linear-algebra

[zinc timber]

you'll get a 0 at some point otherwise

[ionic laurel]

right

wow

that makes so much sense

like a zero vector for one of them yeah

okay that makes a lot of sense

thank you Ryu

[zinc timber]

the first LD vector in your list will turn to 0

[ionic laurel]

@zinc timber i have a question if you don't mind answering

actually you seem to be unavailable

nvm

[zinc timber]

[ionic laurel]

oh

you're here

okay so from a past final

Given this scenario

[zinc timber]

you can ask anyone not just me

[ionic laurel]

The projection of y onto W perp is [-1 -1 -1]

which is asked by a question

the following question is to give a basis for W perp

so i solve the equation A^Tx = 0

transposing A i mean to say

and i get a basis of [1 1 1]

they are both right answers yes?

[zinc timber]

yes

[ionic laurel]

okay thank you ryu just making sure

[zinc timber]

they are a constant multiple of each other

so they'll span the same subspace

[ionic laurel]

true

thats what i was thinking

[zinc timber]

also you can just take the cross product of the cols

[ionic laurel]

they did no work because from the previous question they just saw that since w perp had at least 1 dimension

they just said that the projection of y onto W perp was a basis

[zinc timber]

,w cross product (1,-2,1) and (0,-5,2)

[ionic laurel]

we did not learn cross products

only dot products

thank you though ryu

my exam 3 for linalg is tomorrow

i studied a crazy metric ton but yknow im still scared

[zinc timber]

,w cross product (1,-2,1) and (3, 0,-3)

[stoic python]

Results provided by WolframAlpha

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

Requested by Ryuzaki#5323

[zinc timber]

lol, I was questioning my sanity for a sec

[dull pilot]

how does the line in blue follow from the lines in red

[zinc timber]

that literally gives you 3 of the eigen values

[zinc timber]

yes

[hot swallow]

https://i.imgur.com/fzUPHo2.png

how can i know whether this is LI or LD?

is it LD casue of 1= sin^2 x + cos^2 x ?

[lavish jewel]

that is indeed the definition of linear dependence

[hot swallow]

thank you, also a matrix is diagnolizable if algebraic mult. = geometric mult? right

[wintry steppe]

true

true in both directions, even

[hot swallow]

alright thanks

[raw cloud]

guys who can describe me the column space ?

[lavish jewel]

it's the vector space spanned by the columns of a matrix

[raw cloud]

does the pivot columns span?

[zinc timber]

the col span will be the same as the col span of the cref matrix

[winged prairie]

if i choose eigenvalues as scalars and eigenvectors and a basis of eigenvectors does it work?

[lavish jewel]

yep

dadrunk, i think i would approach this by first picking the canonical basis and letting the matrix be M. then a change of basis is a sandwhich BMB^-1

and the idea is that BMB^-1 = some matrix, say W

and no matter which B you pick, you always get W

then again, inspecting this does go in the direction of simultaneous diagonalization

i guess one could then argue that the EVD is of this form, so W must be diagonal, and all permutations of the eigenvectors should deal the same W

something along thosel ines

[winged prairie]

sorry what is EVD?

[lavish jewel]

eigenvalue decomp, what you proposed

[winged prairie]

ah k

imma try ur plan see if i get somehwere

thanks

[lavish jewel]

but this assumes that the matrix was originally diagonalizable. there must be a way to not have to assume that

[zinc timber]

i was thinking in terms of group theory, like T has to be an elem of the center so constant multiple of I

other than that there is also a theorem states that if T is not scalar then there is a basis wrt which T has 0's in the diagonal, but it's hard to prove

[lavish jewel]

fancy

idk, with what i wrote it's trivial to show that scaled identities satisfy it, but not that nothing else does

[zinc timber]

pretty crazy 🤯

[lavish jewel]

show da proof

[lavish jewel]

sweet

[crystal oracle]

Do I understand correctly that in the inner product space of continuous functions from [0,1] to ℝ, the orthogonal complement of the subspace of polynomials contains only the 0 vector?

[zinc timber]

ig it's true

[wintry steppe]

[zinc timber]

are you gonna say something on the space not being hlibert and then move on to manifold theory?

[wintry steppe]

[crystal oracle]

That was an answer to me, right?

[zinc timber]

ya

polynomials are dense in C[0,1]

[crystal oracle]

Yes, my analysis lecturer claiming this is what caused me to think about it.

[wintry steppe]

i dont think i have any manifold theory stuff to rant about right now, i've mostly been doing commutative algebra recently

[zinc timber]

this is something I wanna stay away from

[crystal oracle]

1. He says polynomials are dense in C[0, 1]
2. The metric topology induced by the uniform norm is finer than the metric topology induced by the inner product
3. Therefore, polynomials being dense w.r.t. uniform norm implies no vector orthogonal to the subspace of polynomials

this is what I thought

[zinc timber]

DG is cool tho

[crystal oracle]

And I am happy whenever I can apply some piece of math I know from another subject in the subject I am currently studying 🙂

[zinc timber]

do you understand what these words mean?

[crystal oracle]

If that was a question to me, then yes. I mean the lecturer only said the thing I listed in 1. 2 and 3 was what I came up with.

[zinc timber]

C[0,1] is not complete wrt induced inner product norm

but complete wrt uniform norm

[crystal oracle]

What does that have to do with anything?

[zinc timber]

nothing

just sayin'

[wintry steppe]

[zinc timber]

[zinc timber]

I do miss your manifold rants tho

[crystal oracle]

Do you by any chance know a simple proof of polynomials being dense in C[0, 1] or of them having only a trivial orthogonal complement?

[wintry steppe]

google "weierstrass approximation theorem"

[zinc timber]

stone-weierstrass is one line proof

[wintry steppe]

ah yes, prove the theorem by appealing to it

[zinc timber]

lmfao corollary comes first

I've seen similar things with adjoint operators, like being able to interchange <Ax, y> = <x, A*y> is the definition, ppl try to prove it

[crystal oracle]

Wikipedia proof seems annoying. I bet I could come up with a simpler proof that would go something like this:

1. Find a way to approximate a characteristic function of a convex set with a polynomial.
2. Therefore, I can approximate step functions arbitrarily well.
3. Since step functions are dense in C[0, 1], so are polynomials.

[zinc timber]

you can also try the proof using Bernsein polynomials

[crystal oracle]

yes that's the one I called annoying

[zinc timber]

[wintry steppe]

how are step functions dense in C[0, 1] if step functions aren't continuous?

[crystal oracle]

Oh, wait

I should've said that C[0, 1] is a subset of the closure of step functions

[vast iron]

What are you doing step-function

[nocturne jewel]

what the fuck

[zinc timber]

couldn't resist

[tropic pebble]

I already knew the span of w and w complement, how do I find the basis in R^4?

W = span ((1,0,1,2),(1,1,-1,0), and W^perp = (1,-1,0,-1/2),(0,1,1,-1/2))

[lavish jewel]

you should be able to just put all of those in a set

[tropic pebble]

You mean all of the four vectors?

My professor gave us a different answer though

[lavish jewel]

there are infinitely many ways of doing this

but what did your teacher suggest?

[lavish jewel]

the 1st 2 vectors are from W, as you see. the last vector is the same as your first vector that spans W^perp, just multiplied by 2

the 3rd vector should be a linear combination of your two vectors that span W^perp

[tropic pebble]

You said I can just put all of the vectors in the span

So, either works?

