linear-algebra

[silent dune]

im confused still

I know how to get one vector

but the 2nd, I don't get what you mean

[lavish jewel]

$\begin{bmatrix} 1 && 2 && 5 && 0 \\ 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 \end{bmatrix}$

[stoic python]

Edd

[lavish jewel]

try to find solutions to this system

[silent dune]

isnt two of them free variables?, x + 2y + 5z = 0

[lavish jewel]

mhm

and well, if you find one, and you want the third to be orthogonal to both the normal and the previous vector

put it back as a row

$\begin{bmatrix} 1 && 2 && 5 && 0 \ -3 && -1 && 1 && 0 \ 0 && 0 && 0 && 0 \end{bmatrix}$

oh oops

[stoic python]

Edd

[silent dune]

why is the 2nd row 3rd column a 1?

oh wait nvm

[lavish jewel]

i had made a mistake earlier

[silent dune]

you picked a different vector

and from there we would?

[lavish jewel]

solve this system

a solution to this system will be orthogonal to the normal and to the vector (-3,-1,1)

[silent dune]

ok so

[lavish jewel]

so the 3 vectors will be mutually orthogonal

[silent dune]

[ 1 2 5 ...]
[ 0 1 2 ...]
[ ... ]

wait oops

i'm confused a bit still

so if I reduce that matrix

I'm confused at how I would choose the third vector from it

[prisma sail]

Can you obtain an identity matrix by multiplying by two non square matrices

I think its no right because it wont be invertible since its not a square..

Is that right?

[lavish jewel]

no

you can take a matrix of size m x n with m >= n

then if the columns are orthogonal, M^T M = I

a simple case could be a truncated identity mat, for example

as a trivial example

[prisma sail]

Oh

[lavish jewel]

this is related to projections onto the row or column space of the matrix

a similar case can be built for MM^T

this is referred to as a pseudo inverse

it works from one side, but not the other

[prisma sail]

Thank you

[noble swan]

So then what would be the simplest way to check if a set of vectors are a basis? Linear independence?

[winter harbor]

If you know the dimension of you space before hand

And if the cardinality of the set you want to check if it is a basis or not happens to have the same cardinality as the dimension of your space, then all you need to check is linear independence.

Otherwise

If you have no information about dimension whatsoever

You have to check that this set is a generating set, i.e spans your whole vector space and is linearly independent too.

This is usually how you go about this.

[noble swan]

Gotcha, thanks!

[ionic laurel]

Hi I have a question

oh wait

wait actually im still confused

If we have this matrix

What would be the integer p such that the ColA is isomorphic to R^p?

Isnt the column space a subspace in R^M which is 4

so p would have to be 4 right?

but why is the answer 3 -> is it because ColA has three dimensions because the basis is made up of 3 vectors?

[sharp trail]

Hi guys, hope you all are doing well

Could anyone recommend me a good book on linear algebra with machine learning exercises?

I'd like to solve problems that have to do with all the layers of machine/deep learning

[marble lance]

Yes

[ionic laurel]

Okay yeah I understand now

I mixed up subspace and isomorphism

Thank you nonetheless

[marble lance]

👍

[errant mist]

Can I just show for this that A has just one L.I column and all other columns of A must be multiples of this column call it $\vec{x}= [x_1 x_2 \dots x_n]^T$ ?

[stoic python]

Fredrikpiano

[wintry steppe]

given x, y, show that uv^T

[boreal spindle]

hi, so I have a Q. Do I need to do any further calculations or is this enough as an answer?

[boreal spindle]

<@&286206848099549185>

[meager steppe]

Could somebody help me prove this, this is what I have so far

[lime zinc]

Replace that I on right with 0

[meager steppe]

oh crap i messed up lol, a times a transpose equals Identity matrix so i think that part is right

[ionic laurel]

hi i have a question

If H was a subspace of V, then would a basis for V also be a basis for H?

[nocturne jewel]

No

[ionic laurel]

i see

what about the converse? @nocturne jewel

[nocturne jewel]

I think it's true.. but I dont quite remember if V is a subspace of V

[ionic laurel]

any vector space is a subspace of itself

that is a true statement i know that

[nocturne jewel]

Then yes:
If a basis of V is also a basis of H, then H is a subspace of V, since H is V

[ionic laurel]

i see

Thank you Mosh

[wispy pewter]

is this legal?

[stoic python]

EmilD

[wispy pewter]

the two equations I have are

[nocturne jewel]

if your equations are $m_1x_1+m_2x_2=35m_2$ and $m_1x_1^2+m_2x_2^2=(35)^2m_2$ then yes

[stoic python]

Mosh

[nocturne jewel]

just not common to write it like that

[stoic python]

EmilD

[wispy pewter]

those are my equations

the x's are all variables

I should add that

when I rref'd the augmented matrix my n-spire gave me a yellow exlamation mark

and the answer was really unhelpful

to solve that system of equations

very messy

[remote hatch]

can i get some help starting this off?

[wintry steppe]

try solving for a

this is the kind of problem where there's literally only one thing to do and it's probably not obvious if you haven't seen anything like it before

so here's a big hint: if x is a real number, x = x * 1

[limber sierra]

solving nonlinear systems by rref is not useful in general

like your manipulations ostensibly work, ie they still represent valid things you can do

but theyre also just fundamentally not manipulations that solve quadratics

[wispy pewter]

so is the only move to solve

for one of the variables

and plug it into the quadratic and solve that?

[limber sierra]

thats probably the best path, yeah.

[wispy pewter]

linear can't help me?

ahh dang

[remote hatch]

Im confused cus it can be any linear map, is the question asking for like a specific number or a general formula like let a = L(x)/x so that L(x) = L(x) whenever x!=0 and when x is zero then L(0) = a(0)=0

[winter harbor]

1 is a basis for R

also

L is a linear map (over R!!!!)

and x is a real number

so L(x) = L(1*x) = ????

use the fact that L is a linear map now

[remote hatch]

1*L(x) = L(x)

[winter harbor]

no

L(x*1) = L(1) * x

[remote hatch]

OH OK

THANK YOU

i swear i had that written down then thought to myself that couldnt be it because you could only factor scalars out

[winter harbor]

but x is a scalar

and that's the reason you can do that

[remote hatch]

just realised thankss

[rotund aurora]

Do the abelianization and determinant have anything to do with eachother

They look similar from the diagrams

[pliant blaze]

hi how do i find a matrix by eigenvalues and eigen vectors?

[winter harbor]

Yes, they are!

[rotund aurora]

It's pd(p^-1) where p is the column eigenvectors in a row and d is the diagonal matrix corresponding to the respective eigenvector

@pliant blaze

Same with gen eigenvectors and Jordan blocks for non diagonalizable matrices

Just look up Jordan canonical form

[wintry steppe]

How?

[winter harbor]

The reason is as follows, for $n \neq 2$ and $\mathbb{K} \neq \mathbb{F}_{2}$ we have that the abelianization of $\text{GL}(n, \mathbb{K})$ is isomorphic to $\mathbb{K}^{\times}$. Moreover, the determinant map:
$$
\text{det} : \text{GL}(n, \mathbb{K}) \rightarrow \mathbb{K}^{\times}
$$
Is precisely the abelianization epimorphism. We can see this because the determinant map satisfies precisely the universal property of $\text{GL}(n, \mathbb{K})$ meaning that for every group homomorphism:
$$
f : \text{GL}(n,\mathbb{K}) \rightarrow H
$$
there exists a unique group homomorphism
$$
f^{\ast} : \mathbb{K}^{\times} \rightarrow H
$$
For which the following diagram commutes:
\begin{displaymath}
\begin{tikzcd}
\text{GL}(n,\mathbb{K}) \arrow[dr, "\text{det}" '] \arrow[r, "f"] & H \arrow[u, dashed, " \exists! f^{\ast}"] \
& \mathbb{K}^{\times}
\end{tikzcd}
\end{displaymath}

[wintry steppe]

Abelianization of matrix groups?

