linear-algebra

[vast torrent]

Find (A^n)v, wasn't it?

[scarlet ermine]

yeah

[vast torrent]

Not (D^n)v

[scarlet ermine]

here's what i got for that

[vast torrent]

A=SDS-¹

[vast torrent]

(SJS-¹)^n=SJ^nS-¹, you can prove this inductively

[scarlet ermine]

so its not particularly useful

though i feel it could be a series?

[vast torrent]

It's a sequence with a closed form

That you will.find if you calculate

(SJ^nS-¹)v and leave it in terms of x1,x2,n

And x3

[scarlet ermine]

i've figured it out via wolfram, but im not sure how to backtrack to the 'mathematically proven' version

[vast torrent]

Hear me out

Do the mtx multiplication in terms of n

You should get a linear combination of (number)^n

That's the closed form

Wait no

Said a stupid thing, nm

[wintry steppe]

Life is [[a, -b], [b, a]]

[scarlet ermine]

@vast torrent sorry im very bad with proofs and that sort of thing

[vast torrent]

No need to apologize

Im done with math for now, good luck

[scarlet ermine]

alright thanks for the help

i think im just gonna call it for todya

[wintry steppe]

thanks btw all functions, dont wanna @ you just in case you're busy

So I'm asked to find the length of the radius of a circle on a graph with the extremities of the diameter being 2,3 and 6,1

The answer is 2,2 although I'm not entirely sure why

[cloud glen]

wrong channel dude

[wintry steppe]

U sure

[sonic osprey]

yes

[wintry steppe]

What even is algebra

Like by definition

[sonic osprey]

this is not linear algebra

[wintry steppe]

What would u consider it to be

That sounded sarcastic it wasn't I'm sorry

[sonic osprey]

You're good

This is more #geometry-and-trigonometry or maybe #precalculus

[wintry steppe]

It's precal but I thought that algebra was variable functions

That's what I was told

[cloud glen]

linear algebra deals with variables in > 2 dimensions

ur algebra is in 2 dimensions

[sonic osprey]

That isn't really accurate

[wintry steppe]

I know cuz it's on a graph not a grid

[jagged saffron]

Let V be a inner product space over R,and let T be an endomorphism of V. How can we relate the minimal polynomial of T, T^* to the minimal polynomial of the operator S(v_1,v_2) = (T(v_1),T^{*}(v_2)) ?

where S acts on V x V

[jagged saffron]

<@&286206848099549185>

[obsidian rapids]

Can I get some help with vectors and axioms? I'm supposed to explain if A subspace of R3: the set of all dimension -3 vectors [x; y; z] such that x+y+z = 0 where x, y, and z are real-valued numbers. holds or doesn't hold using axiom 4: There is a zero vector 0 in V such that u+0=u.

[slow scroll]

The zero vector in R3 is (0, 0, 0). Indeed, for any other (x,y,z), we have (x,y,z) + (0,0,0) = (x,y,z). So, does (0,0,0) satisfy the conditions to be in the subspace you defined in R3? @obsidian rapids

[obsidian rapids]

See that's how I thought about it too. Just wanted to make sure I wasn't overly simplifying it.

[slow scroll]

ah ok.

[north sierra]

when diagnolizing a matrix, you can write the answer in many ways depending on the order of matrix P where P is the matrix of eigen vectors right?

@ me when reply

[vast torrent]

@north sierra the JNF is unique up to permutation of blocks

There's no "best" way to order the blocks, though if all eigenvalues are real we often order them

[north sierra]

whats JNF

@vast torrent

[vast torrent]

Jordan normal form

The diagonalized matrix in this case

[undone garnet]

does imf(A) = imf(B) implies A=B?

or

implies A = kB?

or

exists x, y

Ax = By?

I mean my prob is

AB=BA

imf(BA) = imf(B)

A^2017 = 0

how to prove

imf(A^2017.B) = imf(B)?

[dusky epoch]

wha's imf

[undone garnet]

imf of a matrix

like f

sorry for typo

ImA

[dusky epoch]

no, im(A) = im(B) doesn't even NEARLY imply A=B

[undone garnet]

hm...so how I solve my prob

any hint?

[jagged rock]

$ \int 3x^2\sin(x^{-1})-x\cos(x^{-1}) dx $

how do

oh shit so sorry

i thought i was typing in calculus

[wintry steppe]

how would one construct an iso between V and V**

physicist asking ree

[wintry steppe]

<@&286206848099549185>

[vast torrent]

@wintry steppe whats V? A Hilbert space and V** it's double dual?

[jagged saffron]

Isn't this trivial from definition of min poly

i.e smallest degree poly such that f(A) = 0

[steady fiber]

yes it is

[jagged saffron]

How can i do this?

I can't apply jcf to this

[young urchin]

hey, why is -2pi = 0 ?

Is there a way i can google this problem to educate myself?

[jagged saffron]

uh

[young urchin]

i only know, that sin(-2pi) is 0

[vast torrent]

Ask in one of the questions chat ill help you

[wintry steppe]

@young urchin you can type equation in wolframalpha

[native lodge]

$e^{-2\pi i}=1$ though

[stoic python]

⚡Amphy⚡:

[native lodge]

look into Euler's identity if you are unsure about this

[ionic dust]

Hi guys

I don't havean immediate question in regards to linear-algebra. I actually just signed up for this class and I'm quite familiar with some of the applications of the class.

However, I wanted to get a head start on the proofs, what books (if any) would you guys suggest for how to approach proofing for an undergrad?

(or a beginner)

[sonic osprey]

Velleman's How to Prove It

[wintry steppe]

Revisiting this question I had yesterday. Prove or disprove: Every square matrix is similar to an upper triangular matrix. By similar it means similar over reals. I was given an answer but figure out the answer was wrong since matrixes aren’t similar to their rref.

[ionic dust]

I'll look it up. Was this a personal favorite of yours #advanced-number-theory

oh

[wintry steppe]

?