[lavish jewel]

should be equivalent

[tropic pebble]

Thank you so much!

[viscid lagoon]

might be a bit of a trivial question, but how do i formally show that the showcased set $M$ is a subgroup of $(\mathbb{R}^2, +)$

[stoic python]

lewis

[viscid lagoon]

it's obviously componentwise addition $+ : \mathbb{R}^{2} \times \mathbb{R}^{2} \to \mathbb{R}^{2}$ as usual

[stoic python]

lewis

[stable kindle]

show it's closed and contains inverses

[viscid lagoon]

i mean yeah, it's obvious that it's a subgroup of R2

but how would i write it down, idk how to find a closed form of the set illustrated in the picture

or like how would i formally prove it

it's a bit odd to me

like how do i precisely represent the dots as a set in this case, so it's approximately correct

i don't quite see a way to do it elegantly

[stable kindle]

test

well first define all the dots

like what are their coordinates of the form of

[viscid lagoon]

yeah, that's my problem here

[stable kindle]

oh well that' seems trivial lol

[viscid lagoon]

🥲

[stable kindle]

all of the points are of the form a(2, 0) + b(1, sqrt(3))

[viscid lagoon]

wait

how'd you get to that tho

[stable kindle]

you can get to any point

by starting at the origin and moving left/right

then moving up to the right diagonally

or down left diagonally as it may be

have you heard of a basis

[viscid lagoon]

a little bit of it

not much

[stable kindle]

hmm

well the two-element set {(2, 0), (1, sqrt3)} is a basis for this subset of R^2

[viscid lagoon]

i think it's actually my next lecture

but yeah, i get the jist of a basis

[stable kindle]

i think you can state without explanation that every point there is of the form a(2, 0) + b(1, sqrt3) for some integer a, b

so now you can just apply subgroup test

show closedness, show inverses

[viscid lagoon]

got it, but how'd you get to sqrt 3

or was it just some approximation

[stable kindle]

well it's hexagons

so

60 degrees

pi/3

the height of an equilateral triangle with side length 2 is sqrt3

[viscid lagoon]

oh, i see

okay, yeah, makes sense

thank you

[stable kindle]

np

[hot swallow]

lets say we are given two vectors in R^3, how can we extend v1 and v2 to a basis?

ik what to do after extending it but im not sure ik how to extend it

[lavish jewel]

what you can do is take these two vectors and produce another that is orthogonal to the two of them

(since being orthogonal also means linearly independent, this achieves both goals in one go)

there are versions of the cross product in C^3, for example

or you could directly satisfy the matrix-vector equation for a vector orthogonal to v1 and v2

by making a matrix whose rows are the complex conjugate transpose of v1 and v2, call it M, and finding nontrivial x so that Mx = 0

[old swift]

if you’re generating an orthonormal basis with gram-schmidt

will you get a different basis if you normalize before/after the process?

the answer is no

[glacial terrace]

I am having some difficulties starting this one

[south radish]

Hey I'm looking at hthe proof for why det(A) = product of its eigenvalues. Why is there this term (-1)^n in the char. polynomial?

[wintry steppe]

det(A - λI) is a polynomial with leading term (-1)^n λ^n

[wintry steppe]

it might help to choose bases wisely

if you see "finite-dimensional" in a linear algebra problem then you can usually solve it by picking a basis in the right way

[glacial terrace]

oh I see, like a basis for W and then when I extend to V I can create another basis for W' with the difference of the vectors

[wintry steppe]

it sounds like you have the right idea

[glacial terrace]

thanks man!

[wintry steppe]

take a basis of W, extend to one of V, and use the vectors you added to get W'

neat

[glacial terrace]

and then can I create the direct sum of W with W' to be V ?

Like $W \oplus W' = V$

[wintry steppe]

\oplus

that'll be true, yeah

[stoic python]

mns

[glacial terrace]

I see

[wintry steppe]

and then i think you can get your projection from that

[glacial terrace]

yeah I did an exercise with just that

lol

[wintry steppe]

there you go

[cunning pier]

complex numbers:
z is a complex number and
z =1/(1+i)
find the real numbers a, b with z = a+b*i
am I supposed to write
a+b*i = 1/(1+i) and solve the equation for a and b?

[wintry steppe]

that's one way to do it

it might be a bit easier to use $$\frac{1}{z} = \frac{\bar{z}}{|z|^2}$$

[stoic python]

TTerra

[nocturne jewel]

Answered in the help channel

[cunning pier]

I tend to ask in the thematic channels as well depending if I need quick understanding or explanation

thanks

[wintry steppe]

keep it to one channel only

[cunning pier]

If I write

--> z = 1/z

do I need to define 1/z_1?

[nocturne jewel]

that's shit notation

since z^2 doesnt have to be 1

you can define w:=1+i then say z=1/w

[cunning pier]

gotcha

[cunning pier]

I'm looking for a real number a so why is

a=(1+b-bi)/1+i```?

[oblique prairie]

i’m very new to linear algebra, so don’t use any super complex notation, but why is this not a subfield? i don’t think the book did a very good job of explaining what a subfield is

C denotes the field of complex numbers by the way

someone ping me if you respond

[nocturne jewel]

cause N isn't a field

doesn't have additive identity or inverses, nor multiplicative inverses

[oblique prairie]

oh i see

i’m aware it says that but for some reason when you said it it made sense lol

thanks

[zinc timber]

it's a fucking test

[cyan copper]

Nah

He

[zinc timber]

[cyan copper]

Hw

[zinc timber]

I'm not convinced

[cyan copper]

@zinc timber

[zinc timber]

also it's not linear algebra, try #prealg-and-algebra

[rugged echo]

is eigenvector the same as eigenspace??

[zinc timber]

no

[rugged echo]

what is eigensace?

[zinc timber]

it's the space spanned by the eigen vectors of the corresponding eigen value

[rugged echo]

eigenspace*

it's the geometric multiplicity?

[lavish jewel]

the dimension of the eigenspace is the geometric multiplicity

[rugged echo]

thanks bro 🙂

if i have a diagonizable matrix what happens to the dimension of its eigenspace?

[zinc timber]

think about the geometric and algebraic multiplicity of the eigen values

[rugged echo]

geo and alt mult are equal

[late valley]

Hello!!new here😊

Newbie*

[zinc timber]

then you already know the answer

[rugged echo]

thanks bro

i'm having second thoughts mybad haha

[shy meteor]

Hey, I am currently struggling with this task.

The translation for this task would be:

For which real number does "a" have a range of 3?
Of course there are some other questions, but I am currently only struggling with this one.

I tried figuring out "a" by trying, but for me it doesn't seem like the most effective approach (since I only found out, that "a" can't be 1, but it can be -1,3 or even 5). Could anyone help me with finding a more effective method to calculate this? I have the feeling, I am thinking way to complicated ...

