linear-algebra

[flint jackal]

Right?

[dire thunder]

what

i mean

there is no real difference for points and vectors at least for finite dimensional case

like each vector for finite dimensional case is just n-tuple (x_1, \ldots, x_n)

which can be viewed as point in F^n

[dreamy iron]

Could you explain a bit more please. I’m still not following.

[dire thunder]

what is not clear in that we define span() = {0}?

[flint jackal]

Are you referring to the coordinate space, $$R^{n}$$?

[stoic python]

dldh06

[dire thunder]

F^n

field is not obliged to be R

[dreamy iron]

I get that span() ≡ {0}. I understand that it is defined to be this way.

The span of the empty list is defined to be the singleton set containing the zero-vector, i.e., the most trivial of vector spaces.

What is span({0})?

[dire thunder]

well

apply definition of span of {0}

[dreamy iron]

=={a0 : a \in F }

[dire thunder]

yes

what is a0?

[dreamy iron]

so {0}

[dire thunder]

yes

thus span of empty set and span of set containing only zero vector match

so you are correct here

[flint jackal]

What exactly is the difference between field and R?

[dire thunder]

in your claim of span emtpy set = span {0}

[dreamy iron]

So the span of the two objects we’ve been discussing are in fact equal:

span() = {0} = span ({0}).

[dire thunder]

field is type of algebraic structure and R is one of these algebraic structures

yes

[dreamy iron]

Well that was easy. THANK YOU.

[dire thunder]

simply saying F is any set with two operations such and "zero" and "identity" such that it behaves like R with its addition, multiplication and 0 wand 1

for example Q is also field

and C is field

[dreamy iron]

Some hopefully helpful intuition building. https://thinkzone.wlonk.com/Numbers/NumberSets.htm

[flint jackal]

So I looked up field space and R space. Is field space technically within R space or am I understanding that wrong?

[dreamy iron]

A field is any set, and any two operations on that set, that obeys a selection of 11 axioms: https://docs.google.com/spreadsheets/d/11rri7XubkLaX6ZXxDDhuE0y79u_Of34fSLOXxNebev4/edit

The canonical examples of fields are:

1. The real numbers
2. The complex numbers
3. The rational numbers
4. The set of all rational polynomials.

[flint jackal]

I've only dealt with vector space so field space is a new concept to me

[dreamy iron]

well, the scalars in a vector space have to come from a field, by definition.

so inadvertently you’ve dealt with fields

[dire thunder]

R^n is not a field

well

unless u define vector multiplication meaningfully

[dreamy iron]

They’re easy-ish to get started with. Try and prove zero times any number is zero using the field axioms only. Should take about thirteen lines at most.

how do you define a multiplication on F^n?

and also get a field.....?

[dire thunder]

for F^2 you have something C-like

[dreamy iron]

n>= 3 ??

[flint jackal]

Isn't this basically true all the time because of basic multiplication rules, 0 times anything is always zero?

[stoic python]

Commander Vimes

[dreamy iron]

Yes, of course it’s true. Now prove it using the field axioms.

[stoic python]

Commander Vimes

[dreamy iron]

There’s an analogous proof for vectors:
-1 v =?= -v

[dire thunder]

yes

[dreamy iron]

The basic rules of multiplications are really basic lemmas of multiplication, and lemmas need proof.

[dreamy iron]

Anyway. Thanks for you kind attention #linear-algebra . I will meditate on this result.

[flint jackal]

I don't understand any of this terminology. The most I've dealt with was the basic linear algebra, where we do SVD, eigenvalue decomposition, Gram-Schmidt, orthogonal projection, etc. That basic stuff

[dire thunder]

linear algebra usually does not treats fields much

it just admits them as having nice property

fields are treated in abstract algebra and analysis

[flint jackal]

So all those topics, I stated, I understand real world applications where they could be used. SVD for image compression, Gram-Schmidt to create orthogonal vectors. But the one topic I learned was LU decomposition. I asked my professor how is was useful, but they couldn't think of one on the spot. What kind of applications would LU decomposition be used in? Besides forward and backwards substitution?

@dire thunder Perhaps you could enlighten me on how LU decomposition is useful?

[dire thunder]

it is used is solving linear systems and finding determinant

and inverse

[flint jackal]

Is that the only thing that LU decomposition is useful for?

[lavish jewel]

when it can be used, it is faster than many other methods of solving linear equations

for computational efficiency, you'd use it whenever you can

many decompositions are useful for this reason. when a matrix has a special structure, a particular decomposition makes it easier to handle. it might seem trivial off the top of one's head, but matrix-vector products happen by the truckload when dealing with data, graphics, physics simulations, etc., and one usually has to invert them in some sense

[halcyon pollen]

guys what's numerical analysis?

[lavish jewel]

the study of numerical techniques for obtaining solutions of continuous problems?

[broken sun]

Hello. Is there any example of a matrix in an infinite dimensional vector space such that it is not similar to its transpose?

[dusky epoch]

operators on infinite-dimensional vector spaces tend not to be described with matrices

[zealous junco]

why does the sequence A_k converge? if we assume Q_k converges to plus or minus I

likewhhy does $\lim_{k\to \infty}A^{(k)}$ exist

[stoic python]

Anticipation

[zealous junco]

would something like showing $|A^{(k)}|$ is cauchy work?

[stoic python]

Anticipation

[zealous junco]

have no idea tbh

[broken sun]

Ok. Please replace "matrix" with "operator".

Hello. Is there any example of an operator in an infinite dimensional vector space such that it is not similar to its transpose?

[dusky epoch]

so you mean not similar to its own adjoint i guess

hmm

[broken sun]

No, transpose

[dusky epoch]

you want an operator $A$ that is not similar to $A^$, where $A^$ is the unique operator satisfying $\ang{Ax, y} = \ang{x, A^*y}$

[stoic python]

Ann

[dusky epoch]

(this A* is called the adjoint even in real infdim spaces)

i'm tempted to say that the forward shift operator on $\ell^2$ may fit your description, though i'm not 100% sure...

[stoic python]

Ann

[broken sun]

The space need not to be a product space; I meant "transpose", not "adjoint"

[dusky epoch]

how are you even defining the transpose without an inner product then

[broken sun]

If $T: X \to Y$ is a linear map; its transpose $T^t: Y' \to X'$, prime denotes dual space, such that $T^t(f)=f \circ T$.

[stoic python]

MathPhysics

[dusky epoch]

okay then how are you defining similarity

what does it mean for T to be similar to T^t

they're operations with different domains and codomains

[broken sun]

You can take the domain and codomain the same

[dusky epoch]

an infdim space need not be isomorphic to its own dual

so again, what would it mean for T and T^t to be similar or not

[broken sun]

Assume that they are the same.

[dusky epoch]

what do you mean

[broken sun]

Assume the vector space and its dual are isomorphic.

Why did you delete it?

[dusky epoch]

i'm still trying to figure out what you want, because i'm getting a little lost and tongue-tied

[broken sun]

The meaning of similarity is not clear?

[dusky epoch]

it sounds to me like whether or not two operators are similar an operator is similar to its own transpose may depend on which isomorphism you choose between X and X'

[broken sun]

I am confused.

[dusky epoch]

like, ok, we have two operators, $A: X \to X$ and $B : X' \to X'$ and we wish to define what it means for $A$ and $B$ to be similar

[stoic python]

Ann

[dusky epoch]

as-is, similarity only makes sense (to me) if we take two operators with the same domain and codomain

and ask whether there exists an automorphism phi of the space such that conjugating one operator by phi gives the other

but we cannot ask this same question for our A and B, because they're operators on different spaces

even if we assume X is isomorphic to X', this still requires fixing an isomorphism p between X and X', and asking whether or not A and p^-1Bp are similar

and at the moment i am not convinced that this is independent of our choice of p

[broken sun]

Assume that the vector space and its dual are isomorphic.