[jagged saffron]

@wintry steppe similar over C implies similar over R

I believe schur decomposition implies this

this depends on the entries in your matrix

it is false if your vs is over R

[wintry steppe]

So how would I know if a matrix isn’t similar to any upper triangular matrices? Do I just prove that there is no such S matrix?

Ah nvm I got it

[vast torrent]

@wintry steppe yeah sorry i confused similarity A=SPS-¹ to equivalence A= SPJ-¹

Howd you solve the problem?

[wintry steppe]

@vast torrent V is just a finite dimensional v.s and V** it's double dual

[vast torrent]

so you have some inner product

can't you just use Riesz twice?

or is that circular

okay so you just need the uniqueness of the adjoint

I dont know how to prove uniqueness of adjoint though 😅

[wintry steppe]

what

can't you just construct an isomorphism

that doesn't require an inner prod

[vast torrent]

if V is f.d., V is iso to V*

and V* is to V**

oh, how are you defining V* then

linear functionals?

[wintry steppe]

ye

and then V** is the v.s of maps from V* to K

[vast torrent]

yeah I'm used to dealing with Hilbert spaces so you can write lin.funs. as inner products

I think the strategy is to compose isomorphisms to V* and then to V**

V is iso to V* by the evaluation map

hmm

[jagged saffron]

Number 1

[wintry steppe]

bruh

[jagged saffron]

ok

[wintry steppe]

different channel ty

[jagged saffron]

What

It was the answer

To your question

[wintry steppe]

oh

[jagged saffron]

[wintry steppe]

no ples no

i don't want to see

[jagged saffron]

O ok

[wintry steppe]

i'm going to figure this out

ty tho

[vast torrent]

@wintry steppe are you not satisfied with just taking the composition of the evaluation maps eval1[v](ell1) = ell(v) and eval2eval1[v] = ell2(eval1(v)(ell1))

[wintry steppe]

$\varphi: V \to V^{**}: v \mapsto \varphi(v)$ with $\varphi(v): V^* \to K: w \mapsto w(v)$ should do the trick

[stoic python]

mop:

[wintry steppe]

ty @vast torrent

[vast torrent]

yeah that's what's I was trying to say but your notation is actually coherent

[wintry steppe]

hi again did I prove this right?

[vast torrent]

Why are they smashing two theorems into one theorem

[wintry steppe]

that's me sksk

should i not

[vast torrent]

You should not

[wintry steppe]

other then that?

[vast torrent]

Yes

It's correct

[wintry steppe]

is this right

[dusky epoch]

no

[wintry steppe]

i think i remember seeing it hoffman kunze or something

but i can't really remember it

[dusky epoch]

this seems like an attempt to convey the idea that span(U) is the intersection of all the subspaces containing U but the notation and wording is just wonky af and kinda falls flat on its face

[wintry steppe]

how would i change the wording/notation

(sorry physicist)

[dusky epoch]

span(U) is the intersection of all the subspaces of V which contain U

that's it

[wintry steppe]

oh cool ty n.n

[dusky epoch]

writing it in symbols will only serve to obscure it imo

[foggy dawn]

Let V be ann-dimensional vector space, and letT:V→V be a linear map. The map T is called diagonalizable if there exists a basis (b1, . . . ,bn) of V such that b1, . . . ,bn are all eigenvectors of T. Prove that if T is diagonalizable, then the composition T◦T:V→V is diagonalizable.
Pls help 0.0

[sage gazelle]

matrix representation of the linear map is my intuition 🤔

[dusky epoch]

T is diagonalizable iff there exists a basis in which the matrix of T is diagonal

the matrix of T^2 will be diagonal in the exact same basis

[eternal jasper]

w

[rose grotto]

why am I not getting the right thing

don't I just multiply that matrix

by 24
19

then mod 26

the first 2 letters of the answer are pa

[half ice]

Why (24, 19)?

Oh is it done in pairs?

[vast torrent]

S E N D N U D E S

[half ice]

@rose grotto
Oh I see the problem. That matrix is used to encrypt, not decrypt.

[rose grotto]

oh

so what

do I do with it

inverse or something

[half ice]

Yup. Go backwards

So find the inverse, that's your decrypt

[rose grotto]

1 34/7
2 33/7

I get that

as my answer in that case

but it stil ldoesn't give me

the right stuff

@half ice

idk

this is the inverse

[half ice]

Hmmstv. That's far from an integer matrix

[rose grotto]

wym

@half ice so wat do I do

z

does anyone know what to do

[rose grotto]

can someone help me with this plz

I tried using that matrix * [24,19] and mod(26) the numbers

but didn't get the right thing

I also tried using the inverse matrix * [24,19] and mod(26) and still not right

im using 24,19 cuz its the first two letter

in the answer the first 2 letters

are p and a

[fallow jolt]

Can I get a hint on how to start this proof?

<@&286206848099549185>

[dusky epoch]

this is false

V = [ 2 -1 ; 1 2 ] satisfies the condition they state, but V^TV is 5I, not I.

[fallow jolt]

But from what I've looked up online, orthogonal columns imply ATA = I

@dusky epoch

@dusky epoch Your matrix satisfies the condition

V.T V = [5 0; 0 5]

=[1 0; 0 1]

[dusky epoch]

no

$5 \neq 1$

[stoic python]

Ann:

[dusky epoch]

5I is not the same as I

@fallow jolt

[fallow jolt]

It is

divide the row by 5

[dusky epoch]

no

i don't think you understand what it means for two matrices to be EQUAL.

[fallow jolt]

I promise you this theory holds

It's well known

I just don't know how to prove it lol

See: https://math.stackexchange.com/questions/2202382/when-ata-i-what-matrix-is-orthogonal-to-what

[dusky epoch]

"A is orthogonal" isn't the same as "the cols of A are mutually orthogonal"

it's "the cols of A are orthogonal to one another and the norm of each col is 1"

i don't think you understand what it means for two matrices to be EQUAL.