[winter harbor]

oops

let me correct the diagram

god dammint

[rotund aurora]

Wait what is k x

[wintry steppe]

double damn

units

no wait

[rotund aurora]

I thought it was a map from the exterior to the tensor products

[winter harbor]

multiplicative group of the field K

[rotund aurora]

Or rather

[wintry steppe]

ita everything besides 0

[rotund aurora]

The quotient by the tensor product

*from the exterior product to the quotient sorry

[winter harbor]

I will correct this diagram

But yeah

We can think of the determinant in a bunch of ways

[wintry steppe]

what was a map?

[rotund aurora]

The determinant

[winter harbor]

We can either think of it as a certain group homomorphism from the space of square matrices over a field which satisfies some special properties

Or

[wintry steppe]

Is it in general?

[winter harbor]

We can think of it as follows

[wintry steppe]

in terms of R modules im guessing?

[pliant blaze]

got it, thanks

[west orchid]

The reason is as follows, for $n \neq 2$ and $\mathbb{K} \neq \mathbb{F}_{2}$ we have that the abelianization of $\text{GL}(n, \mathbb{K})$ is isomorphic to $\mathbb{K}^{\times}$. Moreover, the determinant map:
$$
\text{det} : \text{GL}(n, \mathbb{K}) \rightarrow \mathbb{K}^{\times}
$$
Is precisely the abelianization epimorphism. We can see this because the determinant map satisfies precisely the universal property of $\text{GL}(n, \mathbb{K})$ meaning that for every group homomorphism:
$$
f : \text{GL}(n,\mathbb{K}) \rightarrow H
$$
there exists a unique group homomorphism 
$$
f^{\ast} : \mathbb{K}^{\times} \rightarrow H
$$
For which the following diagram commutes:
\[
\begin{tikzcd}
\text{GL}(n,\mathbb{K}) \arrow[dr, "\text{det}" '] \arrow[r, "f"] & H  \\
& \mathbb{K}^{\times}\arrow[u, dashed, " \exists! f^{\ast}"']
\end{tikzcd}
\]

[rotund aurora]

Hm never thought about in terms of modules it should work for free modules though right

Anyway sorry continue

[winter harbor]

Let $V$ be a finite dimensional vector space over a field $\mathbb{K}$ and $T : V \rightarrow V$ an endomorphism of $V$, we define:
\begin{align*}
\bigwedge^{k} T :& \bigwedge^{k} V \rightarrow \bigwedge^{k} V \
u_{1} \wedge \cdots u_{k}& \mapsto Tu_{1} \wedge \cdots \wedge T u_{k}
\end{align*}
Since $\text{dim}_{\mathbb{K}} \left(\bigwedge^{k} V\right) = \binom{n}{k}$, then if we consider the the top power $\bigwedge^{n} V$ it has dimension $1$, so the induced map $\bigwedge^{n} T$ is such that $\forall x \in \bigwedge^{n} T$ we have:
\
\
$\bigwedge^{n} T (x) = \text{det}(T) x$ for some unique constant $\text{det}(T) \in \mathbb{K}$ and this constant we call the determinant of $T$.

[wintry steppe]

lol

[winter harbor]

No, what is wrong with the diagram is the direction of the dashed map

[west orchid]

oh

[winter harbor]

it should be the other way around

it goes from K^x to H

in any case

[rotund aurora]

Wait why is the function labeled as a wedge product

[winter harbor]

It is a notation

[winter harbor]

the exterior power is a functor

so if we have a map T : V -> V

it induces a map between the exterior powers of V

and I am denoting the induced map of T by the exterior power function via \bigwedge^{k} T

this is now an endomorphism of \bigwedge^{k} V

[winter harbor]

thanks 🙂

[stoic python]

MISTERSYSTEM

[winter harbor]

Ok, I hope there are no more typos now

But this is how you are thinking about the determinant map

[rotund aurora]

Doesn't h need to be abelianthough

Wait

How do you connect the two though

Or how does the universality property give the actual form for the determinant

[winter harbor]

yeah, H needs to be abelian

[rotund aurora]

But thanks it was actually pretty cool when I realized the similarity

[winter harbor]

Well, it tells us that the determinant map is precisely the abelianization map between GL(n,K) and K^x

https://mathoverflow.net/questions/102109/a-polynomial-homomorphism-from-gl-to-the-group-of-units-is-a-power-of-the-determ

This might help

I remember seeing this on Lang's algebra

Hold on

Yeah

The commutator subgroup of GL(n,K) for n not equal to 2 and K not equal to the field with two elements is precisely the special linear group SL(n,K)

and by the exactness of
1 -> SL(n,k) -> GL(n,k) - det -> K^x -> 1

this gives us that the abelianization of GL(n,K) is precisely K^x (excpet for GL(2,F_2))

I was also able to find a MSE question related to this

https://math.stackexchange.com/questions/1092804/abelianization-of-general-linear-group

@rotund aurora hope this wasn't too confusing

I apologize mainly for my bad LaTeX typesetting lol

and for the typos

[rotund aurora]

Nw it's been helpful

Wish I hadn't spent the whole semester being depressed, I always feel pretty hype when I learn shit

[winter harbor]

Aw man

Hope you are doing better now

[rotund aurora]

Eh not really but better than the one before, hopefully things look up soon

Thanks for asking though

How about you everything good

[winter harbor]

I guess so

[rotund aurora]

Are you in grad school?

[winter harbor]

Not yet

Hopefuly in the near future

[rotund aurora]

Nice

I was thinking of going for grad school but I think I am actually not good enough tbh to do any productive research

Kind of sad since I would love to do it but that's reality ig

Not sure what to exit into that will give me any joy

But ig that's life for almost everyone

What do you want to study

[wintry steppe]

same and im in junior year

[pallid rampart]

How do I show this?

[lavish jewel]

you could use the induced norm

though i'm not sure if the bars there denote the 2-norm or just any norm in general

[pallid rampart]

Any norm in general

[pallid rampart]

What's induced norm here?

[lavish jewel]

the induced norm is defined like this

from wikipedia

[pallid rampart]

Wait but that's matrix norm though

[lavish jewel]

you can see that, by definition, this means that $\Vert Ax \Vert \leq \Vert A \Vert \Vert x \Vert$, where $\Vert A \Vert$ is the norm that is induced on the matrix by the vector norm $\Vert \cdot \Vert$, whatever it is

[stoic python]

Edd

[lavish jewel]

you can then use submultiplicativity, which is part of the definition, to get the result

might also need some relation to the spectral radius of the matrix to tie it all up

[pallid rampart]

Ah I see

Tho I don't think I'm allowed to use that yet

[lavish jewel]

hmm i see

[zinc timber]

$|T^m| \leq |T|^m$ now norm of $T$ will be < 1 so, there will be a $m\in\mbb{N}$ s.t. $\norm{T} ^m <\varepsilon$

[stoic python]

Ryuzaki

[pallid rampart]

The whole point is I don't know what norm is, and I don't know the norm of T is < 1

[zinc timber]

maybe then try to do something like $\lim_{n\to\infty} T^n = 0$

[stoic python]

Ryuzaki

[lavish jewel]

i thought of that ryu, butthey want a fixed m, not the limit

so you need an extra step in the middle

[zinc timber]

ya this method involves way too theory.

