linear-algebra

[wintry steppe]

But don't I need to let things be elements of the bigger subset

And show that every element in the smaller subset can be written in that form?

[dusky epoch]

im referring to this

[wintry steppe]

Yes me too

[dusky epoch]

write U := {a1(1,0,1,0) + a2(0,1,0,1) | a1, a2 in R} for brevity

youre showing U ⊆ range(T)

[wintry steppe]

Yes

So I need to let x in range T

[dusky epoch]

no you have it backwards

[wintry steppe]

Which is (a+c,b+c,a+c,b+c)

Why?

[dusky epoch]

if you let x in range(T) and show that x is in U

this proves range(T) ⊆ U

which is the other direction

[wintry steppe]

Oh, never mind you are right.

So I have a vector of the form (a1,a2,a1,a2)

I need to show that this can be written as (a+c,b+c,a+c,b+c)

For real a,b,c,

[dusky epoch]

if you wish

(a1, a2, a1, a2) = T(a1, a2, 0)

thats all you need, and no more

[wintry steppe]

Hmm, okay. What if I wanted to do what I wish?

[dusky epoch]

the other direction?

[wintry steppe]

Showing that it can be written in the form above

[dusky epoch]

i mean

if you really wanted to

you could set up a system of equations in a, b and c

and find a particular solution to it

which i did for you by inspection

[wintry steppe]

How? We have too many variables

[dusky epoch]

no we don't?

a1 and a2 are treated as known

arbitrary but known

with a, b and c expressed in terms of them

[wintry steppe]

Oh okay

[dusky epoch]

i mean like

youre overthinking it rly

[wintry steppe]

Wait, no we don't

For example, c = a1 - a = a2 - b

We can't solve it

?

[dusky epoch]

you're overthinking things

[wintry steppe]

How do we solve it?

[dusky epoch]

solve what

[wintry steppe]

The system of equations..

[dusky epoch]

$\begin{cases} a+c = a_1 \ b+c = a_2 \end{cases}$

this?

[stoic python]

Ann

[wintry steppe]

Yes lol

[dusky epoch]

treat c as a free variable

you get a = a1 - c and b = a2 - c

so the general solution will be $(a_1 - c, a_2 - c, c)$

[stoic python]

Ann

[wintry steppe]

Okay

Can we do that? The free variable part?

[dusky epoch]

why couldn't we

are you following some kind of conduct which forbids you from using the techniques and terminology associated with linear systems to solve a linear system?

[wintry steppe]

Let S and T be linear maps from U to V, both of which are vector spaces over C. Prove that null S cap null T = null S + T cap null S - T

We have to show that null S cap null T subset null S + T null S - T

So let x in null S cap null T. Then x has to be in both null S and null T, which implies that Tx = 0 and Sx = 0

Therefore Sx + Tx = (S + T)(x) = 0

Which means that x is in null (S + T)

[lavish jewel]

could you use latex?

[wintry steppe]

Similarly, Sx - Tx = (S - T)(x) = 0, so x is in null (S - T)(x)

I will after this next line. But latex doesn't have a command like \null which makes it annoying.

[gray dust]

we do have \ker

[wintry steppe]

Since we've shown that $x \in \mathrm{null~}(S-T)$ and $x \in \mathrm{null~}(S + T)$ it follows that $x$ is in their intersection, so we have shown that $\mathrm{null~}S \cap \mathrm{null~}T \subseteq \mathrm{null~}S + T \cap \mathrm{null~} S - T$

[stoic python]

n/c

[wintry steppe]

Now we will show that $\ker S +T \cap \ker S - T \subseteq \ker S \cap \ker T$

[stoic python]

n/c

[wintry steppe]

Let $x \in \ker S +T \cap \ker S - T$. Then $x \in \ker S +T$ and $x \in \ker S - T$

[stoic python]

n/c

[wintry steppe]

This implies that $(S + T)(x) = 0$ and $(S - T)(x) = 0$

[stoic python]

n/c

[wintry steppe]

But by definition, $(S+T)(x) = Sx + Tx$ and $(S-T)(x) = Sx - Tx$ and so $Sx + Tx = 0$ and $Sx - Tx = 0$

[stoic python]

n/c

[wintry steppe]

Hence $Sx + Tx = Sx - Tx$. Adding $- Sx$ to both sides, we get $Tx = -Tx$ or $2Tx = 0$ so $Tx = 0$

[stoic python]

n/c

[wintry steppe]

This also tells us that $Sx = 0$ which means $x \in \ker S$ and $x \in \ker T$

[stoic python]

n/c

[wintry steppe]

Does this proof work?

