linear-algebra

[wintry steppe]

I just had too many variables and nothing to do with them

[lavish jewel]

the thing is that in general you can even give counterexamples

[wintry steppe]

Like what?

[lavish jewel]

so these linear maps are in the dual space of V, yeah?

since they are f: V -> F

(do tell me if i'm misunderstanding)

so one could make an example in, say, R3

hmm

oh

is this finite dim?

(i was wrong about the counter examples)

if it's finite dim, the transformation can be written in some basis as a 1 x R^n matrix

since it's rank 1, the orthogonal complement of the null space is also dim 1

so it lies on a line

[wintry steppe]

Yeah it's finite dim

[lavish jewel]

i.e. the row space is scaled versions of the one basis vector that spans it

[wintry steppe]

I can't use those kind of tools

Needs to be more primitive

[lavish jewel]

from which step on?

[verbal pivot]

this is an exercise from axler's right? or my memory is failing lol

[wintry steppe]

@verbal pivot It is

Chapter 3

@lavish jewel I can't use matrices yet

[broken sun]

Should I ask my question again?

[lavish jewel]

ok, but you can use the dual space

[verbal pivot]

A quick suggestion could be to look at the previous excercises, are they useful? (I kinda remember them being a lot in axler)

[wintry steppe]

I know what the dual space is, but it has not been introduced formally

[lavish jewel]

could you wait until this discussion dies out a bit? doesn't seem like anyone here atm knows how to address your question

[wintry steppe]

I don't think so @verbal pivot

[broken sun]

You are answering to me?

[lavish jewel]

yep, i was

[verbal pivot]

kay, assuming finite dim, what can you say about the dimension of the nullspace of f or g?

[wintry steppe]

They're the same dimension

[verbal pivot]

(I don't remember well if it's finite dim tbh)

but

what happens if you apply the fundamental theorem of linear maps?

[wintry steppe]

We know that the dim of the images of f and g are the same

[lavish jewel]

wouldn't you get that the rank of f and g is 1?

and they have the same ortho comp

[verbal pivot]

(they can be 0 too, i think)

[lavish jewel]

right

<= 1

[wintry steppe]

The case where they are zero is irrelevant since any constant c would work when both maps are the zero map

[lavish jewel]

right

and in the other case, they are both the rank 1 orthogonal complement to a dim (N -1) space that is a subspace of a dim N space

[wintry steppe]

I can't use that Edd

[lavish jewel]

after choosing a basis for V, they are in the direction of a normal vector to a dim (N-1) hyperplane so they're parallel?

not sure what you can and can't use

[verbal pivot]

still, what edd says can be useful, think about the dim of range f, can it be any number?

[wintry steppe]

No, they're both 1

dim range f and dim range g

[verbal pivot]

so, for both f and g, their range are spanned by only 1 vector right?

[wintry steppe]

That's true

[verbal pivot]

what can you say about those vectors? (keep in mind that range f is a subspace of F)

[lavish jewel]

i still think this was enough

it's a dim 1 subspace

what can you say about its possible bases?

just replace rank 1 with dim 1

[verbal pivot]

(also, I don't really remember this exercise being a finite dimensional one, maybe you should check that out after )

[wintry steppe]

I know that the range f = range g = F

But I still don't know how to use that

To get our constant

[lavish jewel]

could you not say that since dim = 1 in the nontrivial case, then f and g are linearly dependent?

[wintry steppe]

No?

That's like the whole proof

We're trying to prove that this is the case

[sage ibex]

I mean usual proof that existence of complementary subspaces → choice

[viral flint]

👀

[sage ibex]

It's like you construct an infinite family of free vector spaces involving your set, one over each F_p

[viral flint]

🤔

[sage ibex]

Idr exactly let me try and find

http://matwbn.icm.edu.pl/ksiazki/fm/fm54/fm5419.pdf this is where I read it

actually idk how "usual" this is

[sage ibex]

Theorem 5, left side on page 6/7

ye it just looks so ugly that I dont wanna read it again

but like you prove these statements FSn which says that from any family of non empty sets you can find multi-choice function that for each set in the family gives you a subset of finite cardinality coprime to n

also please tag when you reply lol I dont usually see this channel

[wintry steppe]

guys did I do this right?

[lavish jewel]

,w {{1,0,2},{0,-1,0},{2,0,3}}^-1

[stoic python]

Results provided by WolframAlpha

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

Requested by Edd#5182

[lavish jewel]

you can check with that. as you wrote, X = ACB^-1+ I

[wintry steppe]

thanks

[wispy horizon]

Hey, I am wondering if the adjoint operation is the same as an involution. I dont quite understand what the professor is asking here since I have never used the word involution here.

[wintry sphinx]

I think involution means "inverse" and adjoint is either the adjugate matrix or an adjoint operator

[wispy horizon]

how would you show this

[wintry sphinx]

I'm just giving definitions

the adjoint operator is equal to the inverse operator iff the operator is unitary

[wispy horizon]

yeah so it is not an equivalence?

since it is required it has to be unitary

[lavish jewel]

it's not the same, as saccharine noted

[wispy horizon]

yeah ok, since most of my class mates voted it to be the equivalence.

was just careful here

[wintry steppe]

Does this question make sense?

[quartz compass]

sure, but without info about U, all you can really do is give a kind of generic description

like maybe say n is a normal vector to the plane U to write the formula

[wintry steppe]

what is the difference between projection and reflection ?

[quartz compass]

projection is like a shadow on the ground

a reflection is like in the mirror, it's not in the plane of the mirror but goes through an equal distance to the other side

[wintry steppe]

i meant as in matrix terminology

[quartz compass]

lol well it's something you should try to work out and derive for yourself

I will say you can create a reflection from a projection

draw some pictures to help work it out

[wintry steppe]

Not sure what is meant in this context

[wintry steppe]

A problem asked: "what is the dimension of $\mathbb{C}^n$", I answered n as I thought there are only n basis vectors, but the answer was that there are 2n, they were $(1 +0i,, 0 + 0i,, \dots,, 0 + 0i),, (0 + 1i,, 0+0i,, \dots,, 0+0i),, \dots$ but aren't the first two vectors multiple of each other..? can't I multiply the first vector by i to get the second one?

[stoic python]

Orangus

[weak needle]

Well, when you talk about a vector space, you have to specify the base field. Over C it is n, for R it is 2n

[wintry steppe]

yea, it's a vector space of n-tuples over C, no?

wait I dont get it

[weak needle]

It can be given the structure of a vector space over R as well (in the obvious way) I assume that was what they were implicitly using

[wintry steppe]

Oh wait, so the
so can be like tuples of R over C and tuples of C over R as vector spaces?

I see, I kinda didn't expect that

[weak needle]

I don’t exactly get what you’re saying but i think you probably have the right idea. C^n is a vector space over R because we can multiply vectors with real numbers component wise and it satisfies all the properties of a vector space

[wintry steppe]

I mean that I didn't even think of possibility of existence of such spaces, always assumed R^n is over R and C^n is over C for some reason

[stoic python]

squirtlespoof

[north anvil]

I wish I knew ;-;

I'm only on homogeneous systems of linear equations 😐

Speaking of which, if I'm given a homogeneous system of linear equations, and a non-trivial solution exists, this means that there are infinitely many solutions, right?

[wintry steppe]

I am never doing that

[restive raft]

I mean sure can't you just scale the non trivial solution

[north anvil]

Isn't the trivial solution just all variables are 0s?