[lucid glacier]

so I have 3 affinely independent vectors v1,v2,v3. I wanna define the affine plane spanned by them using a linear functional. I calculate the cross product of v2-v1 and v3-v1, and then to find thr constant for the equality (Since an affine plane is defined by the equation <v,u>=a) I take the dot product of the resulting vector with v1. This should give me an affine plane equation right? But for the vectors (1,1,1),(2,1,1) and (5,2,2) it's not and i'm not sure what i'm doing wrong

[lavish jewel]

you mean the vectors are position vectors of 3 points that are on an affine plane?

(just making sure i get the scenario)

the plane eq should be normal dot (p - p0) = 0

so what you said sounds fine

might be a coding error?

[lucid glacier]

Yes

I tried doing it by hand and using wolfram and got the same thing

[lavish jewel]

and what is it that makes you say the eq does not define a plane?

what error do u get

[lucid glacier]

I mean that the fact that the dot product is zero means that there's a vector plane going through those 3 points

But that shouldn't be possible since they're affinely independent

[lavish jewel]

whoch one gives u 0, normal dot v1?

[lucid glacier]

Yes

The normal is (0,-1,1) and dotting it with (1,1,1) gives 0

[lavish jewel]

oog, i'm not on my pc to check on octave

[lucid glacier]

Geogebra draws an affine plane (just making sure i'm not crazy)

Oog

[lavish jewel]

hmm

idk, looks non affine to me lol

[lucid glacier]

Wdym

It's not passing through 0

Also when I did the process with 5,2,2 instead the dot product was nonzero??

I don't understand, order shouldn't matter here

[lavish jewel]

wait

those are not the same points tho

you said 211 earlier

not 221

make the plane with c = (2,1,1) for my peace of mind

[lucid glacier]

Oh oops sorry

Ok wtf it is a vector plane???

Now i'm even more confused

[lavish jewel]

a what

[lucid glacier]

A plane going through 0

[lavish jewel]

it is, if c is 2,1,1

exactly as we both found

[lucid glacier]

Shouldn't this only happen if the points are affinely dependent tho

[lavish jewel]

show the plot

i have never seen the def of affine independence so i cant comment

but the eq of a plane showed thusly

[lavish jewel]

looks about right

[lucid glacier]

v1,v2,v3 are affinely independent if v2-v1, v3-v1 are linearly independent

[lavish jewel]

i would call that colinearity or smth like that

all planes satisfy your definition

[lucid glacier]

true....

[lavish jewel]

im gonna get dizzy reading on the train, so i leave you with that. maybe look for a dofferent def of affine indep

[lucid glacier]

alright

I was under the impression that the plane had to be affine for some reason

but maybe this is fine

for my purposes

[rich garden]

can you uniquely determine a matrix given A^TA and AA^T?

[lavish jewel]

no, the info about the eigenvalues is lost

the eigenvalues of those two matrices are the same, and they are equal to the absolute value squared of the singular values of the original matrix A

[earnest sigil]

how to approach this sum

[zinc timber]

first try writting b=a+d, c=a+2d. last row gives you a hint in what to do first.

[dark brook]

How would you calculate something like this?

Its something of a mouthful

Wait

[golden reef]

A and B are matrices, i just am not sure how he re arranged from second to third step

how does |B| |A| |A| |B| = |A| |A| |B| |B|

[wintry steppe]

expanded out

[golden reef]

does the order matter?

[wintry steppe]

they are determinants

[golden reef]

since its the determinant of the matrix, can i multiply without the worry of the order?

[wintry steppe]

so they are elements in a field which are commutative

[wintry steppe]

Yes, images of determinants are elements of base field

If there were no determinants this would be false

[golden reef]

ok i see thanks

[wintry steppe]

You compute the determinants and solve the equation for its roots

Its a cubic equation I think

[nocturne jewel]

Multiplication in R/C is commutative

[wintry steppe]

So first term is l((4-l)(8-l)-35)

Linearly independent has same notion, if you can write one function in terms of others multiplied by a constant

@stable kindle

[stable kindle]

yeah i know

[torpid socket]

Hey so

If I have 3 equations in matrixes with 2 variables only but the second variable belong in column 3 do I put 0 if there is no variable for a column ?

Like this

[nocturne jewel]

Yes you put in the placeholder 0s

[torpid socket]

Alright perfect

Also what over R and F3 mean

[nocturne jewel]

x_i is from the field R or F_3

All scalars*

[torpid socket]

I don’t understand

Wait is it the size of the matrix ? @nocturne jewel

[nocturne jewel]

No

R is a field, so solving over R means you're solutions are from R

Same with F_3

[torpid socket]

Hmm

I see

R is real numbers and F is finite right

[dim epoch]

yes F_3 is a finite field

[wintry steppe]

Is this correct?

lambda scalar, a,c in V b,d in W and tensors are in V tensor W

wait huh

[winter harbor]

Wait, if $a,c \in V$ and $b,d \in W$ are vectors on a vector spaces $V$ and $W$ respectively, how are you making sense of $a/b$ and $d/c$?

[stoic python]

MISTERSYSTEM

[wintry steppe]

this feels weird since they are vectors

yeah good point

oh

I know what to do

i think

do i add 0 into the vectors?

[winter harbor]

But since we are dealing with general vector spaces

this doesn't make much sense

[wintry steppe]

yeah

[winter harbor]

What exactly are you trying to prove tho?

[wintry steppe]

reading some article and it just says this is true given two rules

and top equals bottom

starting from right side is easier

tbh

[winter harbor]

Is lambda a scalar?

[wintry steppe]

idk if it is

yeah

[winter harbor]

Ah, so proving that is not hard lol

It's basically a consequence of how we define the tensor product of modules

Meaning

[wintry steppe]

nvm

i got it i think

[winter harbor]

Ah, alright then.

[dim epoch]

by force of will

[wintry steppe]

lol nvm

i was trying to figure out a way to add 0

so a tensor b = a + something tensor b + something

but failed lol

[winter harbor]

Look, this is just a consequence of how we define the tensor product V ⊗ W of two modules over a commutative ring as the quotient of the free module F(V x W) by a certain ideal.

That's basically how multiplication by scalars is defined in V ⊗ W

[wintry steppe]

[torpid socket]

Over here it is showing me that 4 * 1/3 = 7/3 what am I missing

[limber sierra]

you mean here?

that means "add 1/3 times the first row to the third row"

so its not saying 4 * 1/3 = 7/3

its saying 1 + (4 * 1/3) = 7/3

which is, of course, true

[wintry steppe]

I want to show directly through the algebra

[wintry steppe]

I understand it this way easier

but I want to show by hand

[torpid socket]

Yeah I noticed later thanks tho 😊

[lofty topaz]

I'm struggling with an inverse vector. Does inverse vector exist here ?

[nocturne jewel]

what's additive identity?

[lofty topaz]

• (in circle) is multiplication, hence I think that the identity element of this operation is 1 vector

[nocturne jewel]

Yes, 1 is the additive identity

so x has inverse y iff xy=1

is there a positive number y such that xy=1?

[lofty topaz]

im not sure if 1/x is correct notation of inverse vector

[nocturne jewel]

yes, y=1/x is the additive inverse of x

[lofty topaz]

Ok, that's what i thought but I was not sure with the notation. I have not seen it anywhere up to now

[nocturne jewel]

it's the number 1/x

you've not seen fractions?

[lofty topaz]

its 1/vector x, or no ,

?