[dusky epoch]

i am already assuming that

... fixing an isomorphism p between X and X' ...

there isn't just one of these, you know

[broken sun]

You say the definition of "transpose" depends on the isomorphism map?

[dusky epoch]

no

i'm saying that the definition of similarity between an operator on the primal space and an operator on the dual space depends on the isomorphism we choose

[broken sun]

Sorry, I meant similarity.

[dusky epoch]

????

[broken sun]

I meant "similarity".

[dusky epoch]

then see this again

need i repeat myself?

[broken sun]

Why does that depend?

[dusky epoch]

so let's say we have two arbitrary isomorphisms $p, q: X \to X'$, and the same operators $A: X \to X$ and $B: X' \to X'$ as before.
and let's say $A$ is $p$-similar [resp. $q$-similar] to $B$ if $A$ is similar (in the ordinary sense) to $p^{-1}Bp$ [resp. $q^{-1}Bq$].

my question is: is $p$-similarity equivalent to $q$-similarity?

[stoic python]

Ann

[dusky epoch]

even if it is, that's something i would like to see a proof of.

[broken sun]

Why not? Do you have any counterexample?

[dusky epoch]

do not confuse my casting doubt on a statement for an assertion of its falsehood.

though after writing it out it looks like the answer is yes, since $$p^{-1}Bp = r^{-1}(q^{-1}Bq)r,$$ where $r = q^{-1}p$...

[stoic python]

Ann

[dusky epoch]

okay

fine, we can speak of similarity between T and T'

the forward shift operator on $\ell^2$ may still fit your description

[stoic python]

Ann

[dusky epoch]

for clarity, i'm considering $\ell^2$ as the space of all real sequences $(x_1, x_2, \dots)$ such that $$\sum_{k=1}^{\infty} x_k^2 < +\infty,$$ and i define the forward shift operator as $T(x_1, x_2, \dots) = (0, x_1, x_2, \dots)$

[stoic python]

Ann

[crude falcon]

hey I'm calculating the eigenvectors associated to an eigenvalue, after reducing the corresponding matrix i get that rank(A) = nº variables, I computed then the cartesian equations associated to that matrix, is the solution vector associated to this system the eigenvector?

or do I need to compute the parametric equations too?

[crude falcon]

<@&286206848099549185>

[dusky epoch]

it's not exactly clear what you're talking about

[crude falcon]

its about computing the eigenvectors associated to an eigenvalue, for that you compute (A-zId)x=0, and the solution of that system is supposed to be the eigenvector right

(z being the eigenvalue)

[dusky epoch]

so you have a matrix A and an eigenvalue z... yes, the eigenvectors are by defn the solutions of (A-zI)x = 0

that's it

[crude falcon]

this is my problem, I end with a (0,0,0) vector, with can't be an eigenvector right?

[dusky epoch]

are you sure you have the right eigenvalue?

what is your matrix and what eigenvalue are you considering for it?

[crude falcon]

the eigenvalues I got are : 3,2 and 6

in the example I showed the eigenvalue that I'm using is 3

[dusky epoch]

one moment

let me doublecheck your eigenvalues

[dire thunder]

,w eigenvalues {{3,-1,1}, {-1,5,-1},{1,-1,3}}

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Commander Vimes#4407

[dusky epoch]

okay

[dire thunder]

well sorry if i sniped

[dusky epoch]

so you have $A - 3I = \bmqty{ 0 & -1 & 1 \ -1 & 2 & -1 \ 1 & -1 & 0}$

[stoic python]

Ann

[dusky epoch]

how exactly did you get that third row in your reduced matrix?

i think you messed up the arithmetic while doing row reduction somewhere

[crude falcon]

okay I'm seing now the error

I used -1 -1 0 in the 3rd row

instaed of 1 -1 0

nvm

[broken sun]

And why is this operator not similar to its transpose?

[dusky epoch]

its transpose is backward shift which isn't injective

[broken sun]

Why is its transpose the backward shift?

[dusky epoch]

you can do the calculations

[broken sun]

How?

[dusky epoch]

for each natural number $k$ let $e_k$ be the sequence in $\ell^2$ consisting of a 1 at position $k$ and 0 elsewhere, and let $\hat{e}_k$ be the element in $\ell^2$ which maps each sequence in $\ell^2$ to its $k$'th term

[stoic python]

Ann

[dusky epoch]

and consider the isomorphism p between these which sends each e_k to its corresponding ê_k

and consider what $\hat{e}_kT$ might be

[stoic python]

Ann

[dusky epoch]

(where T is the forward shift, of course)

you should get that $\hat{e}k T = \hat{e}{k-1}$

[stoic python]

Ann

[dusky epoch]

and $\hat{e}_1T = 0$

[stoic python]

Ann

[broken sun]

And, an operator similar to an injective operator must be injective?

[dusky epoch]

yes

[broken sun]

Why?

Ok. Thanks.

[dusky epoch]

why are you thanking me when i haven't had the chance to answer your latest question

[broken sun]

I got it.

[dusky epoch]

ok

...wait, so do you need an explanation of why injectivity is preserved across similarity or not

[broken sun]

Yes, but it is obvious, isn't it?

[dusky epoch]

it is

take two similar operators $A$ and $B = \varphi^{-1} A \varphi$ and suppose $A$ is injective (which is equivalent to saying $Ax \neq 0$ when $x \neq 0$)

[stoic python]

Ann

[dusky epoch]

then $Bx = \varphi^{-1}A\varphi x$

[stoic python]

Ann

[dusky epoch]

since phi is an automorphism we have phi x ≠ 0

since A is injective we have A(phi x) ≠ 0

and since phi^-1 is an automorphism we have phi^-1(A phi x) ≠ 0

[broken sun]

We cannot say the composition of injective maps is injective?

[dusky epoch]

or you could say that too if you want

[wintry steppe]

Hey guys, what do I have to do here?

I only know that the rotation counterclockwise is (0 -1 ) (1 0)

[quasi prawn]

U are looking for the function (in this case a linear illustration (not sure about the phrasing here as we call it "lineare Abbildung" in germany)) that will take an input vector v (2 dimensional) and rotate it counter clockwise by 90 degrees. A matrix can be seen as a tool to perform a linear illustration , as when u rightwise multiply the matrix with a vector this will perform a certain manipulation on the vector. Thus in your case you are looking for a 2x2 matrix that, when rightwise multiplied by a 2 dim. vector, will rotate that vector counterclockwise by 90 degrees.
As u stated correctly this matrix is
(0 -1)
(1 0)
which is the solution to the question.

Do you understand why it is exactly that matrix that gets the job done or do you need an explanation?

