linear-algebra

[wintry steppe]

and people say i insult people what is going on???

[pallid rampart]

stfu and don't drag me into your bs

[cursive narwhal]

Don’t start a fight here

And don’t discuss this here

[pallid rampart]

Sorry I'll stop xD

[cursive narwhal]

Take this to #chill

[wintry steppe]

@pallid rampart you insult me though???

[cursive narwhal]

I said stop

[wintry steppe]

i don't insult people lol

[cursive narwhal]

If y’all want to discuss this, then bring it to #chill

[ocean sequoia]

hey this might sound kinda dumb but is the whole point of upper triangle matrices that they have the eigenvalues on the diagonal and its an invariant transformation?

[median forum]

I feel a deja vu

[pallid rampart]

I suppose it's useful in backward substitution when solving a system of linear equations?

[median forum]

ok
a second deja vu wtf

[ocean sequoia]

hahaha its a similar question

[cursive narwhal]

Upper/Lower triangular matrices also lend themselves to easy calculations of the determinant of the matrix

[ocean sequoia]

cool

[median forum]

I remember somebody asking that question like 2 days ago and the first answer being bws sub or somethigng similar

[pallid rampart]

Oh that's true

[cursive narwhal]

So if you can reduce a matrix to upper/lower triangular form, calculating its determinant is easy

[ocean sequoia]

sometimes i just expect something deeper idk

[median forum]

reminder you can use them to do LU

then p much do anything else

doing jacobi for instance

[ocean sequoia]

so it makes things easier?

[median forum]

well yes
its hard to pinpoint the motivation of concepts

sometimes they just appear in a bunch of things

and later somebody generalizes

[ocean sequoia]

im satisfied with that answer I appreciate it

[median forum]

really

I basically gave a non answer

uh

[ocean sequoia]

eh not really sometimes that is th answer i feel like

i wasnt sure if there was some massive theory or importance that I was missing other than "it makes things easier"

[median forum]

but what you mentioned also is a point

theyre a type of matrix with special and easily computable properties

also appear in qr decomp for example

[ocean sequoia]

out of curiousity if you can reduce a matrix into LU form does that mean the original matrix would be invariant as well?

like does it exhbit the same properities im not sure if that makes sense?

[median forum]

dont think so

the answer is probably no, but cant say Im familiar with spectral properties preserved by LU

would have to search

[ocean sequoia]

thanks no worries

[median forum]

[ocean sequoia]

i love your emotes

[median forum]

@ocean sequoia thank you

btw

what is your username about?

[ocean sequoia]

ok honestly?

so did anyone here read Eragon when they were younger?

[cursive narwhal]

Ah yes I miss Aragog the Acromantula

[median forum]

thought br was brazil

[ocean sequoia]

there was a spell i thought was really cool when i was younger called brsingir

and i wanted to make that my xbox name

but i mispelt it and it became brzig

[median forum]

ah

[ocean sequoia]

then it just sorta stuck

[median forum]

unique enough

[ocean sequoia]

yea its only ever been taken like twice

[median forum]

oh

not that unique

when my name is taken I just do
fractal+capuccino = fraccuccino

[ocean sequoia]

hahahha i like that

[median forum]

same

easy enough to remember

[wintry steppe]

what is your username about?
@median forum
mathematics

[median forum]

and very unique

polynomial is actually a cool name

[wintry steppe]

yeah i know

[median forum]

but when I do nicks, its generally more flexy shit

[ocean sequoia]

like what

[median forum]

like paracompact hausdorff fractal
tangend bundle fraccommuter

[ocean sequoia]

i dont know any of those words

[median forum]

neither did I 1-2 years ago when I used them

[ocean sequoia]

haahhahahah

[median forum]

now I at least know paracompact hausdorff
tangent bundle is still kinda hard to grasp
vector field commuter is more general ig

[pallid rampart]

I like the name PCLPCSLSC space

[median forum]

fractal is my known nick

[pallid rampart]

Which stands for path connected locally path connected semilocally simply connected space

[median forum]

so I combine it with other concepts

but path connected is locally path connected

not sure, slim

in fact, Im not familiar with the concept of fractal at all besides the idea of fractal dimension

[pallid rampart]

Path connectedness does not imply locally path connected

[median forum]

but ig itd be nice to have some top structure to fractals

isnt locally path connectedness having a path connected neighbourhood for every point?

not equivalent

if thats the case every pc should be lpc

[dusky epoch]

#point-set-topology y'all

[median forum]

ah mb

[ivory basin]

can someone double check to see if i did part a right?

[wintry steppe]

your indices in checking the "positivity axiom" are wrong

other than that, it looks like you have the right idea (some more words would be nice though)

[marble lance]

For b, you should have $\left<u+v, w\right> = (u_1 + v_1)^2w_1^2 + \dots$

@ivory basin

[dusky epoch]

\left< \right>

[median forum]

or like big brain roka taught \ip{}{}

[stoic python]

Lunasong:

[ivory basin]

ah ok i originally had it in that form but i changed it

what about part d im not sure if its correct or not. I feel like its wrong but idk where i would have messed up

[marble lance]

I only see that you did a and b?

@ivory basin

[ivory basin]

sorry not part d i meant part 4 of part a the last axiom rule

[marble lance]

Positivity? Yeah, any counterexample works

It would be better to write explicitly. If u = (0,1,0) then <u,u> = 0 but u ≠ 0

[ivory basin]

ah ok that makes sense thanks

[mint sentinel]

I have z^4 = something, and have to find all z and give the answer of polar form, how do i do that?

[stoic python]

Commander Vimes:

[daring tundra]

Hey, so I was watching 3blue1brown's series on linear algebra, and I was kind of stuck on the idea of viewing a system of equations as a transformation of a vector, since I'm not sure how to relate the understanding of systems as a way to find the intersection of two lines with this new understanding

does anyone have any resources that "bridge the gap", so to speak?

[dire thunder]

mit strang's lectrues

[limber sierra]

at least in the case of homogenous systems, if you enjoy your "intersection of two lines" intuition, you can view a system acting on a vector as "reframing" the vector to now exist in a new space with the "lines" of the system serving as the new axes

(at least in a very specific, very narrow 2 dimensional case)

[daring tundra]

wait

I'm still kinda lost as to how that happens

ngl

ok so

the way i'm currently seeing it is

a system of equations can represent a linear transformation because it follows the same mathematical rules

and so it's sort of like a tool to model a transformation?

but said tool can also represent the intersection of two lines

because they follow the same rules?

[limber sierra]

finding the solution to a system of equations is represented by an intersection of two lines

[daring tundra]

oh true

sorry it's 5 am and my brain isnt making much sense rn

[limber sierra]

itd be very easy to get into some very sophisticated linear algebra topics here involving kernels of linear transformations and etc but i'll avoid that

lets just use a concrete example

suppose we have a system of equations

[daring tundra]

kk

[limber sierra]

2x + y = 0
x + y + 1 = 0

im keeping it homogenous (equal to 0) here for simplicity

of course then the solution is represented by the intersection of two lines

but each equation here itself represents a line on the coordinate plane

2x + y = 0 is a line (specifically the line y = -2x)

and similarly x + y + 1 = 0 is a line (y = -x - 1)

so now suppose we have a vector (x, y)

here this vector is just representing what values x, y take

the solution to this system is the vector (1, -2)

and indeed:

$\begin{pmatrix}2&1\1&1\end{pmatrix}\begin{pmatrix}1\-2\end{pmatrix}=\begin{pmatrix}0\-1\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

you might be able to see how that relates to the original system i gave

just visually

(rearranging the second equation to become x + y = -1 might be useful)

[daring tundra]

yeah I can see how that works

[limber sierra]

so in a sense, this tells us

if we care about setting the first expression equal to 0, and the second expression equal to -1

or geometrically, if we find where they intersect when we set the first expression equal to 0 and the second equal to -1

the solution to that is the vector (1, -2)

but what if we try a different vector?

say, (1, 0)?