So scaling them wouldn't change it?

Oh non-trivial

[restive raft]

yeh

the zero vector would be like that

[north anvil]

Oh yeah, you probably can

cuz ax + by = 0 for example

if each of the nontrivial solutions are increased by a factor of some real number k

then just dividing by k will yield the original result, proving that non trivial solutions when multiplied by k work

Thanks

:D

[wintry sphinx]

seems right

[north anvil]

Appreciate that :)

[viral flint]

Thanks for this

[viral flint]

Same.
You alone have been making sure I see this channel.

[sage ibex]

[warm kite]

Hey f

Does anyone know how this works

q is a generalized eigenvector if that helps

[native rampart]

Are you trying to solve some recurrence relation?

[native rampart]

No

Take X=span{e_1,e_2} and Y=span{e_3}, Z=span{e_1+e_3,e_2}

[lavish jewel]

how about knowing that its eigenvectors from an orthonormal basis for V?

P being unitary would mean P*P = PP *

over the reals, this would be P^T P = P P^T (e.g. when representing the projection as a matrix)

it does not mean P = P*

also it is P that is diagonalized, not V

symmetric and hermitian mats have that property, but they are not the only ones

those are actually special cases of normal matrices

no?

i don't think so, anyway

normal mats are unitarily diagonalizable

and P*P = PP * is defined over the reals the same way it is over C

i mean matrices. since you're in fin dim,v linear transformations can be written as a matrix using some basis

that looks whonky

self adjoint is a flavor of symmetry

keep in mind i might be wrong, you should always assume i am

i'm pretty sure that is not needed

anyway the spectral theorem is given originally wrt hermitian matrices, not normal ones

but normal matrices are also unitarily diagonalizable

if you have an orthonormal basis, subspaces that don't share elements in their bases are ortho

hmm i think ker P = W^(perp) has to be justified more tho

i couldn't say for sure :x

[grizzled flicker]

How do I solve this?

[broken sun]

Hello. Can an eigenvector $v$ of an operator $T$ with eigenvalue $\lambda$ be in the image of $T-\lambda I$?

[stoic python]

MathPhysics

[dusky epoch]

sure

take lambda = 0, v = [1; 0; 0] and T: R^3 -> R^3 defined by T([x;y;z]) = [y;z;0]

[wintry steppe]

Suppose that $f \in L(V,F)$. Suppose $u \in V$ is not in $\ker f$. Prove that $V = \ker f \oplus { au : a \in F}$

[stoic python]

n/c

[wintry steppe]

So there's two parts two this, and the first is showing that ker f + that set (we can call it W) is equal to V

However I am having trouble showing that V is a subset of W

[dusky epoch]

??

[wintry steppe]

I need to find v in ker f and a in F such that v + au = x, for some x in V

[dusky epoch]

isn't it obvious that W ⊆ V

oh, you mean the sum, my bad

[wintry steppe]

I typo'd

V subset W

W subset V is not hard, since all elements of W are elements of V

So it's a subset

[dusky epoch]

let $c = f(u)$, and consider that $c \neq 0$ and that $f(c^{-1}u) = 1$.

[stoic python]

Ann

[dusky epoch]

(c is a scalar)

consider also how you might construct a vector in ker(f) in terms of your x using this info

take v + au = x and turn it around: v = x - au
can you find a such that x - au ∈ ker(f)?

[wintry steppe]

Wait why are you doing the inverse thing?

[dusky epoch]

i'm too lazy to format the fraction properly

i mean, fine

$f\paren{\frac1c u} = 1$

[stoic python]

Ann

[dusky epoch]

if that makes you happier

did this address your objection to what i said?

@wintry steppe

[wintry steppe]

No?

My question is what made you do that

[dusky epoch]

dunno

it's not strictly necessary

it's just where my thoughts wandered

consider also how you might construct a vector in ker(f) in terms of your x using this info
take v + au = x and turn it around: v = x - au
can you find a such that x - au ∈ ker(f)?

this bit is more important

[wintry steppe]

I don't know

[dusky epoch]

do you know what it means for a vector to be in ker(f)?

[wintry steppe]

Yes, f(x - au) = 0

[dusky epoch]

correct

now do you remember what it means for f to be in L(V,F)?

[wintry steppe]

I forgot to use that part

[dusky epoch]

you're doing linear algebra and you made no use of the fact that f is a linear map...

[wintry steppe]

Right, so

We have v = x - au

fv = f(x - au) = fx - fau

[dusky epoch]

keep going

can you find a such that f(x) - f(au) = 0?

don't overthink it.

it's very simple.

[wintry steppe]

fx = afu, so we have fx/fu = a

[dusky epoch]

perfect

so you see now? you have $x = \paren{x - \frac{f(x)}{f(u)}u} + \frac{f(x)}{f(u)}u$

[stoic python]

Ann

[dusky epoch]

and there you have it, your decomposition of x into the sum of a vector in ker(f) and a multiple of u

[wintry steppe]

Thanks, it makes sense

[dusky epoch]

thank god it does

do you want to ask me anything else while i'm still here?

[wintry steppe]

Yeah, this question

Suppose f and g are linear maps from V to F that have the same null space. Show that there exists a constant c in F such that f = c * g.

I've shown that there does exist a constant that depends on v for v in V, but not that there is a single constant

[dusky epoch]

take some arbitrary vector v such that g(v) = 1, and let c = f(v)
attempt to show that f - cg is the zero map

you may find it useful to have V = ker(f) \oplus span(v) ready to make use of

[wintry steppe]

But that's so weird, that we're looking for g(v) = 1

How could I ever think of that?

[dusky epoch]

if you are allergic to the number 1, any vector outside of f and g's shared kernel will do

[wintry steppe]

Oh okay

[dusky epoch]

obligatory edge case notice that i'm assuming their kernel isnt the whole space, which would make both f and g the zero map and the whole problem would be trivial

however, once you have a vector not in the kernel, you can always scale it so that the value of g at it is 1

makes things marginally more convenient

[broken sun]

Why does the last identity hold?

[wintry steppe]

Thank you Ann

[dusky epoch]

@broken sun any $x \in W$ can be written as $(T - \lambda_kI)(y)$ for some $y \in V$, which makes the left-hand side into $p(T)(y)$, and $p$ being the minimal polynomial of $T$ means $p(T)$ is the zero map

[wintry steppe]

Why?

[dusky epoch]

better if you try it yourself than have me tell you, i guess.

[stoic python]

Ann

[wintry steppe]

But you said arbitrary vector in v such that gv = 1

How is that arbitrary?

We need gv = c_1 and fv = c_2

[dusky epoch]

arbitrary as in there are many such vectors, and i do not care which one

unless, of course, you reject outright the mere existence of vectors v satisfying g(v)=1.

or you reject my decision to select one of those instead of any other vector outside of the kernel.

[wintry steppe]

No, but why don't we need "arbitrary" as in c_1 for c_1 in F

Rather than c_1 = 1?

[dusky epoch]

not every arbitrary choice needs to be abstained from as you're trying to suggest, n/c.

[broken sun]

Thanks.

[dusky epoch]

there is nothing in the world, no law of math, stopping me from basing further reasoning on a vector v such that g(v) = 1.
there is no 'need' to introduce two variables when one does the job just fine.

[wintry steppe]

But how do we know it will hold true even if gv is not 1?