[nocturne jewel]

A is full rank iff det(A) != 0

[zinc timber]

if A is square

[wintry steppe]

[oblique prairie]

i don’t know what’s going on here, but

[shy meteor]

To be honest, I am still a little confused

[zinc timber]

^

[shy meteor]

Yes I got that, but for some reason the faculty is confusing me the more

[zinc timber]

[wintry steppe]

the faculty

[lavish jewel]

emma might be german

or from some other place where the factorial is called "faculty"

they probably refer to the ! in det(A) != 0

[zinc timber]

oh lmao

Now I get what they meant

[shy meteor]

Yes true 😅

Sorry haha, language 😂

[zinc timber]

I thought you are talking about yr professor or something

[lavish jewel]

fun times with falsche freunde

[shy meteor]

But yes also the faculty is confusing me :D

[shy meteor]

The thing is he is ill and doesn't give any classes, so that is also true xD

[shy meteor]

Thanks for helping me out with that :D

[lavish jewel]

yep

so with != they mean "not equal"

this is the notation in some programming languages

$\text{det}{A} \neq 0$ is what they meant

[stoic python]

Edd

[lavish jewel]

if the determinant is nonzero, the matrix is full rank/invertible

[lyric dawn]

does anyone how the name of this matrix and its generalisation to cos(3w) etc... it would like to show it is semi positive definite

it seems to be true from what i have tested

any help would be appreciated

[zinc timber]

idk what you are looking for exactly but it's symmetric, circulant

[lyric dawn]

toeplitz not circulant

i want to show that it is semi positive definite

[zinc timber]

oh yeah, not circulant right

do you have the determinant?

[lyric dawn]

for n=2 it is sin^2(w) and for n=3 it is 0

[zinc timber]

for n=1 it's 1

so you can use Sylvester law of intertia

[lyric dawn]

let me check that

[zinc timber]

ig for for all n>=3 the det will be zero

you can generalize it then

[lyric dawn]

wow it is a strong result

however it does not answer my initial question unfortunatly

[zinc timber]

it doesn't?

did you check sylvester law?

[lyric dawn]

having the det = 0 does not mean that the matrix is semi positive definite, right ?

[zinc timber]

no, that's not it

positive semidefinite means your eigen values are all positive or 0

sylvester tells you that the sign of the eigen values are the same as the determinant of the sub matrices

or the sign of the pivots.

as you have already showed, det of submatrix are given 1, sin^2(w), 0

so eigen values are non-negative

means your original matrix is +ve semidefinite

[halcyon spindle]

I have to prove any finite generating system of vectors in a vector space V contains a basis.

Here my proof. Proof. Suppose for an arbitrary system of $v_1, v_2, ..., v_n \in V$ is a generating system. Since the system is already complete we have to show it is linear independent for it to be a basis for V. Suppose on the contrary the system is linearly dependent. From that we know there exist a $v_k$ in the system that is a linear combination of the other vectors, that is $v_k = \sum_{j = 1, j \neq k}^{n} a_jv_j$. Since $v_k \in V$ and the system is complete it also follows that $v_k = \sum_{j=1}^{n}a_jvj$.

[stoic python]

Kanga Gang Crocodile Plegasus

[halcyon spindle]

It not complete, I been stuck around the end trying to use the two different linear combination of v_k to get a contradiction.

[zinc timber]

@lyric dawn has your doubt been resolved?

[halcyon spindle]

My question is can I say it is not true there exist a v_k in the system that a linear combination of the other vectors in the system, since v_k already has to be equal to that last sum?

[lyric dawn]

i draw the case for n=4. you call what i have circled a "sub matrix" right ? but i dont know its determinant

[zinc timber]

no not that, major diagonal ones

[lyric dawn]

ooooh i see now

[zinc timber]

you can use schur components to show that the det is 0

[zinc timber]

any arbitrary generating set is not independent by default

you have to turn it into one

[halcyon spindle]

I see.

[lyric dawn]

https://www.wikiwand.com/en/Sylvester's_criterion

i understand know, thank's a lot for your help

[halcyon spindle]

quick question For a generating set S that is linear dependent, we know there exist a v_k in S that a linear combination of the rest of the vectors in S. Why is that removing v_k in S we still get a generating set?

if what I am saying makes sense?

[zinc timber]

because the vector is redundant, you can achieve all the vector with other vectors as well

because that vector is already a LC of other

[halcyon spindle]

oh I see thank you.

[wise oriole]

Can someone help me with this one

[stoic python]

Bart

[wise oriole]

Do i just need to reduce it that's it or what's the next step

[quasi vale]

@wise oriole You just need to remove the linearly dependent vectors in the set C

[weak apex]

Can someone help me with some questions in dms?

[zinc timber]

no

[weak apex]

?

[lavish jewel]

you can simply post here, the channel is free at the moment

[wintry steppe]

if (A^T*A)^-1 does not always exist (because it relies on the columns of A being independent) then wouldn't that mean that not all least square solutions exist? if so, what is the purpose then... since it's supposed to solve systems that initially did not have an exact solution? or is it that does A^TA is not always invertible, there are now an infinite number of least square solutoins

[zinc timber]

pseudo inverse is used in that case

[wintry steppe]

in what case.. where A has dependent columns?

[zinc timber]

another eat to deal with it is to remove the linear dependent column from A and try again

[zinc timber]

even if cols are independent,u can still use pseudo inverse

[wintry steppe]

let's say I have a data set to fit a least squares solution Ax = b. And that doesn't work (because the columns of A are dependent), so I move to least squares and try solving A^TAx = A^Tb. Are you saying this will have no solution or infinitely many?

And if infinite, these are a class of points that all have the same "distance" to the column space, and thus are all equally "correct"?

[zinc timber]

yes, the solution will have same dist from the col space, just there will not be an unique description of it

[lavish jewel]

in the case of no sol, you get the projection onto the row space btw

you can always do it with a pseudo inverse. in the invertible case, the pseudo inverse equals the inverse and the sol is unique. in the rank deficient case, you can have either infinitely many solutions or no solutions. when there are no solutions, you get an orthogonal projection onto the row space. when there are infinitely many solutions, you get the solution with the minimum 2-norm

[zinc timber]

I don't understand the "projection onto row space" part

[lavish jewel]

lemme see

let's say that we have a vector w

and we say that Aw + n = y

and that n is in the ortho complement of the column space of A

the LS sol comes from x = (A^TA)^-1 A^T y, yeah

let's replace the inverse with a pseudo inv, cuz were in a case where we're assuming n might be rank def

and for simplicity and jank, use the SVD A = U S V^T

[zinc timber]

meaning null space of A'?

[lavish jewel]

so that we get (A^TA)^-1 A^T = (V S^T U^T U S V^T)^-1 V S^T U^T

yeah, "left null space" as they call it

in this setup, consider only the economy sized SVD, so that S is square and U and V have columns only spanning the col and row spaces, but not their complements

this means S^T = S

we simplify a little and get that (A^TA)^-1 A^T = (V S^2 V^T)^-1 V S U^T = V S^-2 V^T V S U^T

= V S^-2 S U = V S^-1 U^T

now recall that y = Aw + n = U S V^T w + n

[zinc timber]

i think a ^T is left out

[lavish jewel]

so that x = (A^TA)^-1 A^T y = V S^-1 U^T (U S V^T w + n) = V S^-1 U^T U S V^T w + V S^-1 (0)

= V S^-1 S V^T w = V V^T w

and V V^T is an orthogonal projection matrix onto the row space

[zinc timber]

nice

[lavish jewel]

so this includes the case that y is not in the col space, and that w has a component in the null space

the more general flavor

then to wrap it up with a simple example, if you have a matrix
1 0 0
0 1 0
0 0 0
and a vector
1
1
1
the pseudo inverse can recover
1
1
0,

the ortho projection onto the row space

[zinc timber]

tho using the pinv only gives a short formula for LS sol, x* = pinv(A)*y

that's what I used to do anyway

[lavish jewel]

wdym

[zinc timber]

Ax=y to x=pinv(A)*y because A was never rank deficient

in my case

[lavish jewel]

oh lol

[zinc timber]

yeah, i used to pad with a column of 1's

[lavish jewel]

in my usual cases, pinv(A) can't even be computed

[zinc timber]

[lavish jewel]

iterative algorithm life

[zinc timber]

yeah not all optimization are least squares to begin with

[wise oriole]

Determine the base transitionmatrix $P_{B \to \varepsilon}$ from B to \varepsilon ? can someone give me the first few steps?