[fallow jolt]

Perhaps I should have clarified

they gave this to us

the norm of each col is 1

o

o

p

s

[dusky epoch]

[ 5 0 ; 0 5 ] is NOT EQUAL to [ 1 0 ; 0 1 ] because their (1,1) entries are 5 and 1 respectively, which are NOT equal

[fallow jolt]

i forgot about that

m

b

Any ideas knowing that now @dusky epoch ?

[dusky epoch]

$\left< v, w \right> = v^Tw$

[stoic python]

Ann:

[dusky epoch]

the $(i,j)$'th entry of $V^TV$ is equal to $v_i^Tv_j$

[stoic python]

Ann:

[fallow jolt]

Sure

ohhh

so would that mean

<V_i, V_i> = 1

How would I explain that rigorously though

[dusky epoch]

that was the missing bit

the norm of each col of V should be 1

and you should know the connection between inner product and norm

[fallow jolt]

<v, w> = |v|*|w|*cos(theta-phi)

@dusky epoch

[dusky epoch]

no

i mean first off what even are theta and phi

but second no that is not what i meant

the inner product of a vector with itself is the square of its norm

[fallow jolt]

ohhhhh

ohhh

oH

https://media.discordapp.net/attachments/540211747613704221/652410047502221323/183668144404037632.png

Can you explain why this is true @dusky epoch ?

[dusky epoch]

what does V^T look like

v_1^T
v_2^T
...
v_n^T

it looks like this

[fallow jolt]

Ah

Separate question

What are the conditions for a set to span a vector space?

[dusky epoch]

...................

[fallow jolt]

Is it just c1v1 + c2v2 + ... = 0?

[dusky epoch]

no???????????????????

????????

????

???????

?????????

?????

no???

[fallow jolt]

V is a vector space

V = c1v1 + c2v2 + ...

[dusky epoch]

AAAAAAA

NO

AAAAAAAAAAAA

[fallow jolt]

chilllll

im learning

[dusky epoch]

no

[fallow jolt]

😦

[dusky epoch]

a set S spans V iff every vector in V can be written as a linear combination of vectors in S

[fallow jolt]

Isnt that what I just said

[dusky epoch]

"V = c1v1 + c2v2 + ..." is just

so nonsensical

you can't equate a vector space to a single vector

[fallow jolt]

If your set is S = {v1 v2 v3 v4}

And your space is V

[dusky epoch]

let alone an infinite sum

HHHHHH

[fallow jolt]

V = c1v1 + ... + c4v4

If you can do this it spans, right?

[dusky epoch]

NO!!!!

NO!!!!!!!!

A VECTOR SPACE AND A VECTOR ARE TWO DIFFERENT THINGS!!!!!

[fallow jolt]

ah gotcha

How do we know if every vector in V can be written as a linear combination of vectors in S

If V is R^n

In particular, I'm trying to solve this:

[dusky epoch]

the cols of V are guaranteed LI

[fallow jolt]

Yes I agree

[dusky epoch]

and so the dimension of their span is N

the only subspace of R^N that has dimension N is R^N itself

[fallow jolt]

"and so the dimension of their span is N"
How do you know this?

[dusky epoch]

there's N of them!

[wintry steppe]

did i prove this right?

[dusky epoch]

tf is up with that lemma 1.15

you basically just said "all linear transformations are surjective" lmfao

why does lemma 1.15 say "the image of beta under phi spans W"

[wintry steppe]

aaaaaaaaa i'm tired

sorry thank you

anything else

[dusky epoch]

uh

seems so

[wintry steppe]

seems good?

[wintry steppe]

sorry for the beep but idk what that meant @dusky epoch

[dusky epoch]

yes this seems ok to me

at a cursory glance

sorry for being unclear

[wintry steppe]

no np ty for pointing that mistake out

[rose grotto]

Who can help with this

[mint sentinel]

How do i write 6x^2 − 2xy + 6y^2 + 19(√2)x − 9 (√2)y = 37 on standard form?

[dusky epoch]

define standard form

[mint sentinel]

on a form of a hyperbola/parabola/ellipse

[merry shuttle]

is there like a formula to find the distance between a vector and a matrix

[half ice]

There's no usual way to define a "distance" between these two things

What did you have in mind?

[quartz compass]

@mint sentinel it's possible to do by making a symmetric matrix out of the first 3 terms, diagonalizing the matrix to get a change of basis which corresponds to a rotation

after that, just complete the squares

[vast torrent]

@merry shuttle you need to define "distance" first in this context. The regular distance definition is based on the pythagorean theorem. here you don't have that natural geometry definition

[mint sentinel]

@quartz compass im stuck here

i dont know what to do after finding P and D

[quartz compass]

your notation is a little off, you're writing vector components u and v with vector arrows over them @mint sentinel

looks like you normalized it, well the upshoot is the P matrices are orthogonal so their inverse is their transpose, I'll rewrite what you had shortly

$\vec z^T A \vec z + \vec b^T \vec z -37=0$

[stoic python]

Merosity:

[quartz compass]

hopefully it's clear what everything is here, z = (x y)^T this is the vector of x,y, A the symmetric matrix

$A=P^TDP$

[stoic python]

Merosity:

[quartz compass]

this is effectively what you just found so now just plug it in, although I it might be the reverse of how you put your P, I didn't check your work

$z^TP^TDPz + b^T P^TPz - 37 =0$

[stoic python]

Merosity:

[quartz compass]

make sure that all makes sense, I'm not really doing anything except plugging in A = P^TDP and I = P^TP

this gives us the change of coordinates

$Pz =u$

[stoic python]

Merosity:

[quartz compass]

and so your entire thing will simplfy to

$u^TDu + c^T u -37=0$

[stoic python]

Merosity:

[quartz compass]

c=Pb of course

now it's diagonal and your cross terms are gone

geometrically all this has done is given you a rotation of the graph that removes the "cross" terms

at this point you can complete the squares

keep asking questions, I left out some details but I knew it was going to get long so I tried to get to the point as fast as possible

[mint sentinel]

thanks

im just not sure how i complete it

[quartz compass]

complete the square?

it's a trick they teach kids around the time they learn the quadratic formula

[mint sentinel]

i mean

i was looking at this example

i just dont see how the numbers in "9x^2+4x^2...." are derived at

[quartz compass]

explain back to me what I explained a minute ago so I understand what you're thinking

in my mind, I already explained that

[mint sentinel]

don't do this to me... >.<

[quartz compass]

I wrote up a bunch of crap

you're wasting my time

[gray dust]

yes

[native lodge]

quick way to verify with small matrices

try a 2x2

then do a 3x3 using cofactor expansion

[steady fiber]

just think about what happens when you do a cofactor expansion along the bottom row of an upper triangular matrix or along the top row of a lower triangular matrix

[median fern]

I don't understand how this matrix is in reduced row echelon form?