[nocturne jewel]

the vectors are real numbers here

(positive real numbers)

so 1/x makes sense for real numbers, just not x=0, but that isnt a vector in the space so it doesnt matter

[lofty topaz]

yeah thats make sense, thanks

[lofty topaz]

and for example, if we would have working in R^2, 1/x would be still good inverse ?

[nocturne jewel]

no

if you're in R^2, your vectors are of the form [x,y]

[lofty topaz]

ok thanks

[winged prairie]

can anybody help me with this one?

have been stuck on it for a whie

while

[subtle walrus]

what did you try?

[winged prairie]

so i took x = a1v1 + a2v2 + a3v3 + a4v4

then i took a1v1 = x - a2v2 + a3v3 + a4v4

sorry a1v1 = x - a2v2 - a3v3 - a4v4

then im taking a1v1 = a1v1 + a2v2 + a3v3 + a4v4 - a2v2 - a3v3 - a4v4

but idk what to do next or if this even makes sense

[subtle walrus]

what is this supposed to accomplish?

[winged prairie]

im trying to rearrange it so i get v1 - v2, v2 -v3, v3 -v4

[subtle walrus]

🤔

[winged prairie]

but i don't think it makes any sense haha

[subtle walrus]

lets take a step back

what does "span" mean?

[winged prairie]

set of all linear combinations

of v1,v2,v3,v4 in this case

[subtle walrus]

what does "span V" mean?

[winged prairie]

it means that ever vector in v can be written as a linear combination of the basis vectors

[subtle walrus]

ok, so

you can write every vector in V as a linear combination of v_1, v_2, v_3, v_4 and you want to show that this other set of vectors can also do this
one way to do this is to show that you can write each of v_1, v_2, v_3, v_4 as a linear combination of this other set

[winged prairie]

so v_1 = v_1 - (v_2 -v_2)

?

wait no

v_1 = (v_1_-v_2)+ v _2

[subtle walrus]

well, but v_2 is not part of the other set

[winged prairie]

ah

[subtle walrus]

(v_1 - v_2) is, but there is another linear combination that will result in v_1

[winged prairie]

wait so maybe this could work

v1 = (v1-v2) + (v2-v3) + (v3-v4) + (v4)

[subtle walrus]

yeah, that works

so v_1 is in the span of the second set

if now v_2, v_3 and v_4 are also in the span, then also every linear combination of them must be

and hence all of V

makes sense?

[winged prairie]

yes

thanks man

[slender spear]

How do you diagonalize a non invertable matrix?

[winter harbor]

The procedure is the same

The only difference is a non invertible matrix has 0 as an eigenvalue

But checking if a non invertible matrix is diagonalizable (and if so, diagonalizing it) requires the same procedure.

[still lodge]

how might i start this

normally id provide what work i have so far but im just

[slender spear]

Oh wait nvm you just look at the characteristic polynomial even with the zero right

[modern palm]

I have 2 matrices, and I need to show they are not similar. How can I do that?

I already checked that they have the same trace, rank, and characteristic polynomial

[glad acorn]

I don’t understand (b)? Is it a=0 b=1?

[limber sierra]

no

or rather, those values work, but theres more

$\langle S \rangle$ means the \textit{span} of $S$

[stoic python]

Namington

[plain cape]

could someone help me to understand a 3x3 matrix?

not sure in which forum to post it

its actually a problem where you have to solve a system of linear equations with 3 variables using determinants of 3x3 matrices (using Cramer`s rule)

[nocturne jewel]

yeah.. what about it?

[still lodge]

[plain cape]

cool so I'm using this as a guide because I couldn't figure out why my own calculations got it wrong

the equations is at the top there

`-4(11 - 11) = 0
1(-41 - 1-1) = -3
0(-4 * 1 - 1*-1) = 0

0+-3+0 = -3
`

this is my own calc of Dx

but its giving me -3 and I have no idea why

[nocturne jewel]

you negate the middle term

$det(A_x)=-4(1-1)-(-4+1)+0$

[stoic python]

Mosh

[plain cape]

oh so its the last part I'm missing

it takes the x y z there

[nocturne jewel]

No...

you just didnt do the det right

[plain cape]

result - (x+y)+z

oh wait its 1 not -1

[still lodge]

h e lp

not looking for ans ofc, just a nudge pls

[plain cape]

good okey, thanks 😄

I was going crazy there for a while because I redid it like 3 times

btw that thing the Bot did is another way of doing it?

[nocturne jewel]

??

I just wrote out the det you put

[plain cape]

if you got other result instead of 0 there you could still do reuslt - (x+z)+0

[nocturne jewel]

No clue what you're going on about... but you just didn't negate the (1,2) Cofactor

[plain cape]

just thinking, thanks the help

wait how come its -1 when in the guide there its 1?

-4 1 0

looking at this for help, does the signs there still apply ?

[-4 = + ] [1 = -] [0 = +]

[still lodge]

if A is a 5x5 over the complexes, what does A^3 = 0 tell us about A

google (unfortunately, im down bad) told me that it means that it's only eigen vector is 0 but why

[winter harbor]

If we have any matrix $A \in \mathcal{M}{n}(\mathbb{F})$ over any field $A$, the ideal:
$$
\text{Ann}(A) = {p \in \text{F}[x] , \vert , p(A) = 0}
$$
Of all polynomials with coefficients over $\mathbb{F}$ that vanish on $A$ is a principal ideal, since $\mathbb{F}[x]$ is a PID. This means that this set is generated by a monic polynomial of lowest degree $m{A} \in \mathbb{F}[x]$ such that $m_{A}(A) = 0$ and this is precisely the minimal polynomial of $A$.
\
\
With this in mind, we have in your case that the polynomial $p(x) = x^{3}$ vanishes on $A$, i.e $p(A) = A^{3}$ and since the set of polynomials that vanish on $A$ is generated by the minimal polynomial $m_{A}$, this means that the minimal polynomial of $A$ divides $p(x)$, so that the minimal polynomial of $A$ can be only any of the following:
$$
x, x^{2},x^{3}
$$
Since the roots of the minimal polynomial are all eigenvalues of $A$ and in any case the only roots of $m_{A}$ are 0, this means that the only eigenvalues of $A$ are 0.

[nocturne jewel]

$\abs{A}=\sum_{i=1}^n (-1)^{i+j}a_{ij}\abs{C_{i,j}}$ is the Laplace expansion, where $C_{ij}$ is the matrix formed by removing the ith row and jth column of A

Probably something wrong with me writing it from memory but you can just google laplace expansion and find the accurate definition

[stoic python]

Mosh

[nocturne jewel]

yeah missed the ij entry

[stoic python]

MISTERSYSTEM

[still lodge]

ok some of those words just went over my head, my prof is unfortunately not going through it with that much rigor

can u break it down a bit please

the basic stuff first: what do Ann(A) and PID mean

[still lodge]

also in general, i must confess that i dont fully understand the idea of a polynomial "annihilating" a matrix

but that is moreso just me being dense

[winter harbor]

Ann(A) is just a notation, I defined it as the set of all polynomials that when applied to A give 0

In any case

are you familiar with the fact that if we have a polynomial we can ''apply it to a matrix''?