[stoic python]

Bart
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sinful valve]

rip people who get excited about vector spaces

[nocturne jewel]

why rip me

[wintry steppe]

who wrote this

[sinful valve]

sheldon axler xd

[wintry steppe]

sheldon cooper

thriller44

∥u−v∥2=(u−v)T(u−v)
Hi, can someone help me understand why this is true?

[lyric dawn]

you can consider || u ||^2 = u^T u to simplify

express both term using the coordinates of u and you should find your result

[weak idol]

I am currently struggling with c)Give a basis of kernel(f) and d)Determine fB1,B2, it would be nice to get a hint on how to start

[urban egret]

to get the basis of a kernel just find the solution to Ax=0 and use the vectors of the solution

[weak idol]

I'm confused as to how exactly i need to apply that. Do i just have to find out f(x) through the images f(x+2) and f(1)? f(x)=-4x-9 and then do -4x-9=0?.

[gray dust]

write ker f as the span of a linearly independent set

let p be any vector in ker f, ie f(p)=0. use this to get information on the form of p

[wintry steppe]

where does the sqrt(n) come in in this answer? https://math.stackexchange.com/questions/505333/finding-matrix-norm-equivalence-constants

[wintry steppe]

it looks like in the first step, he applies ||x||_1 < sqrt(n) ||x||_2 to ||Ax||_1 in the numerator but not to ||x||_1 in the denominator. in the second step, he applies a similar identity to the ||x||_2 in the denominator but not the ||Ax||_2 in the numerator

[zinc timber]

cauchy schwartz

[lavish trout]

do similar matrices necessarily have the same minimal polynomials?

[quartz compass]

what have you tried so far @lavish trout

[lavish trout]

im just studying

[wintry steppe]

does anyone know about this?

Question (True or false): Suppose A is a 3 ×3 matrix, then det(2A) = 2 det(A).

I feel like its false?????

[lavish trout]

but I'm thinking similar => same rcf => same invariant factors => same minimal polynomial

[wintry steppe]

cause multiplying matrix by a scaler and then taking the determinant is not the same as multiplying the determinant by 2??

[quartz compass]

work out the case when A is the identity matrix

[wintry steppe]

ok

[zinc timber]

yes

converse not true however

[lavish trout]

ok and was my train of thought correct?

[wintry steppe]

if A is the identity matrix, 2|A| = 2 and |2A| = 8

2 does not equal 8 and therefore, it is false?

[quartz compass]

so there you go, you have found a counter example yup

[zinc timber]

can't really say much about rcf

[wintry steppe]

thanks 🙂

[quartz compass]

in general det(cA)=c^n det(A) @wintry steppe

n being the dimension of the square matrix

[wintry steppe]

gotcha

[lavish trout]

what do you mean?

[zinc timber]

I'm not too familiar with rcf

you can try jcf

[quartz compass]

and you can prove that by writing det(cA)=det(cI * A) = det(cI)*det(A) = c^n det(A) @wintry steppe very simple yeah?

just an important thing to know for the future

[wintry steppe]

noted

it came up on my study guide (finals are around the corner 😦 )

[lavish trout]

do similar matrices have the same jcf?

[quartz compass]

you can prove it directly by writing down the polynomial relation

if $A=SBS^{-1}$ then if $p$ is the minimal polynomial of A you have $0=p(A)=p(SBS^{-1})=Sp(B)S^{-1}$ and so you can always turn a minimal degree polynomial for A or B into the other

[stoic python]

Merosity

[wintry steppe]

A,B, and C are all n x n matrices and they satisfy C=AB. If both the linear system $Ax=0$ and $Bx=0$ only has the trivial solution, then $Cx=0$ must have only the trivial solution

[stoic python]

embeddedmonk20

[wintry steppe]

anyone know anything about this?

[lavish trout]

ah yes that's pretty straightforward

[wintry steppe]

it's a true or false question

y'all are so smart in here lol and here I am. Back to the math....

[lavish trout]

...in reply to @quartz compass

[quartz compass]

lol

[quartz compass]

as a first attempt you could try to assume it's false and multiply by x on both sides and see if you get any kind of contradiction maybe

[wintry steppe]

Is this subset a subspace of the vector space? {(3s+t,t+1,-4s+t) such that s,t \in \mathbb{R}}

(3s+t, t+1, -4s+t) are elements of all real numbers (I can't do latex lmao)

[quartz compass]

show what you've tried

[wintry steppe]

It has to be closed under addition and multiplication

since its all reals, what happens if you multiply it by i? It wouldn't be all real numbers since i is complex

otherwise, if you do anything else with it, it seems to be real numbers

I think its a subspace????????

[lavish trout]

Are these properties necessarily defining matrices with 1's along the diagonal and 0 otherwise?

[wintry steppe]

if you're working in finite dimensions, sure

[lavish trout]

so actually im curious about that

my class has mainly stayed within finite dimensional spaces

do the majority of the results fall apart because of the lack of guarantee of a basis?

for infinite dimensional spaces

[wintry steppe]

lack of a finite basis, yeah

stuff like existence of complementary subspaces, injectivity = surjectivity for operators

usually infinite dimensional spaces are studied with some extra structure on them

like a norm

and instead of trying to do things algebraically you do them analytically/topologically

e.g. the entirety of functional analysis

[lavish trout]

ahh I see

wow, that provides a lot of context for this stuff

so i guess now what differentiates analysis and topology

i've taken adv calc so im curious how it's abstracted or goes further

is there a reason why the jordan form is nested twice?

[zinc timber]

yes

[quartz compass]

stop giving answers

[zinc timber]

oh i thought he's fact checking, nvm

[dark brook]

How do I find the algebraic multiplicities? If I've got this one for example

[zinc timber]

try finding the characteristic poly

[dark brook]

I thought of that, but the question is specified such that we should not try and find the eigenvalues or the eigenvectors. I thought that would involve finding the char.poly

[wintry steppe]

you usually find the characteristic polynomial before finding the eigenvalues and eigenvectors

so i think you're in the clear if you do that

[dark brook]

I found it to be $\lambda^3 + \lambda^2 - 2 \lambda$ but I am not sure how I would get alg.mult from that?

[stoic python]

HrJonas

[wintry steppe]

do you know what algebraic multiplicity means

[dark brook]

Actually, not sure.

Just know p(\lambda) = det(A - \lambda I_n)

[wintry steppe]

the algebraic multiplicity of an eigenvalue is its multiplicity as a root of the characteristic polynomial

so you have to find the roots of the characteristic polynomial (you have to factor it)

[dark brook]

Oh factorise it. My nightmare!
When I've done that, what would be the next step?

[wintry steppe]

find its roots and their multiplicities

which should become clear once you factor it

[dark brook]

Ahhh I think it makes sense! Just need more practice in finding the roots. Thanks a lot 🙂

Ohhh when I've found all this informations about the roots, the alg.mult is the sum or something like that right?

like the alg.mult always needs to be equal to n or something?

