy iron]

Yup

[dusky epoch]

anyway no, these are not special cases of each other

[dreamy iron]

There’s a pattern here, seemingly:

$v_1, v_2 \text{ is a basis} \iff v_1 + v_2, v_2 \text{ is a basis} $

$$v_1, v_2 \text{ is a basis} \iff v_1 + v_2, v_2 \text{ is a basis} $$

[stoic python]

ninnymonger is a Physics main.

[dusky epoch]

$v_1, v_2$ is a basis iff $v_1+v_2, v_2$ is a basis

[stoic python]

Ann

[dusky epoch]

in this case however these bases differ by three vectors, not one.

[lavish jewel]

you can use the definition of linear independence and see what happens

[dreamy iron]

and....

$$v_1, v_2, v_3 \text{ is a basis} \iff v_1 + v_2, v_2+v_3, v_3 \text{ is a basis} $$

[stoic python]

ninnymonger is a Physics main.

[dusky epoch]

okay so

[dreamy iron]

Is there a name for this pattern?

[dusky epoch]

no

no there is no name for it

[dreamy iron]

and I also think that the following is true

$$v_1, v_2, v_3, , \ldots, ,v_n \text{ is a basis} \iff v_1 + v_2, v_2 +v_3, , \ldots, v_{n-1} + v_n ,v_n\text{ is a basis} $$

[stoic python]

ninnymonger is a Physics main.

[dusky epoch]

it is true

theres just no name for that particular change of basis

[marble lance]

Linearly dependent?

Yes

The reasoning should be clear based on the defn

[crude falcon]

hmm I have two formulas for obtaining the iterative solution in Jacobi Method

the first one is the wikpedia one, the second is the one I've been taught, they are the same except in the second one you add the L+U matrix

are they equivalent?

[lavish jewel]

no, the second one seems wrong

the method is based on letting A = L + D + U, and then subtracting (L + U)x from both sides

in your second image, this would mean that L and U have a different definition, i.e. they are the strictly upper and lower triangular parts of negative A, instead of A

[crude falcon]

oh true, they are using -A

[wintry steppe]

Assuming the system given by Ax = b is consistent, then it has a unique solution only if A has full column rank. Is that correct?

[lavish jewel]

sounds ok

[wintry steppe]

Thanks

If A has full row rank we can't infer the uniqueness right?

[lavish jewel]

only if you knew the dimensions of A

A would have to be square

[wintry steppe]

Yup, assuming the general (non-square) case

[dreamy iron]

Hi #linear-algebra .

I need to find ALL the vector spaces with only one basis.

Step 1: I can prove that if my basis is the empty set, then my vector space is the singleton set containing the zero vector {0}.

Step 2: If can prove that if my vector space is {0}, then my basis is the empty set of vectors.

does this prove that {0} is the only vector space with one basis?

Step 1: is by definitions of span of the empty set and linear independence of the empty set.

Also by set theory, I know that the empty set is in unique.

[viral flint]

This proves that {0} is the unique vector space with the empty set as a basis

You would still have to show that a vector space has exactly one basis iff that basis is the empty set.

Which BTW I think is false in one case: a one-dimensional vector space over the field with two elements

[dreamy iron]

Curious you mention that: MSE says something along those lines. https://math.stackexchange.com/questions/1258364/what-kind-of-vector-spaces-have-exactly-one-basis

[native rampart]

{0} and span{v} for any vector v under F_2

[dreamy iron]

Lol. I don’t know how to work with F_2. I guess I can make the multiplication table.....

Also this mentions that finite field. https://math.stackexchange.com/questions/468339/necessary-condition-of-a-vector-space-having-only-one-basis

[native rampart]

F_2 is nice in that span{v}={v,0}

[dreamy iron]

Can you be more explicit. I dont see that

[native rampart]

You have v

and v+v=2.v=0.v=0

[dreamy iron]

So that’s the addition table for F_2.

right??

[dire thunder]

F_2 is just field of remainders modulo 2

[native rampart]

Yes

[dreamy iron]

$$v+v = v (1+1) = v* 0 = 0$$

[stoic python]

ninnymonger is a Physics main.

[native rampart]

Well,You will have 0.v and 1.v in span{v}

Yes

[dire thunder]

F_2 = {0,1} and +, * defined as usual operations but taken modulo 2

[native rampart]

Your field has to be over F_2

Otherwise,you can just pick {-v_1} instead of {v_1} to span the space

[dreamy iron]

Okay, so my only scalars in F2 are Zero and One:

$$ span(v) = { 0v , 1 v }$$

Okay, i think I follow.

[stoic python]

ninnymonger is a Physics main.

[native rampart]

Yes

[dreamy iron]

I don’t understand this.

obviously -v is in V over F_2......?

[dreamy iron]

How would I go about doing this?

[native rampart]

In F_2,v=-v

[next vapor]

could still use help with this if possible

i tried it with a basis of just 2 vectors but things got really messy and led to a dead end

[viral flint]

1. If basis is empty, it's the only basis.
2. If basis is non-empty, there's another basis.
   As has been mentioned, 2. is wrong in general, but you can show it if you assume that it's not a basis of size 1 over F_2.

[viral flint]

Was it obtained by the Gram-Schmidt algorithm?
If so, prove that || |w_i|^2 <= |v_i|^2 for every i, (and in fact equality iff w_i = v_i)||.

Hint: ||w_i is defined as v_i - a for some expression. Prove that |v_i|^2 = |w|^2 + |a|^2.||

[next vapor]

idk if im messing up but i get that |w_i|^2=|v_i|^2+|a|^2-2<v_i,a>

which doesnt seem right

[next vapor]

this is just calculating <v_n-a,v_n-a> right?

[short coral]

Any easy proof for Sylvester rank inequality?

[broken sun]

Hello. Can you give me an example of an operator which has no adjoint?