In the following sense

[still lodge]

to be clear, i understand what an eigenvector and eigenvalues are, what they mean for in terms of the matrix theyre obtained from, and how to obtain the characteristic polynomial

[still lodge]

i think so but cant hurt

[winter harbor]

Suppose that we have a polynomial $p(x) \in \mathbb{F}[x]$, where $\mathbb{F}$ is a field, so we can write it as:
$$
p(x) = \sum\limits_{k=0}^{n} \alpha_{k} x^{k}
$$
for some constants $\alpha_{0}, \cdots, \alpha_{n} \in \mathbb{F}$, then if we have a square matrix $A \in \mathcal{M}{n}(\mathbb{F})$, we can define p(A) as:
$$
p(A) = \sum\limits{k=0}^{n} \alpha_{k} A^{k}
$$
where $A^{k}$ is just the product of matrix $A$ with itself $k$ times.

[stoic python]

MISTERSYSTEM

[winter harbor]

Is that fine with you?

so if we have this notion

we can talk about the set of matrices that vanish on a certain matrix A

and I denote it by Ann(A)

And this is the set (In fact an ideal of F[x]) of all polynomials such that p(A) is equal to the zero matrix.

and turns out

that F[x] is a principal ideal domain, meaning that every ideal of F[x] is generated by only one element.

you might know this fact

or might have heard of the fact

that if a matrix is such that p(A) = 0

then the minimal polynomial of A divides p(x)

[winter harbor]

And this is a consequence of this fact,

Although I think most LA courses don't prove this theorem using this language

in any case

if a polynomial vanishes on A

Then the minimal polynomial of A divides this polynomial

[still lodge]

especially not mine, the one im taking is geared towards CS students, im trying to learn the rigor myself

[winter harbor]

https://math.stackexchange.com/questions/2042296/if-qa-0-then-q-is-divisible-by-the-minimal-polynomial-of-a

There you go

This is a really useful result

Are you familiar with Cayley-Hamilton's theorem?

[still lodge]

i know of its existence

prof promised me a proof and never taught it

im not familiar enough w it to recognize an application, i have a lot to do for this class

[winter harbor]

basically

it tells us that the characteristic polynomial vanishes on A

meaning that if we denote by p_A the characteristic polynomial of A

then p_A(A) = 0

[winter harbor]

By this

this also tells us that the minimal polynomial divides the characteristic polynomial

which implies that any root of the characteristic polynomial (i.e, an eigenvalue!) is in fact a root of the minimal polynomial too

with all this information together

we know that since p(x) = x^3 vanishes on A

i.e, A^3 = 0

then the minimal polynomial of A divides x^3

so there are a few possibilities for the minimal polynomial

it is either x, x^2 or x^3

[winter harbor]

from this fact, we also know that the roots of the minimal polynomial are eigenvalues of A

so that in any case

i.e

if the minimal polynomial of A is x, then the only root of this polynomial is 0

so that the only eigenvalue of A is 0

if it is x^2, then same deal, the only root of the minimal polynomial is 0, so the only eigenvalue of A is 0

and the same for x^3

so that we know that the only eigenvalues of A are 0

[still lodge]

holy shit this is a handful

ok im gonna need a bit but im gonna try and get this all understood

tremendous thanks for the detailed explanations <3

[winter harbor]

this is a very important result, I'd recommend you to take a look.

You will also need this for part (b).

for instance

you know the minimal polynomial of A

and by this theorem

[still lodge]

part b

[winter harbor]

you know the eigenvalues of A immediately

[still lodge]

ok that's easier to see i think just bc it's already a polynomial

[still lodge]

i think it's this core idea that i need to get in my head properly bc clearly it's shaky

[winter harbor]

After finding the eigenvalues, you are pretty much done tho.

it's all a matter now to study the possible jordan blocks and so on.

depending on the geometric and algebraic multiplicity of the eigenvalues.

[forest quiver]

Ok so just to make sure I am getting it correctly

In linear algebra

or under vector spaces

[still lodge]

unrelated but how'd ya get that studying role

[winter harbor]

you type '',iam stud'' on #bots

[forest quiver]

There are 3 types of multiplication (products)

1. Inner product/Dot product : a multiplication of two vectors given by |a||b|cos(theta) or a_1b_1 + a_2b_2 + ... +a_n*b_n

2)Cross product: a multiplication of two vectors that has some sort of geometric area interpretation

3. Normal product: a multiplication between vectors and scalars, matrices and scalars, matrices and matrices, and matrices and vectors. (given than they are compatible)

Informal but is the idea there at least?

[winter harbor]

Yeah, it is.

[forest quiver]

Ok this is probably the most confusing thing

to be honest

so inner prod is a projection

[winter harbor]

In some sense...

So

The idea of the inner product can be quite abstract at first

But it basically formalizes some intuitions we have about ''angles and orthogonality''

[forest quiver]

Yeah I know

Oh nice

ok

[winter harbor]

There's an article I was reading a few days ago

and it explains pretty well

[forest quiver]

OK

[winter harbor]

this idea

[forest quiver]

cross product is pretty useful

We used it in class to find area of a parallelogram and parallelopiped

kind of confusing though

[raven badger]

How the h3ll is the dimension of these row vectors 3?

Surely, they're in R^5. I get that some of the vectors can be (in fact are) linearly dependent. Therefore not contributing to the span.

But I do not get, how the basis of these row vectos can be formed using only 3 vectors, when I have another theorem/proof, stating that for S in R^n -> dim(S) = n???

[forest quiver]

Well thanks @winter harbor for not bombarding me in jargon when you explain stuff. Makes it pretty nice to ask questions, to be honest. I can't say the same for my teacher

[raven badger]

I am basing my assumption, that Dim(RowVectors(A)) = 3, on the fact, that the non-0-row-vectors of RREF(A) are the basis for the row-vectors of A.

[winter harbor]

https://www.lesswrong.com/posts/rs2focaRymwvkW2jS/inversion-of-theorems-into-definitions-when-generalizing

This article is REALLY GOOD

Basically

we know what angles and lenghts are intuitively

since millenia

and after René Descartes introduced the notion of coordinates in geometry

people started playing around with these ideas

using what was already known from euclidean geometry

And using stuff like the pythorean theorem and the cosine rule

we were able to conclude stuff about the inner product

that is related to geometry

i.e

<u,v> = |u| |v| cos(theta) for two vectors and so on

or that the lenght of a vector

is equal to the square root of <u,u>

It turns out that nowadays, we sort of do the opposite, we think of angles and lenghts in terms of the inner product.

and not the other way around

[plain cape]

why does this 3x3 matrix solution fail when it comes to a problem like this?

I tried it with another calculator and it gives me the same result

either this is correct or my math book is wrong

[plain cape]

Ah I guess you have to use the inverse approach in this case

[still lodge]

wdym exactly by "vanish"

is this like "matrix roots"

[winter harbor]

well

p(A) is another square matrix when A is a square matrix, right?

so it makes sense to ask for p(A) = 0

where 0 is the 0 matrix

[still lodge]

yeah i suppose

[winter harbor]

and we say that p vanishes on A if p(A) = 0

or that p annihilates A

[still lodge]

ok fair

[still lodge]

this isn't cayley hamilton is it

[winter harbor]

No

It is not

[still lodge]

ok didnt think so

[winter harbor]

Cayley Hamilton tells us that the characteristic polynomial of a square matrix A vanishes on A

meaning that if the characteristic polynomial of A is denoted by p_A(x) = det(A-xI)

then p_A(A) = 0

[still lodge]

i have that previously written as "every matrix is a root of its own characteristic polynomial"

which is suppose is analogous(?)

[winter harbor]

It is analogous, yes.

but ''root'' in a different sense

ofc

polynomials act on matrices in a different way as they do on elements in a field

[still lodge]

just in the sense that plugging in the matrix to the polynomial will "send it" to a matrix of 0's thus annihilating it

[winter harbor]

yeah

[still lodge]

epic

[still lodge]

i know very little about fields, just that R is the elementary example of one - how important is this

[winter harbor]

Oh, if you are not familiar with fields, then you can just consider the case where the field is the reals, the complex numbers or the rational numbers.