[wintry steppe]

I understood it fully, but I dont unterstand the last sentence in my question

[quasi prawn]

@wintry steppe to give you the general formula for rotating any 2dimensional vector by any angle, this would be the matrix needed:
(cos(alpha) -sin(alpha) )
(sin(alpha) cos(alpha))
whith alpha being the angle by which the vector is supposed to be rotated by. With alpha=90 degrees (thus cos(90)=0 and sin(90)=1) this evaluates to the matrix above

[wintry steppe]

(Im also german but translated it myself 😄 )

[quasi prawn]

Meaning of last sentence: Find the 2x2 matrix that performs the linear illustration f (which as described in the first sentence depicts from R^2 into R^2) on a given 2 dimensional vector v

[quasi prawn]

OK:D I dont think german is allowed here so lets keep it in english:D

[wintry steppe]

Sure! Now I understand the question fully, thank you but I think the question is to hard for me to solve and maybe I will skip this one

[quasi prawn]

Well you have already solved it ^^ No problem and good luck

[wintry steppe]

Yes but I dont know what to put in f(u) = Au. But nevermind thanks for your help 🙂

[next vapor]

Let F be a field, A an nxn matrix over F and p(x) a polynomial in F[x], show that there exists a polynomial q(x) with degree less than n such that p(A)=q(A)

so if degree of p is n then i can just use the division algorithm along with cayley hamilton to get the result

but not sure how to go about this if the degree is less than equal to n

[dusky epoch]

just take q=p lol

@next vapor

[next vapor]

oh wow

thank you lol

[digital solstice]

does anyone here know the intuition behind the magnitude of the cross product, why is it equal to the area of a parallelogram? and also, i noticed that it's a necessary defining condition for the cross product to be distributive, what i mean is that if you start with the assumption that the cross product is distributive (along with a couple other properties about the cross products of the unit vectors i, j and k) you end up implying that its magnitude is the area of the parallelogram spanned by the two vectors, and i wonder why thats true exactly

[wintry steppe]

@empty copper Hi, so can you help me with the inequality?

[elfin meteor]

i'm looking for a tutor for Linear Algebra. I'll be paying $30~60 an hour. If you DM me, please turn on the setting to accept messages from nonfriends, or else I cannot reply. Or, send me friend request to accept.

[dawn fractal]

@digital solstice if it still is on ur mind here's a link to a Youtube vid I found helpful

https://youtu.be/eu6i7WJeinw

you'd probably need to watch this vid before that though

https://youtu.be/Ip3X9LOh2dk

[dire thunder]

i guess it would be nice if you would also indicate what level of LA you are interested in

[elfin meteor]

right, thanks

basic to around medium level. fairly simple stuff revolving around eigens, but i need dumbed-down intuitive explanations

[stoic python]

n/c

n/c

[wintry steppe]

Will this proof work for the => direction? Sorry for long paste.

[wintry steppe]

delete the source messages

[wintry steppe]

watching 3b1bs "the essence of linear algebra" might save you a lot of money.you should definitely try it out

[elfin meteor]

i did. loved it

it's hard to explain

for the current assignment im doing, the topic seems to be overly easy. so much that it makes me anxious

[wintry steppe]

@wintry steppe done, sorry

[elfin meteor]

the only challenge is needing to explain it at a super-elementary level, where i'm also anxious that i might miss something

[wintry steppe]

idk but maybe youre just understanding the concepts so good that the task isnt har for you.i found most of my hw in LA1 quite simple too

[nocturne jewel]

Or just post questions you're stuck on and save yourself some money

[wintry steppe]

Can anyone look at my proof and tell me if it works?

[native rampart]

Go on

[wintry steppe]

guys i am not understanding something
When calculating the
Legendre momentos of a MxN array (image in this case)
many sites say it is this way
https://gyazo.com/04423e0600a12dd9513df531eb162b30
So
for the pixel[0][0], the legendre moment is L_00?
i mean, does each pixel have its own Legendre moment?

[wintry steppe]

@azure trout I posted it asbove

[azure trout]

Ummm

You pinged the wrong buncho dragons

@native rampart

[frosty vapor]

lmao

[native rampart]

The assumption is null T_1 is a subset of null T_2. There's no necessity that T_2x=0 implies implies T_1 x=0

[dawn fractal]

in fact, it's the other way around, by definition of subset

more precisely, x in null T_1 implies that x is in null T_2

writing that any map S will work might sound contradictory to what you need to prove

[wintry steppe]

Okay

I'll redo the zero case

What about the nonzero case?

[dawn fractal]

when writing another image of a vector under a map, use another variable, like w_j

it's fine to use a different index (e.g. j) for the second summation

[wintry steppe]

... Why?

[dawn fractal]

because w_i is the same thing all throughout your proof

you don't need it confused with the other w_i

[wintry steppe]

What?

I don't understand your point sorry.

[dawn fractal]

you'd be implying that T2vi = wi = T1vi

when doing by-cases proof u have to state it explicitly
Case 1.
.
.
Case 2.
.
.

and remember to state explicitly what you're proving at the end of each case, or at the end of the whole proof if it's one consequent

.
.
Hence T2 = ST1

[wintry steppe]

I don't understand this notation at all.

Whose entries Aj,k are defined by Tvk = A1,k w1 + ... + Am,k wm

How do I find A1,k and so on?

It seems circular that the Aj,k entry is defined as that

Since I need to use the Aj,k entry to find the Aj,k entry.

[nocturne jewel]

You find the transformed vector under the basis of the output space, then the column of the matrix is just those scalars

so take your v_k and apply T to it and you'll get a vector in W

that means you can write it as a linear combination of W basis vectors

the scalars of the linear combination become the column of the matrix

with the fact that the kth v vector will be placed in the kth column

[wintry steppe]

@nocturne jewel But this doesn't tell us how to find the matrix.

Are we reading the same thing?

Whose entries Aj,k are defined by Tvk = A1,k w1 + ... + Am,k wm

Yeah, this definition makes 0 sense.

uh

Tv_k is an element of W, so it's a linear combination of w_1, ..., w_m with unique coefficients (since w_1, ..., w_m is a basis). we define the kth column of A to be those coefficients.

what's wrong with that?

[nocturne jewel]

I mean what I said is equivalent / uses co-ordinate vectors

[wintry steppe]

That's not what is written there @wintry steppe

?

[nocturne jewel]

The definition is perfectly fine

[wintry steppe]

slightly edited for clarity

we define the kth column of A to be those coefficients.

I seem to be missing this sentence from the definition.

?

Yes, so?

[nocturne jewel]

that's what terra said

[wintry steppe]

do you agree that for each k there exist scalars A_{1, k}, ..., A_{m, k} such that this is true

yes or no

furthermore do you agree that these are unique

Of course, but that's not what the definition says to write the column k of the matrix M as

The word column is literally never used

[nocturne jewel]

$T[v_1]=A_{1,1}w_1+...+A_{m,1}w_m$ means the 1st column of the matrix would be $[A_{1,1},...,A_{m,1}]^T$ since $k=1$ in this instance

[stoic python]

moshill1

[wintry steppe]

A_{j, k} means the entry in the jth row, kth column.

I know that

But then it's written as if the entry of Aj,k is Tvk

????????????????????????????????????????????????????????????????

[nocturne jewel]

it's not

[wintry steppe]

How's it not?

[nocturne jewel]

Cause it never says that

[wintry steppe]

"whose entries Aj,k are defined by Tvk = ..."

[nocturne jewel]

yeah, then the scalars are of the form A_{j,k}

[wintry steppe]

the entries of the matrix ARE NOT the Tv_k themselves. they are the SCALARS in the linear combination in terms of w_1, ..., w_m that Tv_k equals

[nocturne jewel]

whose entries Ajk are defined as the scalars of the linear combination

[wintry steppe]

I know but that's not what it says, it's written badly imo

[nocturne jewel]

It's not

(maybe not how I would personally write it, but that's cause I learned it with co-ordinate vectors)

[wintry steppe]

$A_{j, k} = w^j(Tv_k)$ qed

[stoic python]

狼人

[wintry steppe]

where w^1, ..., w^m is the dual basis to w_1, ..., w_m

there you go

this is as concise as it could possibly be

[nocturne jewel]

in not knowing dual basis

That's my course's note on it

[wintry steppe]

the solution to n/c's confusion is for them to switch to a better book

This is so confusing

If I want to make the differentiation map of polynomials with real coefs from deg 2 to deg 1 into a matrix

A basis for the P2(R) is (1,x,x^2) and a basis of P1(R) is (1,x)

[nocturne jewel]

yes

[wintry steppe]

Then the matrix would be [D(1,0,0) D(0,x,0) D(0,0,x^2)]

Right?