[dusky epoch]

https://math.stackexchange.com/questions/650619/linear-operators-with-no-adjoint

[broken sun]

Thanks. I have already seen that thread. However, it is too hard for me to understand; also, I did not see any specific example there. There is no simple example of such an operator?

[dusky epoch]

well you can pick a particular sequence xi and construct the operator as the first answer writes

[next vapor]

sorry for the ping but i cant seem to get the equality |v_i|^2=|w|^2+|a|^2@viral flint
its just computing the inner product of v_i-a with itself right?

[viral flint]

Yes. Why can't you get the equation?

[broken sun]

I cannot understand such a construction. Can you elaborate?

I know basic math.

[next vapor]

so <v_i-a,v_i-a> = <v_i , v_i-a>+ <-a, v_i-a>= <v_i , v_i>+ <-a , v_i>+ <v_i , -a>+ <-a,-a>

which simplifies to |v_i|^2+|a|^2+2<v_i,-a>

so |w_i|^2=|v_i|^2+|a|^2+2<v_i,-a>

but how do you go from this to that equation

or am i going wrong somewhere?

right? @viral flint

[viral flint]

Right so far

So if you want to get from here to the desired equation, what do you need to show?

[next vapor]

i would need to have |a|^2-2<v_i,a> equal to -|a|^2 right

so |a|^2 would have to be equal to <v_i,a>

[viral flint]

Yes

You have a formula for a presumably

Use that to calculate |a|^2 = <a, a> and <v_i, a> and check that their difference is 0.

[next vapor]

so a is <v_i,w_1>/|w_1|^2w_1+...<v_1,w_i-1>/|w_i-1|^2* w_i-1

wait cant i just say that <v_i, a> -<a,a> =<v_i, a-a>=<v_i, 0> which is just 0?

wait no thats wrong

hm im not seeing any cancellations its just a jumbled mess

[next vapor]

im not seeing this ill come back to it later

i just wanna double check a solution

I have an inner product space and i want to find all vectors v and w such that proj_v(proj_w(v))=proj_w((proj_v(w)

but when i apply the formula for projections i get that v=w

but that seems too simple

[next vapor]

anyone mind confirming or denying this? 😅 <@&286206848099549185>

[viral flint]

Oh sorry forgot about this

[viral flint]

Wouldn't it be <v_i - a, a> i.e. <w_i, a>?

Anyway

[viral flint]

Yes now substitute it into <a, a> and <v_i, a> and try to calculate both

[next vapor]

yea i expanded the inner products using that formula but i dont see any cancelllations, am i missing something?

[viral flint]

Yes
Did you use that w_1, …, w_(i-1) are orthogonal?

[next vapor]

oh right i completely forgot about that

so when i expand <w_i,a> i get inner products in terms of the orthogonal basis

and it equals 0 so that show the equality

and just to confirm that equality happens iff v_i=w_i for all i

if they are equal then equality follows trivially because its already an orthogonal basis

the other direction we have |v_i|^2=|w_i|^2 --> so |a|^2 is 0 meaning all the inner products <v_i,w_i> are 0 meaning theyre orthogonal

so our original basis is already orthogonal

does this look right? @viral flint

[lavish zephyr]

hey guys, why do we need gram-schmidt to obtain orthonormal bases. Why can't we just take basic projections and then obtain them

[viral flint]

No it doesn't show <v_i, w_i> = 0. It shows that v_i = w_i.
Otherwise correct

[viral flint]

What is your construction exactly?
Suppose e_1, ...., e_n is a basis for the vector space.
How exactly would you obtain an orthogonal basis w_1, ..., w_n from it?

[next vapor]

wait why doesnt it show <v_i,w_i>=0?

a is <v_i,w_1>/|w_1|^2w_1+...<v_1,w_i-1>/|w_i-1|^2* w_i-1 and we want everything to be 0 right

and since w_i is a non zero vector that means each <v_i,w_i> has to be 0 for the whol thing to be 0 no?

[viral flint]

Yes (although you need that w_i are linearly independent, not just non-zero)
So <v_i, w_1> = <v_i, w_2> = ... = <v_i, w_(i-1)>
Doing this for all i,
<v_i, w_j> = 0 whenever i > j
which is extremely different from saying <v_i, w_i> = 0 for all i (notice the same i in both subscripts)

[next vapor]

but w_i are linearly independent arent they, since theyre an orthogonal basis

[viral flint]

Yes, I was just pointing out that you need to use that to conclude that the coefficients are 0.

More importantly, did you understand the difference between
<v_i, w_j> = 0 (all i,j such that i > j)
and
<v_i, w_i> = 0 (all i)
?

In any case, this is getting unnecessarily complicated.
If |w_i|^2 = |v_i|^2 = |w_i|^2 + |a_i|^2, then |a_i| = 0 so a_i = 0 so w_i = v_i - a_i = v_i.
Here a_i is the a you used for finding w_i from v_i, labelled with an extra i for clarity.

[next vapor]

yea i see now

thanks alot i appreciate it

[next vapor]

If A is an nxn matrix and i wanted to show that A and A+I have different characteristic polynomials

is there an easier way than calculating the determinant explicitly

[wintry steppe]

char poly of A = det(A - tI)
char poly of A + I = det(A + I - tI) = det(A - (t - 1)I)

[nocturne jewel]

Idk, seems complicated

[merry orchid]

someone helps me in this activity

someone helps me in this activity

[atomic badge]

Hello chat. I would like to confirm if it's possible to exist more than one t which would make phi a linear Transformation with this given vectors? I have found it to be t != 1. thanks in advance!

almost forgot this if this matters at all

[naive skiff]

Can someone help me with this? I saw something like this in a classical mechanics book and I'm bent

I'm looking to understand the steps

Is this true?

[native rampart]

How are you defining derivative wrt B?

[dull rune]

May I get assistance for this question. Thank you so much

[quartz compass]

what have you tried

[dull rune]

I do not know where to start. I've scoured the textbook but no clue where the professor got it from.