[wintry steppe]

the sum of the algebraic multiplicities will always be equal to n

the algebraic multiplicities themselves are the multiplicities of the roots of the characteristic polynomial

[dark brook]

So I found that the roots are 0, -2 and 1

[wintry steppe]

yup

so what are the multiplicities of those roots?

[dark brook]

Good question, sorry if it sounds dumb, but I don't know or see the connection

[wintry steppe]

do you know what "multiplicity" means for a root of a polynomial?

[dark brook]

I must admit I have it hard with understanding the multiplicity and in a different langauge as well

[wintry steppe]

it means the biggest power of (lambda - root) appearing in your factorization, basically

so as a random example, if i have a polynomial like $$\lambda^4(\lambda - 1)^{500},$$ then the root $1$ has multiplicity $500$

[stoic python]

TTerra

[dark brook]

Ohh, I think it makes sense. But I have two parenthesises with each a multiplicity of 1, does that mean I have a multiplicity of 1 or the sum of them both?

[wintry steppe]

a multiplicity of 1

if you're trying to find the multiplicity of a specific root then you don't really care what's going on with the other ones

you just need to read the powers of each factor in the factorization you found, basically

[dark brook]

What if the first parenthesis had a multiplicity of 2 and the other 1?

[wintry steppe]

in my example?

i can't see your steps so i dunno what you're referring to lol

[dark brook]

I mean something like this $\lambda(\lambda + 2)^2 (\lambda - 1)$
I've got that but without the power of 2.

[stoic python]

HrJonas

[wintry steppe]

yeah, you're right. in this example, 0 would have multiplicity 1 and -2 would have multiplicity 2

[wintry steppe]

actually i think you switched up first and second here, but you're right that one of them has multiplicity 1 and one of them 2

[dark brook]

It was just more of an thought example. Just trying to make sense of other cases than my one

[zinc timber]

ok, is there a way to calculate the algebraic multiplicity without finding the chr poly??

[wintry steppe]

dimensions of the generalized eigenspaces

though good luck finding those without knowing the eigenvalues ahead of time lmao

[zinc timber]

ah right, nice

we can try every number in ℂ

[dark brook]

When trying to do an orthonomalbasis, does it really matter where the w_1, w_2 and w_3 etc are placed?

[lavish jewel]

no

the resulting vectors will look different though

it's not a problem, just be aware of it when checking your answers, for example

[dark brook]

Hmm weird. Not sure why matlab is acting like it isn't working

[lavish jewel]

they might not exactly match the ones in your book, but you can check orthonormality by putting the resulting vectors as columns of some matrix M and taking M^T M

[dark brook]

Not sure why its doing it. MatLab says that it can't divide with 0, yet there shouldn't be any

[dusky epoch]

can you show your matlab source

[dark brook]

Sure, here it is.

x3 = sym([0.1e1 / 0.2e1*i i i; -0.1e1 / 0.2e1 1 1; 1 0 1;])

x4 = sym([i -i i/2; 1 1 -1/2; 0 1 1;])

a = orth(x3,'real')
v = orth(x3,'skipnormalization')

k = orth(x4,'real')
j = orth(x4,'skipnormalization')

[dusky epoch]

'real'

try replacing it with 'complex'

[lavish jewel]

also probably call the j variable something else

i dont remember if in matlab modifying j also affects i

[dark brook]

orth is only able to take real or skipnormalization.

x3 works fine with 'real'

[vast iron]

It also appears that one of your imaginary numbers in first row is negative here

(as opposed to the picture you posted, which has a determinant of i)

[dark brook]

Yea the picture is wrong, sorry!

It gets an output without any options, but x3 is quite neat for displaying the result

[lavish jewel]

you can always do pretty(a2)

[vast iron]

That can't be an actual command

[lavish jewel]

it is

[dark brook]

Ah that is neat! I'll have that neat command in my grave. It worked and it looks nice. Though I still wonder why it hates I switch the places

But hey, it works! Thanks a lot 🙂

[dark brook]

What does it mean that an matrix has a rank of 1 and 2?

[lavish jewel]

the that's the number of linearly independent columns (or rows, the number is the same)

[dark brook]

Whops too fast. Does it mean the same for SVD's and doing approximations?

[lavish jewel]

what?

rank means only one thing when talking about matrices

(albeit with several equivalent interpretations)

[dark brook]

I've got this matrix A. I need to find the low rank approximation for rank 1 and rank 2

[lavish jewel]

so you want matrices with only 1 linearly independent row (equivalently col), and 2 for the latter case

[dark brook]

Oh

[lavish jewel]

you can interpret the SVD as a decomposition of a matrix into the sum of rank 1 matrices

[dark brook]

so for σ_1 to σ_r I would discard everything else if I just had rank 1?

[lavish jewel]

you'd keep sigma_1 and set all the others to 0

[dark brook]

Oh okay, thanks 🙂

[pine raft]

Let v= (3,0,4) and w = (2, 4, 4) use this to compute |v x w| = |v||w|sin(theta).
I've calculated this and get the wrong answer!???
for example, |v x w| = sqrt(16^2+12^2+4^2) = 4sqrt(26).
|v| = 5
|w| = 6
sin(theta) = sqrt(1-cos^2(theta))
cos(theta) = 4/5
5 * 6* sqrt(1-cos^2(4/5)) = 21.52
21.52 =/ 4sqrt(26).

Where did I go wrong with this?

[lavish jewel]

where'd u get that cos theta from

cuz i get that cos theta is 11/15

you agree we can get the cosine by normalizing the vectors and taking the dot prod, right?

v dot w = 6 + 16, then divide by (5*6)

[pine raft]

That's what I did 24/30 = 4/5

[vast iron]

Apologies

[lavish jewel]

it's 22/30

[pine raft]

😭

[lavish jewel]

it's bound to happen, doing so many small additions and multiplications

good on you that you're checking with 2 methods

[pine raft]

Appreciate the help in checking 😄

[zinc timber]

[dark brook]

If I've got $\sigma_1 = 10$ and $\sigma_2 = 4$ and I wanted to find the relative errors (?) I would use the formula (one of them) (attached). However I don't see as of why it is that $\lVert A - A_k \rVert}= \sigma_{k+1}$ ?

[stoic python]

HrJonas
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lavish jewel]

i assume that A_k is a rank k approximation of A

what the induced 2 norm does to a matrix is give you the largest singular value (singular values are, by convention, ordered by decreasing size)

if A and A_k have the same first k singular values, subtracting the two matrices will set all k of those singular values to 0

so the k+1 singular value of the difference is the first singular value that can be nonzero

this also means that the notation sigma_k refers to the singular values of A

[dark brook]

Ah, thats weird, but neat! I wonder where that explanation is in my book.
Again, thanks a lot, I appreciate it a lot ❤️

[lavish jewel]

if you know the singular value decomposition, you can easily crack these open

[dark brook]

Indeeed

It is just a lot of training and practice 🙂

[pine raft]

After correcting cos theta, I still find that u * v * sin(theta) =/ 4sqrt(26) by ~ 0.3

[lavish jewel]

,w 56sin(acos(11/15))

[stoic python]

Results provided by WolframAlpha

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

Requested by Edd#5182

[lavish jewel]

4 sqrt(26) is exactly what you got with the other method

[pine raft]

Ah! I had the method wrong

Thanks!

