linear-algebra

[solid sorrel]

there are as many 2x2 rotation matrices as there are rotations on the plane

[torpid barn]

oh wait

im clowning

[wintry steppe]

you either define a rotation by that form, and it's definitional, or you accept that a "rotation" ought to be a linear transformation represented by an orthogonal 2x2 matrix with determinant 1, and then you prove that such matrices take the form you're familiar with

[torpid barn]

nvm

thanks guys

[wintry steppe]

Hey guys, I got a symmetric matrix, I've gotten its eigenvalues, and now I have to get the min based on x^2+y^2 =1

How do I minimise this?

[wintry steppe]

Greatly appreciate any help

[torn stag]

@clear kettle Assume that $av_1 + bv_2 = 0$, and prove that $a = b = 0$.

[stoic python]

IlIIllIIIlllIIIIllll

[clear kettle]

it is linearly independent in math notation

[viscid lagoon]

Do $A \in \mathbb{R}^3$ and $\alpha, \beta, \gamma \in \mathbb{R}$ exist such that:
$$ \left(\begin{array}{lll}
\alpha & 0 & 0 \
0 & \beta & 0 \
0 & 0 & \gamma
\end{array}\right)=A^{-1} M A .$$
If so, evaluate $A$ as well as $\alpha, \beta, \gamma$. Does anybody have just a small hint for this?

[stoic python]

lewis

[wintry steppe]

is M an arbitrary matrix ? if so the solution is obviously yes, any diagonalizable matrix satisfies that by definition

[viscid lagoon]

oh sorry

wait, let me type it out

$$M=\left(\begin{array}{ccc}
0 & 0 & 3 \
1 & -1 & -9 \
0 & 2 & 7
\end{array}\right)$$

[stoic python]

lewis

[wintry steppe]

Ah, seems like they're asking you to determine whether this matrix is diagonalizable

[zinc timber]

that proves v1 and v2 are linearly dependent

[viscid lagoon]

yeah, basically

[zinc timber]

in math notation

[wintry steppe]

typically you would start by calculating the characteristic polynomial

for each eigenvalue found you determine the eigenspace

[viscid lagoon]

I did, the problem here is we haven't introduced

the fact that something is "diagonalizable"

so I have to justify that this conjugation exists

no?

[wintry steppe]

wdym

[viscid lagoon]

this

Assume we don't know what it means for a matrix to be diagonalizable.

Just from this question and the given matrix M, what would the approach be

I've already calculated the characteristic polynomial

and for each eigenvalue, i've determined the eigenspace

Let me type out my results for this:

[wintry steppe]

Ah I see, without that I don't see how we could determine such matrix

[viscid lagoon]

I mean there is a recipe on how to find such a matrix, no?

[wintry steppe]

you found that in which chapter

[viscid lagoon]

I would just have to justify it

somehow

[wintry steppe]

yes there is but it has to do with diagonalization

[viscid lagoon]

What is it then

I mean diagonalization is just a way to present, how do you justify that the given "recipe" works.

[wintry steppe]

you could determine it using diagonalization and just plug in the calculations but that doesn't seem satisfactory

[viscid lagoon]

Yeah

[wintry steppe]

a matrix is diagonalizable iff the sum of the dimensions of eigenspaces is equal to the dimension of the space we're working in.

to prove that we start by noticing that a sum of eigenspaces is always direct, and when the matrix is diagonalizable the dimesion of each eigenspace is equal to the multiplicity of the corresponding eigenvalue in the characteristic polynomial

but not sure if this is what you're looking for

[viscid lagoon]

yeah, this seems a bit much

hmmm

I've been wondering what they want

or expect me to do

[wintry steppe]

to formulate your question differently

they're asking you prove that M is equivalent to a diagonal matrix

[viscid lagoon]

yeah

pretty much this

and I'm sure we'll have to use the results of the eigenspaces and eigenvalues I've determined

[wintry steppe]

Oh so you are allowed to use them ?

[viscid lagoon]

Yeah

I'm allowed to use them

[wintry steppe]

I see

so oncee you determined the eigenspaces

[viscid lagoon]

The problem is just

If I do use the recipe, then I need to justify it

[viscid lagoon]

yeah

?

[wintry steppe]

wdym justify it

this recipe is in any textbook about linear algebra

unless they want you to reproduce the proofs

[viscid lagoon]

yeah

probably

All we have is

[wintry steppe]

well that's quite hard

[viscid lagoon]

the definition of eigenvalues

and eigenspaces

basically

[wintry steppe]

maybe they just want you to determine the eigenspaces properly

justify any intermediary steps etc

did you determine the eigenspaces of each eigenvalue ? what are they

[viscid lagoon]

yea

wait

the eigenvalues are 1, 2 and 3

[rose crag]

hi

@languid wigeon

can u refer to the problem u have posted

[viscid lagoon]

the eigenspaces for each eigenvalue are E_1 = span((-3, -3, 1)), E_2 = span((3, -5, 2)) and E_3 = span((1, -2, 1))

@wintry steppe

[languid wigeon]

i started from #linear-algebra message

[viscid lagoon]

and the three vectors that span each of the eigenspaces are linearly independent

[rose crag]

after taking mod n, new d becomes

mR transpose mod n right?

you missed mod n

[rose crag]

sure

[wintry steppe]

I am back

Okay so the dimension of each eigenspace is what ?

[viscid lagoon]

1

so 1 + 1 + 1 = 3

[wintry steppe]

yep!

so it's diagonalizable

[viscid lagoon]

problem is

why is it diagonalizable then

haha

[wintry steppe]

because the sum of the dimensions of eigenspaces = the dimension of R^3

that's a standard result of diagonalizability

I already gave you the outline of proof if you want to look into it

but not sure you'll need it here

[viscid lagoon]

i'll look more into it

is there an easy proof to that

or i mean, even (b) works

[wintry steppe]

all of them do haha

[viscid lagoon]

yeah

but one of them is easier than the other

i guess

[wintry steppe]

Yes true

that's why we have multiple equivalent conditions

[viscid lagoon]

yeah

[wintry steppe]

you mean proof for (a)=>(e) ?

[viscid lagoon]

yeah

or no

prove (e) => (a)

oooor

(b) => (a)

would help me

i guess

[rose crag]

@languid wigeon why didnt you include mod n?