[quartz compass]

well, what's a transition matrix?

@dull rune

[dull rune]

a matrix that when multiplied with one matrix gives out the a desired output matrix... something like that

so if u multiply the transition matrix by B, itll become C

[quartz compass]

not quite

it's the matrix that lets you change basis

we have two bases, the set B and set C

[dull rune]

oh right

[quartz compass]

so if you have a vector represented with coefficients on the basis B, the transition matrix will take those and give you what the coefficients on C basis should be

fortunately this isn't too bad, let's just build it up one step at a time

[dull rune]

yay

ok so I have an idea how to do soemthing similar had it been simple integers

[quartz compass]

what polynomial does the column vector (1,0,0) represent in the B basis?

[dull rune]

hmm

erm, the first one id have to assume, though i am guessing at this point

[quartz compass]

yeah, write it out

[dull rune]

2+x+x^2

[quartz compass]

perfect

so now what's this polynomial look like in the C basis as a column vector?

[dull rune]

2, 1, 1?

[quartz compass]

yeah exactly

so we just transformed the first basis vector from B to the C basis

and we could write that in a matrix like:

$$\begin{bmatrix} 2 & - & - \ 1 & - & - \ 1 & - & - \end{bmatrix}$$

[stoic python]

Merosity

[dull rune]

ahh

yes i have that great

ok so if i follow that pattern, ill get something like (2,1,1),(1,2,1),(2,0,1)

[quartz compass]

maybe

why would that work

[dull rune]

what do u mean by why?

[quartz compass]

well it's your idea

[dull rune]

uh

[quartz compass]

why would this work

[dull rune]

well following the pattern from the first polynomial

the other two polynomials have similar degrees

[quartz compass]

how does it work

[dull rune]

u map each coefficient from B in a column of a matrix

following the 'formula' given by C

[quartz compass]

why do we do columns and not rows

[dull rune]

now that... uh...

i am not so sure why

is it just a rule? i remember glossing over the definition in my textbook

[quartz compass]

well I just figured it out, it isn't like random

we were trying to construct a matrix that transforms from one basis to the other

[dull rune]

Right

[quartz compass]

let's see, where does the column vector (0,1,0) in the B basis get transformed to in the C basis?

if you take any matrix and multiply it by this vector, what's the output?

[dull rune]

(0,1,0)?

[quartz compass]

$$\begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix}\begin{bmatrix} 0 \ 1 \ 0\end{bmatrix} = ?$$

[stoic python]

Merosity

[quartz compass]

what do I get from this matrix multiplication

[dull rune]

uhhhhh (b+e+h)

wait.

fuck

god i feel so stupid

(b, e, h)? surely im going crazy or i cannot do matrix multiplication

well the result is a 3x1, so that is why it must be a column

[quartz compass]

yeah good

so we got the middle column

we know the transition matrix needs to take the column vector (0, 1, 0) and output (1, 2, 1)

[dull rune]

oh

[quartz compass]

and you just showed that the output of that is the middle column so

that's why that's there

[dull rune]

i see

🧠

[quartz compass]

knowing where the basis vectors go tells you where all vectors go

that's one way to think about what linearity means

[dull rune]

oh wow

[quartz compass]

ok cool, so now you have P, maybe the rest is easier now

[dull rune]

for part b) do i do the same but going in the other direction from C to B

[quartz compass]

that would be hard wouldn't it

[dull rune]

..yeah

[quartz compass]

luckily you don't have to

[dull rune]

i see matrix inversion

ill go and review that

[quartz compass]

yeah that's what you'll do, really quickly think of it this way:

let's say you have the vector v in the B basis then you transform it to the C basis by doing Pv

then if you want to transform from the other way, there's some matrix Q that goes from C to B so you could do QPv

but now you're back in the B basis, so it should be like nothing happened

v = QPv

you could write Iv = QPv

this has to hold for all vectors v, so you can say I = QP and so Q is the matrix inverse of P

[dull rune]

okay i think i understand

[quartz compass]

cool

[dull rune]

lastly for part C, I feel like i get the gist of it but would u be able to offer a brief outline

[quartz compass]

try to explain it to me first and I'll correct you

[dull rune]

basically i have to find 'u' relative to C, and use whatever i got for part B and apply it?

[quartz compass]

yeah what is u relative to C

[dull rune]

(3, 1,1)?

[quartz compass]

yeah easy

[dull rune]

oh

nice

[quartz compass]

so fairly simple steps over all, getting P is easy, writing u in C is easy, since C to B is P^-1 is also easy to get, not much more to say

[dull rune]

so (3,1,1)(P^-1)

ill give it a try

[quartz compass]

well, you multiply P^-1 on the left

and (3,1,1)^T you should probably write it that way to be clearer

[dull rune]

oh ok

[quartz compass]

I am always making sure to say 'column vector' because I don't want to write the transpose

[dull rune]

thank you so much

[quartz compass]

you're welcome 👍

[dull rune]

i got it done and turned it in

u are brilliant i cant thank u enough
'

[quartz compass]

thanks, glad to help, you're welcome

[next vapor]

I have an inner product space and i want to find all vectors v and w such that proj_v(proj_w(v))=proj_w((proj_v(w))

I believe this is only satisfied when v=w and the algebra checks out but im unsure, can anyone confirm or deny?

[quartz compass]

nope

they could be perpendicular

[next vapor]

ah yes

thats the bit i missed in my working

that means <v,w> is 0 and i cant cancel it, thank you

[quartz compass]

yup exactly

you're welcome

[next vapor]

but just to confirm those are the only 2 scenarios right

[quartz compass]

yeah that's all

[next vapor]

show that the square of an average of n numbers is less than or equal to the average of the squares

i know that <v,w> <= |v||w|

so if i take the dot product as my inner product and v=(x_n) and w=(1) this should work

nvm i could just square both sides

[dreamy iron]

Hi #linear-algebra.