The third column contains a -1

I thought the first entry of a 1 in a non zero column, should have zeros everywhere else in the column for a reduced row echelon

<@&286206848099549185>

Sorry what was that?

[sonic osprey]

Read #❓how-to-get-help please

[median fern]

Okay

Sorry. I see. @sonic osprey

[north sierra]

oops i deleted my msg

but thanks @gray dust

[native ore]

I know A is true

but do B and C always follow?

[median fern]

<@&286206848099549185> Sorry to ping, can someone please help explain why the matrix I posted above is in row echelon form?

@native ore They're all true and right

You posted that question in the wrong channel though. Go to #precalculus next time. 👍 @native ore

[native ore]

@median fern

oh my bad

can you help me with something over there

[median fern]

Sure I can try

[wintry steppe]

did i prove this right?

[wintry steppe]

<@&286206848099549185>

[wintry steppe]

in 1.21 $\delta_{ij}$ should be $\delta_{ji}$

[stoic python]

mop:

[wintry steppe]

plz halp

[toxic pendant]

If I have a row reduced augmented matrix of the form Ax=b, what do I do with the b when finding the basis of row(A)?

[native lodge]

the only place b comes in is when you are finding the particular solution

[toxic pendant]

wow free present

Ty, so it's just the row vectors correct?

Then if I was finding the solutions and I had a 0 column vector, how would that affect my solution?

[native lodge]

you can have this then lol

It doesn't tell you how to find the particular solution, but it does say how to find all the bases

[toxic pendant]

Ty

[native lodge]

tell me if anything is unclear or if you don't get it, always looking to improve how to present the material there

[toxic pendant]

Sounds good

[vapid owl]

Question: Is it possible to get multiple answers with eigenvectors?

[toxic pendant]

You mean multiple eigenvectors from the same eiganvalue?

[vapid owl]

I did the following problem and got:
[4 2 ]
[3 -1 ]

Lambda1 = 5, lambda2 = -2
...
5 [2]
[1]
...

-2 [ 1 ]
[-3]

I checked my eigenvalues on symbolab and I think they're correct.

Symbolab says I should be getting [-1] for the second eigenvector.
[3]

It's exactly the opposite of what I have.

@toxic pendant I mean different eigenvectors for the same eigenvalue.

[toxic pendant]

Yes that's possible

You can have multiple eigenvectors from the same eiganvalue

Anyways I just did the question and I got two eigenvectors

[vapid owl]

Okay cool. Thanks. 🙂 I was just wondering if I screwed up.

The last part is kind of arbitrary. We assign a value to x1 or x2 and then solve for the other.

So I guess it depends on which one you choose to solve for first.

[native lodge]

you can multiply an eigen vector by any scalar value and it's still a valid eigenvector, it's just somewhere else along its span now

in your case, you are only different by a factor of -1

[vapid owl]

Thanks. That's kind of what I thought might be happening. 🙂

[wintry steppe]

sorry repost but is this proof right?

[wintry steppe]

<@&286206848099549185> plz help

[vague bronze]

Hey guys can I have help in a lag question

[dusky epoch]

post it

[vague bronze]

How many injective linear maps f :V to W are there for two vector spaces over a finite field Z/pZ, p prime, with V d dimensional and W e dimensional

I know the total amount of linear maps is p^(de)

So it's definitely less than that

[dusky epoch]

it's equal to the number of linearly independent ordered sequences (w_1, ..., w_d) of vectors in W

[vague bronze]

What do you mean by linearly independent ordered sequence

Oh NVM I get you

Oh so that means that if d > e there are no injective maps? Or am I confusing myself

[dusky epoch]

indeed.

there are no injective maps from a higher dimension to a lower dimension space

[vague bronze]

Now that I think about it that can be proved using rank nullity

Can you give me a hint as to how to count them?

And finally is in not the case that for any linearly independent sequence (w_1,...,w_d) we have d! Injective maps using those vectors

[dusky epoch]

no, because we're counting them as ordered

makes it easier

because each ORDERED sequence uniquely determines a linear map

bc you can pick a basis (v_1, ..., v_d) for V

and then send v_i to w_i

and extend by linearity

anyway i believe the total count of linearly independent sequences of $d$ vectors in $W$ would be $(p^e - 1)(p^e - p)(p^e - p^2)\cdots(p^e - p^{d-1})$

[stoic python]

Ann:

[vague bronze]

Ah ok thank you so much

[south sedge]

How do I show that F1 is not a linear transformation, by using addition?

[dusky epoch]

$F_1 = (e_1 x_1 + e_2 x_2) = x_2^2e_1 + x_2e_2$

[stoic python]

Ann:

[dusky epoch]

what on earth is that line supposed to mean

[south sedge]

the line under the e?

[steady fiber]

that entire line

what is it supposed to mean

is that supposed to mean the input to the function is $\left(e_{1}x_{1}+e_{2}x_{2}\right)$ and the output is $e_{1}x_{2}^{2}+e_{2}x_{2}$

[stoic python]

PorosInMyAshe:

[south sedge]

oh, accidentally put an equal-sign there

F1 is not a linear transformation, and i wanna show that by using addition, i already know how to do that by using a scalar.

[dusky epoch]

i mean

if you want to show it ISN'T linear

just give a concrete counterexample

why not... idk let's go with u_1 = 3e_1 + 4e_2 and u_2 = 2e_1 + 7e_2

check that F_1(u_1 + u_2) ≠ F_1(u_1) + F_2(u_2)

[wintry steppe]

I in no way know linear algebra, but is there anyway you can get this from two normal vectors in 3d space?