[dawn ocean]

you made an error in your calculation of the answers weren't the same on both sides.

for this problem you seem to be struggling with what vector addition and subtraction means geometrically, forget the the angles entirely you don't need them.

to start off with draw the origin zero, then draw v and w from the origin.

to add vectors v+w you translate a copy of w so that the it's origin is at the end of v and then draw a new arrow v+w from the origin to the new place the vector ends, vector subtraction is the same thing you just flip the arrow of the second vector and do the same process of adding them together

[stoic python]

AimaneSN

AimaneSN

[languid wigeon]

i used $\mod{n}$ to kill $uvU$

[stoic python]

vin100

[languid wigeon]

for $mR^T$, i proved that that's bounded above by $\sqrt{n}$

[stoic python]

vin100

[languid wigeon]

i mean entry of its entry is bounded above by square root of n

every entry

[rose crag]

ohh

so you mean that mod n wont affect it as its upper bound is

root n

am i right

[languid wigeon]

acutally i found this problematic

the gap $n - \sqrt{n}$ gets larger and larger as $n$ grows

[stoic python]

vin100

[languid wigeon]

$\mod{n}$ only guarantees that an entry has absolute value less than $n$

[stoic python]

vin100

[rose crag]

mm yes

[languid wigeon]

i'm not sure what mod n does actually

if the mod operator % in the computer guarantees nonnegative output then we're fine

i'm afraid that the mod operation might pull an entry near $\sqrt{n} - n \in (-n,0)$

[stoic python]

vin100

[languid wigeon]

i tried in #bots . the % operator in ,calc gives nonnegative results

[rose crag]

oh

your condition is valid only for n>=4 right?

what should we do for n =1,2,3

[languid wigeon]

if we assume this, just like the assumption $0 \le r < q$ in Euclidean Algorithm, then the $nvU$ wouldn't affect $mR^T$.

[stoic python]

vin100

[languid wigeon]

sorry i don't think $n= 1$ is worth considering

[stoic python]

vin100

[languid wigeon]

and $n = 2,3$ have no practical value

[stoic python]

vin100

[languid wigeon]

$m \in {0,1}^n$ is a binary message of length $n$.

[stoic python]

vin100

[languid wigeon]

to see how much an emoticon takes, let's try the bot

,,ℹ️ info icon rendered in unicode code

[stoic python]

vin100

[languid wigeon]

each hex digit is equiv to $2^4 = 16$ bin digit

[stoic python]

vin100

[rose crag]

why

[pearl elm]

You are referring to this right? #linear-algebra message

[rose crag]

no

[pearl elm]

Sorry was trying to reply to @dawn ocean

Shoulda went on laptop

I know I made many failed parallelograms, just trying to figure out which one it is your commenting on

[languid wigeon]

that's the binomial theorem

sorry idk its connection with linear algebra.

[gleaming trench]

I don't know whats linear algebra I just looked for whatever looked the hardest to understand what it was. Thank you.

[static bane]

how do I show this using determinant's properties?

[dawn ocean]

too all of your drawings they're all labeled wrongly

it suggests you don't know how to add vectors geometrically

like you're just drawing random theings and hoping it works

there is a process to add vectors geometrically and i described it in my ping

[languid wigeon]

\begin{align}
& \det\begin{pmatrix} 1 & a & bc \ 1 & b & ca \ 1 & c & ab \end{pmatrix} \
&= \frac{1}{abc} \det\begin{pmatrix} a & a^2 & abc \ b & b^2 & abc \ c & c^2 & abc \end{pmatrix} \
&= \det\begin{pmatrix} a & a^2 & 1 \ b & b^2 & 1 \ c & c^2 & 1 \end{pmatrix} \
&= \det\begin{pmatrix} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{pmatrix}
\end{align}

[stoic python]

vin100

[static bane]

thanks!

[dusky epoch]

3 + 2x + 1?

please don't call me sir.

is it written exactly like that?

not 3x^2 + 2x + 1 maybe?

so it is 3x^2 + 2x + 1?

okay so 2x+4

??

3+2x+1 is literally the same as 2x+4 tho

in any case

whatever the polynomial might be

finding its coordinates wrt a basis is no different than any other "find the coords of this vector wrt this basis" problem

you are to find the constants a, b and c such that 2x+4 = a*1 + b*(1+x) + c*(1+x+x^2)

expand, collect like terms, equate coefficients

a=4, b=2, c=0?

is that what you're claiming?

no

4 * 1 + 2 * (1+x) is 2x+6, not 2x+4

extra credit for 7th grade
uses linalg terminology like "vector space" and "basis"

anyway what do you want to be walked through

my rebuttal to your claimed answer? or how to get the real answer

2x + 4 = a*1 + b*(1+x) + c*(1+x+x^2)

expand and collect like terms on RHS to get:

2x + 4 = cx^2 + (b+c)x + (a+b+c)

do you follow so far?

what comes after will come after. i am stopping to make sure you understand what i've said so far.

ok

now equate coefficients

0 = c
2 = b+c
4 = a+b+c

no

well you got the value of c right, it is 0

but b isn't 4, it's 2

and a is also 2

yes...

if you know how to solve systems of linear equations then this should pose no trouble for you

[nimble wedge]

the answer is 42 as always

[raven badger]

I am reviewing some notes here, and I have a hard time understanding why my solution is wrong.

This is wrt. Finding the eigenvectors of my matrix A.

One of the eigenvalues is $\lambda = 1$, after subtracting this from the diagonal of A and doing RRE I get the RRE(T) as seen in my screenshot.

For me, this tells me that $x = -7*y-> V_\lambda = <1/7,-1>$, because if $x=1 -> y = -7$ and $x=1/7 -> y = -1$.

But apparently this isn't correct; Why is the correct answer $<-1,1/7>$ i.e reversed?

Please tag me.

[stoic python]

CoronaVirus

[pearl elm]

Maybe this is it

[stuck tendon]

You made an error. $x=-7y$. If $x=1$ then $1=-7y$, which gives $y=-1/7$, not -7. Also, the eigenspace $V_{\lambda}$ is a set, not a vector, or an inner product.

[stoic python]

1345631

[halcyon spindle]

The diagonal vector that bisect the angle between v and w should be v+w . The vector up top should be w not v+w. The side vector should be v not v-w. Just responding based on the diagram don’t know the context to this.

[pearl elm]

#linear-algebra message

[pearl elm]

yea thats literally just what i came up with

[pearl elm]

i derped

[dawn ocean]

again, to add vectors you translate a copy of one of the other vectors to the end of the other one and then the new arrow at the end is the end of your new vector. you draw from the origin to the new vector

in this picture y +x you can imagine translating y to the top and then draw from the origin of x an y to get to the end of your coppied x

that new vector is y+x

you can do it with the vector y as well

you just translate it to the right

and you get the same vector

[snow plinth]

how would i even start to do this

[wintry steppe]

check the definition of a subspace

if you don't know it, google "definition of subspace" or "subspace test" or someone here will just tell you

[snow plinth]

ok i know what that is im just kind of confused on how i test for those in this specific problem

[wintry steppe]

why? it shouldn't be any different from checking any other kind of set is a subspace

do remember that trace is linear

[snow plinth]

hmmmm ok

[noble swan]

What do I do with i? Is it just a constant? Am I overthinking it?

[wintry steppe]

constant, you may pull it out of the transpose

[noble swan]

Also, unsure if the ones I've done already are sufficient enough to prove whether it's Hermitian

Gotcha, ty

Are these all sufficient to prove they're Hermitian?

[wintry steppe]

you messed up the first one

(AB)^T = B^T A^T

you need to switch the order, you can't just distribute transpose

i think for these kinds of simple computation questions you should just fully show how to go from e.g. (A + A^dagger)^dagger to A + A^dagger, don't leave out steps

[noble swan]

Oh, woops

Just making sure I understand correctly, A + A^dagger = (A + A^dagger)^dagger, right? Because that's the definition of a Hermitian matrix?

[wintry steppe]

that's what you want to prove

[noble swan]

How would I go from the last step I've written down to the original equation? The question states A may not be Hermitian, so what would be my next step?