Im once again trying to find all the vector spaces with only one basis, but considering only the fields R and C.

Specifically, I’m trying to prove point 1.

That’s what I got...

it doesn’t feel right

[dusky epoch]

any nonzero real vector space necessarily has more than one basis

just take any basis and take any vector in it and scale it by 2, now you have a different basis.

your proof feels empty

[dreamy iron]

I add remarks

[dusky epoch]

this feels bureaucratic

[dreamy iron]

I don’t understand.

Im not doing the finite field portion cuz I don’t have to (and frankly I don’t understand finite fields well enough just yet.)

[dusky epoch]

this proof feels bureaucratically bloated

claim: in a nonzero real vector space, there exists more than one basis.

proof: take any basis of your nonzero vector space. take any vector in the basis. replace this vector with twice itself. now you have a different basis.

that's it

[dreamy iron]

Okay

And how about ** Find all vector spaces with only one basis.**

[vapid vale]

This is true if the field is not of characteristic 2.

[dreamy iron]

Yes.

My proof is only for R or C

[dusky epoch]

i said real vector space.

unless you meant to generalize my thing

[dreamy iron]

If a basis is empty, then it’s the only basis.?

[dusky epoch]

in the zero vector space, the only basis is the empty basis, because the only other set of vectors you could have is {0}, which is linearly dependent.

[dreamy iron]

Is that the proof?

[wintry steppe]

you can also think of the empty sum as being equal to 0

[dreamy iron]

That’s much simpler than what I have.

[dusky epoch]

yes

sometimes things really are simple, ninnymonger

not every proof deserves a seven-page paper

[dreamy iron]

Axler was kind was kind enough to define it outright

[dreamy iron]

Have patience with me

[marble lance]

Lol, I thought Ann just came up with ninnymonger on the spot, but it is your name

[dusky epoch]

ninnymonger (derogatory)

[marble lance]

😂

[lavish jewel]

[dreamy iron]

Lmao

[next vapor]

is the fact that a vector space is a direct sum of a subspace and its orthogonal compliment also true in the infinite dimensional case

ive been told it is but all proofs im finding are for the finite dimension case

[dire thunder]

well suppose it is not direct sum

first of all suppose it is just sum

[stoic python]

Commander Vimes

[dire thunder]

wait i think it is not true @next vapor

ye

here is counterexample

[stoic python]

Commander Vimes

[dire thunder]

then fix point c in [a,b] and take U to be set of all functions in that vector space s.t f(t)=0

@next vapor see that U and its orthogonal complement do not span space

[next vapor]

i see

hmm then how would one go about finding an orthogonal compliment for the subspace of odd functions on V=C[-1,1] given the inner product integral from -1 to 1 of f(x)g(x)

i know that even functions are in the compliment, but how do i show there arent any others

[dire thunder]

prolly by ibp spamming

hmm

i do not remember but it sounds like antiderivative of odd is even and of even is odd

[quartz compass]

I think you can do it by showing every function can be written as the sum of an even and odd function, $$f(x)=\frac{f(x)+f(-x)}{2} +\frac{f(x)-f(-x)}{2}$$

[stoic python]

Merosity

[next vapor]

yea i tried that, doesnt it need to be a direct sum tho?

[quartz compass]

why do you think it isn't a direct sum

[next vapor]

sorry i mean it is a direct sum yea

but the dimension is infinite right

and i can only conclude using the direct sum argument if the dimension was finite no?

as in the vector space is a direct sum of a subspace and its orthogonal compliment is only true in the finite dimensional case

or am i missing something

[quartz compass]

what's 'the direct sum argument'?

I'm just writing a continuous function in a way that makes it the sum of an even and odd function

[next vapor]

how would you go from that sum to the orthogonal complement being exactly the set of even functions tho?

[lavish jewel]

the even functions are the orthogonal complement of the odd ones

[next vapor]

yea thats what im trying to show

[lavish jewel]

the product of an odd and an even function is odd

then use the properties of odd functions?

[next vapor]

hm ig i could just say that if we inner with an odd function we get an even function so the integral is not 0 unless one of the functions is 0

[lavish jewel]

what

inner product of what with an odd function

[next vapor]

sorry i mean taking an element of the odd function subspace

and inner it with another odd function

[lavish jewel]

you wanna show that evens are orthogonal to odds, yeah?

[next vapor]

nah i know they are, now i just have to show that the orthogonal compliment is exactly the set of even functions

[lavish jewel]

then you already have everything you need

just check what merosity wrote above

[dusky epoch]

why tf is everything so thin

[arctic stone]

Hey uhhh isn't this example of finding out B' incorrect?

I was taught that the change of basis matrix from B' TO the standard basis

is actually easy

but here it's the other way around?

[dire thunder]

prolly due to transparent background

[arctic stone]

God these stupid change of basis matrices just confuse me more and more, I've been trying to fully understand them for months now and whenever I think that I get them something contradictory comes up

[dusky epoch]

youre not alone

i personally keep getting confused which way they go even after years of this shit

[arctic stone]

I also hate how there are so many ways of writing them in different parts of the world

[golden ermine]

I also get confused even after drawing the commutative diagram

[arctic stone]

you mean a diagram like this one? or something else?

[golden ermine]

Oh something like this

[arctic stone]

ohhh

[broken sun]

Hello. Can you give me a simple example of an operator which has no adjoint?

[dusky epoch]

take V = the space of all polynomials with real coefficients with inner product given by <f, g> = sum f_k g_k [where f_k and g_k are the coefficients on x^k in f and g respectively]

define a map T: V -> V by Tp = p(1)*x^5

then T will not have an adjoint

[wintry steppe]

When we talk about two vector spaces being isomorphic, in what category are we talking about? The category of vector spaces?