I can also turn them around

I'm trying to find the diagonal line

[dusky epoch]

what am i looking at

[steady fiber]

what are they perpendicular to

I assume that the thing at the end of the lines is the perpendicular angle thing

but to what

[dusky epoch]

uhhh

"they"?

no like

what am i looking at. what are these arrows supposed to be. why are there right angle-like marks at their tails.

[steady fiber]

they as in the two solid arrows

[wintry steppe]

Lol they are walls

I don't really need this anymore though, I found a better way

[rose grotto]

what would be the transition matrix

for this

[north sierra]

if i wanna find the unit vector, why is that i am allowed to scale vector c (to remove the fractions ) and use that

[tidal kettle]

@north sierra scaling it by a positive constant doesn’t change the direction

That’s why you can do that

[north sierra]

thx

[tidal kettle]

No problem

[vital torrent]

is it just me or are none of these answers right?

https://i.imgur.com/12cw9TS.png

i got:

https://i.imgur.com/cFJTe0K.png

[rose grotto]

anyone know how to make a transition matrix

with this

[vast torrent]

$\begin{bmatrix} (\text{case 1 $\to$ case 1}) &(\text{case 1 $\to$ case 2}) \ (\text{case 2 $\to$ case 1}) & (\text{case 2 $\to$ case 2} )\end{bmatrix}$

[stoic python]

gfauxpas:

[vast torrent]

@rose grotto

[steady fiber]

shouldn't it be
$\begin{bmatrix} (\text{case 1 $\to$ case 1}) &(\text{case 2 $\to$ case 1}) \ (\text{case 1 $\to$ case 2}) & (\text{case 2 $\to$ case 2} )\end{bmatrix}$
assuming your case vector is
$\begin{bmatrix} \text{case 1} \ \text{case 2} \end{bmatrix}$

[stoic python]

PorosInMyAshe:

[vast torrent]

let me think about that for a few minutes, I could be wrong

yes

your version is correct

oops, soryr

[rose grotto]

@vast torrent @steady fiber what numbers would I put : O

cuz

won't I need

4 diff numbers

then [1,0]

[steady fiber]

yes, you need four different numbers

you're given 2 probabilities

the probability of something happening must be 100%

so you can easily find the other 2 probabilities

1 other probability for each case

[rose grotto]

wat numbers would I use so I can understand better v-v @steady fiber

[vast torrent]

in a probability matrix

the columns sum to 1

actually wait

rows or columns, I remember there were differng conventions

A common convention in English language mathematics literature is to use row vectors of probabilities and right stochastic matrices rather than column vectors of probabilities and left stochastic matrices; this article follows that convention.[2]:1-8

-WP

right okay but why would you use row vectors

weird convention

if you're using column vectors, the columns sum to 1

yeah I remember they mentioned this when I took probability, that some books use row vectors for stochastic matrices

I dont remember why

[opal plaza]

I'm prolly missing something dumb but a little confused by this?

[slender yarrow]

a lil problem of dimension

yea

[rose grotto]

I understand mod26

encoding

decoding

but

how is that the answer to this

it doesn't make sense

I tried inversing it then mod(26) the determinant and multypling it by that

[quiet bobcat]

encoded how?

[north sierra]

@rose grotto what are you trying to do?

find the inverse?

[rose grotto]

@north sierra I don't know what im trying to find

considering

I can't figure out

how to get the answer

but the decryption matrix

its not just

the inverse

you can tell just by looking at it

[north sierra]

true yeah

dont know what that is unfortunately

what level Linear Algebra is this?

[nimble egret]

@rose grotto you got 25 for the determinant right?

[rose grotto]

ye

and modulo 26

of 25

is 25

[nimble egret]

yes

[rose grotto]

then wat

[nimble egret]

usual 2x2 inverse process

25 * [11, -2][-4, 3]

25 * 11 (mod 26) = 15 right?

25 * -2 (mod 26) = 2

and so on

should mention

25^-1 = 25 (mod 26)

since 25 * 25 = 1 (mod 26)

[blissful vault]

how do i do this?

unit vector means the norm is 1

so it made me think of the formula

u dot v = ||u|| ||v|| cos theta

[torn hornet]

how is the norm defined

also yeah you use that

so think about how || is defined

[blissful vault]

it's the square root of the components squared

[half ice]

"Both unit vectors" is the trick here

[torn hornet]

whats another defination

using dot product

[half ice]

What's ||v||?

[blissful vault]

hmmm

[torn hornet]

whats u.u

[blissful vault]

ohhhh

sqrt(u.u) = norm u?

[half ice]

||v|| is the norm of v

"Unit vector" means norm is 1

[vast torrent]

u.u

uwu

[blissful vault]

so u.u = 1

[torn hornet]

yeah so just use that ivey-kun

[blissful vault]

thanks !

[vast torrent]

New notation proposal: <u,u>=uwu

[half ice]

What's wrong with u.u

It's also pretty

[blissful vault]

u.u is good

what we really need is a toggle to turn off | |

as spoilers

[vast torrent]

Oh thats why

[half ice]

Use \

\||v|| will display correctly

[vast torrent]

\u\

|u|

\\u\\

?

||owo||

Ah

[blissful vault]

ohhh

||u||

[blissful vault]

how do i convert this

equation for a line

nvm

[earnest orbit]

why do we need the absolute value for vectors often?

[gray dust]

context?

[paper egret]

what does it mean for 2 matrices to be similar?

[dusky epoch]

do you know the definition

we say A is similar to B iff there exists a matrix P s.t. PAP^-1 = B

[paper egret]

oh yes

what does it mean intuitively though

kinda struglging with the intuition

[dusky epoch]

well

basically if two matrices are similar you can pick two (potentially) bases on R^n such that the matrices' actions on their respective bases are the same

[paper egret]

uhh do you have a quick numerical example to show it?