$\begin{pmatrix}2&1\1&1\end{pmatrix}\begin{pmatrix}1\0\end{pmatrix}=\begin{pmatrix}2\1\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

this represents another "set of values" our equations can take

the important thing here is that

for any given input vector, we come out with a certain set of values of the equations

visually you can intuit this as

we "remap" the x and y of the vector to instead use the lines generated by the equations (2x + y and x + y) as axes

and then figure out where the vector would end up

under this new system

(in the case of the vector (1; 0) it ends up at (2; 1))

let me try and draw this

it might make more sense

[daring tundra]

that makes sense, though ngl I'm still kind of confused as to how the interpretation of "lines" fits in

like

idk the way we were taught systems was "these two lines intersect somewhere"

[limber sierra]

the orange line represents the expression 2x + y

er wait

i drew it upside down

one sec lmao

ok so the orange line is the expression -2x = y, the blue line is the expression y = -x

and the red line is (the endpoint of) the vector (1; 0)

but when we write (1; 0)

we're thinking of this vecotr in relation to the x and y axes

i.e. the white lines

let's instead try and think of this vector in relation to the blue and orange lines

with the orange line being our sort of "x" (call it x') and the blue being our "y" (call it y')

now if we try and "deform the plane" to make x' end up where the x axis currently is, and y' end up where our y axis currently is

[daring tundra]

Oh I see

wait but ok

if that's the case

say our system was

2x+y=0

3x+y=0

why would our matrix be:
[2, 1]
[3, 1]

instead of

[1, 1]
[-2, -3]

because if the new basis vector were in the direction of the second line

[limber sierra]

this isnt quite a change-of-basis

[daring tundra]

oh okay

[limber sierra]

if thats what you mean

we think of ``applying a vector to this system" as writing [\begin{pmatrix}2&1\3&1\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix} = \begin{pmatrix}0\0\end{pmatrix}]

[stoic python]

Namington:

[limber sierra]

if you actually do the matrix multiplication

to expand the left hand side

you might see why

[daring tundra]

yeah no I get that

[limber sierra]

because when you expand the left hand side you get:

ah

okay

[daring tundra]

I just don't see how those basis vectors that we get are related to the lines on the plane

[limber sierra]

maybe this is helpful

my drawing skills are kinda bad but

the yellow line is supposed to be parallel

to the blue line

and the green lines to the orange line

note that we're making the orange line our "new" x axis, and the red vector is now 2 "green lines" away

and similarly the red vector is 1 "yellow line" away from our "new" y axis

this drawing is very bad since i was imprecise

[daring tundra]

oh i see

[limber sierra]

but you might be able to get the idea here

[daring tundra]

so those lines are like

the old basis? kinda

like

yeah okay I kinda get it, though I'm probably going to need to revisit this once I wake up

thank you so much for explaining!

[limber sierra]

yeah admittedly the connection here is fairly abstract

[split heart]

To find the transformation of [x]B I've to do A.[x]B right?

A is the attached matrix

[wintry steppe]

Well if your transformation is defined by a matrix A, and you want to transform "x" then you do Ax @split heart

[split heart]

Alright and the indirect way to find T([X]B)

Is: [X]B . P . [X]B' . B . [T(X)B'] . P ** -1?

[wintry steppe]

What are the dots and the ** ? I've never seen that notation

[split heart]

Exp

^

[wintry steppe]

and the dots ?

[split heart]

multiplication

[wintry steppe]

And what's indirect

[half storm]

I can't reallly make out what it is your doing.

[wintry steppe]

Do you mean inverse ?

[split heart]

Yes

I speak spanish, I find difficult to explain myself, sorry

[half storm]

Your english is fine it's just the notation

[wintry steppe]

To find the inverse of a matrix I make an augmented matrix with my matrix on the left, and the nxn identity matrix on the right

and row reduce until the left matrix becomes the nxn identity matrix

[A|I]

[split heart]

Oh yes, that's not what I'm talking about. Maybe that scheme is more clear

[half storm]

I see now, you're asking us about that diagram

I see.

[split heart]

Yes

[half storm]

You're trying to see if your doing the matrix multiplication the right way so it follows the diagram.

One sec.

So the diagram is telling you two ways you can find what you want. I.e. you can take two paths $${[T(x)]}{\beta '} = P^{-1}AP{[x]}{\beta '} $$

That's one way to perform the multiplication to get the desired vector

And then there's another

which is ${[T(x)]}{\beta '} = B{[x]}{\beta '}$

[stoic python]

JohntheDon:

[half storm]

That's what the diagram is telling you

[split heart]

Alright

[half storm]

Does that make sense?

[split heart]

yes I've to go through the diagram

[stoic python]

JohntheDon:

[split heart]

but what happens if I wish to go from [X]B to T[X]B without using A[X]B

P^-1 A P = B right?

[half storm]

Yea you can say that.

[prisma pier]

I've seen Riesz rep written as both f(v)=v*f and f(v)=f*v. Is one of these wrong (assuming this is in a complex inner product space) or are they alternate versions?

[dusky epoch]

for a complex inner product space it depends on which argument your inner product is linear in

if it's linear in the first argument it's <v,f>
if it's linear in the second argument it's <f|v>

[prisma pier]

ok I think I see

Thank you very much

[ocean sequoia]

can someone give me a bit of a hint on this

im struggling with eigen/invariant-related proofs

👍

[wintry steppe]

oh I like that problem!

[ocean sequoia]

yea honestly these problems have been really hard

im getting stuck around 3

does anyone know of any other books that might have easier eigen/invariant proofs so i can get some more practice and come back

[half storm]

The text used in at universities is david c. lay and we used jeffrey holt at my school.

holt as a ton of problems.

[ocean sequoia]

perfect thanks

[tardy quail]

Can someone check this? I know this is probably in the wrong channel but I'm very anxious to know if I got this right or wrong on an exam I took

[ocean sequoia]

@tardy quail #prealg-and-algebra

[dreamy iron]

A question regarding Direct-Sum of two Subspaces and it's proof:

Axler makes no mention of the mutual subset inclusion, but I thought that was necessary.

[wintry steppe]

mutual subset inclusion?

what does that mean?

[dreamy iron]

part 1 of my proof

part 2 of my proof

[wintry steppe]

What's the working definition of "direct sum" here? That a vector w in U⊕V can be uniquely written as w = u + v?

[dreamy iron]

mutual-subset inclusion is

$$\mathbb{U} \cap \mathbb{W} = {\vec{0}} \iff \mathbb{U} \cap \mathbb{W} \subseteq {\vec{0}} \text{ & } {\vec{0}} \subseteq \mathbb{U} \cap \mathbb{W}$$

[stoic python]

ninnymonger:

[wintry steppe]

Oh. Extensionality of set equality.

[dreamy iron]

this is lecture video about the same proof from the same book, it also makes no mention of the double subset inclusion: https://youtu.be/wUEmkt7Qxck?t=199

[wintry steppe]

The first part checks out.

[dreamy iron]

of my proof?

[wintry steppe]

Yeah, you proved the one direction just fine.

[dreamy iron]

kk, tyty.

[wintry steppe]

Although I wonder if there's a more direct, intuitive way of doing it. But your proof is valid, for sure.

θ is in U and it's in W, therefore, θ = θ + 0 = is one decomposition into U⊕V while θ = 0 + θ is another.

but since you have a unique decomposition, you get the factors equal to each other: θ = 0 (and on the other side, 0 = θ)

But that might come off as more hand-wavy without taking time to establish that such a unification argument is valid

[hollow finch]

is there some possible simplification to this

$\vec{b}\times(\vec{a}\times\vec{b})$

[stoic python]

nix:

[wintry steppe]

It would be easy enough to check by components, I'd think

Or to look it up on Wikipedia: https://en.wikipedia.org/wiki/Vector_calculus_identities

[hollow finch]

feel free to check my work but this is what i got and i dont see anything immediately

$\begin{vmatrix}0&b_3&b_2\a_1&a_2&a_3\b_1&b_2&b_3\end{vmatrix}\vec{i}+\begin{vmatrix}b_3&0&-b_1\a_1&a_2&a_3\b_1&b_2&b_3\end{vmatrix}\vec{j}-\begin{vmatrix}b_2&b_1&0\a_1&a_2&a_3\b_1&b_2&b_3\end{vmatrix}\vec{k}$

[stoic python]

nix:

[hollow finch]

or this but i dont think its much better lol

$\begin{vmatrix}(b_3\vec{j}-b_2\vec{k})&(b_3\vec{i}-b_1\vec{k})&(b_3\vec{i}-b_1\vec{j})\a_1&a_2&a_3\b_1&b_2&b_3\end{vmatrix}$

[stoic python]

nix:

[wintry steppe]

Probably want to go with the just-look-it-up route then 😮

[median forum]

b×(b×a) = b(b•a) - a(b•b)

[hollow finch]

i guess top row looks like ijk crossed with b which is weird but idk

aha that looks promising

[median forum]

so minus that

[hollow finch]

let me see if i can massage this

[wintry steppe]

My thought if you want to work at it algebraically is that the Jacobi identity might be useful.