(Or some other very concrete number)

[dusky epoch]

i mean, i have not presented my actual line of reasoning yet, but i guess i must

fix a vector v satisfying g(v) = 1, and let c := f(v). we will attempt to show that f - cg is the zero map.
since f and g have the same kernel, (f - cg)(x) will be zero for every x in said kernel.
additionally, (f - cg)(v) = 0 by our definition of c.
since V = ker(f) \oplus span(v), every vector in V can be written as the sum of a multiple of v and a vector in ker(f).
by linearity, we get that (f-cg) will vanish on every vector in V. thus f - cg = 0, thus f = cg as desired.

now, if you really ARE allergic to the number 1, as you seem to be,

you could just pick any vector from outside the kernel as your v

and instead let c = f(v)/g(v)

[wintry steppe]

Okay that's what I did, but this is a false proof

What we're doing here is that, we're first picking v, and then we get c from that, right?

But c needs to be constant

It's not clear that c is constant here

[dusky epoch]

we're picking one particular vector v

and basing our c on that, true

at no point did i claim that c would be the same no matter which v we picked

i just picked one and rolled with it

[wintry steppe]

But we need it to be the same, right?

[dusky epoch]

c is just a number

you're overthinking it

[wintry steppe]

How am I overthinking it?

c has to be the same for every v

[dusky epoch]

c is just a number

we picked ONE v and constructed c from it

the goal is now to prove that this c is the c we are looking for

(i.e. that f - cg = 0)

[wintry steppe]

Oh I see. So let x in V, and then we want to show that fx - cgx = 0, so fx - fv/gv * gx = 0?

[dusky epoch]

...yes, if you insist.

[wintry steppe]

But now what?

[dusky epoch]

(f - cg)(v) = 0

do you agree or disagree with this

[wintry steppe]

I agree with it

But now what about our arbitrary vector x?

[dusky epoch]

our arbitrary vector x can be written as the sum of a multiple of v and a vector in ker(f)

since V = ker(f) \oplus span(v) as you proved in the previous exercise

do you need me to write this out explicitly, or are you able to intuit that if a linear map sends two vectors to zero it also sends any linear combination of them to zero?

[wintry steppe]

I see

Thanks

[wintry steppe]

I think, it's free.

How can one just describe a specific vector whose tail is in a specific point?

Let's say we have a vector $$A = (1,2)$$

[stoic python]

discordi.a

[wintry steppe]

But it's an arbitrary vector that could be found anywhere in the $$R^2$$.

[stoic python]

discordi.a

[lavish jewel]

vectors don't inherently have an associated "location"

[wintry steppe]

Yeah, because of the definition, right? I recognized that.

[lavish jewel]

you would need to parameterize your setting in some way that includes this info

e.g. using two position vectors, instead of just their difference vector

[wintry steppe]

But still the difference vector thereof is anyway an arbitrary vector that has whatever the slope is.

let's we have the difference vector and it's $$D = (4,5)$$

[stoic python]

discordi.a

[wintry steppe]

still can be found anywhere in the R2.

[lavish jewel]

that's why i said you use the two position vectors instead

[wintry steppe]

I mean because of its definition.

[lavish jewel]

then the vector is related to the segment joining the two points the vectors point to

i.e. you cannot do this with just a single vector. to track the position, you need many vectors

and some relationship between them

[wintry steppe]

precisely position vectors?

[lavish jewel]

position or displacement

think of it this way

you have 2 points and a vector that points from one of them to the other

you need at least 2 of these 3 things to find the third one, if it is missing

if you only have one of these 3 things, you cannot find the other two

like having a + b = c

you need two of the three to find the third

this works, e.g., if a is a point, b is a displacement, and c is another point

or if you do a = c - b, then b is the head of a vector and c is its tail

[wintry steppe]

Okay. got what you've said.

Why I asked this question, is that because of this:

"Q sub 0" defines a set of vectors., but I'd say defines a set of "position vectors whose tip on the line".

[lavish jewel]

right, so they're doing a convex combination of 2 vectors

yeah

[wintry steppe]

But not the line itself?

Am I right.

[lavish jewel]

kinda the same thing

it describes a line, and it can be interpreted in two ways

1.) you have a "tail" point at x0 and a direction vector (x1-x0)

so x0 + t(x1-x0) is a ray with source at x0

then the bounds on t specify a portion of the ray

2.) alternatively, let x1 and x0 be position vectors. then their convex combination x0 + t(x1-x0) yields a position vector along the segment joining the points x1 and x0

1.) uses vectors with tail at x0, while 2.) uses vectors with tail at 0, which is more or less what is conventionally done

guess it also depends on whether you wanna talk about x0 and x1 as points or vectors

[wintry steppe]

Actually it's the taste, right?

How you wanna interpret it.

[tough pebble]

hello

guys i have an exam

can someone help me ?

[wicked palm]

I also have an exam! Twinning!

[tough pebble]

omg

[wicked palm]

anyway no academic misconduct, blocked, reported and the police have been called

[strange delta]

kek

is it me or is this solution wrong

for A' isn't the 3rd column supposed to be

(0,-3,3)

phi(x-x^2) is for the 3rd column

[sudden narwhal]

its correct

[strange delta]

why is it (0,0-3)

[sudden narwhal]

x-x^2 is the vector in the basis B

so a vector of R_2[X] is written as
P = a(1) + b(x) + c(x-x^2)

[strange delta]

ah

i think i get it now

since it's with respct to the basis

i can write it as -3(x-x^2)

[sudden narwhal]

yes

[wispy horizon]

I have no idea how to line break on discord so I am terribly sorry for whoever tries to read this

[lavish jewel]

don't leave spaces after nor before the $$

try \ for line breaks

double backslash

[wispy horizon]

i tried. k ill try again

[lavish jewel]

this will look like poop anyway, don't put your text inside the $$

just put $S = ---$, and similarly for the limit

[stoic python]

Edd

[wispy horizon]

lmao ok

[lavish jewel]

\text{ or }

[wispy horizon]

Can someone explain this in a simpler way? (the (b) part):

$S = {\lambda \in C : |\lambda| < 1 \text{ or } \lambda = 1}$
$\$
Theorem 5.13:
Let A be a square matrix with complex entries. Then $\lim_{m \to \infty} A^m$ exists if and only if both of the following conditions hold.$\$
(a) Every eigenvalue is contained in S $\$
(b) If 1 is an eigenvalue of A, then the dimension of the eigenspace corresponding to 1 equals the multiplicity of 1 as an eigenvalue of A

[stoic python]

Eso

[cobalt tartan]

Just want to double check something: Is the reason why closure under addition for a subspace E is sufficient to show that the zero vector is in E because for any x in E, x + (-x) will also be in E, and x + (-x) = 0?

[dusky epoch]

closure under addition alone won't cut it

so it's not sufficient

the set { [a;0] | a ≥ 1 } is closed under addition yet does not contain 0

@cobalt tartan

[cobalt tartan]

hrm

Ah, but if E contains the additive inverses as well as the original elements, then what I said holds?

[dusky epoch]

if E is closed under multiplication by -1 and addition, and is nonempty,

then yes.