[lavish jewel]

remember 11/15 is cos theta, not theta

[autumn kraken]

f is a linear map. I am supposed to find out the basis of ker(f) and im(f) but I'm stuck here. I have ker(f) = {(x,0) in R² | x in R}, and so I think ((1,0)) is a basis of ker(f). To prove that this is a basis I need to show that the vectors in the tuple are linearly independent (which they are, there is only one), and that the tuple is a generating system of R². This is where I'm stuck. I think I need to calculate the linear hull <{(1,0)}> but I don't know how to correctly apply the definition of the linear hull. It's a linear combination apparently? Please someone help 😄

Now that I think about it, ((1,0)) cannot be a generating system of R², right? So I did something wrong when trying to figure out the basis?

But if f is a linear map, then it has a basis right?

oh nvm

it has to be a generating system of ker(f) and not R², right?

then it makes sense to me

[dusky epoch]

a basis of ker(f) has to generate ker(f), yes.

[autumn kraken]

okay thanks 👍

[autumn kraken]

I wonder

is it the same thing if you have a basis ((1,1,1)) and ((1,0,0), (0,1,0), (0,0,1))?

do they span the same spaces?

I am not sure what to prefer

[lavish jewel]

they do not

the former can only span a 1 dimensional space

the latter spans R^3

[autumn kraken]

I have a subspace {(x,-x) in R²} and need to find a basis. I feel like I can either go with ((1,-1)) or ((1,0), (0,-1))?

hm

[lavish jewel]

the former, not the latter

[autumn kraken]

okay

[lavish jewel]

to be fair, they both span the subspace, but the latter doesn't span ONLY the subspace

it spans all of R^2

[autumn kraken]

what I always do is (x,-x) = x * (1,-1) pull out the variables like this. So you only want to have 1 vector per variable with this method?

hope that makes sense

[lavish jewel]

yep that's right

[autumn kraken]

ok thanks

but that's not a proof that it's actually a basis right

[lavish jewel]

sure it is

[autumn kraken]

I still need to show that it's linearly independent and a generating system?

oh

[lavish jewel]

x is an element of the field, so you have written all vectors in the subspace as a linear combination of vectors

a single vector at that, so since it's nonzero, it's trivially a linearly independent set

[autumn kraken]

ah right

but then I need to show that it's a generating system or not?

we defined a basis as a linearly independent generating system

[lavish jewel]

i would say that was already included in what you did

[autumn kraken]

so this is what I wrote down before to prove that, but dunno if it's time wasted <{(1,-1)}> = {(x,-x) in R²} (with extra steps)

[lavish jewel]

you showed directly that all the vectors in the subspace are given by a unique linear combination of a set of vectors

maybe someone can reword it differently so that it matches your nomenclature better though

[cedar crescent]

does equality occur when v is a linear combination of those vectors...

i'm uncertain about it

[lavish jewel]

extend the orthonormal set to an orthonormal basis for R^n and express v in that basis

[cedar crescent]

nice

i thought i was wrong

[lavish jewel]

you should achieve equality when v lands in the span of the original set

[zealous lagoon]

i dont understand this

[lavish jewel]

do you know how to compute a determinant using minors?

[zinc timber]

what kind of LA question is that

[lavish jewel]

not really LA but

this sort of stuff does show up when doing computational linalg and you have to use a specific method for whatever reason

[zinc timber]

yeah like can't we manipulate rows/cols and get a upper tri angular giving us 0 3x3 minor calc?

these questions don't make sense to me

[lavish jewel]

it can be an interesting thing to do for oneself

to compare the computational complexity

[zinc timber]

aha,

[lavish jewel]

cuz getting it to triangular form has the complexity of row reduction, right

and that can be finicky sometimes

[zinc timber]

basically LU decom

~n^3 I think

[lavish jewel]

in this case, for example, you can do minors twice to reduce it to 3x3, then write the determinants explicitly for those, which is roughly O(1)

so that's around n^2

it's sounds dumb, but it could actually work

[halcyon spindle]

Just learned about linear transformation properly in my book. I find it amazing the fact we only need to know how the transformation acts on a given basis for us to be able to get any vectors in the target space.

I don't know why but it feels so nice.

[lavish jewel]

it's pretty cool

[noble swan]

I found the least-squares solution, but my answer has the signs swapped. Where did I mess up?

[zinc timber]

yes it should be -4,3

how are you writting (12 8 \ 8 10) = (6 4 \ 4 5)

also A'b is not (12 1)

[noble swan]

I shouldn't have changed the signs on A^T*B

Other than that, I'm just reducing the matrices

[zinc timber]

u don't just reduce matrices like that

[zealous lagoon]

Yes

[viscid lagoon]

Let $\mathbb{K}$ be a field such that $#\mathbb{K} \notin \mathbb{N}$ and $V$ be a vector space over $\mathbb{K}$. Let also $n \in \mathbb{N}$ and $U_1, \hdots, U_n$ be subspaces of $V$ such that $U_i \neq V$ for all $i \in {k \in \mathbb{N} \mid 1 \leq k \leq n}$. Show that
$$\bigcup_{i=1}^{n} U_{i} \neq V$$
I'm not really sure on what to do, my first intutition is to prove it by contradicition. So assume that $$\bigcup_{i=1}^{n} U_{i} = V$$
holds true. Maybe inducing over $n$ is what I should be aiming for next? But I have no idea on how to approach this.

[stoic python]

lewis

[modern palm]

If a matrix is in jordan normal form, does it mean it cannot be diagonalized?
Since no other matrix is similar to it

[chilly shard]

well, a diagonal matrix IS in jordan normal form, so the answer is no, trivially.

if you're asking if thats true when a matrix is jordan normal form and it isn't diagonal, then the answer is yes, since jordan normal forms are unique up to reordering.

[modern palm]

yeah, something like $J_3(0)$

[stoic python]

Kurama

[chilly shard]

i take it this is the 3x3 jordan block with eigenvalue 0?

[modern palm]

Yep

[chilly shard]

since there is only one jordan block in its normal form (itself), the only possible jordan forms it has are just itself, which is clearly not diagonal

[modern palm]

oh yeah

thats right

thanks

[chilly shard]

np

[chilly shard]

are you assuming V is finite dimensional?

[wintry steppe]

Is there an algorithm called "unit column algorithm" which helps you find the unknowns in a system of equations by using elementary row operations? It is like Gaussian elimination, however, you don't have to form a triangular matrix

[nocturne jewel]

Yes.... just do Gauss Jordan...

or since RREF is unique, any set of EROs until you get to RREF

[teal grotto]

induction might be a viable approach

then splitting into cases, U_n+1 intersects all of the spaces trivially, or it is a subspace of one of U_1,…, U_n

[still lodge]

found some slides on gram-schmidt

and i have to apply it to find orthonormal basis of the standard P2 basis

inner product defined as integral from 0 to 1 of the product of the two polynomials

this is my work so far, my question is which polynomial i should be using for p_1(x) in finding p_2(x)

like should i use the one i find before (x-1/2) or after (2sqrt(3)x - sqrt(3)) dividing by the magnitude to obtain a unit vector

[still lodge]

update i dont think it matters

[rigid sorrel]

If we are given rank of a linear transformation, how do we find nullity?

say we have a linear transformation T:R^3----->R^3 and rank(T)=3. How would we find nullitiy(T)?

[still lodge]

look into the rank nullity theorem

[rigid sorrel]

it says in google that rank nullity theorem gives us the property that rank(A)+nullity(A)=dimension so in this case Rank(T)=3 and we know its transformation with regards to R^3 so 3+nullity(T)=3 so nullitiy(T) just equals 0?

is my logic correct or am i missing something?