[median forum]

a(b•b)-b(b•a)

[old flame]

Question : Suppose V is finite dimensional. Prove that if $U_1,…U_m$ are subspaces of B such that $V=\bigoplus_{i=1}^{m}U_i$, then $dim V = \sum_{i=1}^{m} dim U_i$.

Here is my proof : Since $V = \bigoplus_{i=1}^{m}U_i$ ,every $v \in V$ can be written as an unique representation $v = \sum_{i=1}^{m} u_i$ for some $u_i \in U_i$.

Then choose a basis from each $U_i$ and express $u_i$ as a linear combination of those basis vectors, hence there are scalars, $a{i1},...,a_{in} \in F$ such that $u_i=a_{i1}v_{i1}+…+a_{in}v_{in}$ for each $i=1,…,m$. Substituting this into the unique representation of $v$ yields, $v=a_{11}v_{11}+…+a_{1n}v_{1n}+…+a_{i1}v_{i1}+…+a_{in}v_{in}+…+a_{m1}v_{m1}+…+a_{mn}v_{mn}$. So the list $(v_{11},…,v_{i1},…,v_{mn})$ spans $V$. Consider the linear combination of 0 from the list, $0=a_{11}v_{11}+…+a_{mn}v_{mn}$, as each list (v_{i1},...,v_{in}) is a basis of $U_i$, they are linearly independent, implying that the only combination of 0 is when all the scalars $a_{11}=...=a_{mn}=0$. Hence is a basis of $V$.

Therefore, since the cardinality of the basis is the sum of the cardinalities of the bases thus $dim V = \sum_{i=1}^{m} dimU_i$. Any problems ?

[median forum]

dont you have to prove LI?

[wintry steppe]

for each some u_i \in U_i.

in the second sentence, did you mean to write "express u_i as a linear combination of those basis vectors"?

[old flame]

yes

[stoic python]

Otoro:

[wintry steppe]

as fractal says

you should probably give some more words of justification for why that list of vectors is linearly independent

the last sentence is kind of nonsense. although it's clear what you mean (to someone who's done this stuff before, at least), you'd benefit a lot from using proper terminology

[old flame]

ah I see, Im working on the lienar independence part now. May I ask what are such proper terminology ?

[median forum]

youre better off saying the cardinality of the basis is the sum of the cardinalities of the bases

[wintry steppe]

^

[old flame]

thank you, I shall correct it

[median forum]

np
thank you for making the community wholesome

[old flame]

is that sufficient for a proof for LI ?

@median forum

[median forum]

I dont see it

where o.o

[old flame]

oh in the original text sorry, Im not sure why the text doesnt pop up

I made the edit

[median forum]

after a while it stops reacting to edits

[old flame]

so what do I do

[median forum]

edit
copy
paste

dont forget to edit first cause discord ommits some symbols

[old flame]

alright

Since $V = \bigoplus_{i=1}^{m}U_i$ ,every $v \in V$ can be written as an unique representation $v = \sum_{i=1}^{m} u_i$ for some $u_i \in U_i$.

Then choose a basis from each $U_i$ and express $u_i$ as a linear combination of those basis vectors, hence there are scalars, $a_{i1},...,a_{in} \in F$ such that $u_i=a_{i1}v_{i1}+…+a_{in}v_{in}$ for each $i=1,…,m$. Substituting this into the unique representation of $v$ yields, $v=a_{11}v_{11}+…+a_{1n}v_{1n}+…+a_{i1}v_{i1}+…+a_{in}v_{in}+…+a_{m1}v_{m1}+…+a_{mn}v_{mn}$. So the list $(v{11},…,v{i1},…,v_{mn})$ spans $V$.
Consider the linear combination of 0 from the list, $0=a_{11}v_{11}+…+a_{mn}v_{mn}$, as each list $(v_{i1},...,v_{in})$ is a basis of $Ui$, they are linearly independent, implying that the only combination of 0 is when all the scalars $a_{11}=...=a_{mn}=0$. Hence $(v_{11},…,v_{i1},…,v_{mn})$ is a basis of $V$.

Therefore, the cardinality of the basis is the sum of the cardinalities of the bases $dim V = \sum_{i=1}^{m} dimU_i$

[wintry steppe]

oof

you might want to fix the _{indices}

[median forum]

seems like you didnt edit before copying

copy while editting is what I meant

[old flame]

omgosh Im so sorry lol, didn't know that all the _ will disappear

[median forum]

@old flame I think its ok otoro

the since kinda misplaced below

but it looks nice

[stoic python]

Otoro:

[old flame]

@median forum "since" is deleted, thank you so much. Final question of the chapter, now I can take a break 🙂

[median forum]

np. anytime : D

[old flame]

and thank you too TTerra

[wintry steppe]

now that you've done it, here's a neat way of doing it (it's basically the same, but shorter):
pick a basis b_i of each U_i. every vector in V uniquely decomposes as a sum of vectors u_i from each U_i, each of which decomposes uniquely as a sum of vectors from b_i. then v decomposes uniquely as a sum of vectors from union b_i, so union b_i is a basis. then you're done by your final sentence

that's just for fun though. what you've done is perfectly good (and if you unravel the definitions, identical)

btw you can use \dim(V) to get a nice looking dim in your tex

[stoic python]

TTerra:

[old flame]

@wintry steppe damn that really summed up my proof, very nice indeed

[prisma pier]

I'm trying to show that given:

the change of basis for the reciprocal bases is given by:

I've been trying for quite a bit. This is what I have so far:

though I can't figure out how the conjugate of T comes in here

I'm really unsure where to go next with this proof

[prisma pier]

ok phew I got it

[open pivot]

Hey guys if I have two subsets with dim1 and dim2, can I say anything about the dimension of the intersection of the subsets without any additional info?

[marble lance]

It is less than or equal to both dim1 and dim2

I am assuming you mean subspaces, and not just subsets

[open pivot]

Yeah I'm working with two bases for two subsets

Which are subspaces

Whats the explanation for that fact?

[dusky epoch]

for any vector spaces U and V, U ⊆ V implies dim(U) ≤ dim(V)

and A ∩ B is a subset of A and of B (and is in fact the largest set with this property)

[marble lance]

What Ann said, and if you want to know why dim(U) <= dim(V) realize that a basis for V also spans U since any vector in U is also in V

[open pivot]

Thanks, im trying to make sense of the definitions. I have two subspaces of R^3 and I need to explain the dimension of their intersection. It's not obvious to me if one subspace is a subset of the other.

[marble lance]

What we are saying is that the intersection is a subset of both of those spaces

[open pivot]

But wouldn't the intersection of two subspaces be larger than either subspace on its own?

Wait im getting confused with the union aren't I

[marble lance]

Yes

Union of subspaces is not a necessarily subspace precisely because it can be larger than either subspace

[stoic python]

whzup:

[gray dust]

most of the time i prefer to write that as dim(U+U')=dim(U)+dim(U')-dim(U cap U') which reads as dim(sum)=sum of dims-dim(overlap)

[dusky epoch]

hehe dim sum

[gray dust]

that was intentional 我真喜欢吃点心

[wintry steppe]

Hello

If matrix A is a 6x5 matrix, then T is from R^5 to R^6 correct (T(x) = Ax)?

Also, Given T: R^5 to R^7, then A is a 7x5 matrix correct ?