[wintry steppe]

that's for you to figure out

[noble swan]

Ah, shucks...

[wintry steppe]

you're just working with conjugation and transposition, use some properties of them or something

[tall hound]

is the adjoint of a matrix literally just another word for the transpose of a matrix?

[wintry steppe]

in the real case, yes

[tall hound]

cus my linear algebra course is nearly over and Ive never heard of adjoint before lmao

[tall hound]

oh ok thx xD

[wintry steppe]

not to be confused with "adjugate" or something, which is sometimes used in some old treatments of determinants/cramer's rule/etc.

[prisma nest]

mr. Ttera can you help me understand a proof in the analysis channel

sorry for the cross posting

I will be forever indebted

[wintry steppe]

i can not

[prisma nest]

that is really a tragedy

for me

[noble swan]

#linear-algebra message

So um...My prof wants me to do this w/ Einstein notation

I don't know how to do that at all lol

[wintry steppe]

doing it with einstein notation sounds like a waste of time

[spare widget]

It's prob practice for working with tensors

[wintry steppe]

all you need to know is that transpose of transpose and conjugate of conjugate return the original matrices

[spare widget]

You have to use $(AB)^i_{j} = \sum_k A^i_{k}B^k_{j}$ also $(A^T)^i_{j} = A^j_{i}$ and you have to drop the sum symbols.

[stoic python]

criver

[spare widget]

swapping the indices of A^T like in the above is wrong unless you assume orthonormal basis btw

As otherwise you need the metric tensor and its inverse to lower and raise indices

[thin stratus]

can someone explain what a basis is

[tribal willow]

a basis is a set of vectors which span a vector space

gonna make an analogy here to painting that i found on a website a while ago

the basis is the set of colors of a palette such that

1. the palette spans the set of all colors
2. each color is independent

for example,

{red, yellow, blue, white, black} is valid because they span the set of all colors (you can combine them to make whatever color) and they are independent (you cannot combine two colors in the set to make another one in the set)

{red, blue, white} is not valid because they do not span the set of all colors (you cannot make yellow with red, blue, and white), but they are independent

{red, blue, purple, white} is not valid because they do not span the set of all colors (same reason above) and they are not independent (red + blue = purple)

[thin stratus]

thank u so much

the example u provided really help

[tribal willow]

👍

[fair wren]

Can you please explain eigenvalues and eigenvectors ?

[keen sierra]

what aspect?

[tribal willow]

cant say i can, sorry

[wintry steppe]

there is a lot to say about eigenvalues and eigenvectors, you should be more specific

do you want to know the definition? do you want to know how to find them? do you want to know why we care about them?

[fair wren]

Last one ye

[wintry steppe]

studying eigenvalues and eigenvectors tells us a lot about the structure of linear operators

for example, finding canonical forms for operators (diagonal, jordan form, rational canonical, etc.) is both of theoretical and computational significance

a lot of problems are simplified by assuming at the outset that some operator takes on a particularly simple form

studying how and when such forms can be achieved is typically done through studying its eigenvectors

here's a theoretical example where assuming that some matrix is diagonal helps us greatly: one has the matrix exponential identity det(exp A) = exp(trace A). the easiest way to prove this is to use the fact that any A with complex entries can be approximated very well by diagonalizable matrices, and, using some algebraic properties of the operators here, you can further assume A is diagonal (where the formula just reduces to the fact that e^{x + y} = e^x e^y).

i'm handwaving some analysis there but whatever, that's not important

that's the first example that came to mind

if you don't understand the example, that's okay

it's kind of difficult to tell linear algebra students why their class is important without just saying "vector spaces, operators, eigenstuffs, etc. show up everywhere in math", but i hope i justified at least a little bit why someone might care about this stuff

[noble swan]

Thanks a bunch! I'm assuming you're right, since this is a mathematical physics class

Does the notation change in any way for the conjugate?

[wintry steppe]

the complex conjugate of a matrix is just the entry-wise complex conjugate, so no

[noble swan]

Gotcha, thanks

[stoic python]

sean

[tacit pelican]

actually

my bad

i'm dumb

[subtle gust]

Alr i want to make sure i understand basis, dimension and coordinates correctly

So i'mma say the things ik and plz sm1 correct me if i say smthng wrong

For R^3

If dim(set of vectors we have)> 3 then the set is linearly dependent and we would be able to express any vector in R^3 in infinitely many ways (infinitely many coordinates)

If dim(set of vectors we have)<3 then there is no way those vectors would span the entirety of R^3

And there would be vectors in R^3 that we wouldn't be able to express using the vectors we have

If dim(set of vectors)=3 and the set is linearly independent then those vectors are basis for R^3 and the coordinates of each vector in R^3 are unique

So coordinates are unique only when we're taking them with respect to a basis

That's pretty much everything ik :)

[native rampart]

You don't call that dimension

You call that cardinality of the set

[subtle gust]

Huh

If we have S={v1,v2,v3}

Dim(S)=3 right?

[native rampart]

dimension is the minimum number of basis vectors required to span the same subspace as S

dim(S) can be 0,1,2 or 3

[subtle gust]

Oh

[floral siren]

the entire vector space has a dimension, not a set of vectors

[subtle gust]

Isn't it just

The number of elements....

[native rampart]

Not true,Take {(0,1,0),(0,2,0),(1,0,0)}

You can never express (0,0,1)

[subtle gust]

Yeah cuz the set is not linearly independent

[native rampart]

Well dim(S) is 2 here

dim(S) can never be >3 in R^3

[subtle gust]

What do we call the number of vectors in a set tho ...

[floral siren]

do you know the concept of span

[subtle gust]

Yeah

[native rampart]

Well,I guess just cardinality of spanning set? There is no specific term for it

[subtle gust]

Yeah i've never heard "cardinality" b4

Weird tbh . I though i read that notation somewhere in the book

Dim(set(

[native rampart]

cardinality of a set is the number of elements in it

[subtle gust]

Yeah that seems to match what i want

So we would express it as

Cardinality(S)?

[native rampart]

If the vector is in span,it can be written in infinitely many ways

[subtle gust]

If cardinality(S)=dim(R^3)

[native rampart]

Or just |S|

[subtle gust]

Only if the set is not a basis

Or if the vectors are not linear indep

If a vector belongs to the span and the vectors in the set are lin indep

[native rampart]

Well,if |S|>3 it won't be linearly independent

[subtle gust]

There would be a unique way to express this slvector

[subtle gust]

True

I don't think i understand dimension well :)

I'm so confused rn

We've been using it

For a few weeks now

To express the number of elements in a set

[native rampart]

dimension is if you start with a spanning set

[subtle gust]

Dim(RS(A))

[native rampart]

Remove all the redundant elements

[subtle gust]

Dim(CS(A))

[native rampart]

And find cardinality of that set

[floral siren]

the number of elements in a basis, or the number of elements in the space?

[subtle gust]

Like

If we have S={v1,v2,v3}

Dim(S)=3 (that's what i thought)

[native rampart]

So for example
{(1,0,0),(2,0,0),(0,1,0)} has dimension 2 because (2,0,0) is redundant

You can express any vector in the span with just {(1,0,0),(0,1,0)}

[subtle gust]

Yeah. So the dimension is the number of linearly independent vectors in the set?