[nocturne jewel]

No

1,0,0 isnt a function

[wintry steppe]

I know?

[nocturne jewel]

$D(1)=0=0\cdot 1+0\cdot x$

[stoic python]

moshill1

[nocturne jewel]

so the first column is [0,0]

$D(x)=1=1\cdot 1 + 0\cdot x$

[stoic python]

moshill1

[nocturne jewel]

so the 2nd column is [1,0]

[wintry steppe]

But you need to input a basis vector into the map

[nocturne jewel]

Yeah, and I did that

[wintry steppe]

So what did I do?

[nocturne jewel]

I put in x which is a basis vector of P2

you put in a co-ordinate vector

[wintry steppe]

We defined a polynomial to be a list

[lavish jewel]

i think their D(...) notation was more like a divergence sorta thing

D(0,0,x^2) = [0,0,2x]^T, for example

giving the benefit of the doubt

weird notation, but ok

[wintry steppe]

Okay wait

We could do it like this instead

D(1,0,0), where (1,0,0) represents 1 * 1 + 0 * x + 0 * x^2

[nocturne jewel]

right

[wintry steppe]

D(0,1,0) where 0 * 1 + 1 * x + 0 * x^2

[nocturne jewel]

yes you can write the functions as their co-ordinate vectors

[wintry steppe]

And D(0,0,1) where 0 * 1 + 0 * x + 1 * x^2

But now, how do we get a list of length 2 back? If we define a polynomial to be a list

[nocturne jewel]

cause it outputs to a 2 dimensional space

[lavish jewel]

it doesn't HAVE to be length 2, anyway

it'll behave the same way tho

[wintry steppe]

@lavish jewel ?

[lavish jewel]

you will get a 2 dimensional subspace regardless

[wintry steppe]

Yeah but how do we know which place to truncate in the list

[lavish jewel]

it doesn't matter

[wintry steppe]

For example, D(1,0,0) = D(0,0) but how would you know?

[lavish jewel]

the dimensionality or rank or whatever you wanna call it won't change

[nocturne jewel]

cause $\mathbb{R}[x]_{\leq 1}$ has a basis ${1,x}$

[stoic python]

moshill1

[nocturne jewel]

so your output vectors will be written as a linear combination of basis vectors, in which there are 2

[wintry steppe]

So why is what I said wrong?

The matrix would be [D(1,0,0) D(0,1,0) D(0,0,1)]

[lavish jewel]

yeah, that was already correct, if each of those is a column

[wintry steppe]

Yes

[lavish jewel]

just make it clear your D(...) is a vector-valued function that puts the input vector as coefficients of a 2nd order poly and then takes its derivative in some special way

[nocturne jewel]

$D[f]=\dv{x}f \to D[(f)_B]=A(f)_B$ where $B={1,x,x^2}$ and A is the desired matrix

[stoic python]

moshill1

[nocturne jewel]

that's how I was taught it

(f)_B being the co-ordinate vector of f under basis B

[wintry steppe]

But I don't see how the basis of the codomain impacts it

[lavish jewel]

you want the output to be made up of coordinate vectors in the codomain's basis

you'll have to do an extra transformation into the correct basis

[wintry steppe]

Look, if T(x,y) = (x + 3y, 2x + 5y, 7x + 9y)

So T in L(F^2, F^3)

What is the difference if we use the standard basis for F^3 or some other basis?

[lavish jewel]

the same way any change of basis works

[wintry steppe]

Haven't learned about that

[lavish jewel]

e.g. you have a vector [1,0,0] in the canonical basis

but for whatever reason you are told to use the basis [5,0,0], [0,10,0], [0,0,-1]

the coordinates of that vector whose coords were [1,0,0] in the canonical basis are now different w.r.t. the new basis

[wintry steppe]

But what about the matrix?

[lavish jewel]

you need to transform it

either directly when you write it, or by multiplying it from the left by a transformation of some sort

[wintry steppe]

How do you do it directly?

[lavish jewel]

you look at it and go "oh, i have to transform it"

i don't have a good example because it's the sort of stuff one should see immediately

if you don't, you'll have to take an extra step

[acoustic zodiac]

I'm a bit confused with matrix notation

Can I get some help?

[nocturne jewel]

post your question if you want help

[acoustic zodiac]

Alright

So this is the way basis are transformed according to my notes

Where

And the same for the ones with the hat

But in the case of vectors

So I have two questions

[wintry steppe]

@lavish jewel How would the differentiation map from P2(R) to P1(R) look like if P1(R) had basis 2,2x or something instead of 1,x

[lavish jewel]

you'd have to divide by 2.2

for example

in more complicated cases, you have to multiply by an inverse mat

[acoustic zodiac]

1: In the first case, each basis vector is written by itself, without components. But how would you write that up in the case of a real basis? Do you just make every column a basis vector, like you would with vector components?

2: Why is the inverse being used in the case of transforming vectors?

[mortal locust]

hello

I have a question

if u have a matrix A and find it’s eigenvalues, and then u plug in one of the eigenvalues (call it lambda) into A-lambda*I

that matrix should be non-invertible right?

since det(A-lambda*I)=0

[brisk fractal]

yes

[mortal locust]

am I making a stupid mistake then

[brisk fractal]

also, the kernel of A-lambda*I is just the eigenspace corresponding to lambda

[mortal locust]

wouldn’t the matrix at the bottom be invertible since it has 3 pivot positions

[brisk fractal]

If the eigenspace is nontrivial then the map cannot be surjective

Because you calculated the eigenvalues wrong I believe

I don't think you wrote down the correct characteristic polynomial

[mortal locust]

oh right

I forgot to subtract 1

ok so stupid mistake was true ok got it thanks needed another pair of eyes

[brisk fractal]

I believe it's (2-a)((3+a)²-1)

[mortal locust]

yeah

thank u

[brisk fractal]

👍

[elfin meteor]

does anyone know if it's true that, if a matrix has conjugate complex eigenvalues, the matrix is a rotational matrix?

[odd kite]

This is false

[elfin meteor]

thank you

[odd kite]

none of these are rotation matrices $$\begin{pmatrix}0 & -2\2 & 0\end{pmatrix},;;;\begin{pmatrix}
e^{-i\theta} & 0\0 & e^{i\theta}\end{pmatrix},;;;\begin{pmatrix}0 & -1 & 0 & 0\1 & 0& 0 & 0\0 & 0& 2 & 0\0 & 0& 0 & 2\\end{pmatrix}$$

[stoic python]

Timon

[digital solstice]

i already saw that video but i dont think it answers my question in particular

[dawn fractal]

My questions might be better answered on here
I was trying to find a connection between this theorem and 3b1b's vid on change of basis:

https://www.youtube.com/watch?v=P2LTAUO1TdA&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=15

at 11:03, he discusses "How to translate a matrix"
ofc, u have to watch the vid to know what im gonna talk about

1. did i make all the connections between the vid and Theorem 10 correctly?
2. does any change of basis matrix (B1, B2 are bases of R^2) correspond to some linear operator on R^2?

[weak blaze]

Hello everyone. Is there anyone to prove?

[weak blaze]

Thanks a lot👍

[dawn fractal]

<@&286206848099549185>

[dreamy iron]

Hi #linear-algebra 😄

I have an Axler QQ: why does the lemma about spanning lists being reduced to form a basis say nothing about finite dimensionality?