[half ice]

Yes

[slow hedge]

I was reading the proof of the spectral decomposition theorem from a Quantum Computation book about the Normal Operators. I get the entire line of reasoning of how the operators PMP=λP, QMP=0, PMQ=0 and how QMQ is a normal operator. What I don't get is the last line of the proof-
"By induction, QMQ is diagonal with respect to some orthonormal basis for the subspace Q, and PMP is already diagonal with respect to some orthonormal basis for P"

Usually, in induction proofs, one says that the argument is true for d=1, then we assume that it is true for d=k and using that prove that it is true for d=k+1. I don't understand the type of induction argument used.

Definitions of Projector and Orthogonal Complements
•Projector: A projector operator defined on some vector space V (n dimensional space) with orthonormal basis {Vi} (i=1...n) projects onto a subspace W (m dimensional space, m<n) with orthonormal basis {Vj} (j=1...m).
P= Σi |i><i| (i=1....m).

•Orthogonal Complement: Q is defined as Q≡ I-P i.e., the projector onto the space spanned by the basis {Vi} (i=m+1....n)

[half storm]

@median forum To show that the derivative operator is a surjective linear operator on the set of smooth functions. Let $f \in C^{\infty}.$
Since $f$ has derivatives of all orders $\implies$ f is continuous on $\mathbb{R}$ and $F(x) = \int f(x) dx$ exist. You then need only to show that $F(x)$ is a function with derivatives of all orders which is pretty easy via induction.

[stoic python]

JohntheDon:

[median forum]

@half storm yes ?

[half storm]

Lol it was a question I forgot question marks

[median forum]

@half storm seems ok

[wintry steppe]

What's the matrix for a rotation around the y-axis ?

Does that mean I have to translate by x, leaving y unchanged, and rotate, then translate back ?

[gray dust]

@wintry steppe is the context a map on R^2 or R^3?

[wintry steppe]

R^3

[gray dust]

so a rotation in y is a rotation in y, nothing complicated, its matrix representation is the 2nd here https://i.stack.imgur.com/VJNzh.png

[wintry steppe]

Cool, thank you

[gray dust]

no prob, and to gut check it IS a rotation in y, see where it maps the standard basis

[wintry steppe]

For $A^{k} = P * D^{k} * P^{-1}$

[stoic python]

Septim VII:

[wintry steppe]

My prof got

And I got

Are they equivalent ?

[limber sierra]

is n odd?

[wintry steppe]

No, any integer > 0.

But I realized that in P, I used only "k"

But my prof used @n1 instead of "k"

I'm not sure why

I think "@" tells the calculator that n1 is an integer

[vague cedar]

wait you can do that on a CAS?

woah that's weird

[honest notch]

I haven't been able to wrap my head around subspaces. How would I approach solving this

[half storm]

I'll sort of start you off and help you with showing that $dim U = 1$. What you need to do is take an arbirtary vector in $U$ and think about how to express that in terms of one vector. But you'll have to make use the defining properties of the set.

So think about an arbitary vectors $ (a,b,c) \in U$. Now, for any vector in $U$ , then we know that $a - 2b + 3c = 0$ and $b + c = 0$. $ b + c = 0 \iff b = - c$ And using this implication $a - 2b +3c = 0 \iff a + 2c + 3c = 0 \iff a + 5c = 0 \iff a = -5c$ Now, with all this $ (a, b, c) = (-5c, -c, c) = c(-5,-1,1)$

So hopefully now you can see that the single vector $(-5,-1,1)$ generates the entire subscape of $U$.

[stoic python]

JohntheDon:

JohntheDon:

[half storm]

Now you need to explain why it is a basis? Can you do this?

[honest notch]

i must prove that all vectors in the set have 3 elements for it to be a basis?

[half storm]

If you're asking that you need to show that all vectors in the set have 3 coordinates, then the answer is no. All vectors in the set will have three coordinates because the set U is a subset of $\mathbb{R}^3$.

[stoic python]

JohntheDon:

[half storm]

The definition of a basis for a vector space is a linearly independent subset of the vector space that generates the space.

But before we move further tothere, do you see why $(-5,-1,1)$ generates the subspace U based on the previous algebra?

[stoic python]

JohntheDon:

[honest notch]

i believe i do from before

[half storm]

O.k. good.

It merely follows from the previous algebra.

Now you need only to show that $ {(-5,-1,1)}$ is a linearly independent subset of U.

[stoic python]

JohntheDon:

[half storm]

Do you remember what it means for a set to be linearly independent?

[honest notch]

apologies for the delay, yes i do

I need to prove that none of the vectors within the set are linear combinations

[half storm]

I need to prove that none of the vectors within the set are linear combinations
@honest notch *of one another

Good. This is correct, but there is a equivalent statement to this and that's what is going to be useful for showing this set is linearly indepdent.

$$
A set of vectors v_1, v_2, \cdots v_n \text{ are linearly independent if and only if there exists a finite set of scalars }a_1, a_2, \cdots a_n \text{ and } a_1 v_1 + a_2 v_2 + \cdots + a_n v_n = 0 \implies a_1 = a_2 = \cdots = a_n = 0.
$$

[stoic python]

JohntheDon:

[half storm]

Ugh I'm tyring to TeX this

[honest notch]

I see, this is equivalent to the statement from before

It's okay (:

[wintry steppe]

get rid of the \text

and it should look fine

[pallid rampart]

A set of vectors $v_1, v_2, \cdots v_n$ are linearly independent if and only if there exists a finite set of scalars $a_1, a_2, \cdots a_n$ and $a_1 v_1 + a_2 v_2 + \cdots + a_n v_n = 0 \implies a_1 = a_2 = \cdots = a_n = 0$

[stoic python]

Whoever:

[wintry steppe]

||sniped lmao||

[half storm]

lol good. It makes the process faster than me fixing it.

O.k. so we're going to use the definition that @pallid rampart TeX'd up.

[wintry steppe]

(they copied your tex lol)

[pallid rampart]

Yeah

[half storm]

In particular we want to show that ${(-5,-1,1)}$ is linearly independet so we use the definition

lol i see.

[stoic python]

JohntheDon:

[half storm]

So we suppose there exists a $a \in \mathbb{R}$ such that $a(-5,-1,1) = (0,0,0)$ and we have to show that $a = 0$ to show that we have a linearly independent set. Do you see how to do this?

[stoic python]

JohntheDon:

[half storm]

Do you see why a has to be zero?

[honest notch]

Yes

Do I need to prove why in my answer?

[half storm]

It wouldn't hurt. It's a very short proof.

$a(-5,-1,1) = (-5a,-a,a) = (0,0,0)$ We know that coordinate vectors are equal if and only if their coordinates are equal yea? So $(-5a,-a,a) = (0,0,0) \iff -5a = 0 , - a = 0$ and $ a = 0$.

[stoic python]

JohntheDon:

[honest notch]

I see

[half storm]

perfect.

So now we have show that $ {(-5,-1,1)}$ is a linearly independent set. We now just have to show that ${(-5,-1,1)}$is a subset of $U$. Do you think you can show that?

[stoic python]

JohntheDon:

[gray dust]

whatcha thonking tera

[honest notch]

I feel like i've missed something, I'm not sure how to prove this

[half storm]

I messed up lol but I fixed it.

Well, we go back to the definition of U right? $$U := {(a,b,c) \in \mathbb{R}^3 | a - 2b + c = 0 = b + c }$$

[stoic python]

JohntheDon:

[gray dust]

@wintry steppe don't thonk me or

[half storm]

So we just have to show that $(-5,-1,1)$ satisfies those equations that define the set right?

[stoic python]

JohntheDon:

[half storm]

Can you see why $(-5,-1,1)$ satisifes them?

[stoic python]

JohntheDon:

[honest notch]

Yes

so proving why this satisfies the equations for the set is the last proof i need?

[half storm]

Yup

Because you have shown that that ${(-5,-1,1)}$ is a linearly independent subset of U and it generates U.

[stoic python]

JohntheDon:

[half storm]

So we're done.

[honest notch]

Thank you for your help

I really appreciate it

[half storm]

Np.

I kind of got to go but I'm sure someone else can help you with the next set.

[floral thistle]

Hello

What's the problem?

[prisma pier]

You need to explain that when one or more of the Bs is not 0, the As cannot be all 0 to prove that last point. Though otherwise your method looks good

[viscid kernel]

Is there a special name for matrixes whose inverse is the matrix itself ?