[still lodge]

epic pt2

ok now that i think i understand what that SO post is stating in the question that feels so random/contrived

[winter harbor]

I'd recommend you to take a look at the definition tho

It's pretty simple

It's basically a set with a multiplication and a sum

Which are commutative, associative, have an indetity, distribute over each other and etc...

And that every non zero element has a multiplicative inverse.

[still lodge]

im unfortunately going to have to leave that for another night, im spending a lot of time trying to understand all the other stuff, too much in fact, and i should get to actually doing my problems soon

though i do get that it's just a general set of things with a set of rules to oversimplify it

mildly related - the minimal polynomial of a matrix A is obtained by first finding the characteristic polynomial and then finding what divides that, and guess+check...?

^probably wrong, going off memory there

wait

that just makes sense

[winter harbor]

Yeah, one way to find the minimal polynomial is by first calculating the characteristic polynomial. And then, using the fact the minimal polynomial has the same irreducible factors as the characteristic polynomial, you basically does a smart guess+check with that.

[still lodge]

the q(x) discussed in the stack overflow post is just something "between" (in terms of divisibility) the minimal polynomial and characteristic polynomial

[winter harbor]

It's just some other polynomial that vanishes on matrix A too.

[still lodge]

hmmm

[winter harbor]

Are you talking about this post?

[still lodge]

yeah

or at least the proposition in the title

[winter harbor]

I think you meant "q" right?

p is the minimal polynomial of A in the notation of the post.

[still lodge]

oh yeah my b

q is something divisible by p

does it follow that q divides the characteristic polynomial of A?

that might be in the post, i havent finished

[winter harbor]

No

[still lodge]

ah okay

[winter harbor]

Not necessarily

For instance

If A^3 = 0

Then the characteristic polynomial of A is x^5

This follows from the fact that the degree of the characteristic polynomial is the same as the size of the matrix

And since your matrix has size 5×5

But notice also

That q(x) = x^6

Also vanishes on A

But q(x) does not divide x^5

Which is the characteristic polynomial of A

[still lodge]

ok understood

[still lodge]

i think i fully understand this now - why do i care though, is it applicable only in questions like the one from the beginning of the discussion?

oh wait

so does cayley hamilton follow directly from that...?

[winter harbor]

No, it doesn't.

It tells us that the minimal polynomial divides the characteristic polynomial tho.

since cayley hamilton tells us that the characteristic polynomial of a matrix A anihilates A.

which is a very useful fact

[still lodge]

ohhhhh

see my next question was about to be regarding the usefulness of the characteristic polynomial, besides being something that we get eigenvalues from

but i suppose that's the usefulness right there isnt it

[winter harbor]

It can give us a bunch of information about the matrix, its eigenvalues (geometric and algebraic multiplicity of the eigenvalues, its jordan canonical form and etc)

It can even tell us stuff about the size of the matrix

if we don't know it beforehand

[still lodge]

well yeah ig i meant in the context of these other polynomials

i think im just used to LA being about what's covered in the 3b1b videos lol, so now that it's getting more numerical my brain has to ~adjust~

[winter harbor]

Yeah, I don't have much a good intuition for this stuff ''geometrically'' too.

It's easier to get some algebraic intuition tho

the minimal and characteristic polynomial are mainly algebraic tools

[still lodge]

i didnt expect it, youve already been immensely helpful just in discussing this

[winter harbor]

I think you may be able to solve the questions now

post the solutions if you get them

[still lodge]

not using latex unfortunately but will do

gonna take a few more notes first tho

[still lodge]

shouldnt this be that the roots of the characteristic polynomial are our choices for the roots of the minimal polynomial

im realizing i probably need more practice just finding the minimal polynomial

speaking of, side question: why do i care about what the the minimal polynomial is anyways - i get that it's the "smallest" polynomial that annihilates A, but why do i care which polynomial is specifically the smallest

[winter harbor]

I'd go as far to say that the minimal polynomial is the natural notion to consider rather than the characteristic polynomial.

[lime zinc]

Looks like it will be help full to write. Matrix jordan form

[winter harbor]

Like, reason why the characteristic polynomial is useful is because it is computationally easier to find it.

But the minimal polynomial is a much more intrisic notion.

Ah and ofc, Jordan Canonical Form.

It's the best we can do to a diagonalization in some cases.

And we need the notion of the minimal polynomial to compute it.

Also

A matrix is diagonalizable iff its minimal polynomial splits into distinct linear factors, i.e has no repeated roots. (This works over any field)

which is really useful

[still lodge]

god there's so much overlapping stuff

dont get me wrong it's cool

but piecing this together is a task

[still lodge]

what about this tho - i was thinking like what if you have a characteristic polynomial (x-3)(x-2), then isn't a potential minimal polynomial (x-2)..?

oh well ig there's an (x-3)^0

but that's lame

[winter harbor]

yeah, the possible minimal polynomials are (x-3),(x-2) or (x-3)*(x-2)

I found a worked out example

https://www.math.uh.edu/~klaus/08spt03_solu.pdf

Maybe this can be useful

[still lodge]

will check it out later, mucho thanks

going wayyyyy back to the original question

i now understand that just reading part a), we know that the eigenvalues are all 0

so then based on this we can just write all the jordan forms w blocks of varying sizes in a 5x5 where the diagonal is 0's right

but those are all just 0 on the diagonal and 1 on the diagonal above it...? so it's just one JC form...?

might be pushing my understanding a bit there

wait no

the 1's change

[winter harbor]

Yeah, you may have different jordan blocks with different sizes.

but they all have 0's on the main diagonal

[still lodge]

👉👈

sorry for orientation

but i think this is it

[still lodge]

order of the blocks doesnt matter right

otherwise this gets really out of hand

[winter harbor]

Yup, it doesn't matter.

[stoic python]

ImperfeKt

[frank yacht]

this transformation is not linear right?

it doesnt follow T(c*v) = c * T(v)

[still lodge]

wait are those forms right then

[teal grotto]

yes, they are not linear. T(cv) = c^2 T(v)

[frank yacht]

but book says they are linear

probably a misprint but I wanted second opinion

thanks

[teal grotto]

unless the field you're working over is the field with two elements, then it fails to satisfy T(cv) = cT(v)

[winter harbor]

That looks pretty good, I will actually give a better look tomorrow.

Since it's pretty late now.

[frank yacht]

input is vector actually,
It's T(v) = v1 * v2 where v = (v1, v2)

[winter harbor]

now try to find the elementary divisors and the invariant factors in each case.

[still lodge]

gotta learn what those are but yeah

[teal grotto]

yes, i understand, but if the vector space is defined with scalars in the field with two elements, then c^2 = c for all scalars c

[still lodge]

yeah it's 2am here tho, again thank you so much for discussing this

[winter harbor]

Has your professor covered the rational canonical form and the primary and invariant factor decomposition?