Furthermore, what linear algebra books talk about things like the question I just asked? Mine doesn't ever use the word 'category'

[native rampart]

Category of Vect,ig

Morphisms are linear transforms

[wintry steppe]

Do you know what book talks about it, Buncho Dragons?

[native rampart]

not very familiar with the category theory approach

But I guess it's similar to how groups or sets work

[wintry steppe]

Are you an undergraduate?

[native rampart]

Kinda but I don't study math

[wintry steppe]

I see, CS or something then

[native rampart]

Aluffi is good,ig

[dusky epoch]

if we are talking about two vector spaces being isomorphic then we are talking about them being isomorphic as vector spaces unless explicitly stated otherwise

[native rampart]

Also,Category Theory isn't necessary to do intro algebra

[dusky epoch]

what are you talking about it obviously is

didnt you take a year of cat theory before you saw 2+2? /j

[faint lintel]

What's worse is that in my school different abstract Lin alg classes use different notations for which way the direction goes

[wintry steppe]

this is why people like being coordinate-free

[crude ridge]

What would be a quick way to check if two sets of vectors are basis for the same space?

[wintry steppe]

if the space is finite-dimensional and you have your vectors in coordinates you can put them together into a square matrix and compute its determinant (i.e., check linear independence)

[crude ridge]

I don't understand how you would do that

Say I have two sets, each having 2 vectors that are in R5

[wintry steppe]

oh

you're asking for subspaces

nvm

hmm

[crude ridge]

Yeah not the entire R5 space

But yeah I'm struggling to think of a way to see if the first set of two vectors span the same space as the other set of two vectors

[wintry steppe]

it'd be enough to check that each of your two sets is linearly independent, and to write the vectors in one set as a linear combination of the vectors in the other one, i believe

[crude ridge]

Oh that's interesting

Visually, that makes sense if I'm thinking of subspaces in R3

[daring ravine]

I'm reading this paper on optimising the "Normalised Cut" for a similarity graph

Just confused by what they've written as the "Eigenvalue problem":

[stoic python]

orm

[daring ravine]

I'm confused by the RHS

Is it saying that $\lambda \mathbf{D}$ are the eigenvalues of $(\mathbf{D - W})$?

[stoic python]

orm

[wintry steppe]

no

what's lambda D?

if D is a linear transformation, no. linear transformations cannot be eigenvalues

[lavish jewel]

looks more like a generalized eigenvalue problem

[daring ravine]

So D-W is a "Laplacian", i thought it would be like:
$L y = \lambda y$

[stoic python]

orm

[lavish jewel]

right, so L is another linear transformation

seems you have $Ly = \lambda D y$

[stoic python]

Edd

[daring ravine]

yeah exactly

which is not an eigenvalue problem i've really seen before

[lavish jewel]

this is a generalized eigenvalue problem

from wikipedia

[daring ravine]

ok awesome thank you!

I was spooked by it and thought i'd misunderstood previous parts of the paper

Thank you for cleaning this up for me ^^

[wintry steppe]

[lapis fern]

HI ALL. is there a name for this change of basis?

I've seen a discussion of something similar a few days ago.

[wintry steppe]

I’ll do anyone’s homework for money dm me

[wintry steppe]

done with a, my characterstic eqation is 0 = x'' + (a+d)x' + (bc-da)x

my question is how do I get x(t) and y(t)

for part B

[dusky epoch]

well you can get x(t) as the solutions of your 2nd order DE

what substitution did you make to arrive at it btw?

[wintry steppe]

i differentiated the first equation and substituted the second one

and then i solved for y and substituted again

[dusky epoch]

so you solved for y in terms of x and x'?

[wintry steppe]

yea

and then plugged it into the third equation, the differentiated one

[dusky epoch]

do you have that written out somewhere

cause thats what youll be using to recover y

[wintry steppe]

yeah sure i could send a picture

@dusky epoch how would I go about solving the second order differential equation

[dusky epoch]

isnt it literally just a linear differential equation with constant coefficients...

[wintry steppe]

it's second order

[dusky epoch]

and?

it'll have two solutions $x = e^{\lambda_1 t}$ and $x = e^{\lambda_2 t}$

[stoic python]

Ann

[wintry steppe]

right

[dusky epoch]

where $\lambda_1, \lambda_2$ are the roots of your characteristic equation, and also the eigenvalues of $\bmqty{a&b\c&d}$

[stoic python]

Ann

[wintry steppe]

yes so I differentiate x twice and plug in to the characterstic equation

and solve for lambda

?

[dusky epoch]

your lambdas are facing the wrong way

but yes

[wintry steppe]

oops

[grizzled sphinx]

https://media.discordapp.net/attachments/490557019623915520/844238677689630761/IMG_20210518_1549322.jpg?width=1440&height=357

[wintry steppe]

i couldve sworn my teacher drew them that way lol

[grizzled sphinx]

how do I do the last part goddammit, it hurty my brain

[lavish jewel]

looks like an exam

lol they left the server

[dusky epoch]

whomst?

[lavish jewel]

pickleman68 posted an exam, and as soon as i asked about it, they left the server

[dusky epoch]

ah

[frosty vapor]

i banned them i thought

yeah i did ban

[lavish jewel]

[crude falcon]

in the second paragraph, when they refer to lambda as dominant eigenvector, that is a typo and they refer to an eigenvalue lambda right?

same with the last paragraph

[lavish jewel]

yeah

[dusky epoch]

actually it's the lambda that should be changed instead

[wintry steppe]

Is there a proper infinite dimensional subspace F of a inner product space V where every vector in V has a orthogonal projection in F?