[native rampart]

Yea,vectors that cannot be written in terms of other vectors

Like (2,0,0)=2(1,0,0)

So (2,0,0) has to be taken out

Or well,you could remove (1,0,0) and keep (2,0,0)

[subtle gust]

Yeah i'm fam with the def of linear independence

Well that brings us to another question

How can the dim of the range of a linear transformation be zero

[native rampart]

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

Range is just {(0,0,0)}

[floral siren]

the empty set spans 0

[subtle gust]

Cuz i was told to check 8f the resulting vectors are lin indep

After finding basis for the range

[wintry steppe]

the "dimension" of a vector space is the "cardinality" of any of its bases, to be pedantic about language in the previous discussion

[subtle gust]

Yeah i wasn't familiar with the word "cardinality" till a few mins ago lol

[native rampart]

Well,In a basis you never include the 0 element by default

[subtle gust]

Yeah obv

Cuz that would make it a lin dep set

But like i can't remember the context tbh

[native rampart]

The vector space {(0,0,0)} is interesting because we just let the empty set be the generating set

[subtle gust]

But i was WARNED by my professor to check if the resulting vectors are lin indep

[native rampart]

Because span(Set) is the smallest vector space that contains the Set

[subtle gust]

When exactly is finding a basis for the range impossible?

[native rampart]

Finding the basis for range is always possible in the finite dimensional case

[subtle gust]

If we have a transformation matrix A

The basis for the CS of A are the basis for the range of the linear transformation defined by A right?

[native rampart]

Then range will be the space generated by column vectors of A

Yea

[subtle gust]

But then why do i need to check for linear independence 🐸

Wouldn't the resulting vectors ve automatically lin indep

[native rampart]

Your transformation matrix could have redundant columns

Take
[1 2]
[2 4]

[subtle gust]

Yeah but we find basis for the CS tho

How could there be redundant elements

[native rampart]

How do you find basis without checking for linear independence

[subtle gust]

We reduce to RREF

And check the columns that have pivots

[native rampart]

RREF will give you the dimension

[subtle gust]

Those would be the ones that are lin indep

And would form a basis for the CS

Obv we don't get the columns from the reduce we only get the pivot columns

The columns that form a basis for the CS are the columns in the original matrix that correspond to those pivot locations

[subtle gust]

It would also give me the lin indep vectors i think

Like

The lin indep column vectors

And those would form a basis for the CS

Which would also form a basis for the range

So question is.. why do i need to check that the resulting column vectors are lin indep if they already form a basis for the CS

[native rampart]

Ok it does

You don't need to

[subtle gust]

Yeah. I rem my prof sayin gsmthng abt it tho

Idr the context but he said it was imp 🥲

Mind if i ask a question abt eigen values and eigen vectors?

[native rampart]

Sure

[subtle gust]

I'm still confused as to what the number of eigen values and eigen vectors corresponding to an n×n matrix should be

So what ik

Is that n×n matrices have at most n distinct eigen values

Or like

Every n×n matrix has n eigen values (although some may be repeated)

Now wb the eigen vectors

Every n×n matrix has exactly n eigen vectors or what?

[native rampart]

Not necessarily

[native rampart]

Only over C

Over R there could be 0 eigenvalues as well

[subtle gust]

Nah we're only dealing with R

[subtle gust]

0 real eigen values

But there's always n eigen values

[native rampart]

Take
[0 -1]
[1 0]

Yea

[subtle gust]

So can we find a similar relationship for eigen vectors

[floral siren]

if the linear map is on a real vector space, then there is no such thing as a complex eigenvalue

now that being said

there is a process to turn a real vector space into a complex vector space called complexification, and thus maps over real vector spaces becomes maps over the new complex vector space, but that probably goes beyond the scope of your course

[subtle gust]

Very true

[subtle gust]

Talking abt the last sentence lol

I just want to know the relationship between the size of the matrix and the number of eigen vectors

And possibly the relationship between the number of distinct eigen values and the number of eigen vectors corresponding to them

And the dim of eigen spaces corresponding to eigen values🥲

And algebraic and geometric multiplicites

I understand everything but i'm failing to see a relationship among them

[floral siren]

first of all, it's important to master dimensions, spans, and bases first because if $v$ is an eigenvector of a linear map $A$, then any scalar multiple of $v$ is also an eigenvector, so

$$
Av = \lambda v \Rightarrow A(kv) = (\lambda k) v
$$

for any scalar $k$

[stoic python]

lucid

[floral siren]

so for any eigenvector $v$ corresponding to an eigenvalue $\lambda$, there is at least a $1$-dimensional subspace of such vectors generated by $v$

[stoic python]

lucid

[subtle gust]

Ok so there is at least one eigen vector corresponding to each eigen value?

Or more like

The dim(eigenspace corresponding to this eigenvalue) is at least 1

[floral siren]

yes

[subtle gust]

What if we have repeated eigenvalues tho

For ex if lambda1=lambda2=3 for ex

[floral siren]

a repeated eigenvalue is simply your eigenspace having more than 1 dimension

[subtle gust]

The dim(eigenspace corresponding to those to eigenvalues) would be 2 ?

[subtle gust]

Exactly but is it related to the algebraic multiplicity?

Meaning

Is the dim(eigenspace) necessarily = algebraic multiplicity

Oh

[floral siren]

geometric multiplicity*

[subtle gust]

Yeah just realised

[floral siren]

the algebraic multiplicity is the dimension of the generalized eigenspace

[subtle gust]

And remembered that they don't have to be equal /:

From our lecture on diagonalisation

[floral siren]

what textbook are you using btw

[subtle gust]

Howard anton i believe

[floral siren]

if only that was linear algebra done right...

anyways sorry for the mini rant

haha

[subtle gust]

Elementary linear algebra

[subtle gust]

Lmao np

So from what i've understood

The geometric multiplicity of an eigenvalue is always less than or equal to its algebraic multiplicity

[floral siren]

yes

but do you know why that is the case

at least intuitively

[subtle gust]

Hmmm i don't have intuition for anything lin alg related unfortunately lol

[tall hound]

When a matrix is orthogonal it's determinant is equal to 1 right?

[dusky epoch]

no, not necessarily.

[tall hound]

why not?

my thinking is that since inverse of a matrix is given by

inverse = 1/determinant * transpose

and orthognal is when transpose = inverse

hence for orthogonals 1/determinant has to = 1

hence determinant = 1

what am I missing in that line of thinking?

[dusky epoch]

no

if you assume 1/det(A) = 1 then of course det(A) = 1

[dusky epoch]

this isnt true either

anyway, $\bmqty{1&0\0&-1}$ is an orthogonal matrix and yet its det is $-1$ and not $1$

[stoic python]

Ann

[dusky epoch]

did somebody ping me here?

[tall hound]

oh ye my b, I replied to something then realised I did something wrong

[frail ember]

If U is a unitary matrix, how are the eigenvalues of U + U^* related to the eigenvalues of U?