[dusky epoch]

not really

but such matrices are precisely the matrices which satisfy A^2 = I

[prisma pier]

symmetric orthogonal? (edit: oh I guess that doesn't account for all cases nvm)

[modern palm]

would this be a proof for U being a subspace?

[dusky epoch]

not as laid out, no.

(ii) is troublesome

=> does not mean "next step"

[modern palm]

right. it should mean implies, i'll change it

other than that, would it be fine?

[dreamy iron]

while we're on subspaces, what are the full subspace requirements, written out with all the quantifiers?

i know theirs names as

1. must contain the zero vector
2. closure under addition
3. closure under a scalar multiplication.

[modern palm]

yep, fixed it. thanks

[dreamy iron]

even if the alpha were distributed out. how does that shows it's in the subset of interest?

[stoic python]

whzup:

[modern palm]

maybe divide both sides by alpha?

order?

[stoic python]

whzup:

whzup:

[dreamy iron]

is it not the following:

1. $\exists, \vec{0} \in \mathbf{U},$ such that $\vec{0}+u = u + \vec{0} = u \$
2. $\forall, u, v \in \mathbf{U}$, if $u, v \in \mathbf{U}$ then $u+v \in \mathbf{U} \ $
3. $\forall, u \in \mathbf{U}, , \forall, \gamma \in \mathbb{F}$, if $\gamma \in \mathbb{F}, u\in \mathbf{U}$, then $\gamma u \in \mathbf{U}$

[stoic python]

ninnymonger:

whzup:

[modern palm]

so start by saying vector x in U, meaning x1 = x2 + x3.

then ax in u, means ax1 = ax2 + ax3 => ax1 = a(x2+x3), then i can divide both sides by a, then use the fact that x is in U

to say that ax is in U

[stoic python]

whzup:

[modern palm]

i see, makes sense

thank you

[stoic python]

whzup:

[dusky epoch]

or using 3 with alpha = 0

[dreamy iron]

the subspace's being non-empty and there being at least the zero-vector in the subspace are consistent with each other. i have a proof of this.

[wintry steppe]

I'm having some trouble with this:

5 students are sitting next to each other. They are numbered 1 to 5, and the professor, watching them, is #6.

Every minute, the student in possession of the answer sheet tries to give it to his neighbor. There is a 1/2 probability that the prof. sees him and takes the answer sheet. If not, the student gave it to one of his 2 neighbors if he is in the middle of the row, or his only neighbor if he's at the end of the row.

What is the state transition matrix for this situation?

If anyone has hints or like the dimension of the transition matrix ?

[wintry steppe]

Would this be right ? Does 1 have 50% chance to pass it to 2 since the teacher might see him, or does he have 100% of passing it to 2 since it's the only person he can pass to

[prisma pier]

50%

[dusky epoch]

the former.

your diagram doesn't even have a node for the professor

and why write 0% arrows explicitly when you can just... not?

[prisma pier]

tho you can't pass to the professor

why is the professor a node?

[dusky epoch]

is "the prof has the sheet" not a valid state for the system

[prisma pier]

oh nvm ic

[wintry steppe]

would every node be linked to the professor node in the middle ?

or instead of a pentagon I make a figure with 6 nodes instead

but then I don't see how that would work 😅

[prisma pier]

I don't know enough about this stuff for my answer to be reliable, so I wouldn't trust what I say unless someone else confirms it

but this makes the most sense to me:

[wintry steppe]

That seems like it makes sense

Each one adds up to 100%

thanks for taking a look

[hallow kite]

Quick question about unitary matrices:

if my matrix U is unitary
does that mean U* = U^-1?

[half storm]

Yea by definition.

[hallow kite]

ty! 👌🏾

[half storm]

👍

[formal remnant]

I need help with the following problem, i have to prove that if 3|17a and 3|(2a-7*b) then 3|b. Thanks in advance.

[pallid rampart]

#elementary-number-theory

[latent ledge]

Suppose U and W are subsets of V with $U \subset W$. Prove that $W^0 \subset U^0$.

[stoic python]

Konoha:

[latent ledge]

Let $\varphi\in W^0$, for all $w\in W$, we have $\varphi(w)=0$. Since $U\subset W$, for all $u\in U$, we have $\varphi(u)=0$. So, $\varphi\in U^0$; hence, we can deduce the conclusion since $\varphi$ is arbitrary

[stoic python]

Konoha:

[wintry steppe]

if you are asking someone to check your proof, then it looks fine

[latent ledge]

thanks, I also want to know whether we can use basis on this problem

[wintry steppe]

the vector spaces arent guaranteed to be finite dimensional

so unless you like working with infinite bases (hamel bases?), no, no basis stuff can be used here

the proof here is plenty nice anyways

[half storm]

I thought he was answering someone question that I couldn't see 😆

or like someone asked him a question in another channel and he was answering it in here basically.

[latent ledge]

so unless you like working with infinite bases (hamel bases?), no, no basis stuff can be used here
@wintry steppe i see

[wintry steppe]

Is W^0 the annihilator?

[half storm]

Yea

I can tell by the context of the question.

[wintry steppe]

First time I've seen that notation

[half storm]

or the context of his proof.

It's the notation I'm famaliar with.

[prisma pier]

wait, about this question earlier:

the book im reading defined unitary matrices as anything satisfying:

in which case A doesn't need to be square

is this an uncommon definition?

[wintry steppe]

it's true but it's definitely uncommon, I wouldn't define unitary matrices that way

that's more of a property

[prisma pier]

hmm ok ty

[wintry steppe]

(Ux, Uy) = (U*Ux, y) = (x, y)

[latent ledge]

$$A=\begin{bmatrix}0&0&0&a_1\0&0&a_2&0\0&a_3&0&0\a_4&0&0&0\end{bmatrix}$$
for which complex number $a_1,a_2,a_3,a_4$ the matrix is diagonalizable.

[stoic python]

Konoha:

[latent ledge]

Currently, my idea is to compute $A-\lambda I=0$ to find all eigenvlues, and show them they are distinct

[stoic python]

Konoha:

[spice storm]

IS already diagonalizble becuase is upper and lower triganlar

@dusky epoch

[dusky epoch]

...

[spice storm]

@latent ledge

[wintry steppe]

why did you ping ann lol

that matrix is not upper and lower triangular. "upper and lower triangular" means "diagonal" and that's definitely not the case here

[dusky epoch]

hes probably gonna pull some bullshit like calling me "really smart" or something to that effect as if that was an excuse to ping me ootb

[wintry steppe]

has that happened to you before

nvm polynomial probably did it a few times

[dusky epoch]

KW likes to pester me in DM and then have me try to battle through his combination of dyslexia and general ineptitude and lack of reading comprehension

thank god poly is no longer here tho

[latent ledge]

@spice storm the problem asks to determine the value a1,a2,..

[wintry steppe]

they might be trolling, but just in case they weren't i explained why they are wrong

[spice storm]

Sorry I was confsued diagonalizable with diagonal..

its been a long weekend.

[wintry steppe]

anyways @latent ledge one big case you can get done quickly is if all of the entries are real (why?)

that matrix would be called antidiagonal @spice storm since the entires are on the off diagonal

diagonal always refers to top left to bottom right

[spice storm]

@wintry steppe Oh okay. Thanks I'll remamber that put it on my notebook

[latent ledge]

anyways @latent ledge one big case you can get done quickly is if all of the entries are real (why?)
@wintry steppe all eigenvalues are real

[wintry steppe]

not necessarily, take (0 1 \ -1 0)

do you know that real and symmetric matrices are diagonalizable?

[latent ledge]

not yet

[wintry steppe]

ah

i thought of it cause it's a cute way to get rid of a large number of cases quickly

anyways

computing the characteristic polynomial is probably your best bet

at least, that's the first step i'd take if faced with a problem like this

[latent ledge]

yea, that is way I am doing, it is just tedious to compute the eigenvalues

at least to me

[wintry steppe]

i haven't sat down and tried computing it yet

maybe you can try a smaller case to see if there's a trick?