[cobalt tartan]

Yea, bc specifically this E is vectors over binary numbers (I don' tknow if I'm phrasing this right)

And uh 1 + 1 = 0 so the additive inverse exists in E

[nocturne jewel]

So $\mathbb{Z}_2$ as a vector space

[stoic python]

Mosh

[strange delta]

why did they express the image as (a+3c) and (b+3d) isn't it supposed to be term by term like a,b,c,d?

origonal question

[lavish jewel]

they're showing that the transformation cannot give you any 2x2 matrix, only a subset of them

which is precisely what you were asked for

[hidden mantle]

Could somebody help me understand the solution for this problem?

https://cdn.discordapp.com/attachments/359052604149465088/851817675570413609/linalg5.png

Note that A ∈ R{m×r} and Y ∈ {r×m} for some r. Given that rank(AY ) = rank(I) =m then r ≥ m.
Furthermore, m = rank(AY ) ≤ rank(A) and rank(A) ≤ m given that A is m by r. Therefore rank(A) = m and m ≤ r.

so I don't understand how Rank(AY) = Rank(I) = m implies that r>=m

[strange delta]

would it be incorrect if i wrote it in terms of a,b,c,d then

[lavish jewel]

it IS written in terms of a,b,c,d

they just took it further to find a basis

[hidden mantle]

(sorry I'll continue once you guys are done 🙂 )

[strange delta]

i mean like single terms a,b,c,d

[lavish jewel]

wdym?

[nocturne jewel]

are you expecting a linear combination with 4 vectors?

[strange delta]

^ yeah

[nocturne jewel]

You could do that sure, but then you'd have excess basis vectors

[lavish jewel]

duk, the rank of a matrix product is bounded by the rank of the matrices you multiplied. you cannot get a result that is higher rank than the components. if r < m, then A and Y have rank at most r, and you could not get Im as a result

for redd, a clearer way to try and see it is to make the matrices into long vectors, and rewrite T as a larger 4x4 matrix. then you'd see that you have linearly dependent columns in the transformation. idk if that helps you

[hidden mantle]

ah thanks!

[lavish jewel]

should be something like rank (AY) <= min(rank A, rank Y). you can wikipedia it

[hidden mantle]

yep we have a proof of it in our lecture notes - i'd forgotten about it

thanks!

[lavish jewel]

at any rate, for redd's problem, the point is to show that a basis for the image of the transformation has only 2 vectors, not 4

[strange delta]

just making sure the "give an example of a linear map" is a trap right?

[lavish jewel]

wdym?

[strange delta]

o

it's not

[lavish jewel]

what about a matrix with columns [0,0,0,1], [0,0,1,0], and then 2 with all 0s?

[strange delta]

hmm

can i just have the [0,0,0,0] vector as the map

[lavish jewel]

no

[strange delta]

hmm actually might need to think a bit more

[lavish jewel]

the one i gave you works for sure, but you should indeed think about it more

if the vector [0,0,0,0] is the map, then you don't have V -> V

[hidden mantle]

how do i do b

i've been trying it for like an hour now

and it's only 2 marks :/

i'm probably missing something very obvious?

[lavish jewel]

should be pretty easy

multiplication of a vector in R^n with a scalar in R is commutative

so (q_i^T x) q_i = q_i (q_i^T x)

then associate from the left

(q_i q_i^T) x

you can expand these operations as sums if you wanna be very thorough

[warm kite]

Yes! Sorry for the late reply but I still don’t understand

[hidden mantle]

Can't believe I missed that

Thanks!

Any idea about what the summation of Pi part is?

Oh nvm

It's as obviously lmao

[marble lance]

[wintry steppe]

what is a row operation?

[native rampart]

This makes no sense except in the context of reccurence relations

[nocturne jewel]

scale a row, swap 2 rows, add a multiple of 1 row to another

[wintry steppe]

how can that result in a matrix with different amount rows than the original?

[nocturne jewel]

it cant

[wintry steppe]

is there a more general sense where it can?

@wintry steppe could you please give a bit more details about what you are asking cause I can't figure it out

i think i figured it out now thanks

ok great

Can anyone help me with this: how do I show that a matrix with only 1's and -1 'sas its eigenvalue is always similar to its inverse?

hmm well they have same eigenvalues that's a start

wait why do they have equal eigenvalues

but that is not enough anyways

inverses of matrixes have multiplicative inverse eigenvalues

oh ok your right

but that's not going to be enough

I though I'd try to show their jordan forms are similar, but I still don't know how to do that

maybe you can try showing their generalized eigenspaces have same dimensions too

that will imply they have "same" jordan

ohh good idea i didnt think of that

[pale spindle]

For those of u who took linear algebra in high school, were you able to skip, test out, or get credit for the class in college?

[hidden mantle]

what's the best way to prove ii without it becoming 10 pages long?

[wintry steppe]

I would say if it can be written as a product of elementary matrices it can be transformed with elementary operations to identity matrix (multiplying with inverses of those matrices), which has rank n (and elementary operations do not change the rank)

For the other way its similar

[hidden mantle]

hmm thanks

[hidden mantle]

dim(range A)<dim(RN) => Not an onto transformation

i don't understand why

is there a theorem which states that the minimum number of linearly independent vectors spanning Rn is n?

[lavish jewel]

isn't that precisely the definition of the dimension of a vector space?

[wintry steppe]

exactly!

[hidden mantle]

oh oops

[wispy horizon]

can someone actually help me with this?

[twilit anvil]

i think it says the dimension of the nullspace of (A - 1I) is the multiplicity of 1 as a root of the characteristic polynomial of A.

im not 100% sure though, because i thought the statement i just gave is true for any eigenvalue of A

[copper heath]

uhhh...quick question, if i have 3 matrices that all describe individual rotations (roll, pitch, yaw)
and i want to combine them into a single matrix which gives you a general direction/orientation
do i multiply all of them together?
or do i add

[limber sierra]

composition of linear functions is matrix multiplication

so you multiply.

this is the whole reason matrices exist, btw

[wintry steppe]

For this problem would a11 be 8 and would b21 be 1?

[latent ledge]

yes

show $R^n=V\oplus V^{\perp}$ where admits a basis $v_1,v_2,\dots,v_p$ such that $L_Av_i=\pm v_i$ and $L_Av\neq v$ for all $v\in V$. $A\in SO(n)$

[stoic python]

亜城木 夢叶

[stray granite]

is it true that the span of any nonzero vector in R^2 is all of R^2

[latent ledge]

no unless they are unparallel

[stray granite]

yeah thought so

[stray granite]

thank you

[balmy finch]

I'm in highschool gr 11 so it's kidna hard to find someone to talk to about this. How does the dot product work? I know the actual formula, but not how it really works.

Can someone help? The internet isn't really helping lol

[sudden narwhal]

yeah what is wrong with the formula ?

[limber sierra]

if you know the formula, you know everything about it

its not some super deep convoluted construction

perhaps youre wondering why its useful? or why connects to "angle"/orthogonality?

but its hard to tell from your question

[vital swallow]

Question. I am trying to prove (or find a counterexample). I'm working with a vector space over a function field....
If U is a unitary matrix (whose entries are functions of x), and A is a vector whose entries are rational functions of x, and if the product UA is a vector whose entries are also rational functions of x, does U have to have rational entries?

[lapis river]

can someone here hope on call with me perhaps later tonight to do this fourier series problem with me and show how to graph on it on Matlab or some software. I'll DM you the problem if you are down.

[hollow ruin]

Does anyone know how to make a dual basis with this?, I am quite stuck

should I sub (w2, w1'' + w2'') into the integral?

[sudden narwhal]

I dont understand your question

[hollow ruin]

me neither

do you know the best way to find the dual basis of this, I tried getting the dual map but am confused since it's 2 dimensional

[wispy pewter]

Does anyone have a professor leonard style series for linear algebra they recommend? if it has a cost thats ok.