[native rampart]

He implicitly assumes VS is finite dimensional

[dreamy iron]

The partner of that particular lemma, the one about linearly independent lists makes a point to call out the fact that the vector space is finite dimensional.

[native rampart]

For example ,R is spanned by the set of irrational numbers along with 1 as a Q vector space

[native rampart]

Ok, it's more like he assumes choice

[native rampart]

It's not obvious that R as a Q vector space has a basis if you don't assume choice

[dreamy iron]

Wait, what‽

There a A LOT of irrationals...

[native rampart]

Yes

[dreamy iron]

So that list is like VERY LONG?

[native rampart]

That list is infinitely long

[dreamy iron]

good lord

[native rampart]

It's not obvious there's a basis for that list

There are people who think there is

[dreamy iron]

But there is?

Oh...

[native rampart]

and there are people who think there is not

It's called axiom of choice

[dreamy iron]

So if you assume choice, any infinite dimensional vector space can have a SPANNING list reduced to a basis ??

[native rampart]

Yes

[dreamy iron]

Okay. Well that’s super neat.

[dreamy iron]

But that basis is still infinitely long?

I don’t know enough functional analysis to keep up with this....

[native rampart]

Yes

[dawn fractal]

bumpu

[soft abyss]

Maybe a dumb question but, is there a generic matrix I can multiply another Matrix by to make all the lower diagonal elements 0

?

[native rampart]

0

[soft abyss]

(and leave the upper diagonal elements alone)

Obviously

[wintry steppe]

Can a spanning list really be reduced to a basis in infinite dimensional case? The proof i know for existence of a basis does not involve reducing a spanning list

[wintry steppe]

No. Suppose there is such M = [x y] [z w], then in particular works for A= [1 0][1 1] but then can show M = [1 0][-1 1]. But that M doesnt work for every matrix (this is the case for right multiplication)

actually even easier: you would have IM = I So M = I

[soft abyss]

Right yeah

I should have known that 😥

Thanks

[wintry steppe]

and i should have taken less time to solve it

[strange delta]

for part v wasn't this already shown in part ii because the null is kernel and the col is the image?

[spiral star]

well, almost. you just have to take the isomorphism back to R_3[x] and R_2[x]

[strange delta]

wdym

[spiral star]

for the matrix you have basis in R³ and R⁴

but F maps between poly spaces

[lavish jewel]

one of the two things deals with coordinate vectors. the other deals with polynomials

[strange delta]

do you know where i can find this on the book linear algebra done right?

well read more about it

[spiral star]

you already applied this when you solved (i)

you apply such an isomorphism every time you write vectors in basis coordinates

now you do the opposite and convert from basis coordinates back

[dawn fractal]

what does "the rank of AB is at most the rank of A and B" mean in inequality form?

[wintry steppe]

rank(AB) <= min(rank(A), rank(B))

in case you're asking "how do i write this as an inequality"

[strange delta]

why is it saying col(b) = r^4

[dawn fractal]

oh ty

[strange delta]

is it because the basis spans R^4 at this point because we can literally make anything out of the basis ?

[wintry steppe]

B has rank 4 meaning it's column space is a 4-dimensional subspace of R^4, i.e., all of R^4

the rref tells you this

[strange delta]

ah yeah

and how was he able to deduce the nullity(b) = 0

well i can kinda just eye it and see any possbile x will just be equal to zero

and hence the null(B) = {0}

[wintry steppe]

sounds good

[agile lion]

How to solve this please?

[wintry steppe]

a nice way to check linear independence in this case is to put the vectors together into a 3x3 matrix and take the determinant

[gentle briar]

so for the 1st part i got r can be anything but 0 in the real numbers

for part 2 and 3 i am a bit confused

[jagged pendant]

Jordan normal form

3 to n get killed

(0 1)
(0 0)

^2

= 0

(0 1)
(0 0)

n=5 example (we're using VC as well if this makes no sense)

[wary sinew]

Using Cayley Hamilton theorem

[flint jackal]

It's because the matrix is full rank and based on the Rank-nullity theorem rank(B) + nullity(B) = n, where n is the number of columns in a m x n matrix

[jagged pendant]

characteristic polynomial

input the matrix A into it and u get 0

[strange delta]

is this correct

[nocturne jewel]

$0=a_1v_1+a_2v_2+...+a_nv_n$ when $a_1=...=a_n=0$

[stoic python]

moshill1

[nocturne jewel]

$f(0)=f(a_1v_1+...+a_nv_n)$

[stoic python]

moshill1

[strange delta]

yes agreed

[nocturne jewel]

wait no, that's wrong

[strange delta]

i'm wrong or your wrong?

[nocturne jewel]

My answer is wrong cause it doesnt show that only the scalars all 0 make 0

[strange delta]

is my answer right

?

[nocturne jewel]

you never write 0

so hard to show independence without the defining feature

[strange delta]

what i meant to show there is the transition from the first set to the second

[nocturne jewel]

well there clearly is a transition

but you havent used independence

[strange delta]

and since we know the first sets a_i are equal to zero

[nocturne jewel]

only when writing the 0 vector

[strange delta]

isn't the defintion of linear independent just

$0=a_1v_1+a_2v_2+...+a_nv_n$ when $a_1=...=a_n=0$

[stoic python]

redd

[nocturne jewel]

yes, you never wrote that

you wrote for a general element in V, which is span not independence

Assume {f(v1),...,f(vn)} is a dependent set, show that's a contradiction

[strange delta]

but in the question we assume the first set is linearly independent don't we

so i can automatically assume a_i = 0

[nocturne jewel]

yes, and that will show that all scalars 0 will make the 0 vector

but that doesnt meant they're indep

${v,2v}$ is clearly dependent, but $0=0(v)+0(2v)$

[stoic python]

moshill1

[strange delta]

hmm

the way i normally show independence is showing it's under addition and multiplication

[nocturne jewel]

what?

[strange delta]

nvm, do you know a way to do this question?

[nocturne jewel]

.

[strange delta]

when showing that would it be similar to how i attempted the question

[nocturne jewel]

it'd show that only the set of scalars {0,0,...0} would work

which is what is desired

[acoustic zodiac]

What is the reasoning behind defining a conjugate transpose as (where $A^\dagger$ is the conjugate transpose):

$<u|A^\dagger(v)>=<A(u)|v>$

[stoic python]

rcatalang

[acoustic zodiac]

Assuming it's a linear application and not a matrix

[wintry steppe]

I would like to understand this idea for the matrix of a vector x in V in basis B, where x = x1 v1 + ... + xn vn. xi in F,=

And we can view any vector x in V as defining a linear map x : F -> V, where 1 gets mapped to x, and lambda gets mapped to lambda * x

Where 1 is the standard basis element of F

[agile lion]

What to do after taking the determinant?

[wintry steppe]

well, the determinant will tell you whether or not it's linearly independent

det A = 0 if and only if the columns of A are linearly dependent

[wintry steppe]

guys, acording to this

https://gyazo.com/19a613f189ecfa165692dab99f4dc99b

what is L_mn??

like, mn loop through the pixels of the image?

Or it is the degree?

[odd kite]

no, the image is f(x,y). m n are indices which identify a particular moment

@wintry steppe

[wintry steppe]

so every pixels needs its moment?

[viral isle]

Completely new to linear algebra. How do I convert this problem into linear algebra equations: (1/2)at^2 + v0 + c0 is distance of object from observer at time t, where a is acceleration, v0 is initial velocity and c0 is initial distance from observer.