I just think that the eigenvalues of U + U* would be \lambda + 1/\lambda where \lambda is an eigenvalue of U

[cold temple]

Hi, can anyone tell me how to find the linear transformation here. The message is too implicit and I cant solve it.

[native rampart]

Yea that's correct

[frail ember]

yeah and see as how theyre unitary both \lambda and 1/\lambda have a modulus of 1

[native rampart]

Indeed

But the sum need not have a modulus of 1

[frail ember]

@native rampartif i write now the eigenvalue of U + U* as lambda_k, is there a relationship between \lambda_k and the \theta in exp(i theta) where exp(i theta) is the eigenvalue of U?

assuming \theta is in 0 to \pi

[native rampart]

I mean just expand and substitute in \lambda + 1/ \lambda

[frail ember]

I think all i can say is that as \theta is minimized (towards 0) we have larger eigenvalues associated?

yes but you get

$e^{i \theta_k} + e^{-i \theta_k}$

[stoic python]

*-algebra

[frail ember]

rofl

right

i got it

[native rampart]

So I guess you can conclude that value lies between -2 and 2

[frail ember]

yeah, nice

[floral siren]

Let $U$ be unitary and therefore normal. Per the spectral theorem, there exists an orthonormal basis of eigenvectors with eigenvalues $\lambda_1, \dots, \lambda_n$, all norm $1$. Then

$$
U = \begin{pmatrix}
\lambda_1 & \dots & 0 \
\vdots & \ddots & \
0 & &\lambda_n
\end{pmatrix}
$$

Then the matrix of the adjoint map under the same basis is simply the conjugate transpose

$$
U^* = \begin{pmatrix}
\bar\lambda_1 & \dots & 0 \
\vdots & \ddots & \
0 & & \bar\lambda_n
\end{pmatrix}
$$

Adding the matrices together:

$$
U + U^* = \begin{pmatrix}
\lambda_1 + \bar\lambda_1 & \dots & 0 \
\vdots & \ddots & \
0 & & \lambda_n + \bar\lambda_n
\end{pmatrix}
$$

[stoic python]

lucid

[native rampart]

I mean that's overkill

Uv=cv \implies U*U v=c U*v \implies U*v = 1/c v

[frail ember]

well U isn't equal to that

you just mean the diagonal post decomposition

[peak lodge]

if $T: A \rightarrow B$ is a bijection, does that mean that $dim(A) = dim(B)$ ?

[stoic python]

Vince Tafea

[native rampart]

Yea

[peak lodge]

thought so

[native rampart]

Well T also has to be a linear map

[peak lodge]

ah yeah T is a linear map

do you think in a proof that is fine to say? or is there a theorem or something associated with this

[floral siren]

you should prob call it an "isomorphism" just to be precise (bijection that is also a linear map)

[keen sierra]

could use rank nullity

bijection = trivial null space = full rank

so the image has the same dimension as the domain

[floral siren]

if this is for a course, you should probably ask your instructor or ta if you really want to be sure, and get their general policy on how you can use prior results in assignments/tests which is useful

but in a proof of anything that isn't very basic that would probably just be treated as a given

[rose crag]

hey @floral siren

i d like to ask a qn

[floral siren]

ye

[rose crag]

c = v · Lˆ + m.
here, if i know any one of the basis vector of L^, is it possible for me to get m by knowing only c?

L^ is nxn matrix

v is 1xn matrix

[keen sierra]

what do you mean by basis vector of a matrix? do you mean column of the matrix?

[floral siren]

are there like entires of L that you'd not know in this problem

also do you know the value of v

[rose crag]

wai

[floral siren]

the description is kinda confusing, all i can say is that if n=3 for example and you know v is something like (1,0,0) for example, then v . L is completely determined by the first column of L i guess

[rose crag]

one min

i ll send you the qn

[floral siren]

you only have to know the columns corresponding to the nonzero entries of v (but this might not be what was asked)

[rose crag]

this is the thing

we have to analyze the security of the cryptosystem. In particular, we have to show that if any orthogonal basic
of Lˆ can be found, then the security is broken

[rose crag]

@floral siren

[rose crag]

https://www.researchgate.net/publication/350369335_Encryption_and_Decryption_of_Images_using_GGH_Algorithm_Proposed

does anyone knows GGH algorithm?

<@&286206848099549185>

[wild elm]

Then just got stuck there and idk if what i did is even on the right track lol

[slim cedar]

This might be basic but im struggling to find an answer, does anyone know of a general form solution to the equation $X\otimes A=B$ where only $X$ is unknown and $\otimes$ is the kroenecker product?

[stoic python]

Fro

[slim cedar]

I know of the inuitive solution which is simply dividing out the contents of A in B, but is there a nice way of performing this

[native rampart]

UV^-1 will be a upper triangular matrix as well as a rotation matrix

Then Note that forces sin \alpha to be 0

and taking cos \alpha = 1 and -1 gives you all the solutions

[wild elm]

hey drake i am not really sure about how you got UV^-1 and how it forces sin alpha to be 0? I thought matrix multiplication wasn't commutative so we cant have Ralpha R-beta * UV^-1

[native rampart]

$U = (R_{\alpha}^{-1}R_{\beta}) V \implies UV^{-1} = R_{\beta - \alpha}$

[stoic python]

Drake

[native rampart]

RHS is a rotation matrix and LHS is upper triangular since U and V^-1 are both upper triangular

[wild elm]

Thank you so much was stuck on that step for the longest time makes so much sense

[native rampart]

np

[native rampart]

I guess you can do index shenanigans. like b_11 will be x_11 a_11 b_{m+1} will be x_12 a_11 and so on where m is the number of columns in A

Well Start with b_11,consider only terms of form b_{km+1}{ln+1} where size of A is mxn, that will give you x_kl a_11

[slim cedar]

Yeah that was what i thought, i was just hoping there was some nice operation that acted as an inverse of kroenecker

[native rampart]

Well I guess this is nice enough

If $X \otimes A = B$,then
$[X]{ij}= \frac{b{{(i-1)m+1},{(j-1)n+1}}}{a_{11}}$

[stoic python]

Drake

[vital drum]

i'm trying to prove this theorem in linear algebra by friedberg et al. 4th edition page 139

[native rampart]

Well,That reduces to finding the solution to differential equation with auxillary polynomial
(t-c)^n=0

[vital drum]

and i think it uses on at least two other results in the section

i'm stuck on the induction step of the hint

[vital drum]

you might wanna read again

[native rampart]

Well I know but I think that gets the point across

[vital drum]

what i'm trying to prove is that the set given there forms a basis for the solution space

i.e. the null space of the differential operator associated with that polynomial

i've already proven that the set (let's name it beta) is a subset of the solution space

i only need to prove that it's linearly independent which i'm trying to do with induction

because that's what given in the hint

we need not worry about the span of beta because the dimension of the solution space coincides with the cardinality of beta

[native rampart]

So you know {e^ct, t e^ct, t^2 e^ct...} Is a LI set?

[vital drum]

yes

that's the lemma

[languid wigeon]

i found the invertibility of $B$ irrelevant to the conclusion that $U = \pm V$

[stoic python]

vin100

[languid wigeon]

we're given $$R_\alpha U = R_\beta V.$$

[stoic python]

vin100

[languid wigeon]

denote $U = [u_1 u_2]$ and so on for $V$

[stoic python]

vin100

[languid wigeon]

so the above matrix identity can be translated into

``$\vec{u}_j$ is rotated $\alpha-\beta$ to form $\vec{v}_j$, $j = 1,2$.''