[vital swallow]

.. who's professor leonard?

[nocturne jewel]

Math youtuber who does flipped classroom-esque videos

[vital swallow]

I'm not sure what a textbook in that style would look like then

[nocturne jewel]

a textbook can't look like a video...

[vital swallow]

oh good point. I expected textbook, so that's where my brain went

[lapis river]

can someone explain why my answer isn't panning out for part c of this question?

I'll show my work

I'd extremely appreciate it if someone took a look

*verify that those equations are true

[vital swallow]

Your vector w_1 is not part of any orthonormal basis

[lapis river]

how so?

from textbook

[vital swallow]

Well, you set w_1 to be v_1, but v_1 isn't normal

[lapis river]

isn't it good enough that it v1 to v3 is linearly independent though?

[vital swallow]

that's orthogonal, not orthonormal

normal (in this setting) means length = 1

[lapis river]

how would i do orthonormal?

[vital swallow]

just divide the vectors by their length. v/|v| is a new vector in the same direction as v but the length is |v|/|v| = 1

[lapis river]

can you help me out and show me?

[vital swallow]

I'm serious. just divide your vectors by their length. So w_1 should be defined as v_1 / |v_1|. You had v_1 = [0, 1, 1] (I think), so |v_1| = \sqrt(0^2+1^2+1^2) = \sqrt(2). So define w_1 as [0, 1, 1]/sqrt(2), or [0, 1/\sqrt2, 1/\sqrt2].

[lapis river]

why wouldn't you divide w1 instead by it's magnitude?

[vital swallow]

It's been since I've needed to actually construct an orthonormal basis, but I'm pretty sure that you can just take the three vectors that you found w_1, w_2, and w_3, and divide each of them by their respective lengths. You constructed them to be orthogonal, and rescaling them shouldn't affect that

[lapis river]

yeah

ohhhh I got you

[vital swallow]

but it will change the pythagorean relation you were supposed to check at the very end

part of the reason that works out is because | av+bw |^2 = |a|^2 |v|^2 + |b|^2 |w|^2 if v and w are orthogonal. if {v,w} is also orthonormal, then |v|=1 and |w|=1, too.

[lapis river]

@vital swallow should me making the vectors normal also affect the decomposition of x as a linear combination

[vital swallow]

yeah. it'll change the coefficients by a factor related to the rescaling of your vectors

[lapis river]

sorry to bother again but could you @vital swallow , or maybe someone else, take a look at my work and tell me where it went wrong?

In the Pythagorean identity

[vital swallow]

your |w2| = sqrt(4^2+1^2+1^2) is incorrect

that's the first error I see @lapis river

[lapis river]

ahhhhh

thank you

[hidden mantle]

Given m by n matrices A and B that have the same column space C ⊂ Rm, is it necessarily the case that A + B also has column space C?

[hollow finch]

no

[native rampart]

Take B=-A

A+B=0

[hidden mantle]

Ok yeah that's smart

Thanks

[lapis river]

how much error is expected if using the gramm-schmidt method?

[lavish jewel]

wdym, how much error?

[lapis river]

oh actually nevermind

im trying to prove a pythagorean identity, but it comes 1 under the actual length of the vector?

can i get a heads up of what im doing wrong

my orthonormal basis has all lengths equal to one i checked

sorry for illegibility xd

the identity

[dusky epoch]

you don't because that's not true as stated

S = { (4,3), (8,6) }

S contains only two points but (S^perp)^perp = {(x,y) in R^2 | y = 3/4 x}

man, that wording

yes

[lapis river]

are there any fourier expansion calculator tools or software out there?

[dusky epoch]

no

to be in (S^perp)^perp a point needs to perpendicular to everything in S^perp

[jagged granite]

I want to revise linear algebra by solving some problems. Is there a good pdf of problems or a recommended list of problems from hoffman kunze ?

from some course webpage maybe

[marble lance]

I think I understand your argument. If the subspace has dimension 2, then it has a nonzero vector x. So ||x|| is nonzero, and you can choose c large enough so that ||cx|| = |c|||x|| is larger than any given integer.

No, I think this is the best way

[lavish jewel]

you can put what both of you said together to simply say "it's not closed under scalar multiplication"

[native rampart]

How would you define degree

[half ice]

T² often means "T composed with itself"

Obv T needs to have the same input and output space

And we can see that in matrix multiplication, A² is A×A

[lapis river]

if u is orthogonal to projx then u dot product projx = 0 right?

[spiral star]

yea thats the definition

[lavish jewel]

yep

[scenic fulcrum]

(p-1)(p+1)(p-3)(p+3)

1920 = 2^7 * 3 * 5

You can check that one of the factors is divisible by 3 and 5 reasoning modulo 3 and 5
(p=/=0 mod 3 and p=/=0 mod 5)

[lavish jewel]

try the questions channels or elementary number theory, this isn't linalg

[wide hemlock]

hello

[spiral star]

note that ker L = Sym and im L = Skew so you are looking for dim(Skew). skew symmetric matrices are fully determined by the entries above the diagonal, for 2x2 matrices there is only one such entry.

[strange delta]

if you have an inner product space, does the cardinality of the vectors match the orthogonal vectors?

[lapis river]

can someone guide me on this?

[dusky epoch]

you have conveniently cropped out the actual question(s)

[lapis river]

oopsies

[wide hemlock]

hello

thanks for previous answers

can you help for this too

is it 4 ?

[robust marsh]

Can anybody teach me anything of linear algebra? I'd like to learn it.

[native rampart]

This is a system of linear equations
2x+y=4
x+y=2

Linear algebra is all about solving linear equations in a sense although that is very not obvious as you go more abstract

(and mostly irrelevant)

[lapis river]

can someone please guide me through this question?

[native rampart]

What is the question?

[lapis river]

it literally just says this

heres an extra line but its arbitrary

[native rampart]

Yea,The last line is very important

[lapis river]

is this good enough?

[native rampart]

What have you tried?

[lapis river]

I'm actually pretty lost on where to begin

some kick in the right direction would be nice

[native rampart]

Start with the defintion of orthogonal

So,You will pick some random n(>=0) and random m(>=1) first and then consider the set {c_n(x),s_m(x)}

Here,c_n(x) and s_m(x) are both functions

Apply the definition of orthogonality on this set

It's not special enough to have a name,ig

[nocturne jewel]

Lemma 7.321 \s

[native rampart]

Just do it bro

Do you have know a vector space has a fixed number of basis vectors?

That is,a vector space has a dimension

So,Do a contradiction

Suppose the subspace is proper,then you can extend that basis of subspace to a basis of the parent space

Since the subspace is proper,this implies the new basis will have atleast dim(V)+1 elements

Proper subspace as in subspace that is not equal to vector space

Parent=vector space

Do you know if u is not in Span(S),where S={a_1,a_2,a_3...a_n} and S is Linearly independent implies {u,a_1,a_2...a_n} is Linearly independent

So,Since The subspace is proper,you can find elements not in it

You will adding elements not in The subspace to complete forming a basis

Your basis will have atleast one more element compared to what you started with

[native rampart]

It says this

Which is wrong because dim(V) is constant

Hence, Contradiction

[native rampart]

This

If it's not a subset ,you will never get a basis of the vector space

You will always have some extra element

[nocturne jewel]

If T is a linear map, then $0\in\ker(T)$

[stoic python]

Mosh

[nocturne jewel]

so consider non-zero v in the kernel, then $v\in ker(T^2)$ since $T[T[v]]=T[0]=0$

[stoic python]

Mosh

[nocturne jewel]

repeat for any positive integer power of T

trivial if T is injective, so assume T is not injective then show that ker(T^5) is also a subset of ker(T^4), then you get set equality

[native rampart]

Do you know what your T is?