[civic anchor]

if i row reduce a matrix do i still get the same cofactors from co factor expansion

[native rampart]

Row reduction does not preserve determinant

So,I don't think so

[civic anchor]

oh okay ty

[wintry steppe]

<@&286206848099549185>

maybe i should rephrase

Is there a pair (V, F) where F is an infinite dimensional subspace of a vector space V such that for every v in V, the orthogonal projection of v in F exists?

if F was finite dim, this projection always exists. and its easy to construct a infinite dim F for which some v dont have

[soft abyss]

maybe a really basic question, but here goes: if you allow column operations from the right to left (i.e. adding a column to a more rightward column) and row operations from the top to bottom, then every matrix can be turned into a permutation matrix (well, a permutation matrix with some columns all zeroes)

what can we say if you only allow row operations from bottom to top, and only allow scaling of columns by some nonzero (real or complex ) number

?

[wintry steppe]

hi

[wintry steppe]

If A² = - I, Identity matrix

How to know that the eigenvalue is not a real number?

[red prawn]

A row operation corresponds to multiplication on the left by the appropriate matrix. You can find out which matrix by applying the row operation to the identity matrix.

As for scaling of columns:
A column operation corresponds to multiplication on the right by the appropriate matrix. Scaling a column would be multiplying on the right by the diagonal matrix with all 1's except for the scaling column. Again, you can find out which matrix corresponds to the column operation by performing the operation on the identity matrix.

[quartz compass]

consider an eigenvector v, $Av=\lambda v$ then $A^2v = \lambda^2 v$ so we have $-v = \lambda^2 v$ that means $-1 = \lambda^2$

[stoic python]

Merosity

[soft abyss]

Sure, I guess I’m just asking about if there is a nice representative for each class (note that we can’t do any row operations, only upward ones)

For example if we allow all row and column operations then every class is represented by a partial permutation matrix

[red prawn]

Maybe find matrices representing these allowable operations, and look at the set of those

[wintry steppe]

How can A* lambda be lambda squared?

[soft abyss]

Yeah, the matrices are the set of upper triangular matrices B, and the algebraic torus T

I was just trying to phrase is in linear algebra terms, but that’s essentially the problem I’m working on

[lavish jewel]

remember the definition of eigenvectors and eigenvalues

A^2 v = A (Av) = A lambda v = lambda A v = lambda^2 v

[soft abyss]

B^+ M B^-1 gives you the permutations a where the B’s are upper and lower triangular

I’m thinking about B M T, where T is like you described

[wintry steppe]

right, thanks

[soft abyss]

It might not have an obvious answer but just making sure I haven’t missed something obvious

[broken sun]

Why does T have not any adjoint?

[red prawn]

You could think of it as orbits of left/right action

[soft abyss]

Ha, that’s exactly what I’m working on 😛

[dusky epoch]

why don't you try finding out yourself?

[broken sun]

If I could, I would not have asked you.

[red prawn]

So "from bottom to top" meaning upper triangular?

[soft abyss]

Right

[broken sun]

Also, "why don't you try finding out yourself" can be an answer to any question asked here.

[dusky epoch]

it's worth knowing what the question-asker has tried, if anything.

anyway, try considering where T* would have to send monomials (1, x, x^2, ...)

[red prawn]

And you're looking for representatives of equivalence classes given by orbit under (b,t).m = bm(t^-1)? Does this sound like appropriate phrasing?

[dusky epoch]

if i am not mistaken, you will find that T*(x^5) would have to be a non-polynomial

[soft abyss]

Yeah I think so

Possibly b ^ + m t but I think it’s similar

[red prawn]

Yes, something to guarantee left action even though you're multiplying on the right

[soft abyss]

I’m not massively familiar with this stuff yet

So most likely yeah

[red prawn]

(also I think of it as pre-composition, so you're moving the coordinates of the system rather than the body, if you're thinking of the maps acting on a vector space)

[broken sun]

T* would have to send those monomials to where?

[wintry steppe]

How could - I v = λ² v imply -1 = λ²

Can we remove v on both side, even though it's vectors? And how could identity matrix I, be 1?

[soft abyss]

Right l think the T is doing the re scaling

[dusky epoch]

do you want me to do the algebra for you

[soft abyss]

And the B is just a normal action of multiplication

I’m not 100% clear either lol

[red prawn]

pre-composition by an inverse, it's like translations in hs algebra. f(x-a) shifts to the positive direction

[broken sun]

I do not think that algebra is as straightforward as basic algebra.

[quartz compass]

Iv = v is a vector so it simplifies to just -v = lambda^2 v

[soft abyss]

Yep ok, I should have remembered this!

[quartz compass]

you can't "cancel" the vectors that's a good point, but what we do know is that eigenvectors are nonzero and that the 0 vector is unique so we could add v to both sides to get 0 = (lambda^2 + 1) v. Since v is nonzero, we know lambda^2+1 = 0

[soft abyss]

I’ve played about with it, and I think it’s like a permutation matrix, but you can have repeated 1s in a row

As well as all zero columns

[broken sun]

People may not be as smart as you expect.

[soft abyss]

Does that sound believable?

[red prawn]

Well you're not allowing column additions, so you wouldn't be able to "get rid" of the repeated ones if you were doing column ops

[wintry steppe]

Thank you very much 🙂

[quartz compass]

you're welcome

[dusky epoch]

let $T^(x^5) = \sum_{k=0}^{\infty} a_k x^k$, where the right-hand side has to have finitely many terms (else it is not a polynomial). consider $\ang{x^m, T^(x^5)}$, then you have:

$$a_m = \ang{x^m, T^*(x^5)} = \ang{T(x^m), x^5} = \ang{x^5, x^5} = 1$$

for \textbf{all} $m$, which would imply $T^*(x^5) = 1 + x + x^2 + x^3 + \dots$

[soft abyss]

But we can get rid of repeated 1s in the B m B case

Which allows column operations

[stoic python]

Ann

[dusky epoch]

@broken sun here

[soft abyss]

I’m 95% sure anyway

But yes they have to stay in our case

[red prawn]

I think you got a clear picture, just a matter of explicitly writing the eq classes by this point

[soft abyss]

I think I do yeah thanks for the chat anyway!