At 1 second, object is 20 metres from observer, 2 seconds it is 55 metres away, 3 seconds it is 110 metres away

Is it simply:
0.5a v0 c0 20
2a v0 c0 55
4.5a v0 c0 110

[dawn fractal]

v0t or v0? @viral isle

[viral isle]

v0

I think I'm wrong because v0 and c0 all being 1 means I can't get it into Reduced Row Echelon Form since they'd all end up being 0s or 1s

@dawn fractal

[stoic python]

!superficialsicko

[viral isle]

oops, I actually had 4.5a

Edited

Thanks

I'm still not sure it's correct because when I try to get it into reduced row echelon form my v0's and c0's are always identical

[dawn fractal]

use Gauss-Jordan elimination method

[viral isle]

Thanks for your help, I am trying that

I can't actually find a solution that works

[dawn fractal]

i found one myself

what's the rref you got?

[viral isle]

I currently have:
1 2 2 40
0 3 3 25
0 8 8 70

[dawn fractal]

then what

[viral isle]

I'm unsure how to proceed tbh

[dawn fractal]

do the same thing u did from the first step

[viral isle]

ok

[dawn fractal]

where u reduced the other entries in the first column to 0

do that for the 2nd column

[viral isle]

My first step was to take the first row from the second and third to get the 0s

but if I do that now won't it just but numbers back in the first column's 2nd and third rows?

[dawn fractal]

that's basically what we're doing

now u need to make sure column 2 has a leading 1 in row 2

but we're working with a smaller matrix now

ignore the first column and row for now

then consider the matrix obtained from that

[viral isle]

oh

[dawn fractal]

essentially make row 2 equal to 0 1 ... ...

and row 3 equal to 0 0 ... ...

and after that, make row 3 equal to 0 0 1 ... (or 0 0 0 ..., otherwise )

[dawn fractal]

do this by dividing row 2 by its 2nd entry

[viral isle]

1 2 2 40
0 1 1 8.33
0 8 8 70

[dawn fractal]

i suggest u write in fractions

but yes, something like that

[viral isle]

Then eight times row two from three, I guess?

1 2 2 40
0 1 1 25/3
0 8 8 70

[dawn fractal]

yes, subtract row 3 from that

[viral isle]

1 2 2 40
0 1 1 25/3
0 0 0 (200/3)-70

[dawn fractal]

yep

[viral isle]

So dumb question, but how can 0 0 0 = anything other than 0?

[dawn fractal]

we're actually searching only the first 3 columns for leading 1s

yes

this is why i asked if u were sure about v0

im familiar with the formula (1/2)at^2 + v0t + c0 = distance

not (1/2)at^2 + v0 + c0

[viral isle]

yep, it's definitely the latter

[dawn fractal]

it means the system has no solution, so there's something wrong with your setup

probably a typo

and forgot to type t after v0

try it

replace v0 by v0t

and the system will be consistent

[viral isle]

Hmm, that might be it

[dawn fractal]

it's a typo

where did that come from

[viral isle]

wow

It's one of my exam prep questions

[dawn fractal]

lmao

i have only seen (1/2)at^2 + v0t + c0 in my physics life

[viral isle]

we have only been shown the elimination steps, not how to put formula into linear algebra equations form. I thought it was a simple as what I was doing but assumed I was wrong because I kept getting an answer on the right in the third column with all 0s, as we have done here

Guess it's good to know I was on the right track

Feel like I just wasted your time though

Thanks for your help, I'll check with the teacher on Monday

[rough aspen]

Hi I feel like this is a really easy question but I'm getting it all wrong. Could I get some assistance thank you

[dreamy iron]

Hi #linear-algebra.

If I have a vector space V, with subspace U. And I have a linearly independent list of vectors is U, how do I know the linear independence is preserved in V.

This feels intuitively true, but I don’t actually know how to prove it.

Like obviously if ONLY the trivial combination is the only way to get the ZERO vector in U....I need it to be the case that the trivial combination is also the ONLY way to get the ZERO vector in V.

And the two, V and U are over the same field.....

[dusky epoch]

i mean

the defn of linear independence doesn't quantify over vectors of the space you're in

but rather over tuples of coefficients

like, any linear combination of your list (of vectors in U) will still be in U

and it's not like U and V have different zero vectors anyway

[dreamy iron]

So a list of linearly independent vectors forms a span. The span is a subspace of the ambient vector space....

Yeah....i don’t think that does it....

[limber sierra]

why is span coming up?

[dusky epoch]

any list has a span

[limber sierra]

you dont care about span, you care about linear independence

[dreamy iron]

I’m shooting in the dark

[limber sierra]

ie the equation $a_1v_1 + a_2v_2 + \dots + a_nv_n = 0$

[stoic python]

Namington

[limber sierra]

if $v_1, v_2, \dots, v_n$ are linearly independent, then this implies $a_1 = a_2 = \dots = a_n = 0$

[stoic python]

Namington

[dreamy iron]

Im still following

[limber sierra]

this is preserved regardless of whether you talk about a subspace or the whole space

[dreamy iron]

Do I need to prove this?

[limber sierra]

since a subspace is over the same field by definition

its immediate

your possible coefficients (ie possible values of a_1, a_2, ... a_n) are the same

since subspaces have the same base field as the vector space

[dreamy iron]

Yes. Since same field

[limber sierra]

so the equation carries over without any changes

and has the same solutions

namely, just a_1 = a_2 = ... = a_n = 0

hence we still have linear independence

[dreamy iron]

I believe every thing you are saying. It’s just crazy that I’ve not seen Axler prove it.

I don’t know. Maybe Im just being paranoid

[limber sierra]

its kind of an obvious fact

considering a different subspace does nothing to the linear independence equation a_1v_1 + ... + a_nv_n = 0

(assuming v_1, ..., v_n are still in your subspace)

so itll have the same solutions

hence linear independence is preserved.

[lavish jewel]

you can think of the space as giving some properties to the vectors. if changing the subspace does not change the vectors, linear independence is shown via exactly the same equation, no changes at all.

[dusky epoch]

what if you operate on an absolute "no such thing as obvious" maxim

[dreamy iron]

If I tried to show that a set B is linearly independent in U and linearly Dependent in the ambient V....i should get a contradiction, right?

[limber sierra]

you would struggle to show that

[dreamy iron]

Lol. Okay

[limber sierra]

if you managed to show it, that would be a contradiction, yes

if im understanding you correctly

[dusky epoch]

just as you would struggle to show 2+2=5

[lavish jewel]

i'm pretty sure such a scenario would imply that U isn't a subspace of V

[dreamy iron]

i mean...i could try and contradict that 2 plus 2 equals 5

[limber sierra]

this is the face of someone unenlightened to the glories of characteristic 1

[dreamy iron]

Any way. Thanks for your kind attention. I will write down “it is obvious”

[limber sierra]

truly tragic that they dont know what theyre missing

[dusky epoch]

F_1 moment

[limber sierra]

alas

such is the harshness of life

man now i wanna make an arxiv april fools paper

"Applications of Commutative Algebra to the Ring of One Element"

which just proves a bunch of comm alg theorems in the special case of the trivial ring

(each proof, of course, is "Trivial. ")

[dusky epoch]

all modules over the zero ring are isomorphic

[snow jetty]

did I understand correctly what does the Gram matrix mean?

[lavish jewel]

the gram matrix, for a given matrix M, is M^T M (over the reals)

[dusky epoch]

@snow jetty yes

[lavish jewel]

yep that looks ok

[snow jetty]

Oh yes, indeed, it works if the scalar product as the abstract "bilinear application" and the M lines as the basis. thanks. didn't think about that

[dusky epoch]

bilinear form with extra properties (symmetry and posdef)

[snow jetty]

Yes, in fact the dimension of bilinear applications for R^3 must be "9", but how can they be constructed?