[strange delta]

if i want to prove something through contrapositive like

a+b =<c

would the contrapostive then be

a+b>c

[nocturne jewel]

I believe contrapositive is only for implications

[nocturne jewel]

that would be the negation

[strange delta]

ah ok so through negation, would my conclusion be to show that negation is true?

i haven't worked this these stuff for a while lol

[nocturne jewel]

if you use negation you're doing contradiction, so assume the negation and draw it to an absurd/false conclusion

[lapis river]

is a vector of (1,1,0,1,1) in r4 or r5?

[nocturne jewel]

5 entries, so R5

[spiral star]

sure. looks like you are working towards generalized eigenspaces :) this reminds me a lot of fitting's lemma

you know that the kernels are increasing and that it must stabilize because its an increasing sequence that is bounded above

for the images you get the result from rank nullity

[native rampart]

No

You have imposed a max value on dim(T^n) and it is non decreasing

So,it has to get constant at some point

[spiral star]

there are some more properties you can get out of this

that are very useful for working towards generalized eigenspaces

in particular, the dimension of ker T^d will be the algebraic multiplicity of 0 in the characteristic polynomial of T

and V = (im T^d) + (ker T^d) as a direct sum

[hidden mantle]

how do i solve B?

I think i've made some progress

by reducing the middle term to

(yT)y where y = Bv

[zealous junco]

hm i feel like passing to inner product makes it easy but the notation seems weird

lambda_min refers to minimum eigenvalue of B i think?

[hidden mantle]

yeah

[zealous junco]

ok u dont need inner product, hint is write v = c1v1+...+cnvn for orthobasis v1,..,vn

i mean dont need inner product wrt B, just standard

[hidden mantle]

hmm

[lapis river]

what exactly is this question asking me?

[lavish jewel]

show that the integral of c_n c_m is 0 if m != n, and that the integral of c_n s_m is 0 always

oh, also that the integral of s_n s_m is 0 if m!=n

[lapis river]

yeah i was gonna sk

thanks!

[peak walrus]

does one only look at the coefficient matrix instead of the whole augmented matrix when deciding if a matrix is in ref or rref?
basically the constant matrix does not affect whether a matrix is in either form?

[lavish jewel]

that's right

the definitions for ref and rref don't really place any restrictions on the rightmost column of the augmented matrix

[exotic wedge]

I was trying to prove this theorem from Linear algebra done wrong

can someone verify my proof ?

[marble lance]

This looks fine. I have never seen the word complete used to mean the basis spans the space, but I will assume this is a term you use.

The second half I think is cleaner when done directly and not using contradiction, but your proof is fine.

Lastly, you are easily jumping between A as a mapping and as a matrix, which is fine ig, so long as you understand why you can do this

[twilit anvil]

is the second half of this proof necessary? if you already have that your set of length n is linearly independent, then it is definitely complete.

from the context i can see that it is given V,W have dimension n

[exotic wedge]

I guess if I make this claim I should prove it too

[exotic wedge]

The book uses complete, spaning, and generating. I like complete

[twilit anvil]

i agree, if it was not proved in your course. just something to keep in mind though.

you can view a basis as a linearly independent set of maximal length

(not sure how much that holds in the infinite dimensional case though)

[exotic wedge]

I've just started self-learning abstract linear algebra so i am not sure if i can comprehend this, but thanks for help !

[twilit anvil]

what i meant by maximal length is this. say you have a set of linearly independent vectors. can you add another vector to it so that the resulting collection is still linearly independent? if not, then your original set is a 'linearly independent set of maximal length'.

[exotic wedge]

oh, this makes sense

[marble lance]

I would consider this the proof that isomorphic spaces have the same dimension

Since here you construct a basis of the same length

[fathom lichen]

If A,B are linear operators and are self adjoint, would the product AB also be self adjoint?

[nocturne jewel]

I believe only if A and B commute as well

[north anvil]

If this is the row-reduced form of a matrix A, and I want to know whether or not the columns of A span R3, how can I reason this out?

[hollow finch]

how many pivots does this matrix have?

[wintry steppe]

the columns span R3 because they are linearly independent

[north anvil]

3 -> does this indicate that it spans R(the number of pivots the matrix has)

[wintry steppe]

what does row-reduced mean exactly (im not english) ?

if you have n linearly independent vectors in an n-dimensional vector space, they will always span that entire space

[hollow finch]

so that's pretty close. it means it spans a 3 dimensional space. since these vectors are in R3, the only 3 dimensional space in R3 is R3 itself.

[north anvil]

[hollow finch]

if you had only two pivots for example, it would span a plane (two dimensional space) in R3

[north anvil]

So in more general terms, if a matrix A has k pivots, then it would span a k-dimensional entity?

[hollow finch]

yeah

[north anvil]

Alright then - appreciate that

🙂

[hollow finch]

in R^m (if its an mxn matrix)

[hollow finch]

no problem

[north anvil]

In R^(number of rows)

yeah

alright then 🙂

cool 👍

Oh, the columns aren't multiples of each other

If they're all multiples of each other, then the span is a line

If exactly one of them is a multiple of exactly one other, then the span is a 2D plane

If none of them are multiples of each other, then they span a 3D plane

[wintry steppe]

not necessarily a multiple of one of the other two, it can be a linear combination of the others and it will still only be a 2D plane

[north anvil]

ohh

i see

i'm gonna bank on the fact that row reduction will take care of it

speaking of row reduction, is a matrix considered a row-reduced matrix if it's in REF form, or RREF form, or either?

[hollow finch]

REF is enough to see the column dependencies usually

RREF is the most convenient form when looking for a homogeneous solution though

[north anvil]

the non-trivial solution(s)?

[hollow finch]

yeah. RREF sets you up conveniently to have all the leading variables in separate equations, so its just the leading variables in terms of the free variables

[north anvil]

👍

[wintry steppe]

Can someone help me with this problem?

[strange delta]

do linear maps have to be injective/surjective?

[nocturne jewel]

no

[hollow finch]

so is your particular solution (your p vector) the only part thats incorrect?

[wintry steppe]

No, idk if I’m approaching this problem the correct way

[hollow finch]

because if so you only need one particular solution. Hint:||you could just find p3 such that the equations are satisfied. for example, that second equation looks simple. you just need to find p3 such that 1p1+0p2+1p3=3. since you have p1 and p2 then you can get p3 easy enough||

[hollow finch]

do you know about homogeneous and particular solutions? they can have different names too like complementary

[wintry steppe]

Yeah I know what they are

Would I have to add on a column of zeros to the matrix that they gave me?