[native lodge]

a diagonalizable matrix

[dusky epoch]

no unfortunately i don't

[paper egret]

outside of diagonalizablematrix plz

i know it in the context of eigenvectors/eigenvalues

what about outside of that context

[dusky epoch]

ok uhh

idk

let V be a vector space of dimension 3

wait no

this is gonna be too abstract nvm

[paper egret]

rip

[half ice]

Honestly no need for intuition.

A ~ B if there's a P such that A = PBP'

This just ends up being a natural way to group matricies together, and similar matricies have a lot of the same properties

[paper egret]

i'm not sure if this is the right way to think of it, but i just think of it as language translating

P is your translator

[dusky epoch]

P'

you mean P^-1

not P^T

[half ice]

Oh yeah change of basis is a great example of this

But this is larger than change of basis

[paper egret]

how is it larger than change of basis?

[half ice]

I used P' as P^(-1) was too lazy to write it

In a change of basis, you'd have to think of a basis, a new basis to express it in, and a transformation

[paper egret]

yes

[half ice]

Too much machinery for this, where we can say that two matricies are similar, and therefore share properties

[paper egret]

oh wait that's a great way to put it that i never thought about

basis A<-> some transformation <-> basis B

makes A ~ B

[half ice]

Also change of basis ignores the grouping part, which is important to think about. Similarity is a great way to group matricies together

[paper egret]

this is such a mindfuck

[half ice]

Yes, if you can express a matrix A as a matrix B using a change of basis, A~B

You'd expect something like this, since both transformations are the same and should have the same determinant/eigens

[paper egret]

yup

that makes each and every diagonalizable matrix unique, if they are even different by one entry in the diagonal

is that true?

[half ice]

That's true! Diagonalizations are unique

[paper egret]

and i'm assuming

each diaognalizable matrix has its own defining eigenvectors

or can two identical diagonal matrix have different corresponding eigenvectors

or does a diagonal matrix always have a set and only one set of corresponding eigenvectors

oh wait nvm i thik i figured it out

eigenvectors = P

you chose the right P to have a diagonal only entries matrix

if you chose any other P, they wouldnt be eigenvectors anymore, and ur matrix aint gonna be diagonal

so does that mean

diagonal iff eigenvectors/eigenvalues?

i hope my question made sense lol

[half ice]

A matrix of size n×n can be diagonalized if you have n eigenvalues with multiplicity

[paper egret]

does it go the other way

[half ice]

Or, that the characteristic equation splits over the field

[paper egret]

if you have n eigenvalues with multiplicity, a matrix of size n*n can be diagonalized

[half ice]

Yes

[paper egret]

ah so its an iff relationship between a diagonal matrix and eigenvectors/eigenvalues

[half ice]

Wait that's what I said lol

[paper egret]

eigenvectors/eigenvalues ALWAYS yield a diagonal matrix and
a diagonal matrix always implies eigenvectors/eigenvalues

wait hold on

sorry

how about

if a matrix of size n*n can be diagonalized, u have n eigenvalues

i'm trying to understand the relationship between diagonal/eigen

[half ice]

If your matrix can be diagonalized, your characteristic equation splits

And vice versa

[paper egret]

ah so diagonal matrix is strictly unique for eigenvectors/eigenvalues and the other way around too

eigenvectors/eigenvalues is strictly unique for one diagonal matrix

and only one

[half ice]

Yeah no matrix has two characteristic equations

[paper egret]

oh SHIT

characteristic equation is the way u group matrices together

HOLY CRAP

IT"S ALL CONNECTING TOGETHER

!!!!!!!!!!!!!!!!!!!

bRO

this is big brain

damn this was basiclaly my whole semester in 30 mins

i feel like a god

[half ice]

Oh yeah true. If A~B then they both have the same characteristic

I think? That one I'm not sure about

[dusky epoch]

you mean charpoly?

[half ice]

Yaya

[dusky epoch]

yeah that's not an iff. goes only one way.

[paper egret]

oh wat?

not an iff?

[dusky epoch]

two matrices can have the same charpoly yet not be similar

[paper egret]

HOW

?????????

[dusky epoch]

[ 1 0 ; 0 1 ] and [ 1 1 ; 0 1 ] both have (x-1)^2 as charpoly

yet one's diagonalizable and the other isn't

[half ice]

^^ The bae example

[dusky epoch]

well the former is the identity matrix so it like

is diagonal

[paper egret]

u cant diagonalize the second one?

seriously?

[dusky epoch]

yes

[paper egret]

😦

[half ice]

The identity matrix is similar to nothing else, since
P'IP = P'P = I

[dusky epoch]

it has one eigenvalue, 1, with multiplicity 2

but only one eigenvector for that eigenvalue

[ 1 ; 0 ]

[half ice]

But two similar matricies will have the same charpoly

[paper egret]

so outside of identity matrices, i.e exlcluding identity, does the statenet still hold true? same characteristic implies A~B

[dusky epoch]

no

i could give you a more substantive example

[paper egret]

oh damn that wouldve been perfect if it went both ways lol

[dusky epoch]

$A = \begin{bmatrix} 2 \ & 2 \ & & 3 & 1 \ & & & 3 & 1 \ & & & & 3 \end{bmatrix} \ B = \begin{bmatrix} 2 & 1 \ & 2 \ & & 3 \ & & & 3 \ & & & & 3 \end{bmatrix}$

[stoic python]

Ann:

[dusky epoch]

$\chi_A = \chi_B = (x-2)^2(x-3)^3$

[stoic python]

Ann:

[paper egret]

yes

[dusky epoch]

but the dimension of their eigenspaces are different

[paper egret]

this has to do with multiplicity right?

two things can have the same char poly

[dusky epoch]

and that's something that gets preserved with similarity

dimension of eigenspace = geometric multiplicity

anyway

[paper egret]

but their multiplicity gets a little sketch

[dusky epoch]

for λ=2, the dimension of A's eigenspace is 2 while that of B's is 1

and for λ=3, the dimension of A's eigenspace is 1 while that of B's is 3

[paper egret]

so would this be true then

A~B iff char poly, and eigenspaces are identical for each corresponding eigenvalue

[half ice]

Nop. That's not true either

[paper egret]

what the

[half ice]

Well, wait, eigens are identical?