[still lodge]

sounds about right

can you think about why that might make sense

[rigid sorrel]

hmmm gimme a sec

no i haven't gotten anything could I get a hint?

[still lodge]

what is nullity

try not to look it up

[rigid sorrel]

its just whats in the nullspace right

[still lodge]

well in literal terms it's the dimension of the nullspace

what is the nullspace

[rigid sorrel]

I'm gonna be honest.... i don't really remember

is it okay if i serach it up?

[still lodge]

then how about this: what does a linear transformation do

[rigid sorrel]

it just changes the input to a different vector space right

[still lodge]

alright sure

well what happens in a linear transformation from say R2 to R1

a dimension is "lost" so to say

[rigid sorrel]

we go from a 2 dimensional vector space to a one dimensional vector space?

[still lodge]

yeah

what kinda matrix can perform that linear transformation

[rigid sorrel]

uhhhh i dont remember

my bad

[still lodge]

in order to go from R^2 to R^1 we need a 1x2 matrix

[rigid sorrel]

ah yes

[still lodge]

what will be the rank of a 1x2 matrix

[rigid sorrel]

2?

[still lodge]

defend your answer

sorry im being hardass but this is how i like to help lol

[rigid sorrel]

nah it's best this way I actually learn. But do you mind if I quicly submit my hw first and then you can explain?

[still lodge]

aye bro do whatever u need idc

im a random person on the internet

[noble swan]

How do I do this problem? I have Au & Av, but I don't know how I'm supposed to use them to get an answer

[noble swan]

<@&286206848099549185>

Dang, thanks for looking

[hazy obsidian]

How do I calculate the first determinant?

[slow scroll]

"x" is a least squares solution if |b - Ax|^2 is the smallest it can possibly be.

i.e, you need to compare |b - Au| and |b - Av|

[fringe fjord]

Use Leibniz expansion. To construct a term that doesn't vanish choose one of the a and b in the first column -- that forces your chose in the last column too. Proceed by induction.

But is the lower right entry really supposed to be b?

[noble swan]

Ooh, I see it

[hazy obsidian]

I don't know, those were exercises from my teacher

[fringe fjord]

And the lower left is inexplicably an a rather than apparently b's everywhere else on the antidiagonal. I think my answer would be "I cannot figure out exactly which pattern of entries the dots stand for ..."

If it \emph{had} been "b 0 0 . . . 0 0 a" in the bottom row, my hint from before would lead to an answer of $(a^2+b^2)^{n/2}$, but ...

[hazy obsidian]

Yeah, it makes sense

Thank you

[fringe fjord]

Beware the signs, though -- I think I got them wrong.

[hazy obsidian]

Wouldln t Gaussian Elimination also help ?

Till now I only learned Laplace and Gaussian Elimination

[fringe fjord]

Hmm, right. It would be simple to make the matrix upper triangular by a few selected row operations.

If you want the easy way out, you can take the typo at face value and say "the first and last rows are identical, therefore ..." :-)

[hazy obsidian]

:)))))

[dim lotus]

quick questions, how can I find the Euclidean matrix norm of a matrix?

is it equal to the largest eigenvalue of the matrix?

[torn stag]

@dim lotus What do you mean by Euclidean matrix norm?

[dim lotus]

as in this question

[fringe fjord]

There must be a definition of the phrase earlier in the text.

[dim lotus]

ok

[torn stag]

Is it the square root of sum of squares of the entries? What is the definition of it?

[dim lotus]

are they asking about the Frobenius norm?

I'll check that rn

[torn stag]

I'd guess Frobenius or the operator norm

[dim lotus]

this is the definition

I think it is the operator norm?

this is the definition + example

[fringe fjord]

Yes, that's the operator norm.

[torn leaf]

<@&286206848099549185>

Can anyone please help

Ping me too

[torn leaf]

Anyone please?

[zinc timber]

looks like a test

[rough jackal]

@torn leaf bro show assignment title, to prove its not a test please

[torn leaf]

Ok

[wintry steppe]

Okay so my teacher confirmed for this problem that I just need to show y^t y > 0 now

Im thinking that means each entry of Y^t * y ? is always > 0

[zinc timber]

=0 yes

that's just the statement of dot products saying <a, a> \geq 0

and 0 iff a = 0

[torn leaf]

Can anyone please help with 2nd picture

@zinc timber sorry for the ping

But could you

[zinc timber]

that is just the properties of determinant

write them out and answer will be clear

[torn leaf]

How

[torn leaf]

is the answer a?

[golden monolith]

https://www.math.ucla.edu/~yanovsky/Teaching/Math151B/handouts/GramSchmidt.pdf

I am trying my best to program this in Python, but I really am starting to become insane ... Every time I think I get something, all of the sudden something doesn't match up

[still lodge]

is saying vectors of a basis are linearly ind. enough justification to show that the linear transformation is one to one

[fringe fjord]

If the images of a set of basis vectors under a linear transformation are linearly independent, then the transformation is injective, yes.

[nocturne jewel]

basis implies independence

[still lodge]

part of the question was showing it's a valid basis

but i did that with wronskian

it's in polynomial space

[nocturne jewel]

if you show they span and are indep. then it's a basis

[wintry steppe]

@zinc timber Is this better

[still lodge]

i did that already

tryna show injective

[nocturne jewel]

bases arent injective

[still lodge]

hol up i keep forgetting there isn't conntext

7 here

[nocturne jewel]

,rotate

[still lodge]

holy shit that's cool

texit based

[nocturne jewel]

You want to find all v st T[v]=0

ie $[T]_B^E v_B=0$ in matrix form

[stoic python]

Mosh

[still lodge]

boom

https://tenor.com/view/obama-mic-drop-largar-o-microfone-gif-7413222

[wintry steppe]

wtfs a qed

[still lodge]

the vine boom of math

[wintry steppe]

mom

[gray dust]

phrase to end a proof. 'im done proving what i wanna prove'

[still lodge]

here's to hoping my prof takes the def of injective as sufficient proof

[zinc timber]

if every LI set is mapped to a LI set, then it's injective

[fringe fjord]

Heck, if every single-element LI set is mapped to a LI set, then the linear transformation is injective :p

[golden monolith]

Is there someone who can see how you would program the R values of this factorisation ?

[zinc timber]

#computing-software

[golden monolith]

Woop

[gray dust]

at that point its clearer to say nonzero vectors are sent to nonzero vectors

[fringe fjord]

Indeed. :-)

[zinc timber]

[ionic laurel]

Hi I have a question

oh my bad

[torn leaf]

Hi

Can anyone please help how to solve this

Ping me too please

[zinc timber]

@torn leaf #geometry-and-trigonometry or #prealg-and-algebra

[torn leaf]

This is linear algebra

Vector topic

[zinc timber]

looks more like a coordinate geometry to me

[torn leaf]

Nvm I’ll figure it out but can u help with this

@zinc timber

[zinc timber]

have you calculated the row operations?

[torn leaf]

Yes but i dont know what values to use in this question

[dusky epoch]

do you have the row echelon form ready yet

if yes can you show what you got

[torn leaf]

This is all I have

[dusky epoch]

so, you have not yet done the necessary row operations to put the matrix in the necessary form...

in fact you haven't even written down the original matrix fully, and have not done any row operations.

why did you put a blank space where k^2 - 7 had to go?

[torn leaf]

That’s what I’m not getting how to do

[dusky epoch]

what do you mean?

what are you not getting?