[stoic python]

vin100

[hollow tangle]

Greetings, can anyone recommend a good book or series of lectures that covers vector spaces and subspaces. Im kind of stuggling with them a bit.

[languid wigeon]

insel's lin alg book is a great classic

[vital drum]

any book on linear algebra

yea, friedberg-insel-spence is a good book

[languid wigeon]

[about isaiah's question]
the assumption that U and V are upper-triangular simply means that u₁ and v₁ are parallel to (1,0)ᵀ. this forces the angle of rotation α-β to be a multiple of 180°. that proves the claim that U = ± V. we don't need any assumptions on their invertibility.

[native rampart]

Multiplying with that will make all the $t^m e^{c_{k}t}$ terms vanish

[vital drum]

with what

what is "that"

[native rampart]

(D-c_kI)^n_k

[vital drum]

it's a linear operator

also it's strange how you used the index m lol

because that's what a solution manual i was looking into also wrote

a badly written one

[stoic python]

Drake

[native rampart]

Well By multiplying,I mean applying it in both LHS and RHS

[vital drum]

yea, i get that much

the linear operator reduces those terms to zero

yea

what do we do afterwards?

[native rampart]

So,you know if we exclude those vectors,The remaining set is LI

[vital drum]

wait a minute

are linearly independent subsets mapped to linearly independent subsets

[native rampart]

(D-c_kI)^n_k will just scale all other terms

[vital drum]

iff injective i believe

[vital drum]

also this is a really unclear argument to me, could you elaborate further

apply the linear operator and some terms vanish yes

but the linear operator is also applied on the other terms

what happens to them?

[native rampart]

Ok That doesn't behave nicely

[vital drum]

yea, that's the part i'm stuck on

it's a lot of messy algebraic manipulations that don't seem to lead anywhere

i'm supposed to find a pattern in it according to friedberg

[native rampart]

So $(D-c_k I)^{n_k} (t^x e^{c_m}t) =( (D-c_m I) + (c_m-c_k)I ) ^{n_k} (t^x e^{{c_m}t})$

[stoic python]

Drake

[vital drum]

OH

wait i got too excited there, i still don't see the solution

[native rampart]

If you manipulate you get this will be a linear combination of {e^c_m t , t e^{c_m t} ... t^{x} e^{c_m t} }

Not a clean way of doing it

[vital drum]

oh huuuh

what do you have in mind exactly because

i can't see a way

[native rampart]

Binomial expansion

[vital drum]

i was thinking we didn't have to come to that but jesus

maybe there is no way out

[native rampart]

This gets very ugly,I think

[vital drum]

yea, i might try thinking of other ways before giving up because i really don't wanna be doing binomial expansions on maps

[native rampart]

Do you know primary decomposition?

[vital drum]

no

but i'm pretty sure it should be doable without that since there hasn't been much complicated machinery introduced in friedberg up until that point

[native rampart]

without prim decomposition,I guess binomial expansion on maps is the only way

[vital drum]

actually i think there's going to be one map

$(c_m - c_k)^{n_k} I^{n_k}$ which is just going to be $(c_m - c_k)^{n_k} I$

[stoic python]

pecfex

[vital drum]

when we expand it, right

[native rampart]

How did you reduce this to that?

[vital drum]

i didn't

it's gonna be in the expansion

one of the terms

[native rampart]

Yea

[vital drum]

let's see if we could do something with it i guess lmao

[torpid barn]

could I get some help with part ii?

this is Ra

which is a rotational matrix

[quasi vale]

Write the equation as $\begin{bmatrix} b \ d \end{bmatrix} = \begin{bmatrix} \cos(\alpha) & - \sin(\alpha) \ \sin(\alpha) & \cos(\alpha) \end{bmatrix} \cdot \begin{bmatrix} u_{12} \ u_{22} \end{bmatrix}$

[stoic python]

Sup?

[quasi vale]

and then u know what to do

[torpid barn]

ok thx

lemme try

umm

how do you rewrite the equation as what you have posted?

[quasi vale]

try multiplying R_a with [u12 u22] using usual matrix multiplication

you'll see you end up with what is given

so just work backwards

[torpid barn]

okay thanks

😄

[clear kettle]

Why does the property have two u's?

[dusky epoch]

wym by "the" property?

there are four properties listed here

do you mean property (a)?

@clear kettle

[torpid barn]

umm, how would I start part iii?

part ii is just above in the chat

[nocturne dock]

havent learnt matrices yet

[torpid barn]

i just started learning it lol

[nocturne dock]

._.

doesnt look like it

[torpid barn]

it wasn't taught in a levels

[nocturne dock]

depends on u

[torpid barn]

ive been watching a bunch of vids and 3b1bs linear algebra vids

but im so lost on this question lol

[dusky epoch]

.....

[wintry steppe]

maybe start using parts i and ii to rewrite the matrix B

[dusky epoch]

okay so yeah i got ghosted and then the person who ghosted has had their question get posted over

[torpid barn]

lol

how could I use parts i and ii to rewrite part B?

[dusky epoch]

B isn't a part, it's a 2 by 2 matrix.

think of B as being made of two column vectors stuck together.

[torpid barn]

yep sorry thats what i meant

[dusky epoch]

also think about what it means for U to be upper-triangular.

[torpid barn]

how do you stick two column vectors together to be a matrix tho

i guess thats what this question is asking

[dusky epoch]

wym

you just do

sticking things together is meant to be taken as literally as possible

[torpid barn]

oh

so not adding them together or anything

[dusky epoch]

using the notation of the problem, $B$ is made of $\bmqty{a\c}$ and $\bmqty{b\d}$ stuck together in that order

[stoic python]

Ann

[torpid barn]

yep

[dusky epoch]

yeah ok so now

you know what an upper-triangular matrix is, yes?

[torpid barn]

not rlly

[dusky epoch]

?!

[torpid barn]

a matrix with values only in the upper corner

[dusky epoch]

no

not just "in the upper corner"

[torpid barn]

from the diagonal onwards

[dusky epoch]

an upper-triangular matrix is a matrix all of whose nonzero entries are on or above the main diagonal

in particular, in your case, $U$'s first column will be of the form $\bmqty{x \ 0}$

[stoic python]

Ann

[dusky epoch]

i.e. it will be a vector parallel to the x-axis

[torpid barn]

right

[dusky epoch]

the first column of $R_{\alpha}U$ will be $\bmqty{x \cos(\alpha) \ x \sin(\alpha)}$

[stoic python]

Ann

[dusky epoch]

match this to the first column of B

you'll fix both x and alpha

[torpid barn]

ok

[bright current]

In each of the following cases, exhibit a basis for the given space, and prove
that it is a basis.
11. The space of 2 x 2 matrices.
12. The space of m x n matrices.
13. The space of n x n matrices all of whose components are 0 except possibly
the diagonal components.

please help

[wintry steppe]

you know how (1, 0), (0, 1) is a basis of R^2? (1, 0, 0), (0, 1, 0), (0, 0, 1) of R^3? try emulating that

[bright current]

so 1 0 and 0 1 is basis of 2*2 matrices?