[hollow finch]

nah

the homogeneous solution usually comes out of solving it normally

[slim gyro]

one quick question yall, what's the rank of the orthogonal complement of a matrix A, is it equal to the nullity of A

[wintry steppe]

Oh okay, because I put the matrix in echelon form, but I’m kind of stuck from there

[hollow finch]

rref is the best form when theres a nontrivial homogeneous solution

that should make it easier, assuming you meant you put it in ref form

[wintry steppe]

Ohh okay, thank you

[lapis sparrow]

So using Bernstein basis functions you can compute some abritrary point Q(u,w) on a surface

If you took a partial derivative of this WRT to u or v you would get something like this

I'm trying figure out the surface normal, so I've taken partial derivatives with respect to both

and then performed the cross product on those two vectors

I cant figure out whats wrong about my algorithm

njin and nkjm is supposed to be the result of taking a derivate of a bernstein polynomial

[misty storm]

boys

what the heck is that symbol?

also, while we're at it

what is that weird parenthesis?

[atomic hound]

looks more like a big S to me

[unkempt ridge]

it's an integral?

[misty storm]

I think it is

[unkempt ridge]

Yea i know i can't see it be anything else lol

:)

[misty storm]

what? I mean those surrounding the "f,g"

those parenthesis with sharp angles

sharper

ah, I get ya

I was asking 2 different things

[atomic hound]

oh yah i responded to the wrong one, and yes it's an integral

https://en.wikipedia.org/wiki/Bracket#Angle_brackets

[misty storm]

thx

[hallow lodge]

Hi, I'm calculating A=PDP^-1 and my result is showing me a flipped matrix, can someone help me figure out where im going wrong?

i just tried switching eigenvalues and their vectors and recalculated back to the same flipped matrix

[atomic hound]

do you mind showing the original question?

@hallow lodge

[atomic hound]

can you check your eigenvalue / eigenvector that resulted in [1 -2], lambda = 1?

[hallow lodge]

if i let x1 = 1, then x2 = -2, seems right

or should that be reversed

[hallow lodge]

if x2=1, then x1=-2

[atomic hound]

this was the clue that gave something away for me

[atomic hound]

i think [-2, 1] should work

[hallow lodge]

thanks i'll try it out

worked thx

[atomic hound]

awesome, np

[nocturne jewel]

it's inner product

[toxic pendant]

does the P_a(x) here mean polynomial

with a representing the power

[dusky epoch]

without context, it is impossible to tell

[dusky epoch]

...if i had to guess, $P_{\mathbf{a}}$ is the projection operator onto some fixed vector $\mathbf{a}$

[stoic python]

Ann

[toxic pendant]

makes a lot more sense thanks

[wintry steppe]

hey could anyone help me understand what a simplex is

I'm reading Peter Lax's linear algebra, chapter on determinant

and he says a simplex in R^n is a polyhedron with n+1 vertices

wikipedia, on the other hand, says it is a generalization of the notion of a triangle or tetrahedron to arbitrary dimensions

then also Lax says: "An ordered simplex S is called degenerate if it lies on an (n-1)-dimensional subspace", which I also cannot comprehend

[merry imp]

use R^2 for intuition. a simplex in R^2 has 3=2+1 vertices and a degenerate simplex would be a line (or even a point) which would lie in the subspace R, having dimension 1=2-1

[wintry steppe]

but is a vertex necessarily a "generalization of triangle", because I can't see how Lax says that?

@merry imp

[merry imp]

that's a nice way of thinking about it, but the actual definition is as lax gives it

[wintry steppe]

If I understand correctly, for R^3 in example, you can choose any four vertices, not necessarily such that they give you tetrahedron?

[merry imp]

(btw, since you're using lax you might like this lecture series which follows it: https://www.youtube.com/playlist?list=PLwV-9DG53NDwKJIwF5sANj6Za7qZYywAq )

[wintry steppe]

oh awesome, i'll definitey take a look at it

Another thing I cannot comprehend is this: "An ordered simplex (0, a1, ... , an) =S that is nondegenerate can have one of two orientations: positive or negative. We call S positively oriented if it can be deformed continuously and nondegenerately into the standard ordered simplex (0, et,... , en), where ej is the jth unit vector in the standard basis of R^n."

[half ice]

@wintry steppe
So for example, in R² a simplex that is "going counter-clockwise" is considered to have positive orientation

The same simplex "going clockwise" would be negative

[wintry steppe]

ok thank you

my next question is why is this true?

@half ice ?

[wintry steppe]

Hello, I have a conceptual question for linear algebra

https://www.slader.com/textbook/9780201658590-introduction-to-linear-algebra-5th-edition/337/exercises/9/

Does order of eigenvalues matter????

When I use octave, it gives me a different order than the key here.

I'm concerned whether the result I get will be very different from the key or not.

[native rampart]

It does not

Also,You will never be finding the change of basis matrix for finding the diagonalised matrix

If you have a different order of eigenvalues,your S will be different

[wintry steppe]

Thank you

How do I determine the correct order or eigenvalues?

Eh, never mind. You just said it doesn't matter.

I just don't understand

The octave program gives one of the values a number that cannot be rationalized...

The only difference is one of the columns for S is different

I am so lost.

[native rampart]

What was the D you got?

[wintry steppe]

I can't even get a D because the S inverse is this

[native rampart]

Yea,S^-1 could be bad

[wintry steppe]

I hate linear algebra

All because octave gives the eigenvalues in a different order

Driving me insane

Like, how am I supposed to know the order?

[native rampart]

You don't need to

[wintry steppe]

I don't get why octave won't work if a different order.

[native rampart]

Have you defined S

[wintry steppe]

Frustrating

Yes, that is S

[native rampart]

That's what the problem looks like

[wintry steppe]

When I invert it, it ends up being the result

[native rampart]

Share a SS of the part where your program outputs S

[native rampart]

No,I mean the matrix S such that SAS^-1=D

[wintry steppe]

Oh, i didn't get D yet.

I can't get D if I can't even get S^-1

[frigid crow]

sry but what is octave

you could directly use MATLAB for this

:/ idts

[wintry steppe]

Are symmetrical matrices orthogonal??

There's so many mini definitions I keep forgetting

[marble lance]

Do you know the definition of orthogonal and symmetric?

[dusky epoch]

no, a symmetric matrix need not be orthogonal

and an orthogonal matrix need not be symmetric

[marble lance]

What the fuck

<@&268886789983436800>

[cursive garden]

Wy r u using such words @marble lance

[marble lance]

🤡

[lavish jewel]

delete that too

[gray dust]

found the deleted msgs

[marble lance]

Yes, just want mods to see my proof, then they can delete it 😂

[lavish jewel]

👍

[dusky epoch]

what happen

[honest token]

just that guy saying nsfw things

[marble lance]

Yes, dw Ann, I wasn't saying it to you

[stable kindle]

can the basis for U just be {(1, 6, 0, 0, 0), (0, 0, -2, 1, 0), (0, 0, -3, 0, 1)}?

[native rampart]

Yes

[stable kindle]

ok ty i was 95% sure but paranoid

[twilit anvil]

how did you find it? if youre finding it in a deterministic way, you should be more confident than paranoid

unless you made a mistake in the calculations or something

[hollow finch]

A lot of problems in linear are easier than they seem, and one can often (and should try to imo) solve them by inspection. Sometimes that seems too easy so it can be fair to question if it's actually a valid solution. I think solving it by inspection, which it seems like Kaisheng did, is the right approach.

[fringe burrow]

If V is a vector space and X,Y are subspaces of V and dim(V)=dim(X+Y), does it follow that V=X+Y?

I guess it's not true but if we add that X inter Y={0} then it is true

[tropic turret]

normal opertator is always diagonalizable?

[tropic turret]

?