[dusky epoch]

? what

je ne comprends pas ta question, désolée

[snow jetty]

I mean: according to the theory, the set of bilinear applications over a vector space V (Bil(V) aka "f:V->V, bilinear") is also a vector space. And its dimension is dim(Bil(V)) = (dim(V))²

[dusky epoch]

...yes? what does that have to do with your previous question?

[snow jetty]

Like suppose I have R³, so how can I define the basis?

[dusky epoch]

basis of what

[snow jetty]

of Bil(R³)

[dusky epoch]

well

you could define nine bilinear forms $p_{ij}$ ($i,j \in {1,2,3}$) by requiring that for all $i,j,k,l \in {1,2,3}$ you have $$p_{ij}(e_k, e_l) = \delta_{ik}\delta_{jl},$$ where $\delta$ is the kronecker delta

in essence, each form sends one of the 9 possible pairs of basis vectors to 1, and all the other pairs to 0

does this make sense?

[stoic python]

Ann

[dusky epoch]

alternatively, you could also write this definition as $p_{ij}(x,y) = x_iy_j$

[stoic python]

Ann

[snow jetty]

yes probably, there will be like the "dual basis" of {e1, e2, e3}, and then I make the product of the "covectors", for every possible combination, and it will be like the basis of (Bil(V))

[dusky epoch]

if you wish...

[lavish jewel]

ngl, idk why we ended up here when we started at Gramian

[snow jetty]

😅 because for every bilinear we have its own Gramian...

The scalar product is just a particular case like M^T * M

[lavish jewel]

bilinear form or bilinear map?

i guess bilinear form

[snow jetty]

Yes, indeed, I didn't even consider the distinction before 😮

The ones that are VxV -> K

[snow jetty]

they said that should be the basis. trying to understand the ⦻ sign...

[lavish jewel]

looks like outer product/kronecker product

but this Bil(v) seems to be a map, not a form

is it now V x V -> V?

(i might be wrong)

[snow jetty]

They still mean a form VxV. But I think I understand, omg it's so easy in fact 🤯

The basis will be just {x1y1, x1,y2, x1y3, x2y1, x2y2, x2y3, x3y1, x3y2, x3y3}

[lavish jewel]

tbh it looks exacty like what ann wrote earlier.

here #linear-algebra message

[snow jetty]

Yes but... I thought like it should be a composition of matrices or something like this... but in fact it is impossible to define a matrix for a bilinear form

Because it takes** two damn vectors** and it is a linear map only if one of them is fixed if I understand correctly

this stuff is way too abstract

[faint lintel]

Can someone give me a good resource on determining when an operator is normal vs self adjoint vs unitary

So it's normal if TT* = T*T

If we have an orthonormal basis of V consisting of eigenvectors of T then T = T* (self adjoint)

And then if it's normal and the eigenvalues for all those eigenvectors are have absolute value 1 then it's unitary?

[zealous junco]

pretty sure

if theres an ON basis of V with eigenvectors of T, then T can still be normal

[faint lintel]

Is the only self adjoint unitary operator the identity matrix?

Cause we'd need such an operator to be TT* = T*T = I

But also T = T*

So then T^2 = T*^2 = I

Oh so then either the identity or negative identity matrix are the only unitary self adjoint operators

I think?

[zealous junco]

no i dont think so

take for example householder transforms

visually u see its a reflection so its its gonna be diagonalizable, and unitary since it doesnt stretch anything

and its easy to verify

[faint lintel]

Householder Transform?

[zealous junco]

Basically a reflection on any hyperplane

[faint lintel]

I see

[faint dune]

proofing positive definit of the L2 scalar

I'm not sure if I see it right, but the p(x)^2 is always positive, so the integral is only 0 if p(x)^2 is 0.

But now I have a slightly different scalar product definition

Can't I argue with the same way, and just say p(x)^2 is positive and has to be 0 therefore ?

Because the proof in the script is a little bit weird and unnecessarily complicated, not sure where my problem is.

[wintry steppe]

well, you'll get p(x_i) = 0 for each i. what are the assumptions on p and on the x_i's?

[rain drum]

hi, can somebody tell me, do u know some kind of book, that has exercises for jordans canonical form with results? i would appreciate it thank u ❤️

[wintry steppe]

friedberg insel spence

[rain drum]

it seems good, thank u

[twilit minnow]

can someone explain row echelon form to me? apparently this is in REF but i read that there has to be a row of all zeros below entries

[odd kite]

you read wrong

[faint dune]

Its about Lagrange Polynoms, interpolation. p is element of Polynom Space of degree n

[faint dune]

This was the proof in the script, which even looks wrong to me at line 3.10

the coefficient aj is missing before the lj and the result would be two sums, one over i and one over j at the end of the aj^2.

The proof is based on the idea that the lagrange polynoms lj are a basis, so linear independant.

[minor bluff]

is the row space of a matrix a span of all its row vectors?

[sand apex]

Is strang or axler better in your opinions?

question really boils down to why isnt axler listed in resources when it seems to be a popular choice in schools

[odd kite]

@sand apex depends on what your goals are

@minor bluff yes

[minor bluff]

ah okay ty

[sand apex]

what is the difference in goals here?

would axler be more theoretical?

(abstract)

[odd kite]

if you're looking more in the applied/computational direction Strang may be better, and may be better overall, put if you're looking for pure math maybe look at Axler

yeah

Axler has a bent against determinants which isn't great so you'll have to learn about those elsewhere

[sand apex]

didnt axler come up with his method of determinants?

[odd kite]

idk, his book is famous for not using determinants

[sand apex]

lol

why would he do that

[odd kite]

but otherwise is has a nice clean presentation that's good from a pure math perspective

[sand apex]

yeah i appreciate the rigorous foundations of spans, directsums and such as presented in his texts - it builds off from these concepts

[dawn fractal]

this is the function defined

i don't get the 1st sentence

in particular, the 2nd equality

[lavish jewel]

in this definition, you have f(x,y)

[lavish jewel]

here, f(y,x), so the x is conjugated instead of the y

after that, just use your regular complex conjugate properties

it distributes over products and sums

and a_i is real, so it is not affected

[dawn fractal]

by "complex conjugate of a complex number" (x+yi), do they mean x-yi

?

[lavish jewel]

yes

[dawn fractal]

for example

cconj of f(xi,yi)

cconj of a_1(xi)(-yi)

[lavish jewel]

what

[dawn fractal]

hmmm

[lavish jewel]

i think you mixed up the subindex i with the imaginary unit

[dawn fractal]

there

[lavish jewel]

still

[dawn fractal]

wdym

[lavish jewel]

$\sum_n a_n x_n y^*_n$

[stoic python]

Edd

[lavish jewel]

the star there is complex conjugate

[dawn fractal]

i literally meant $xi$

[stoic python]

!superficialsicko

[dawn fractal]

and $yi$

[stoic python]

!superficialsicko

[lavish jewel]

like multiplied by i?

[dawn fractal]

yes

[lavish jewel]

well, are x and y real or complex?

i think in this scenario they are in general complex

[dawn fractal]

i mean

in the form of a+bi

[lavish jewel]

so $\text{conj}(i\cdot x) = -i \cdot \overline{x}$

[dawn fractal]

x and y are both in the form of b

x* ?

[lavish jewel]

x_i is of the form x + i b

[dawn fractal]

i literally meant 0+xi and 0+yi

[lavish jewel]

ok, sure, but that's not what you have in the text above. just so you know.

[stoic python]

Edd

[dawn fractal]

in the scenario where n = 1

and we're simply considering CxC

[lavish jewel]

this is always valid btw, even if x is real

sure

sure, if n = 1 you have $a_1 x_1 \overline{y_1}$

[stoic python]

Edd

[dawn fractal]

so is f(0+xi,0+yi)=a_1(0+xi)(0-yi)?