They appear in this context

[still lodge]

i'm staying up anyways so ¯_(ツ)_/¯

[winter harbor]

In any case, if you know the characteristic and the minimal polynomial

finding them is not hard

[winter harbor]

Just follow this

[still lodge]

nah this is stuff im doing on my own lol

prof gave me problems to look thru on my own

[winter harbor]

https://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain

Anyway

It's 3am now

I gotta go to bed oof

[still lodge]

yeah i got 9 more problems i wanna finish im staying up the night

but thank you, good night :)

[still lodge]

are invariant factors the same as elementary divisors

[slow scroll]

nah

[still lodge]

what are each then :(

[slow scroll]

elementary divisors are powers of primes and invariant factors satisfy the divisibility condition d1 divides d2 divides d3 etc..., but both the elementary divisors and invariant factors are unique for a given f.g. module over a pid

[still lodge]

what about when something asks me for the invariant factors and elementary divisors of a matrix

im doing this

i already found JC

[slow scroll]

hmm okay, what did you get for the minimal polynomial?

[still lodge]

well about that

the characteristic is -(x-1)^3

using x for simplicity

from that do i just guess the minimal...?

but ig the only options are (x-1) and (x-1)^2

or ig (x-1)^3 too...?

it's gonna take a bit for this all to sink in lol

[slow scroll]

right, this is because by cayley hamilton, the minimal polynomial divides the characteristic polynomial, so the minimal polynomial could be any of those three options. Whichever one has smallest degree and still annihilates A

i don't know of any good ways of figuring out which is the minimal poly tho.

better than just directly checking*

[still lodge]

well why do i need it in this case anyways

[slow scroll]

in this case there is only one elementary divisor which is equal to the minimal polynomial since the minimal polynomial is a prime power

[still lodge]

it's (x-1)^2

[slow scroll]

so you need it for that

alright so (x-1)^2 would be the only elementary divisor

[still lodge]

is that the case in general? a question asking for elementary divisor(s) is just asking for the minimal polynomial...?

actually ig that just

makes sense

[slow scroll]

no, not quite. Let's say your minimal poly is like $p(t) = (t-x_1)^{n_1}\cdots (t-x_k)^{n_k}$. Then since i've written $p$ in its prime factorization, the elementary divisors must be $(t-x_1)^{n_1}, \dots, (t-x_k)^{n_k}$

[stoic python]

kxrider

[slow scroll]

idk how much generality ur working in, but the general procedure would be to take your minimal polynomial, and factor it as a product of prime powers (primes need not be linear!). Then those become ur elementary divisors

If you're working over the complex numbers, then primes are always linear (of the form t - a)

[still lodge]

so ig just the bare "elements" of the minimal polynomial

[slow scroll]

yea kinda

[still lodge]

but they retain their powers

[slow scroll]

like x^2 + 1 would be a prime polynomial over R, because its roots are +sqrt(-1), -sqrt(-1), and we'd have to leave the reals to factor it any further

[slow scroll]

yea, so like (x^2+1)^5 could be an elementary divisor of some matrix over R

now uh lets see for invariant factors i'd have to think.

[still lodge]

what are those anyways

[slow scroll]

they are another set of polynomials u get by the structure theorem, except these are polys q1, q2, ..., qn that satisfy q1 divides q2 divides q3 etc..

wrong channel

and this channel is occupied

[still lodge]

this is just something ive seen but does it have to do with when you create like a chain of divisibility...?

[slow scroll]

yea, but it can't be just any chain of divisibility either

[still lodge]

i mean wouldnt my only option here be (x-1) tho...?

[slow scroll]

it would be something like x-1, (x-1)^2, but im not confident enough on this to say for sure. I've only dealt with these concepts in a very abstract setting
maybe someone else can explain it better

[still lodge]

im just writting (x-1)|(x-1)^2 as the ans

i think that's it but i also wanna keep moving w these problems

merci :D

different thing: if a question asks me to "describe up to similarity all 4x4's s.t. [insert rule here]" then it's just asking me for the JC forms right

[slow scroll]

if [insert rule here] has a char poly or something then yea

[still lodge]

something like A^5 - A^3 = 0

so it translates to a char poly yeah

[slow scroll]

so this says that t^5 - t^3 is a poly that annihilates A, so t^5 - t^3 might not be the char poly, but anything dividing it could be the min poly (including t^5 - t^3 itself!)

and the min poly is what determinate the JCF

[wintry steppe]

let's say you have R^2 and take the x-axis as a subspace U = {(x, 0) | x ∈ R} and also take the y-axis as a subspace Z = {(0, y) | y ∈ R}

now the sum of subspaces U + Z would be R^2 right?

[still lodge]

so ig it doesnt really matter what the char poly is, just that from this info we can find the min poly

[slow scroll]

we can find all possible minimal polynomials, yea

[slow scroll]

yes

[still lodge]

oh god this is annoying

so since i have x^5-x^3

and that factors to x^3(x-1)(x+1) every combination of those are the possibilities right

[slow scroll]

anything that divides x^3(x-1)(x+1) yea

.... which has 16 divisors. hmm i wonder if there is a way to narrow it down

maybe instead of writing every single matrix, just compute the possible jordan blocks

and find some clever way to write down the answer with as little ink as possible lol

[still lodge]

bleh alright i suppose

[lofty topaz]

Is it subspace ? Should I prove it using mathematical induction ?

[still lodge]

dawg

also okay there's 5 possible jordan blocks

and it's a 4x4

and three of the jordan blocks are from x = 0

so does that mean there's only 6 possible JC forms

[still lodge]

also i cant really read that but i see recursively defined sequence i default to induction or something something monotonicity

[still lodge]

i realize people aint active rn but next someone is: some help understanding how you can literally pull an eigen vector out of your ass at will for some problems would be very nice

https://www.youtube.com/watch?v=dPxAha-W9j8 like it's done here near the 12 min mark

[slow scroll]

im going to bed in like 5 minutes but wdym by this?

[still lodge]

*8 min mark

im doing this now

did the eigen value and nullspace stuff

eigen values are 0 and 1 but you only get one eigenvector for each

peyam does an example where something similar happens and just fucking generates an extra vector and i dont get why/how it works

i might sleep soon too

i tried me darndest

holy i just realized what you were replying to

[still lodge]

i just wrote 6 JC matrices im chillin but ty

[slow scroll]

okay so the idea for this is you want to find a basis of B which is in JCF

so what peyam is saying at the 8min mark doesn't make sense?

[still lodge]

it's the part right after it where he pulls the eigenvector out of his ass

it's the generation of that new one that i dont get

i can follow the process it's just weird to understand ig

[slow scroll]

okay where is the exact point he pulls the eigenvector out of his ass? He starts with two eigenvectors, and computes the last vector (not an eigenvector) he needs for a Jordan basis if im understanding correctly.

[still lodge]

also i have no idea how he makes the conclusion he does at 13:30

ohhh it's not an eigenvector?

that computation ig

[slow scroll]

so lets say you have a JCF like this. Then the columns of each of those entries i circled corresponds to an element of your jordan basis which is an eigenvector

does that make sense?

[still lodge]

yeah

[slow scroll]

so lets take the second column for example now. Lets say v2 is the second basis vector. Then if you call that matrix "A", then
Av2 = v1 + lambda1 v2
so (A - lambda I)v2 = v1, so to find the basis vector u need for the second column, you have to find the solution to (A - lambda I)x = v1

in your particular case on the problem of computing $B^n$, you are going to have one nontrivial Jordan block which looks like $\begin{pmatrix}\lambda & 1 \ 0 & \lambda \end{pmatrix}$ where $\lambda$ is one of your eigenvalues. So pick one of the eigenvectors you got, call it $v_1$. Then you need to solve $(B - \lambda I)x = v_1$, where $\lambda$ is the eigenvalue of $v_1$

[stoic python]

kxrider

[slow scroll]

the solution x is the second basis vector you need to write down the change of basis matrix stuff for the jordan block

so your final jordan basis will be something like v_1, v_2, x, where v_1 and v_2 are the two eigenvectors you got, and x is the additional vector you got from solving the equation from above. And you would use this basis to make a change of basis matrix etc...