[smoky lagoon]

Just use a calculator to get the eigenvalues honestly

That’s what I always did

[wintry steppe]

[smoky lagoon]

On homework at least just because it was tedious

I would only recommend doing that once you understand the process though

I just used it as a timesaver after I was able to do it consistently

[wintry steppe]

@latent ledge i sat down and computed the characteristic polynomial

the roots should have a somewhat nice form

somewhat

nvm they still kind of suck (i mightve made an error tho)

there has to be a smart way to do this

ok, there's definitely a nice form for the eigenvalues

so i'll say you can go the route of finding the roots (heh)

[dusky epoch]

shouldn't the charpoly be $(\lambda^2 - a_1a_4)(\lambda^2 - a_2a_3)$

[stoic python]

Ann:

[wintry steppe]

yes

i made a calculation error somewhere

[dusky epoch]

i have a hunch the eigens are just $\pm \sqrt{a_1a_4}$ and $\pm \sqrt{a_2a_3}$

[stoic python]

Ann:

[dusky epoch]

and i think the matrix will ALWAYS admit a diagonalization

[hollow finch]

$\vec{v}=\vec{p}+k\vec{n}$

[stoic python]

nix:

[hollow finch]

is there a way to minimize the magnitude of v without calculus?

[dusky epoch]

sure, |v|^2 is quadratic in k

[hollow finch]

i assume youre indicating to find the vertex?

$\vec{v}=\vec{p}-\frac{\vec{p}\cdot\vec{n}}{|\vec{n}|^2}\vec{n}$

[stoic python]

nix:

[hollow finch]

thats what i ended up getting

[dusky epoch]

yes

that sounds exactly right

[hollow finch]

$\vec{v}=\vec{p}-\operatorname{proj}_{\vec{n}}\vec{p}$

[stoic python]

nix:

[hollow finch]

is that right?

[dusky epoch]

yes

[hollow finch]

i suppose i could go even further with

$\vec{v}=\operatorname{proj}_{\vec{n}^{\perp}}\vec{p}$

[stoic python]

nix:

[hollow finch]

pretty cool. now i just have to think about why that is.
thanks for the help, Ann.

[split heart]

Hi, it's i^715 = -i?

[gray dust]

yes

[split heart]

Thank you.

[wintry steppe]

How did they know to only row-reduce to that, instead of putting it in rref ?

[marble lance]

How is that not in rref?

[wintry steppe]

Ah my bad

I put P in rref instead of P - I

So it didn't match up

All good

[sullen pollen]

I try to find the eigenvectors of this problem but it doesnt work. I found the first 2 (for 2 and -1) but im stuck at when the eigenvalue is 1

[wintry steppe]

If it's 1 you subtract by the 3x3 identity matrix

You didn't say what your problem is

[sullen pollen]

Its solving a diffeq by using the method of eigenvalues and eigenvectors but im stuck at the eigenvectors part

[wintry steppe]

Subtract A by the 3x3 identity matrix and then put that into rref and then augment it with a 0 column

then solve for the general solution, then write the general solution in parametric form and you will have your eigenvector

That's for the eigenvalue 1

[half storm]

Are you solving a system of differential equations? I'm trying to remember why you would find the eigenvalues of a matrix for anything related to DE.

It's been a while.

[sullen pollen]

Yes John

and I will try septim

[half storm]

I got you I can see it in my head now.

Does this method only work if the differential equations have constant coefficients?

[gray dust]

yes & if coeff matrix is diagonalizable

[latent ledge]

shouldn't the charpoly be $(\lambda^2 - a_1a_4)(\lambda^2 - a_2a_3)$
@dusky epoch I have lambda^4-a1a2a3a4, what are the reasons that you split char-ploy into (\lambda^2 - a_1a_4)(\lambda^2 - a_2a_3), is it we are seeking real symmetry matrix?

[dusky epoch]

??

[wintry steppe]

ann did that to find the roots, i.e. eigenvalues

if you havent learned "real symmetric implies diagonalizable" then forget about making things real and symmetric

We consider the origin O (0,0,0). Let the vector OP where P(2,1,3). Find the image of this vector if we apply a rotation of 45 ° around the y axis followed by a reflection with respect to the x axis. Give the matrix of this transformation.

Can someone check my work please? And the range of the vector is just the OP vector ?

Image = range I think ? I translated from french.

Also here u is the original vector (2,1,3) and w is the vector with just a 45 degree rotation around the y-axis

Why does it look like it's been rotated clockwise ?

I thought rotations were counter clockwise ?

lol

had not considered that

thanks

Do you know if the range is just the new vector itself in this case ?

It says "Trouvez l'image de la matrice"

Find the image of the matrix

I gotcha

So if the image is the set of vectors than can be expressed as a linear combination of it's columns

and it only has 1 column with no variable, then the image must be that column ?

Well if I had let's say [r - 2w, r, w] - then the image would have 2 vectors right ?

Yeah

Like a solution of a linear system ?

Okay, I will try to say it more clearly

No

If I solve a linear system for it's general solution, I can get the vector form of the general solution: [r - 2w, r, w]

Do we agree ?

And if we say that the image is defined as "Given some matrix A, which vectors can be expressed as a linear combination of its columns?"

Then the image of the general solution would be easily found if we rewrite it in parametric form:
r * [1, 1, 0] + w * [-2, 0, 1]

The image is {[1, 1, 0], [-2, 0, 1]}

How would you have written the image then ?

Ok, I gotcha

So now in the question, they ask me for the image of the new vector

Unlike [r - 2w, r, w] - the new vector has no variables

Therefore the image is simply span{[new vector]} ?

I see, and A is the transformation matrix right ?

Therefore, the image of v under A, is Av because A * v only has one result ?

Oh I think I get what you mean

The meaning of image in this case, is simply the vector resulting from the transformation ?

Ah I see 🙂

Right, that was ambiguous

But now I know

Thank you

[half storm]

lol

[limber sierra]

a basis of a vector space is a linearly independent, spanning subset of that vector space

do you know what all those words mean

"spanning" means "every element of the vector space can be written as a linear combination of the subset's members"

[half storm]

A basis for a vector space $V$ over a field $F$ is a linearly independent subset $\beta = {x_1, x_2, \dots x_n } \subseteq V$ such that $\forall v \in V,$ $\exists a_1 ,a_2, \dots , a_n \in F$ s.t. $v = \sum_{i = 1}^{n} a_ix_i $

[limber sierra]

theres a lot of jargon involved here admittedly

;

\exists

[half storm]

sorry namington I already know you answered but I was still already in the middle of TeXing and wanted to practice.

[limber sierra]

but uh theres a couple issues

your sum should be over a_i x_i

not a_i x_1

[half storm]

yup

[limber sierra]

and its not clear where the scope of the \exists ends

[sick tartan]

welp, gotta start at vectors again i suppose

[half storm]

I think that's a neat TeX

How do you guys format multiple existence quantifiers like that

[subtle walrus]

use words instead

[stoic python]

JohntheDon:

[subtle walrus]

yes, same

if i have to use them i add spacing

[half storm]

Oh yea I use words on exams and stuff

[subtle walrus]

and maybe colons

[half storm]

I don'treally use logic symbols

But sometimes I like for them to look pretty with my symbols.

I don't really. One of my professors didn't like them.

[subtle walrus]

quantifiers look really ugly for some reason

a mathematical text is an english text primarily

[half storm]

Yea I got you. I use them in my notes because sometimes I do get tired of using for all and stuff.

Yea yea.

[sick tartan]

i use snipping tool to do my bidding

[half storm]

Symbols and notation are a god send honestly.

Thank both India and the Middle East for our numerical system.

and thank whatever mathematicians and logicians came up with symbols. Sometimes I get little overwhelmed with so many words.

lol you ever get points off for handwriting

like they can't read your work?

If you guys have ever tried reading an old math paper without any sort of the conventional notation we have now it is literal hell to read.

I remember in school they made us read some stuff on the study of algebra. An arabic mathematician was doing some work on algebra ; his name escapes me. But basically he was doing something with the quadratic formula I think. And it was like "the roots of a polynomial are given by the negative of it's second coefficient added to square root of the second coefficient squared ... " so on and so forth.

[subtle walrus]

i mean that is really old then

math papers from the past century are understandable

[half storm]

Yea, lol. It was small excerpt but it was super old.