[dusky epoch]

you might want to look into something called jordan normal form

[paper egret]

uh i mean

[half ice]

If the eigens are identical they're the same matrix

[paper egret]

each corresopnding eigenvalue, has same geom mult

[dusky epoch]

yeah no that's insufficient still

here lemme

[paper egret]

ok

[dusky epoch]

hh

nvm

[half ice]

This idea extends pretty well. In abstract algebra, two elements a,b in a group are conjugate if for some g, g'ag = b. Two conjugate elements have similar properties.

[paper egret]

so what'll make it go both ways? what condition for A~B

[dusky epoch]

same JNF

[paper egret]

lemme look that up

[dusky epoch]

(jordan normal form)

[paper egret]

ooo that's cool

math can be so perfect, yet at thes same time not

[gleaming topaz]

If I want to find a vector orthogonal to two lines, do I just construct a vector on each line and take the cross product of them?

[steady fiber]

yes

in 3 dimensions at least

[gleaming topaz]

Yeah my bad missed to include that

[blissful vault]

the proprieties of subspace i learned was

1. subspace must not be empty

2. given x y in subspace, x+y also in subspace

3. given k in Real, kx also in subspace

i don't see how i can prove that abc is/isnot a subspace...

[quartz compass]

if abc=0 what do you know about a,b,c?

[blissful vault]

wait no

if abc is 0 then one of them is perpendicular to the other?

[quartz compass]

well, I'm just asking about the 3 numbers not talking about vectors yet

[blissful vault]

then one of them is 0

[quartz compass]

ok good

so write some examples of what your vectors would be like

with the idea of trying to see if you can find an easy contradiction to the properties of a subspace that you just listed

[blissful vault]

hmmm

so i can simplify that into (a,b,0)

[quartz compass]

that would be one possibility yup

I was thinking more concretely like (1,1,0) would be one

since 1*1*0=0

similarly (1,0,1) would also be one cause 1*0*1=0

[blissful vault]

k(a,b,0) = (ka,kb,0) -> ka(1,0,0) + kb(0,1,0)

and n(a,b,0) + m(a,b,0) = na(1,0,0) + nb(0,1,0) +ma(1,0,0) + mb(0,1,0)

[quartz compass]

what about (a, 0, c)

[blissful vault]

i will just change it to ka(1,0,0) + kc(0,0,1)

[scarlet hamlet]

hey, was just wondering if the converse of If hermitian, then unitarily diagonalizable. is true?

[quartz compass]

you have to have both of them

simultaneously

(1,1,0) and (1,0,1) are both in the set

you can't pick and choose

[blissful vault]

ah so the set is actually empty?

[quartz compass]

haha no it's not empty

you're on the other end of it

it's not a subspace

[blissful vault]

yeah it's not a subspace because it's an empty set?

[scarlet hamlet]

you know the 3 conditions of a subspace
so attempt to show

1. non empty
2. closed under addition
3. closed under scalar mult

[quartz compass]

it's not an empty set

it contains (1,1,0) and (1,0,1)

[blissful vault]

so is it that if i add 110 to 101 it becomes 212 which doesn't satisfy abc=0

so it's not closed under addition

[quartz compass]

bingo

[scarlet hamlet]

yup

[blissful vault]

ay thanks

[scarlet hamlet]

go step by step!

try and go through each condition

[rose grotto]

hi I don't understand how it goes from this

to this

its not just the inverse

they do something else to it

but im not sure what

when im decrypting something

you get the determinant

and do mod 26

[slender yarrow]

it's probably mod 26 or some shit like that

[rose grotto]

and times it by the inverse

I tried that

maybe the answers wrong

even if you did do mod 26

it makes no sense how they are all positive

and how the 2 and 4 stay the same

[slender yarrow]

they took the coeffs mod 26 i suppose also

[rose grotto]

what u mean

[slender yarrow]

well what have you got?

[scarlet hamlet]

if matrix A is unitarily diagonalizable, does it mean its hermitian?

unitarily diagonalizable means A* A = AA *

[quartz compass]

and hermitian means?

I know what it means, I'm just saying write it

[scarlet hamlet]

A* = A

idk i recall the dude sayign something in class about a certain converse

being not true

idk if its this one though loool

prolly not ?

[quartz compass]

I wouldn't have defined unitarily diagonalizable as what you wrote down

but it's clear to see (A*A)*=A*A so it's certainly hermitian according to your definition

[scarlet hamlet]

U*AU = D -- oh i could compare it with A * and find out that its the same thing

[quartz compass]

in fact it's independent of A*A=AA*

[scarlet hamlet]

hm lemme play with it

[quartz compass]

if the entries of D have imaginary numbers, then it's not hermitian

since hermitian matrices have all real eigenvalues

so if you can write 'unitarily diagonalizable matrix' as U*AU all the diagonal entries should be real

[scarlet hamlet]

ok ill try and construct a proof of 'if A is unitarily diagonalizable, then it only has real eigenvalues'

[slender yarrow]

@rose grotto rip

[rose grotto]

?

when I do it regularly

I forget

but its not that answer

its

det 25*the inverse

mod 26 of 25 is 25

so its 25*(3,-4,-2,11)

[slender yarrow]

it should be $$25\begin{bmatrix} 11 & -4 \ -2 & 3\end{bmatrix}$$

[stoic python]

emeric75:

[slender yarrow]

but yeah

(i'll just use the fact that 25 is -1 mod 26 to simplify a bit)

so it becomes

$$\begin{bmatrix} -11 & 4 \ 2 & -3\end{bmatrix}$$

[stoic python]

emeric75:

[slender yarrow]

and then reduce it mod 26 (get the coefficients in the matrix between 0 and 25)

$$\begin{bmatrix} 15 & 4 \ 2 & 23\end{bmatrix}$$

[stoic python]

emeric75:

[slender yarrow]

which gives the answer they have @rose grotto

[rose grotto]

ty

[slender yarrow]

[rose grotto]

the 2 values that z are equal to

are both positive

so how do you maximize it

usually you take the lowest negative number

and then do the pivot and all that

but what do you do

if its already all positive

[vast torrent]

I'd start by introducing a slack variable and turning the inequalities into equalities

but I'm not very experienced in linear programming 😦

[opal plaza]

asked yesterday but still a bit confused on this one

[slender yarrow]

i mean yeah their thing clearly misses a row

[quartz compass]

what's the first thing you learn when multiplying two matrices, when are you allowed?