[dire thunder]

no

consider infinite dimensional spaces e.g

[dire thunder]

yes

[tropic turret]

NM I got it's not true

[dire thunder]

wym

of fuck ye

[tropic turret]

Yeah but thats true just if the operator is complex

[dire thunder]

self-adjoint operator is always diagonalizable

[tropic turret]

alright thanks! 😁

[tropic turret]

How can I prove that if T is a normal operator with some natural k s.t T^k=0 so T=0?

[wintry steppe]

What is this symbol

[dusky epoch]

looks like an equals sign with a hat on top

do you have any more context for it

cause idk about anyone else but i have no idea what this might mean

[atomic hound]

does this help?

https://tex.stackexchange.com/questions/103408/symbol-for-corresponds-to-equals-sign-with-hat

i'm not familiar with it either

[fringe burrow]

I was speaking about V finite dimensional in particular

[limber sierra]

my guess is that thats not actually a hat, and is instead a very sloppy $\cong$

but without further context, hard to tell what it means

since that symbol can mean many things

(isomorphic? congruent modulo n? approximately? etc)

[stoic python]

Namington

[tropic turret]

<@&286206848099549185>

[native rampart]

If T is normal,just find the basis in which the matrix of T is diagonal

Since T^k=0 you get that matrix to be 0

And you are done

The spectral theorem is amazing,ik

[rose umbra]

this means S( S(x)) = S(x) ?

[native rampart]

Yes

[rose umbra]

ty

[native rampart]

@tropic turret

[tropic turret]

Wait but who says T is diagonal? it's true just if T is complex

[native rampart]

It's true if T is normal

It's called spectral theorem

[tropic turret]

Yeah but this theorem correct above the complex numbers and not necessarily above the real

[tropic turret]

I can give you counterexample of real normal matrix that is not diagonalizable

[native rampart]

mb

[tropic turret]

what? 😅

[lavish jewel]

show me an example, i'm curious

[tropic turret]

take $T:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ that represented by $\begin{pmatrix}2 & 3\
-3 & 2
\end{pmatrix}$

[stoic python]

go

[tropic turret]

@lavish jewel

this is not diagonalizable because your'e above R

the spectral theorem correct just for the complex field

[lavish jewel]

aha, i always forget that

[tropic turret]

I mean I can see how it's work but idk how to correct this problematic situation

[wintry steppe]

Can I get help on number 12? it's already in echelon form but idk how to parmaterize it

[cursive ginkgo]

Hey ! Any idea how to find this determinant ?

[torn hornet]

what did you try?

[cursive ginkgo]

I tried to find an induction relation by developping with the first collumn but without sucess

[torn hornet]

induction/recursion is the way to go

how far did you get

[cursive ginkgo]

If we name this matrix A_n then I get
A_n = (1+x+x²) A_n-1 but i don't know if that's enough

[torn hornet]

first I assume you mean A_n is the determinant of the nxn version

and second that is wrong (although if it was true it would be enough)

[cursive ginkgo]

it's wrong ? could you explain ?
I'm developping with the first column so
A_n = (1+x²) A_n-1 + x A_n-1

[torn hornet]

so the second cofactor is not A_(n-1)

$A_n = (1+x^2)A_{n-1} -x \begin{vmatrix} x& x& 0&\dots \ 0 &1+x^2&x& \dots\ 0& x& 1+x^2& \dots \ \dots \ \dots &\dots &x& 1+x^2 \end{vmatrix}$

[stoic python]

JohnDS

[torn hornet]

not best made matrix but w/e

anyhow clearly not the same cofactor right

[cursive ginkgo]

Ok I understand why the minus but why do you still have a x on the top left ?

[torn hornet]

hold on lemme just send a picture lol

[wintry steppe]

when you do gauss-jordan elimination its forward elimination then backward elimination right?

[torn hornet]

its this cofactor

and remember you do alternate sum of the cofactors

thats why you have -

[cursive ginkgo]

That yes.

[torn hornet]

anyhow can you find the determinant of the matrix i showed you interms of some A_k

[cursive ginkgo]

But I did it with the 1+ x^2 on top left and then and the x below, to develop with the column

[torn hornet]

i see

you still would not get A_(n-1)

did you figure out the determinant of the second cofactor @cursive ginkgo

[novel shuttle]

hey guys what is a good visual textbook for linear algebra?

[azure knoll]

Hello

[mild current]

can anyone elaborate what this means.

[azure knoll]

Im struggling on dis homework assignment and I just need it so that I dont have to do summer school It probably gonna be hot asf in that classroom and my mom making me get a job during summer any help would be appreciated

[mild current]

Especially whats up with the notation

[mild current]

@novel shuttle Linear Algebra A geometric Approach- S.Kumarsean
Linear Algebra Done Right-S.Axler

[nocturne jewel]

The set of real valued functions from X to R is a vector space with the typical pointwise operations

[dreamy iron]

https://math.stackexchange.com/questions/2287698/books-on-linear-algebra/2288250

[glad finch]

Hello everyone, I don't really understand what this means, anyone has any idea about this?

[dire thunder]

point of intersection means that there is vector (x,y,z) that is on both planes and line simultanoeusly

[glad finch]

so it means that the line that parameterized is x, and the other equation are y and z?

[dire thunder]

no

(x,y,z) can be vector parametrized by line

that is (x,y,z)=(-t, 5t,1)

or (x,y,z) may be just vector s.t. its coordinates are on planes

[glad finch]

ahh okay I got it. But then what are we going to do with that equation that given?

[dire thunder]

find if there is such point (x,y,z) that satisfies all three equations

to be precise

find if there is t s.t (-t, 5t,1) satisfies both equations of planes

[glad finch]

ohhhh okay okay I understand now. Thank you!

[soft coral]

this is kinda a geometry question, but what does it mean for the determinant to be the signed area of the parallelpiped created by the columns of the matrix

like i geometrically get orientation in 2d and 3d, but my ideas dont really generalize to higher dimensions

the best idea i have rn is (and assume that the determinant isnt 0), if u take the first n - 1 cols of the matrix, this creates an n - 1 dim subspace of the vector space ur in, splitting it in half
if u stand at whereever the nth col vec is and "look at" the subspace, if it looks positively oriented from that direction then ur det > 0, and otherwise its < 0, and this corresponds to which half of the vector space the nth col vec is in

this makes sense for 2d and 3d, but i cant really see it beyond that

u can say a 2d subspace looks positively oriented from a position depending on whether the first basis vector is counter clockwise or clockwise to the second

but idk how ud say something similar for 3d subspaces in 4d

[glad finch]

wow it's more complicated than I thought. My brain is tryna understand your explanation. Thank you tho!

[heavy crown]

Prove or disprove;
If A is matrix of order m x n, and A*A^t is invertible then m ≤ n

Is it true?

I know it means based on given that det(a)≠0 and rank(A*A^t)=rank(A)=m but how can I conclude if m ≤ n ?

[dusky epoch]

how are you talking about det(A) when A is not square?

[heavy crown]

I was thinking because det(AA^t)≠0 then it means det(A)²≠0 and then det(A)≠0 ?

[dusky epoch]

username.

A is not a square matrix.

[heavy crown]

I'm not allowed to?

[dusky epoch]

A is not a square matrix.

non-square matrices do not have determinants.

determinants are only something you can talk about for matrices that are square.

det(A) is literally nonsense.

[heavy crown]

Oh okay I needed that clarification

[dusky epoch]

anyway,

AA^T is a matrix of size m × m

so rank(AA^T) = m