Might have been during Euclid's time or something.

Well maybe not that old.

I don't think they new of polynomials or roots of them.

That wasn't the exact example though, I can't remember exactly what it was but it was definitely and like elemntary algebra at that.

[north sierra]

I remember in school they made us read some stuff on the study of algebra. An arabic mathematician was doing some work on algebra ; his name escapes me. But basically he was doing something with the quadratic formula I think. And it was like "the roots of a polynomial are given by the negative of it's second coefficient added to square root of the second coefficient squared ... " so on and so forth.
@half storm https://en.m.wikipedia.org/wiki/Muhammad_ibn_Musa_al-Khwarizmi this guy?

[dreamy iron]

My level is "I'm self-studying Axler". I'm just worried I'm over reaching trying to prove that "the set of all univariate polynomials forms a valid vector space."

I know what the zero polynomial is.
If you give me two specific examples of polynomials, I can take their linear combination with two arbitrary scalars. (I don't exactly know how to do this in a general case.)

So what Axler is having me do is to show that the set of all polynomials of at most degree m is a valid subspace of the vector space of all polynomials.

1. Am I not supposed to show that all univarite polynomials forms a vector space?
2. How do I add two polynomials together, as a general case?

okay, some definitions:

$p \equiv \sum^{m}_{i=0} \alpha_i x^i \equiv \alpha_0 x^0 + \alpha_1 x^1 + \cdots + \alpha_m x^m$

$q \equiv \sum^{m}_{i=0} \gamma_i x^i \equiv \gamma_0 x^0 + \gamma_1 x^1 + \cdots + \gamma_m x^m$

[stoic python]

ninnymonger:

[dreamy iron]

what does it mean to add two polynomials p+q?

like what if deg(q) < deg(p)?

and that's like just okay?

some of those \gamma terms could be zero, (or maybe they're all zero) and it's still valid to just smash the two functions together?

What is 5x plus 1-x^2
@wintry steppe

-x^2 + 5x + 1

[half storm]

@north sierra Yup that's the dude!

[north sierra]

cool cool

he's a legend

[dreamy iron]

$\mathcal{P}(\mathbb{F}) \equiv {\text{set of all polynomials over a field F}}$

[half storm]

@north sierra He was perisan not arabic though. Messed that up

[pallid rampart]

Specify the coefficients

[dreamy iron]

$\mathcal{P}_{m}(\mathbb{F}) \equiv {\text{set of all polynomials of with coeffs in F of degree at most m}}$

[stoic python]

ninnymonger:

ninnymonger:

[dreamy iron]

okay, so if i have polynoms p and q

and i add these to together

p+q \in P(F)

but how to show this, just smash the summands together?

just to satiate my curiosity, can you walk me through the closure axiom for polynomial addition?

$(\alpha_0 + \gamma_0)x^0 + (\alpha_1 + \gamma_1)x^1 + \cdots +(\alpha_m + \gamma_m)x^m$

[stoic python]

ninnymonger:

[dreamy iron]

okay, and does that prove:

1. closure under the subspace polynomials of at most degree m?
   OR
2. closure under the vector space of all univariate polynomials?

(i can't tell if i'm over complicating this or if i'm actually confused about something)

okay, yeah, how do i do that? how do i account for different degree sizes?

intuitively i know that adding two polynoms of at most degree m will not create a new polynom of degree exceeding m, but how to state that explicitly?

i feel like there's a inequality here i need to prove:

$deg(p) + deg(q) \geq \deg(p+q) ??$

[stoic python]

ninnymonger:

[quartz compass]

I prefer to write it more strongly as deg(p+q) <= max( deg(p), deg(q) )

with equality when deg(p) != deg(q)

this is called the strong triangle inequality

[dreamy iron]

https://math.stackexchange.com/questions/2027287/the-degree-of-a-sum-of-two-polynomials-proof-question

[quartz compass]

or also the ultrametric inequality, although that's not relaly important right now

[dreamy iron]

1. is this outside the scope of axler?
   or
2. is this something i should be able to do?

[quartz compass]

idk, how obvious does it seem to you

if I add x^3 and x^2 I get x^2+x^3 so the max degree wins out

if they're the same degree they have a chance of competing and cancelling out to smaller degree

like x^3 and x^2-x^3 both degree 3

but added they're degree 2

[dreamy iron]

i mean, intuitively it's very obvious, having worked with a lot of polynomials.

but the intuition of the argument is not a 1 to 1 translation into the intuition around the proof

[quartz compass]

I can't answer this question for you, you have to decide how deep you want to go in anything you learn

it's ultimately a bottomless pit and you must draw a line somewhere

if it interests you, learn it

if you think it's obvious enough and not interesting, don't

[dreamy iron]

okay, i guess what i want to know here is "in what class is this proof regarding degrees" usually encountered?

the MSE article says "rings" which is an algebra topic

(my skill points in abs. algebra are minimal)

the lecture series i'm following that uses Axler as the main text for the course just says "it should be obvious that you can't get a larger degree, just try a few examples"

[half storm]

lol I used to have this problem and still do sometimes. Sometiems I'm like "this is intuitively obvious" but there really is no way to be super formal about it.

[cobalt tartan]

How do I prove that if $Ax = b, Gx \leq f$ has no solution, then there exist y, z such that $z \geq 0, A^Ty + G^Tz = 0, b^Ty + f^T < 0$?

[stoic python]

Liria ^(;,;)^:

[cobalt tartan]

So far I have'nt been able tod o anything with it

I tried wwriting it as a single matrix but that hasn't beenn going anywhere

[median forum]

@cobalt tartan you sure y and z multiply from the right?

[cobalt tartan]

Yea

[median forum]

ah

[cobalt tartan]

Wait I forgot a bit

It's supposed ot be $b^Ty + f^Tz < 0$

[stoic python]

Liria ^(;,;)^:

[median forum]

is that the full question?

I can see that being the case if y and z multiply A and G from the left

[half storm]

So that's a dot product.

[median forum]

ye

[half storm]

I'm just curious about the quesiton.

I was looking at it like wtf

[median forum]

dot doesnt matter the order anyway

[half storm]

Yea

Yea if it's left multiplication then I might be seeing where you're going with it.

I took a look and immediately my first thought was to take the transpose of matrix equation so you end up $x^TA^T = b^T$

[stoic python]

JohntheDon:

[half storm]

But then I got back to looking at what I was trying to prove and I was like "this might be out of my depth"

[median forum]

ye

also if $xA^T \neq b^T$, you can always choose x st. $xA^T > b^T$

[stoic python]

Fractal:

[half storm]

Like if $\forall x (xA^T \neq b^T)\implies \exists x ( xA^T > b^T)$

[median forum]

ye

because R is a group

[stoic python]

JohntheDon:

[cobalt tartan]

Pardon?

Yea it should be ad ot product, b^Ty + f^Tz < 0

but iunno what to do with A^Ty + G^Tz = 0

[dusky epoch]

My course's lecture notes allow for the basis of {0} to be {0} itself
that's wrong because {0} is a linearly dependent set and hence cannot be a basis

1*0 = 0

to be clear, the zeroes here denote the zero vector while 1 denotes 1 the scalar

i.e. the zero vector can be written as a nontrivial linear combination of {0}

[viscid kernel]

Does every matrix that has a null space also have a left nullspace ?

[half storm]

I think you mean a right nullspace. The usual thing that we refer to when we say the "nullspace of a matrix A" is the same thing as the "nullspace of the left-multiplication of matrix A" i.e. the set of vectors such that $L_A(x) := Ax = 0$

[stoic python]

JohntheDon:

[half storm]

The fact that A has a left nullspace wouldn't affect whether it not it would have a right nullspace though. You would just define the right nullspace to be the set of vectors $x$ such that $xA = 0$.

[stoic python]

JohntheDon:

[half storm]

@viscid kernel

[viscid kernel]

Aight I got my answer thank you

Sorry for the late reply btw @half storm

[half storm]

np

[stoic python]

slimvesus:

[half storm]

At first i was confused but your basically considering $\mathbb{R}$ as a vector space over itself.