[opal plaza]

oh so the solution is wrong?

[slender yarrow]

1000%

[opal plaza]

lol

I was so confused

thank you

[north sierra]

if im trying to show that a set of vectors is an orthogonal basis, what's the point of showing that each vector is a unit vector?

[native lodge]

that would be if you wanted an orthonormal basis

[north sierra]

oh snap i kept reading orthonormal as orthogonal

[rose grotto]

would this be the transition matrix

[rose grotto]

it doesn't seem so

answer is like 15%

so idk

[dense holly]

Can the spectral radius of a matrix be negative
Like if -5 is my largest eigenvalue in magnitude is the spectral radius 5 or -5

[dusky epoch]

no the spectral radius cannot be negative

if -5 is my largest eigenvalue in magnitude is the spectral radius 5 or -5
it's +5

[violet field]

Hello everybody...
Could I get some help on how to begin this problem?

I’m specifically confused about the latter half (computing basis of hull space)

Prob 4

[dusky epoch]

null*

well

what is the null space of d^2/dt^2

i.e. what polynomials in P4 become 0 upon differentiating twice

[violet field]

@dusky epoch I see 1 and t become zero, so that is the basis?

And the range as well

[jagged saffron]

its the basis for the null space

[violet field]

Also is the range then just 1 t and t2 since that is the column space

[jagged saffron]

is p_4 polynomials of degree 4?

[half ice]

Yus

[violet field]

Basis of the range*

[jagged saffron]

yea then thats right

[violet field]

Thanks wow that was pretty simple 😅

Looked at it for a good hour haha

[dusky epoch]

is p_4 polynomials of degree 4?
at most 4*

the polynomials of degree exactly 4 don't form a vector space.

[north sierra]

so this is a True/False question

I was wondering what normalized means?

it's not clear in my book

[native lodge]

normalized means they are unit vectors

[north sierra]

ohh okay

[native lodge]

so magnitude of 1

[north sierra]

thx that makes answering the question much easier now

[native lodge]

orthonormal basis means your basis consists of orthogonal unit vectors

[north sierra]

yup!

[slow scroll]

u dont have to delete all of your messages when you figure something out lol

[violet field]

Lol I’m still tryna figure it out 😅 just need time if I wanna ask something again, don’t wanna flood

Repost:
Given a basis B w/ 3 vectors and T:R^3–>R^3 so that I have T(v1) T(v2) and T(v3). Once I calculate [T]B how do I find the standard matrix of T?

[slow scroll]

So you know what T does to beta. And you already have that T outputs in standard coordinates. So basically, T is a map that takes vectors in beta to the standard basis.

To find what T does to the standard basis, you should find a matrix that sends vectors in the standard basis to vectors in beta. In other words, something like this: $$ [T]{\mathcal E\beta} [I]{\beta \mathcal E} $$

[stoic python]

kxrider:

[slow scroll]

So question is: what does [I]_betaE look like.

The tricky part about change of basis, is that anytime you are given a basis (like beta), by writing down coordinates you are assuming the standard basis convention (unless told otherwise). So v_1 = (2, 1, 1) is actually v_1 in the standard basis (while v_1 in the basis of beta is just (1, 0, 0)).

[violet field]

Hm @slow scroll

I thought I would call those vectors Matrix A and A^-1*[T]B*A to get the standard matrix of T

[native lodge]

$A^{-1}[T]_{B}A$

[stoic python]

⚡Amphy⚡:

[slow scroll]

@native lodge you think that is the case here? In the problem, it says that T outputs to the standard basis already.

[native lodge]

I was just writing it in Latex so it looked better

since the _ messed up the formatting

[slow scroll]

Hm, @violet field lets say you were told what T does to the basis beta, in terms of the basis beta. So T takes inputs in beta and outputs in beta. THEN your change of basis matrix would look like that. However, the problem is saying that T already outputs to the standard basis, so the left multiplication at the end is unnecessary.

[violet field]

Oh ok thank you! So it’s just A^-1*[T]B?

Thxxxx

[slow scroll]

well, hold on

looking at part b looks wrong to me

[violet field]

Yeah it is

[slow scroll]

take note that T(v2) = 2v1 So you should expect that in the basis of beta, you have T((0,1,0)) = (2, 0, 0) as an example. Does that make sense?

[violet field]

Yeah, I’m still figuring this out. I think I did It incorrectly. I should have created a linear combination of the basis and set them equal to each T(vi) and the solutions to the augmented matrix would be the columns of the matrix TB

[slow scroll]

part b is sort of the opposite of part c. The input part is done for you i.e. we know what T does to inputs in beta, but we need the outputs in beta as well.

So you need a change of basis matrix that sends vectors in the standard basis to vectors in beta. i.e. (-4, 2, 2) would map to (2, 0, 0)

[violet field]

[0 1 0]^t [2 0 0]^t [-1 1/2 1]^t should be the columns of [T]B

[slow scroll]

nope. What do I get when I plug v1 into T?

[violet field]

I should have created a linear combination of the basis and set them equal to each T(vi) and the solutions to the augmented matrix would be the columns of the matrix TB

?

[slow scroll]

Its given in the problem that T(v1) = v3. So that means in the basis of beta that T((1,0,0)) = (0, 0, 1), or in other words, the first column of T_B is [0,0,1]^T by the definition of matrix multiplication with a vector. We did something similar with v2 earlier, so check the second column.

Then the third column is not so obvious, but luckly, you did the work in part a already