[wintry steppe]

"emulate", not "copy"

those aren't even 2x2 matrices

[dusky epoch]

well

property (a) concerns the inner product of a vector with itself

[wintry steppe]

"unit length" always means length 1

it's only that v_1, v_2, v_3 have length 1

[rare hazel]

Hi everyone, I’m having some trouble understanding the solution to eigenvectors for this problem

Here is the matrix

I’ve calculated the eigenvalues as 10, 6 and 2

I’ve got the first eigenvector as (sqrt(3),1,0) which is correct

But the second eigenvector is (0,0,1) I’m struggling to understand why the first two values are 0

[viral olive]

Can i assume you got (0,0,1) from the eigenvalue 6?

[rare hazel]

Yes

[empty hemlock]

You can just multiply the matrix by (0,0,1) to verify that it has 6 as an eigenvalue.

[rare hazel]

But how do I get to that vector in the first place

[empty hemlock]

S (x,y,z) = 6 (x,y,z). Now solve for x,y,z.

Replace y1 with x1 and y2 with x2.

[viral olive]

Guys a more phylosophical question. Today i got the question why are motivated to explore the relationship between Scalarproducts and norms.

More accurately asked what advantage does a norm that was induced by a scalarprodukt in comparison to a norm that has no scalarprodukt as its origin.

[empty hemlock]

Yes.

[viral olive]

One thing i told was in numerical Analysis at least we have a more well conditioned function in form of the Condition Number in the case of Norms that are induced by Scalarproducts in comparison to the counter part. Which is why i wonder if similiar advantages are know in linear algebra

[empty hemlock]

The expression (a|a) only has one variable.

[empty hemlock]

Scalar products are useful/necessary when you need to consider orthogonality. If you just need to consider lengths of vectors or distances, you don't need a scalar product - the norm will do just fine.

Yes.

[frail hemlock]

not sure if this is the right place, but i am doing mesh simplification with quadric error metrics, and the fundamental math part of this algorithm is solving a linear system Ax = -b where A is a 3x3 matrix, x is the new optimal position of a vertice i am trying to solve, and b is the cost of moving the vertice (i.e. how much does the mesh change if i collapse two verts into this vert). Now, i can usually solve the system with a combination of LU decomposition and SVD, but sometimes that does not work. I was thinking it might be possible to solve this (3d) system on a line along the edge of the two vertices instead of full-on 3d as a fallback, but i am not sure where to find resources on how to do this, i haven't been able to find anything

geometrically speaking, the A matrix is the quadric error matrix, its called that because it forms a quadric surface, and if visualized it would be ellipse-like, and it tells you where you can move a new vertice in such a way where the error introduced is some value (usually 0-ish), except sometimes this matrix can be non-invertible because the possible positions are infinite, either on a plane or a line (such as in the case of a flat plane mesh)

[stoic python]

alexisreen

[night wren]

Are the vectors (cv,w) and c(w,v) the same vector in VxW?

[crystal iris]

Where do they get the simultaneous equations from?

[grand hare]

how to find values of c such that 3c^2t^2 + ct + 4 ∈ Span{t^2 + 4t − 3, 2t − 2, t^2 + 6t − 5}

[dawn ocean]

L(p(t))= tp(t)+p(0)

p(t)=t in our case

L(t)=t*t + 0 =t^2

transpose cause you need to express the representation of L in terms of column vectors with numbers not polynomials

i personally haven't seen the [L(t)]_T notation before

but i assume that's what's going on

[dawn ocean]

yeah i said column vectors

[wintry steppe]

$(1, 0, 0)^T$ is $$\begin{pmatrix}1 \ 0 \ 0 \end{pmatrix},$$ it's just a compact way of writing it that doesn't break the line

[stoic python]

TTerra

[fair wren]

look at T

[wintry steppe]

"Consider the ordered bases S = {t, 1}..."

[fair wren]

p(t) is in S

so its either 1 or t

[wintry steppe]

to find the representation of L with respect to S and T you need to compute how L acts on the elements of S and then write the results in terms of the elements of T

i am referring to the question

are you asking how to find the representation?

okay

it's called a subspace

note that "is a span" implies the set is linearly dependent

it is not, it's just notation for another basis (completely unrelated whatsoever to S, T)

[dawn ocean]

no it's just a symbol that means a different set theres only so many letters we can use

lots of things in math use the same notation with completely different meanings

you get used to it

[wintry steppe]

"is a span" means "equals span(S) for some set S" means "every element is of the form \sum c_i v_i for some scalars c_i and vectors v_i in S" and i leave it to you to see why such a set is going to be linearly dependent

[dawn ocean]

it says, they're already in standard basis

[wintry steppe]

i don't understand what you mean

also, please don't ping me with every message

you asked what to call a set that is linearly dependent and a span of vectors

of course, any subspace can be written as a span of independent vectors

any set of vectors, regardless of linear dependence or independence, spans some subspace. it's called their span

[dawn ocean]

it's wrong

[wintry steppe]

alright i don't want to decipher what your book means through you anymore

can you screenshot the part of your book you're confused about?

[dawn ocean]

basis vectors span a space

i'm so confused

the definition of span is given a set of vectors the span of that set is the set of all finite linear combinations of the vectors

[wintry steppe]

same thing

transpose is just a compact way of writing a column vector

inside a line without breaking it

[dawn ocean]

the definition of span says nothing about independence or dependence of vectors

[wintry steppe]

no, i mean in a typesetting context.

writing a column vector as a row vector transposed is just a neat way of doing it that saves space. there's nothing mathematical going on

[dawn ocean]

?

there's nothing about basis vectors in here

no

you don't

be respectful if you want my help

[wintry steppe]

what is L?

is this a continuation of the previous question?

if you're talking about the vectors on the top left in blue, those were just obtained by matrix multiplication by A.

the vectors

try computing $$A\begin{bmatrix}1 \ 0 \ 0 \end{bmatrix}, A\begin{bmatrix}1 \ 1 \ 0 \end{bmatrix}, \text{ and } A\begin{bmatrix}1 \ 1 \ 1 \end{bmatrix}$$

[stoic python]

TTerra

[wintry steppe]

each of these should be a column vector with two entries

A is a 2 x 3 matrix. multiplying a 2 x 3 matrix and a column vector with 3 entries results in a column vector with 2 entries

... the product of an m by n and an n by p matrix is an m by p matrix

so the results shouldn't just be scalars, which is what you had

[fringe burrow]

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

[stoic python]

øst

[dusky epoch]

this is not a vector space this is an expression

do you mean ${ x_0 e^t + x_1 te^t \mid x_0, x_1 \in \bR }$?

[stoic python]

Ann

[dusky epoch]

(a subspace of C^1(R) presumably)

[wintry steppe]

Yes, that's what i mean

[dusky epoch]

well then this is literally span {e^t, te^t}

by defn

[wintry steppe]

Lmao that's the basis?

[dusky epoch]

yes

it's clearly LI

[wintry steppe]

Yeah obviously so, in matrix form would you just write it (0 1)?