[stoic python]

JohntheDon:

[half storm]

@wheat shoal most people would say that the empty set generates the null set. Though in another channel, I was speaking with someone who was a TA and a certain professor - who specialized in LA - was teaching his graduate students that the zero subspace is not generated by any set.

pretty nonstandard.

I feel like you can't do that though because of Hausdorff's maximal principle that says every vector space has a basis.

but they know more than me and there is probably some cavaet there.

Yea, it makes sense to me.

but he was doing something weird.

I don't know. What you just said is the way it was taught to me.

I tried to get more information out from the guy because I was geninuely curious but he said if he revealed anymore he would doxx himself lol.

Must be this one dude that does it and is known for doing it in the community; no clue though.

[split heart]

Hi, what complex is defined by z=e?

[tribal nebula]

the number e?

[split heart]

Yes

P=1 as I see

[tribal nebula]

if z = e, then z = e. That's the complex number

e + 0i

[split heart]

Alright so if there is no exponent the angle is 0

Thank you

[half storm]

Yea

[tribal nebula]

no problem

[half storm]

That's it?

I mean, I'm not sure if you can really deduce anything from just that alone.

[dusky epoch]

I mean, I'm not sure if you can really deduce anything from just that alone.
yes you can

are you allowed to assume V is findim?

[half storm]

From that yea.

You can use rank-nullity theorem?

[dusky epoch]

then the dimension of the kernel will be dim(V) - 1, simple as that

i invite you to prove that a nonzero linear functional has as its image the entire number line

[half storm]

I see.

But i was just like if there are no other assumptions, then you can't really say anything can you?

Probably could have asked for the context though.

[wintry steppe]

How do I write [2r+2] in parametric form ?

r * [2] + [2] ?

[half storm]

Yea

I was a bit confused because i usually seen anyone write one-dimensional matrices i.e. numbers like really

[modern palm]

hello, can anyone help me with the following T/F question?

If X is a spanning set for a subspace U, and V is any set that contains X, then V is a spanning set for U.

I think it should be True, would that be correct?

[half storm]

Yea it's true.

[modern palm]

Ok, my logic is

Since X = {x1, x2,..., xk} and U = span(X) = span{x1, x2,..., xk}
Suppose V = {v1, ..., vk, x1, x2,... xk}, which contains X.
Which means U = span(V) = span{v1, ..., vk, x1, x2,... xk}

[half storm]

Yup that's pretty much it.

[modern palm]

Alright. thanks

again

[wintry steppe]

1. unless you're guaranteed that the spaces are finite dimensional you should be more careful with the notation
2. if V = X union {vector that isn't in span(X) = U} then span(V) will be larger than span(X) = U, so unless "spanning set" means "span contains the space" and not "span equals the space" then the answer should be false
   @modern palm

Find the basis of the subspaces associated to the eigenvalues

That just means find the eigenvectors of an eigenvalue?

yes

more accurately, find a basis, and find linearly independent eigenvectors forming a basis of the eigenspace

[modern palm]

@wintry steppe I don't think we are at the level to take that into consideration yet, at least I'm not sure what you mean by be careful with the notation because it doesnt say that the spaces are finite dimensional.

and yep I think "spanning set" means "spans contains" not "spans equal" to the space for us.

[wintry steppe]

i can't respond to this right now, in a lecture

[half storm]

Yea that's more nuanced. I guess i was just thinking about the fact that span(X) is a subset of the span(V) so U is contained in span of V but yea if you're taking a "spanning set" to mean the spans are equal then you have no real idea whether U = span(V)

[wintry steppe]

ill let someone else answer this, unless you want to ask again in 30 minutes

[half storm]

I was thinking of span as generates the set; and I think that's how I've always learned it.

[modern palm]

sure

@half storm yep, that was how its stated in our notes as well.

[wintry steppe]

does "generates" mean span equals or span contains

[half storm]

I think I've learned it as the latter.

I'm looking up what's standard right now.

[modern palm]

we defined it as span(x1, x2, ..., xk) = {c1x1 + c2x2 + ... + ckxk} where c1,c2,...,ck are constants. So I was thinking the v1, v2 v3 vectors can be multiplied by 0.

[half storm]

That's the definition of span, but TTera is asking a bit of a question regarding the terminology that you've been taught. Like if you have a vector space V, if I say that "S spans V" does that mean that the span(S) = V or that V is a subset of the span of S.

[modern palm]

the latter

[half storm]

because the terminology is important to determine whether the question is correct or not actually.

Then it's correct.

or true.

[wintry steppe]

so you say that {x_1, ..., x_k} generates V if V \subseteq span({x_1, ..., x_k})

i don't like it, but if it's the convention you're following then the T/F problem is indeed true

the LA book i used took "X generates V" to mean span(X) = V

i'm sure there's some standard meaning of generates, but it feels very wrong to me to say that X generates V if span X contains (but is not necessarily equal to) V

[half storm]

Yea mine does too.

[wintry steppe]

friedberg :)

[half storm]

Yup

lol

Friedberg has some weird stuff though. Like it calls the rank-nullity theorem "The Dimension theorem" and the exchange theorem the "replacement theorem"

[wintry steppe]

also uses N(T) for null spaces / kernels lol

[half storm]

Yea

I only knew that the dimension theorem was known as rank-nullity from my first class in LA where we used a different book and I had to look up exchange theorem.

I saw someone in the channel use it in here and I was like "is that the replacement theorem?".

[wintry steppe]

yeah, friedberg has some strange conventions and notations

otherwise it is a solid LA book

has a shit ton of examples and actually covers determinants unlike another popular choice

[half storm]

lol.

Yea I like it. It's definitely a textbook geared towards people who really want to do math for the sake of math.

Not like a super applied approach like some other ones

It's proofs-based.

[wintry steppe]

Give the relation between a, b, and c so that [a, b, c] is a linear combination of v1, v2, v3

Wut ?

[half ice]

There's gotta be more info than that

[wintry steppe]

I just have the values for vector 1, 2, and 3

But it says a, b, c out of nowhere

[half ice]

Basically:
a = mv1 + nv2 + ov3
Find m,n,o

But if you don't know anything about a then that's not reasonable haha

[wintry steppe]

hmm

[half ice]

Or actually it doesn't look like a is a vector

So I dunno. You'll need more info

[wintry steppe]

a, b, c are variables

the vector is [a, b, c]

and we want that as a linear combination of v1, v2, v3

weirdness

we want the relation between a, b, and c

so that [a, b, c] is a linear combination of other vectors

ugh lol

[half ice]

What relation? [a,b,c] is just [a,b,c] there's no room for a relation there haha

If v1, v2, v3 span R³ then [a,b,c] is definitely a linear combination of them

Given that a,b,c are real I suppose

[wintry steppe]

Yeah, that makes the most sense

[cursive crest]

Linear algebra is easy

Y=mx+b

[hollow finch]

So say I have a 2x2 matrix A with a repeated eigenvalue. That means I can decompose it into the form PJP^-1

$P\begin{bmatrix}\lambda&1\0&\lambda\end{bmatrix}P^{-1}$

[stoic python]

nix:

[hollow finch]

Does the second column of P even matter as long as P is invertible?

[tribal nebula]

Linear algebra is easy
@cursive crest lmao that's not linear algebra

apologies, I don't mean to interrupt

[cursive crest]

Yeah I’m joking @tribal nebula

😉

[tribal nebula]

Oh ok

[pallid rampart]

@tribal nebula hey but that’s algebra and it’s a linear equation, so it must be linear algebra

I don’t see anything wrong

[tribal nebula]

🤔 if you think about it, x is just a 1x1 matrix...

[cursive crest]

Math is cool

[tribal nebula]

Sure is

does that mean you can't square a matrix??? Cause then it's not linear... 😢

[quartz compass]

technically squaring a matrix should be defined as A^T A

since it makes it square and is well defined while A^2 isn't

[tribal nebula]

Well A² works for square matrices

[quartz compass]

so what

I addressed what you said before you said it

[dusky epoch]

mero

[wintry steppe]

[tribal nebula]

ngl mero, A^T A ≠ A² If A is a square matrix

[quartz compass]

so what

A^T A literally makes a square

what's the squared length of a vector?

v^T v

seems like the right definition to me