[torn leaf]

How to solve it

What numbers to use

[dusky epoch]

if there was no k and it was just numbers, would you be able to do it?

[torn leaf]

For ref

[dusky epoch]

if there was no k and it was just numbers, would you be able to do it?

[torn leaf]

Yes

[dusky epoch]

okay then do it the same as if there was no k.

you won't even need to involve k in any of the coefficients in the row ops here.

[torn leaf]

Then what values

[torn leaf]

Of this?

[dusky epoch]

$\begin{array}{ccc|c}1&1&1&2 \ 3&2&2&5 \ 3&2&k^2-7&k+7 \end{array}$

you start with this

your first step is to zero out the first column leaving only one entry equal to 1 at the top

how would you do that?

i am nearly CERTAIN i've seen you do basic gaussian elimination before.

[torn leaf]

I have to do ref of this right?

[stoic python]

Kanga Gang Annihilator Ann

[dusky epoch]

yes! yes you need to put this in REF!

as the problem tells you!

i'm surprised this even needs to be repeated

[torn leaf]

Well u didn’t reply so I asked again

[dusky epoch]

you said that if there were no k then you would be able to put the matrix in REF

and i am saying that the presence of k does not impede that in any way whatsoever

[torn leaf]

Since there is k

U said not to include

Then what to replace with?

[dusky epoch]

no i didn't say not to include it

i didn't say to replace all instances of k with some number

are you intentionally misreading the things im saying??????????????

it's infuriating

[torn leaf]

Ok wait

Is it

[dusky epoch]

the operations are "add <number> times <row> to <row>", "multiply <row> by <number>" and "swap <row> and <row>"
what i said is: it is possible to put the matrix in REF without putting k into any <number> slots in these operations

[torn leaf]

(3 2 2 5)
(0 1/3 1/3 1/3)
(0 0 k^2-9 k+2)

Is it correct?

Sry of it’s hard to read

[dusky epoch]

,w row echelon form {{1, 1, 1, 2}, {3, 2, 2, 5}, {3, 2, k^2-7, k+7}}

[stoic python]

Results provided by WolframAlpha

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

Requested by Ann#0413

[dusky epoch]

...okay, WA went too far

this looks correct at a glance but i do not know why you chose to put the row [3, 2, 2 | 5] at the top

[torn leaf]

I swapped row 1 and row 2

[dusky epoch]

yeah, and i don't know why you did that

[torn leaf]

It’s wrong?

[dusky epoch]

it's not wrong

it's just needlessly complicated from my point of view

[torn leaf]

I thought that’s what we have to do

[dusky epoch]

you think that a sequence of row operations is invalid if it never involves any row swaps?

[torn leaf]

No

[dusky epoch]

I thought that’s what we have to do

you did a row swap because you thought you always had to do at least one row swap

yes or no

[torn leaf]

Yes

[torn leaf]

Oh yes I do

I misread it

[dusky epoch]

well then you are wrong

it is perfectly possible for a sequence of row ops to never swap any rows and remain valid

[torn leaf]

Oh ok

But it’s not wrong to swap right?

Coz that’s what i was told

[dusky epoch]

it's not wrong to swap

but it's not wrong to not swap either

[torn leaf]

I see

Ok

After that what do we do

1 question

Since the question says 1

But I got 3

[dusky epoch]

you have a system in REF where the last equation reads (k^2 - 9)z = k+2

[torn leaf]

It doesn’t matter right?

[dusky epoch]

you got 3?

oh

[torn leaf]

The first value

[dusky epoch]

okay now i see what you meant

technically you will have to divide the first row by 3 to make it fit the question's requirements

[torn leaf]

How to do that

[dusky epoch]

do what

[torn leaf]

Ohh just the first row dived by 3?

[torn leaf]

The next step

[dusky epoch]

WHY SO MANY REPLY PINGS

[dusky epoch]

this

[dusky epoch]

is

[dusky epoch]

very

[dusky epoch]

annoying

[torn leaf]

Ok

I was just replying to the specific message

[dusky epoch]

you can disable pings

[torn leaf]

So no confusion would be there

[dusky epoch]

anyway, what i was trying to say is that you made things more complicated for yourself with the swap.

[torn leaf]

Fine I won’t ping

Why

[dusky epoch]

if you had not swapped anything you could have just kept the first row as is

anyway, now that you have the equation (k^2-9)z = k+2 you need to understand two things:

• x and y can be unambiguously expressed in terms of z using the other two equations
• thus whether or not (k^2-9)z = k+2 has a solution completely determines the number of solutions for the whole system

[torn leaf]

??

[dusky epoch]

yes you divide the first row by 3

[torn leaf]

Ok

Done

Then

How do u mean by 2 equations?

[dusky epoch]

the other two equations

(1/3)y + (1/3)z = 1/3

you can isolate y in this one to express it in terms of z

then in 3x+2y+2z=5 you can isolate x to express it in terms of y and z and since you know the expression of y in terms of z you can plug that in too to get x in terms of z only

this relies on the fact that the coefficients on y and x in these equations are nonzero

[torn leaf]

Ok I’ll try

[dusky epoch]

all this should be obvious with the slightest bit of algebraic proficiency

[torn leaf]

Ok

[sleek sundial]

https://cdn.discordapp.com/attachments/540211747613704221/919200321266401340/unknown.png

[vast iron]

This is so much easier if you know Dirac's notation

Sum of <• ,e_i><f_i, •> up to m is the identity matrix if m is the dimension of V.

[sleek sundial]

thanks I'll try to work with that

[vast iron]

Well I guess imagine that the inner product of <x,e_i> = c_i, <x,f_i>= k_i, the sum becomes sum of |c_i k_i| = |c_i| |k_i|, now if m=n, these are just components of x in some basis and must add up to | |x| |^2. In case m<n, you would get the inequality.

[vast iron]

What I mean in a more mathy language is that imagine you find n-m vectors that are orthonormal to {e_i}, {f_i}. Then these new sets of orthonormal vectors form bases for V, let's call them {e'_i}, {f'_i}. Then
Sum |<x,e'i> <x,f'i>| from 1 to n. But this is just norm square of x. Now,
|c_1 k_1| + ... + |c_m k_m|+ |c(m+1)k(m+1)| +...+|c_n k_n| = |x|^2
=> |c_1 k_1| + ... + |c_m k_m| < |x|^2
So in case m=n, you get equality, in case m<n, you get the inequality.

[cedar crescent]

If so, can I apply just apply Bessel's Inequality for this

[cedar crescent]

i mean, this

smh

Is it true that $| \textbf{u} |^2 + | \textbf{v} |^2 \ge 2| langle \textbf{u} \cdot \textbf{v} \rangle |$

[stoic python]

usr/bin/kannaaa3777777777

[zinc timber]

try $\ip{u-v, u-v}$

[zinc timber]

we know it's >= 0

[cedar crescent]

ah yes

so I wonder if I could apply Bessel's Inequality if I could prove it

I mean,...

[zinc timber]

i mean it's not the only proof so

[cedar crescent]

ok i just realize the more i study, the more my brain fu*k up

[zinc timber]

also I think you have miswrote the inequality

[cedar crescent]

i lost my ability to know that the heck is this

[zinc timber]

it should be

[cedar crescent]

yea i just realize

[zinc timber]

$\norm{u}^2+\norm{v}^2 \geq 2|\ip{u, v}|$

[cedar crescent]

yea