Or (1 0, 0 1)?

[dusky epoch]

what matrix form

[wintry steppe]

Nevermind you've helped me a lot

I'll figure out the rest

Thank you Ann :) ❤️

[torpid barn]

could i get some help with this?

This is B

and this is Ra

So i found that u11 and u22 cannot be zero as otherwise B would not be invertible

but where do I go from here?

[raven badger]

How is the eigenvector for the eigenvalue 1 equal to <0,0,1>, if the RRE of A - Eigenvalue*E is equal to diag(1,1,0)???

[viral olive]

@torpid barn So lets suppose you know following information B is invertible so and you know the form of $R_a$ and know the form of U

[stoic python]

Cho

[viral olive]

What can you say about U first things first

[torpid barn]

U has to be invertible aswell?

[viral olive]

well cant say that yet

Let me ask you this is $R_a$ invertible?

[stoic python]

Cho

[torpid barn]

yes

and its inverse is the same as its transpose

making it rotational

[viral olive]

For every a $\in \mathbb{R}$

[stoic python]

Cho

[torpid barn]

wait

ye it should be right?

[viral olive]

Perfect so letssuppose you have following information \begin{itemize}
\item $B=R_aU$
\item $B=R_bV$
\end{itemize}

[torpid barn]

cos of the identity cos^2 + sin^2 = 1

[stoic python]

Cho

[viral olive]

You know both Rs are invertible

thats all the tips i am gonna give you for now

[torpid barn]

huh

[viral olive]

the solution is more or less straightforward

except maybe i am missing a point here

[torpid barn]

im not following sorry

[viral olive]

Try to formulte such that $V=R_aU(R_b)^-1$

[torpid barn]

right

[stoic python]

Cho

[torpid barn]

ye

[viral olive]

can you try to work from there on

[torpid barn]

but does Ra Rb^-1 = I?

it doesnt right

or it doesnt have to be

[viral olive]

Depens on your choice of a and b

[torpid barn]

yea

[viral olive]

if a =b yes it does

if not then no

[torpid barn]

but a doesnt have to equal b

I guess the thing i have to proof is that Ra Rb^-1 has to either equal to I or negative I

[viral olive]

I can give you one more tipp

[torpid barn]

yes pls

[viral olive]

Rotation does what ?

[torpid barn]

it rotates the basis vectors

[viral olive]

It rotates the Matrix or in this case the spann of a matrix if the rotation is fitting all rotation could possibly do is change the direction of a basis vector from start point to positive values to start point towards negative values

Have to go now though

this should be more than enough for you to smash your head against it

good luck

[torpid barn]

ty

sm

[raven badger]

Can you help me with my question aswell

[viral olive]

I have 5 percent on my phone pray its enough

gonna give you little tios

Do you knoe hoe you find out about the eigenvector of an Eigenvalue?

Because your question kinda implies you didnt quite got the fundamental idea behind solving homogenous Linear equation system

Which is why look at it this way

\begin{itemize}
\item $x_1+0+0=0$
\item $0+x_2+0=0$
\item $0+0+0=0$
\end{itemize}

[stoic python]

Cho

[viral olive]

Try to solve this and think why this works

[torpid barn]

@viral olive So, since Ra and Rb^-1 are rotational matrices, their product is also a rotational matrix

and because rotational matrices are only able to rotate U, either in the positive of negative direction

U = +- V?

is that explanation correct?

[viral olive]

Well that would be if it was VRaRb

but its RaVRb

[torpid barn]

but there is no difference between the two right?

[viral olive]

Well can you explain why there is no difference?

[torpid barn]

from the original equation

[viral olive]

Because multiplicative kommutativity is not a given thing in Vectorspaces

?

[torpid barn]

RaU = RbV

[viral olive]

Ok

[torpid barn]

that works right?

[viral olive]

nope

[torpid barn]

oh

[viral olive]

Like i said you are taking it as a given thing

there is a reason why i specifically said the formular RaVRb

because it doesnt have to be VRaRb

Except you have an explanation why it works

well phone charge is dying

2 percent left think about what i said

[torpid barn]

wait

so u meant in the latex earlier U = RaVRb^-1 right?

damn

im still so lost

[raven badger]

This is the exact equation I have written (except x,y,z ofc.).

x1 = 0
x2 = 0

And then there is no x3

[spare widget]

@torpid barn Study the two ways of writing of $BB^{-1}=I$

[stoic python]

criver

[spare widget]

once using that $B = RU$ and $B^{-1} = V^{-1}Q^T$, and once that $B = QV$ and $B^{-1} = U^{-1}R^T$

where R and Q are orthogonal matrices

[stoic python]

criver

[torpid barn]

Right

that makes sense

since its rotational Q^t is the same as Q^-1

[spare widget]

When they tell you that a matrix is invertible, and that it has two representations B = QV and B = RU, then if you have no idea what to do, just form properties involving those, as in the above.

[torpid barn]

and then try to compute it to make U = +- V?

[spare widget]

try to study this first

see what you get

[torpid barn]

right

RaU U^-1 Ra^T == I

oh wait

one sec

[spare widget]

use R and Q for simplicity

so you don't confuse the matrices

[torpid barn]

ok

[spare widget]

R = Ra, Q = Rb

[torpid barn]

so we have BB^-1 == I

and that is the same as RUV^-1 Q^t == I

[spare widget]

yes, now try the other way to write this, such that the rotation matrices end up on both sides

i.e. B = QV, B^{-1} = U^{-1} R^T

[torpid barn]

yep

okay

[spare widget]

wait, nvm, I am not sure that will help you, sorry about that

[torpid barn]

oh ok

[spare widget]

I wanted to show that Q^TR = R^TQ which would imply that this is either the identity or a 180 degree rotation

but I think I missed something

without using that B is invertible or that U and V are triangular you get RU = QV -> U = R^TQ V

and I think you also need to somehow get that U = QR^T V from the fact that they are triangular and invertible

then QR^T = R^TQ would mean that the rotation is either 0 degrees or 180

[torpid barn]

right

the fact that B is invertible means u =

but u11 and u22 cant be zero

otherwise U wouldnt be invertible

[spare widget]

wait

rotations in 2D commute

then R^TQ = QR^T

am I tripping

[torpid barn]

lol

[spare widget]

then it doesn't even matter whether U or V is triangular or whether B is invertible

I was thinking of n-d

but your problem is 2D

[torpid barn]

but like the question says it should matter tho

[spare widget]

and if this is true: R^TQ = QR^T

no wait that's not enough

I needed that R^TQ = Q^TR

i.e. that (R^TQ)^{-1} = R^TQ then it's either 0 degrees or 180 degrees

[torpid barn]

wait I dont get why its either 0 or 180 degrees

I meant i get that it needs to be that

but why is it 0 or 180?

is it because if the inverse is is equal to the original then that would mean the rotation either doesnt rotate at all or rotates in a direction where it doesnt matter which way it rotated?

[wintry steppe]

Is it possible to find order basis so that matrix representation of linear operator is diagonal matrix provided it has two eigenvalues ? Dimension of vector space is 4

[stuck tendon]

You can if there is a basis of eigenvectors

[wintry steppe]

Is there a method to do this ?