[broken sun]

The basis is orthonormal?

[dusky epoch]

yes, with my choice of inner product the monomial basis is orthogonal

[broken sun]

Thanks. But, why did such a simple example not come into my mind? Because I am not smart enough to study math?

[dusky epoch]

i will not indulge your self-deprecating remarks

i do not know how much experience you have with linear algebra (and higher math in general), though now i think it's too little

so it's not that you're not smart enough, if anything it must be that you're not experienced enough

[broken sun]

I usually do not solve math problems; I usually read only main texts.

[wintry steppe]

sounds inefficient

doing the problems is a great way to get better

[quartz compass]

reading about math is to math like going to an art museum or watching a sports game are to painting and playing the sport

[dusky epoch]

ngl this is no way to learn math.

[broken sun]

What is "ngl"?

[dusky epoch]

math isnt a spectator sport; if all you do is passively read texts without sitting down and doing some exercises / solving problems / proving results then your retention will be almost nil

ngl stands for "not gonna lie"

[stable kindle]

yeah, you gotta go through the exercises

you really really gotta actually rederive the proved results and then apply them to the problems, it's the onyl way

[wintry steppe]

good morning guys

if anyone can explain this that would be nice :)

i dont know what trA means

I understand that detA means the determinant of A

[stable kindle]

trace

[wintry steppe]

what is that

😐

[stable kindle]

sum of the elements on the main diagonal

[wintry steppe]

mmm

ok

[wispy thicket]

hello can anyone explain how to find an orthogonal basis that contains a set?

given an inner product, I found that the set is orthogonal. Not sure if that's part of the solution

[vale plover]

well, they should be orthogonal to be part of the basis

what i would do is applying vectorial product to v1,v2

so you obtain a third vector v3, that will be orthogonal to v1 and v2 at the same time

that can be a solution

[wispy thicket]

I technically found said v3 in a previous subproblem

so those 3 vectors would form the basis, then?

[vale plover]

another solution would be proposing a generic vector v3=(v3x, v3y, v3z). Then it should be orthogonal to v1 and v2, so it has to satisfy <v1,v3>=<v2,v3>=0

[vale plover]

yup, an orthogonal basis of R3

well more formally what you have found with those solutions is a group of 3 orthogonal vectors in R3

if you want to go further, you should prove that those vectors generate R3

[wispy thicket]

I see, thanks. I wasn't sure if I just had to do that since some of my notes regarding this was just Gram-Schmidt

[vale plover]

i didn't put it beacuse is easy to see that they will generate R3

[vale plover]

well i can't remember well, but i think that G-S is a more general way of finding a base from a group of vectors

in this particular case you know that v1 and v2 are orthogonal

if they weren't you can use GS

[wispy thicket]

ah I see, thanks!

[wintry steppe]

by checking that it sends π to 0

It's already the feature of the set.

what?

I mean that it's already f(pi) = 0.

that's the condition for a function from R to R to be in the set

you check that to show that something is in it

[wintry steppe]

Jeez.

?

ok

It's the FUCKING CONDITION.

not the feature.

[lavish jewel]

it says V is a vector space of real functions. check if the functions for which f(pi) = 0 are a subspace of V.

[wintry steppe]

calm down lol, sorry if i offended you

[wintry steppe]

Are you German?

[lavish jewel]

no, but i know some german

[wintry steppe]

Oh, okay.

edd knows everything

Edd is God himself.

[native rampart]

Ergo God is Engineer

[lavish jewel]

tteppa is a considerable tteppa

[wintry steppe]

considerable is tteppa a tteppa

[wintry steppe]

I checked the closed under multiplication and addition.

they're all good.

kinda stuck at the first point.

showing that zero function is an element of T.

[lavish jewel]

well, i would sure hope that z(x) = 0 is 0 if you evaluate it at z(x=pi)

[wintry steppe]

f(pi) yields surely 0.

but the zero vector must map every element of the domain onto zero in the co-domain.

Am I right?

[lavish jewel]

that's what i wrote up there

let z(x) = 0, the zero function

this satisfies that z(pi) = 0, so it's in T

[wintry steppe]

Jeez.

[wintry steppe]

It's the FUCKING DEFINITION.

[lavish jewel]

oh, another foreigner studying in germany

i feel you

[wintry steppe]

When you can't finance your studies, you have to go to DEUTSCHLAND.

It's cheap you know.

[lavish jewel]

hey, i'm doing the same

😛

[wintry steppe]

WHAT THE.

Where are you now?

[lavish jewel]

somewhere in thüringen

[wintry steppe]

from the States?

[lavish jewel]

somewhere in latinamerica

[wintry steppe]

hmm.

Masters?

[lavish jewel]

did masters, now doing phd

[wintry steppe]

viel Glück dude.

[lavish jewel]

gleichfalls

[wintry steppe]

(afd appears)

[lavish jewel]

oof

[wintry steppe]

have you ever encountered one of them?

Jeez, they're so fucking dumb.

one of them just said to me: verpiss dich Amerikaner. Hahaha.

i assume that does not mean anything good

[lavish jewel]

classic. yeah, not the nicest encounters... this is a better convo for the discussion/chill channels tho

[wintry steppe]

yes, exactly.

[wintry steppe]

sheesh

can you guys help me with my linear algebra final haha

[dusky epoch]

help in what sense?

we are not going to help you cheat on it

[wintry steppe]

i need a sense of direction

[nocturne oracle]

most cellphones nowadays should have a compass built in

they also usually have some type of map

[wintry steppe]

lol

[nocturne oracle]

🙂

[wintry steppe]

i think im facing lake michigan right now

should be facing north then

[charred shell]

could anyone please help me

I have
W1= {x(1,2,3) where x is in R} C R3
I have to find dimension of W and Wt(orthogonal complement)

I found it as W= 1 and Wt=3-1=2

I'm not sure how to find Wt

[short coral]

https://math.stackexchange.com/a/438762

Please help me understand this explanation

I didn't get which elementary operations were applied

[novel crescent]

hey guys i need help with linear equations, who can help me? dm

[wintry steppe]

or you can just post your question here

[novel crescent]

okay

hold up

someone pls help

i been stuck on this problem

[lapis fern]

,rotate

[lapis fern]

https://www.desmos.com/calculator/fr6sxdas2p

the point of intersection needs to be a solution to both, because that point is on the red line, as well as also being on the blue line.

also, with those empty boxes.....plug in your value for y and x

you'll see that the equalities agree. it's like a sanity check.