[lavish jewel]

sure

[dawn fractal]

ok

[lavish jewel]

just to reiterate though, that is not the form of the x and y up there

[dawn fractal]

i was asking about the specific case

[lavish jewel]

ok

[dawn fractal]

so real conjugate of a+bi is -a+bi?

[lavish jewel]

real conjugate

[dawn fractal]

?

[lavish jewel]

i have never heard of that

maybe it's defined in some scenario, but i have never seen that

[dawn fractal]

i don't get the point of adding "complex" to conjugate of a complex number

maybe

[dawn fractal]

as in "complex" conjugate

[lavish jewel]

maybe to denote that it only makes sense for complex numbers

[dawn fractal]

hm

[dawn fractal]

is this by definition?

[lavish jewel]

you can prove it

it isn't a definition

it's a consequence of the definition and properties of the complex conjugate

[dawn fractal]

let's say all x_i's and y_i's are respectively of forms a+bi and c+di

then f((x_1,...,x_n),(y_1,...,y_n)) = a_1x_1(c'-d'i) + ... + a_nx_n(c^(n)-d^(n)i)

[lavish jewel]

i don't understand your notation there

[dawn fractal]

c',c'',c''',c^(4),...,c^(n) are the real parts of the y_i's

yes a bit cringey

[lavish jewel]

[wintry steppe]

[lavish jewel]

sure, looks ok i guess

but wth lol

[dawn fractal]

wait

basically the first sentence follows from f being an inner product?

since f(x,y) = conjugate of f(y,x)

[lavish jewel]

isn't it backwards?

[dawn fractal]

wdym

[lavish jewel]

you're trying to show that this is an inner product

[dawn fractal]

ah

[lavish jewel]

so they're proving this

[dawn fractal]

i still don't follow from this

i'm not much acquainted with complex numbers

[lavish jewel]

just do it yourself

i * (a + bi)

go on

expand that and take its conjugate

[dawn fractal]

ai-b

conj is -ai-b

[lavish jewel]

now factor out a -i

[dawn fractal]

-i (a-bi)

[lavish jewel]

and if x = a + bi, what is a - bi?

[dawn fractal]

conj x

[lavish jewel]

there you go

[dawn fractal]

but i don't see any i in the summation

[lavish jewel]

me neither. this is for the expression you made up

[dawn fractal]

the imaginary number i mean

[lavish jewel]

this is the general form of the example you made up, for whatever reason

[dawn fractal]

so how do i show the first equality

[lavish jewel]

i didn't know if you meant to use $x, y \in \mathbb{C}^n$ as in the original problem, or if $x, y \in \mathbb{R}^n$

[stoic python]

Edd

[lavish jewel]

you show the first equality the same way we just did

first, show that the complex conjugate distributes over sums

then, show it distributes over products

and that's it

[dawn fractal]

so to do that i need to add the terms of the summation first?

to get a complex number

[lavish jewel]

use an ersatz sum that is simpler

let w_n = a_n x_n y^*_n be a generic complex number a_n + i b_n

work with the sum of w_n's first to show what happens if you complex conj the whole sum

and then show what happens if you conjugate one individual w_n

the way they present this just in passing kinda implies these are things one should've learned before. i guess now is as good a moment as any for you to do so 😛

[dawn fractal]

that's just my prof assuming xD

i haven't learned complex numbers in-depth before college, only a bit

sum of w_n's vs only one w_n?

[lavish jewel]

yeah

if you show the conjugate goes into each summand individually, you don't need to bother with the sum anymore

so show it goes into the sum, and then apply it to one element to see what happens when you conjugate a product

past this point i would suggest you go to #real-complex-analysis

[dawn fractal]

i see

[lavish jewel]

what type of "modulus" are we talking

you could use an inequality

if you absolutely must split it, all you can do is provide a bound

since the result depends on how x is represented as a linear comb. of the matrix's row space/domain and its orthogonal complement

smth like |Ax | <= |A|_(induced 2 norm) |x|

otherwise, decompose x in the basis of the row space and factorize A somehow

but if you have A and x, the easiest is to multiply first, take norm after

[wintry steppe]

Let T : R^3 -> R^4 be a linear map such that T(1,0,0) = (1,0,1,0), T(0,1,0) = (0,1,0,1), and T(0,0,1) = (1,1,1,1). Find a basis for null T and range T

I think it's clear that a basis for the range of T is (1,0,1,0),(0,1,0,1)

And a basis for null T would be (1,1,-1)

(I think)

How would I make sure these answers are correct, if they are?

[native rampart]

Use rank nullity

[wintry steppe]

I don't know what that is

[native rampart]

rank(T)+null(T)=3 here

[wintry steppe]

Yeah, so that's one sanity check

[dusky epoch]

well the range of T is spanned by the values of T at (1,0,0), (0,1,0) and (0,0,1)

[wintry steppe]

Which is good

[dusky epoch]

for the nullspace you can just write out the explicit formula for T and find the nullspace as the solution set of T(x)=0 [treating this as a linear system]

[wintry steppe]

That's what I did I think

So to write the matrix for Tv, where v in R^3, I had this

$\begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 1 \ 1 & 0 & 1 \ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} a \ b \ c \end{bmatrix} = \begin{bmatrix} a + c \ b + c \ a + c \ b + c \end{bmatrix}$

[stoic python]

n/c

[wintry steppe]

Because 1,0,0, 0,1,0 and 0,0,1 form a basis for R^3

I just looked where they are sent by Tv and put them as the columns of the matrix

And a,b,c is an arbitrary vector in R^3

So then for the null space, we want a + c = b + c = 0, which gave me (1,1,-1)

For the basis of the null space

[native rampart]

Yes

[wintry steppe]

But how do I make sure it's correct?

[native rampart]

So you need a+c=0 and b+c=0

[dusky epoch]

doublecheck your work? idk

[native rampart]

Implies a has to be -c and b has to be -c

[dusky epoch]

triple check

[native rampart]

That's it

[dusky epoch]

quadruple check

N-fold check for N as high as needed to satisfy you

[wintry steppe]

How do I double check it"

[dusky epoch]

you doublecheck a piece of your own work by going through it again and reading every step out loud making sure you know exactly why each step was taken and why it is all correct

this should be done only a certain time after writing it, so as not to lull yourself into a false sense of security like "oh i know this is ok"

[lavish jewel]

if you got the spaces right, you should be able to check they satisfy all the definitions.you can also pseudo invert stuff in the image back onto the row space

[wintry steppe]

So basically

I need to check that ${ a_1(1,0,1,0) + a_2(0,1,0,1) : a_i \in \mathbb{R} } \subseteq \mathrm{range~}T$ and $\mathrm{range~}T \subseteq { a_1(1,0,1,0) + a_2(0,1,0,1) : a_i \in \mathbb{R} }$

[stoic python]

n/c

[wintry steppe]

And similarly for null space?

And then also that they're linearly independent, but that's obvious

?

[dusky epoch]

i mean yeah you could do this

neither check is really all that hard

it's more bookkeeping than anything

[wintry steppe]

I am having a dumb moment

How do I show the first subset?

[dusky epoch]

every vector of the form (a1, a2, a1, a2) is the value of T at some point

[wintry steppe]

I need to let x = (a + c, b + c, a + c, b + c) right?

[dusky epoch]

?

[wintry steppe]

Tv is that

For v = (a,b,c)

[dusky epoch]

you are showing that every vector of the form a1(1,0,1,0) + a2(0,1,0,1) is the value of T at some point

more concretely in this case it happens to be equal to T(a1, a2, 0)

thats all you need

[wintry steppe]

What?