[still lodge]

jesus christ my brain is gonna melt

im gonna take a break

but ty for the help <3

will read this when im functioning again

[slow scroll]

Aight sounds good!

[still lodge]

<@&268886789983436800> i hope this is the right time to ping but this is in every channel

[subtle walrus]

i think they got the memo, more pings won't help

[wintry steppe]

is there much calculus/diff eq involved in linear alg?

[lavish jewel]

there can be, since many operations you deal with in calc and diff eq are linear

[marble lance]

^ Examples can come from calculus and diff eq, but you don't need it for the theory

[dusky epoch]

it usually goes the other way around

diffeq typically involves a sizeable amount of linalg

[still lodge]

had study role didnt see other channels getting spammed too before pinging

[lavish jewel]

all good, the spammers were taken care of

[still lodge]

kinda LA but is the conjugate of a real number just the same real number

[lavish jewel]

yes

[still lodge]

im being asked to prove this and ive gotten to <v,u>/2 + <u,v> /2 = <v,u>

but that feels like it should just be true for a real inner product space

(question specifies that)

[lavish jewel]

what are the brackets and || supposed to represent? the usual inner product and 2-norm in C^n?

[still lodge]

only instruction is "Prove that if V is a real inner-product space, then"

and then that

but we've only really acknowledged the presence of complexes in the class, havent really worked with them

<> is inner product, || is norm

but judging by "real" im assuming it's in R^n

[lavish jewel]

yeah the extra info is needed because those two things can be defined in several different ways

there is no "THE inner product", nor "THE norm"

[still lodge]

that's true my b

i just saw a sunrise im very tired lol

[leaden tide]

i think it's an inner product and the associated norm ||x|| = √(<x,x>)

so really, just calculate and it'll work

[sick sandal]

Let $\mathbb{F}$ be a field of characteristic zero and let $V$ be a
finite-dimensional vector space over $\mathbb{F}$. If
$\alpha_{1},\ldots,\alpha_{m}$ are finitely many vectors vectors in
$V$, each different from zero vector, prove that there is a linear
functional $f$ such that $f(\alpha_{i})\neq 0, i=0,\ldots,m.$
i tried using induction but not sure how im supposed to proceed of if thats even the correct approach

[stoic python]

james_ash_.

[sick sandal]

all i need is some lead and i can solve it

im also unsure how having char zero field will assist here

[dusky epoch]

i think you might want to pick a LI subset from your alphas

hm wait no

this might not work as cleanly as i might've imagined

[slow scroll]

@sick sandal Lets say you have $\alpha_1, \alpha_2, \dots, \alpha_{m+1}$ for $m = dim (V)$. Then there exists $1 \leq j \leq m+1$ such that $$\alpha_j = \sum_{i \neq j} a_i\alpha_i.$$ Also,
$$f(\alpha_j) = \sum_{i\neq j} a_i f(\alpha_i) $$

Lets say you have $\alpha_1, \alpha_2, \dots, \alpha_{m+1}$ for $m = dim (V)$. Then there exists $1 \leq j \leq m+1$ such that $$\alpha_j = \sum_{i \neq j} a_i\alpha_i.$$ Also,
$$f(\alpha_j) = \sum_{i\neq j} a_i f(\alpha_i) $$

[stoic python]

kxrider

[slow scroll]

spark any ideas?

wrong channel try #groups-rings-fields

(that "wrong channel" was not to you ofc, james)

wait actually what i have in mind might not work, didn't think through this very far

ok yea there is something simple you can do here. If f(a_j) is zero, you can just modify one of the f(a_i) so that f(a_j) is nonzero. Can you see how?

the particular way you have to modify one of the f(a_i) depends on the characteristic of the field being 0

also the list of alphas being size m+1 is not important here. I had a different idea before

[sick sandal]

not sure i see it but i think it should be enough so i can figure it out from here tysm

[slow scroll]

aight np, let me know if u still have any trouble

[winter harbor]

This is called a polarization Identity.

[wintry steppe]

a banach space is a normed vector space with a metric right?

[winter harbor]

A Banach Space is a normed vector space whose induced norm is complete, i.e every Cauchy sequence converges.

And by induced metric of a normed vector space $(V, | \cdot |)$ I mean that the function:
$$
d(u,v) = | u - v |
$$
For $u,v \in V$ is a metric.

[wintry steppe]

thanks

[stoic python]

MISTERSYSTEM

[winter harbor]

No more typos now

[wintry steppe]

lloll

[winter harbor]

Np

[winter harbor]

Let $(V, \langle \cdot, \cdot \rangle)$ be a real inner product space and denote by:
$$
| u | = \sqrt{\langle u, u \rangle}
$$
Its induced norm on $V$.
\
\
Now, let $u,v \in V$, then we have:
$$
\langle u+v, u+v \rangle = \langle u,u \rangle + 2 \langle u,v \rangle + \langle v,v \rangle
$$
Thus, this implies that:
$$
\langle u,v \rangle = \dfrac{| u + v|^{2} - | u|^{2} - |v|^{2}}{2}
$$
Moreover, we also have that:
$$
\langle u-v, u-v \rangle = \langle u,u \rangle - 2 \langle u,v \rangle + \langle v,v \rangle
$$
And thus
$$
\langle u, v \rangle = \dfrac{|u|^{2} + |v|^{2} - |u-v|^{2}}{2}
$$
And by summing both equations together, we prove that:
$$
\langle u,v \rangle = \dfrac{|u+v |^{2} - |u - v |^{2}}{4}
$$

[stoic python]

MISTERSYSTEM

[marble lance]

All that langle rangle must have taken forever 👀

[winter harbor]

Yeah, I am on my cellphone too oof

[marble lance]

Just do < , > and hide when the latex purists come to yell at you

[winter harbor]

But doctor... I am a LaTeX purist...

[winter harbor]

Anyways, this is called a Polarization Identity because it let's you recover the inner product from its induced norm.

And is pretty much just a fancy name for the parallelogram law.

[earnest sigil]

is there a way to find determinant for any matrix whose order is greater than 3

[lavish jewel]

there are many. most are not suitable to do by hand

you could try using minors and cofactors

[hard drum]

row reduction usually best in my experience for nxn matrices, n > 3

(by hand, i mean)

[lavish jewel]

you have to keep track of the operations in case any alter the det, tho

but yeah, triangularizing the matrix

[wintry steppe]

What's -x/9=x/5?

Answer is 90 and no idea how

[lavish jewel]

try either #prealg-and-algebra or a math help channel

[wintry steppe]

Did it in pre calclus

but how on earth is it 90

[lavish jewel]

are you just trolling?

[wintry steppe]

No lol

I been getting 14x/45

And the teacher said it's A which is 90

Literally

[earnest sigil]

how to do this

[sick sandal]

can you send the question as its presented in your book/paper @wintry steppe in #precalculus

[wintry steppe]

Gotcha.

[wintry steppe]

I'll show the question. Also bold of you to assume I'm trolling

[lavish jewel]

i have to make sure

i'll go take a look as well

[wintry steppe]

#precalculus

[stoic python]

MISTERSYSTEM

[winter harbor]

I forgot to square everything

[lavish jewel]

my main remark was rather that you were using | instead of \Vert

i didn't even know that worked

[winter harbor]

\Vert\Vert just takes so much work to type every goddam time

\ | is the way to go

[gray dust]

do $\norm{x}$

[stoic python]

RokettoJanpu