---

.

from Axler page 42.

my question: an immediate corollary of this is that dim U + dim W = dim (U ⊕ W )?

we have a span of W, and the w_i's are all linearly independent

so its a basis for W, of length n. hence dim W = 5.

also $$v = a_1u_1 +\, \ldots \,+ a_m u_m + b_1w_1 +\, \ldots \,+ b_nw_n$$ demonstrates a basis for V, so dim V is m+n ?

[stoic python]

Videlicet

[lapis fern]

basically i need a justification for $$dim(U \oplus W) = dim V = dim U + dim W$$

[stoic python]

Videlicet

[wintry steppe]

uh

idk if you guys are busy

but im wondering what it means for the origin to be stable/unstable

just using Matrix A with a system of equations

[novel crescent]

Can someone lmk if this is corrrect?

[lapis fern]

its wrong. are you doing it by eye?

[wintry steppe]

this belongs in #prealg-and-algebra, see the pinned message here

[wintry steppe]

to say the origin is stable is to say that any solution to the ODE starting close enough to the origin stays close to the origin.

you can check stability at the origin by looking at the (real parts of the) eigenvalues of A

if they are nonpositive, then the origin is stable, and conversely

[nocturne oracle]

i wish i had terra 2 help me on my la final 🥺

[lapis fern]

TTerra, while youre here might i trouble you with a question, please?

#linear-algebra message

[sonic quartz]

@wintry steppe help pls exam due in 10 minites

[frosty vapor]

@wintry steppe be considerable

[frosty vapor]

Doubt

what are the conditions on U and W

[lapis fern]

U is a subspace of V, and so is W.

and $$U \oplus W = V$$

[stoic python]

Videlicet

[frosty vapor]

so this forces U intersect W to be ?

well

no

[lapis fern]

zero vector?

[frosty vapor]

no it isnt necessarily

this is what i was getting at

it needs to be backwards

we need intersection to be zero vector first

then it follow

U sum W could be V sure

but that isn't enough i think

@wintry steppe wat u think

my enlgihs

i am brain worm

[lapis fern]

i guess what i want is : $$ U \oplus W = V \implies dim U + dim W = dim V$$

[stoic python]

Videlicet

[lapis fern]

is that true?

[frosty vapor]

ye no that is false pretty sure

[wintry steppe]

this is the internal direct sum right

dim V = dim(U + W) = dim U + dim W - dim(U cap W), and this last term vanishes

[lapis fern]

internal? what's that?

[wintry steppe]

ok nvm, ignore that comment

[frosty vapor]

monkers

why the last term gonna vanish

er

oh this comes from U, W subspace V

[wintry steppe]

U \oplus W = V usually means U + W = V and U cap W = 0

like by definition

[frosty vapor]

ye i see

okay yeah this works

i was more worried about the U cap W going to zero

[lapis fern]

so its a corollary?

[frosty vapor]

if you want to call it that ye

[wintry steppe]

linear algebra

[lapis fern]

i do too, lol

[frosty vapor]

i never knew that the oplus had that condition tho

i thought it was just take linear combo of everything

cool

[wintry steppe]

that's how it's defined in friedberg

[lapis fern]

axler calls it that.

[frosty vapor]

my class didn't do oplus in my linear algebra class 🐲 no wonder

makes sens now tho

excelent

the only thing i know from oplus was from the group theory

[wintry steppe]

same thing

vector space is group wrt addition

well

i guess you have to account for scalar mult

[frosty vapor]

vector space is vector space

😼

[sonic quartz]

oplus is just coproduct in Ab

obv

[wintry steppe]

die

[nocturne jewel]

math is math motel

[frosty vapor]

hsct

[lapis fern]

#linear-algebra message

sorry, just one last clarification, my intuition about this is correct, right?

[frosty vapor]

.

[sonic quartz]

im being sullied

Where Is The Lie

[wintry steppe]

you're correct, but also, bruh

[frosty vapor]

[sonic quartz]

wdym this is the correct pedagogical approach

[wintry steppe]

[sonic quartz]

lol every time I join a question channel I just turn it into #chill

or #discussion

[wintry steppe]

ok, quick, everyone give me a cool linear algebra fact

[sonic quartz]

🚶

[wintry steppe]

the prize is my recognition

[frosty vapor]

Fredholm alternative

[wintry steppe]

ok ultra

[frosty vapor]

what

[wintry steppe]

lol

"fredholm" makes me think of C* alg stuff

[frosty vapor]

ooga

[wintry steppe]

[frosty vapor]

i learned about fredholm alternative in my diffeq class 💀

its so cursed

[sonic quartz]

was that the class that blackboxed differential forms

[frosty vapor]

no that was the first diffeq class

this is the "intermediate" one

but uh

this is how it should always have been

with linear algebra

first diffeq class had none

anyways time to read friedberg because i never learned direct sum

[sonic quartz]

[frosty vapor]

in group theory class we learned about internal and external product

which was cool

but

now i can return to linear algebra

[verbal pivot]

i find the fact that null spaces stop growing pretty cool to be honest

[wintry steppe]

elaborate

[verbal pivot]

Consider an operator $T \in L(V)$ where $dim(V) = n$, for all $k > n$ we have that $null T^{k} = null T^{n}$

[stoic python]

b2unit