I'm not sure why the field C was mentioned..

[gray dust]

@wintry steppe it works

[wintry steppe]

Thanks

Why was the field C mentioned?

Just as a distraction?

[dusky epoch]

this works over any field of characteristic not 2, i think...

the only place where it matters that our base field is $\bC$ is when you divide by 2, because you don't want $1+1=0$ to be true in your field

[stoic python]

Ann

[gray dust]

@wintry steppe late reply but it affects the logic in the 2nd to last tex. as said above, a field having characteristic 2 means 2=1+1=0, so if x is a vector then 2x=0 doesn't necessarily mean x=0

[wintry steppe]

Okay

So basically if Sx = 0, but the field is Z/2Z

Then it could just be the map x^2 + x

But x = 1 in this field

When calculating the best approximation to vector b, in a subspace V. Do you have to use the standard basis or the orthogonal basis of V, in your proj_v(b) formula?

[native rampart]

Should be orthogonal

[wintry steppe]

ok, thanks

[versed hearth]

for a matrix $X$ and a vector $v$, how can $\norm{X^Tv}^2 = Tr(XX^Tvv^T)$ ? there Tr is the trace

[stoic python]

lyinch

[native rampart]

What is the inner product?

I think you use tr(AB)=tr(BA) smh

and tr(ABC)=tr(BCA)=tr(CAB)

[carmine dune]

Hey what can we say about the ratios of a triangle's altitudes?

We know that the altitudes would be concurrent

but why is it that the ratios of lengths FA/CE = BC/BA holds?

[versed hearth]

for a vector $v$ it makes sense that $\norm{v}^2 = Tr (v^Tv)$ but when I plug in $v = X^Tw$ then I get that $\norm{X^Tw}^2 = Tr(w^TXX^Tw)$ which is close, but still not what I need

[stoic python]

lyinch

[carmine dune]

oops sorry

[native rampart]

You use Tr(ABCD)=Tr(BCDA)

[carmine dune]

didn't knwo there was a discussion going on

[versed hearth]

but $w$ is a vector, doesn't this identity require that ABCD are square? Or is it only the requirement that their product is square

[stoic python]

lyinch

[native rampart]

Ok, doesn't work

Is X square?

[versed hearth]

no

[native rampart]

It requires that both products are well defined

A(BCD) and (BCD)(A) should both be well defined

i.e,should have the correct number of rows and columns

[versed hearth]

X is nxd and w is nx1

okay I'll check if it works out with the dimensions

seems to work work with Tr(ABCD)=Tr(BCDA), thanks

but I've never seen that rule 🙂

[native rampart]

https://math.stackexchange.com/questions/252272/is-trace-invariant-under-cyclic-permutation-with-rectangular-matrices

That's like the defining property of trace

trace is the only linear functional on matrices that does that

[versed hearth]

oh very nice

[hoary venture]

math pros

[carmine dune]

hi

so in this type of triangle we know that the altitudes are concurrent, but what can we say about the ratios?

why is it that the ratios of lengths FA/CE = BC/BA holds?

where FA & CE are altitudes, and BC is a side of a parallelogram and BA is a diagonal of a parallelogram

[nocturne jewel]

#geometry-and-trigonometry

[coral skiff]

hey this question stumped me and @wintry steppe

nobody can figure it out so far

I have two points, p1 and p2 at (x1, y1, z1) and (x2, y2, z2)
I have rotation matrices R1 and R2 describing the orientation at each of those points
i extend a line of length l1 from p1 along the x axis at p1
and a line of length l2 from p2 along the x axis at p2
i then join p1 to p2 with a straight line, and the ends of the two lines with another straight line, forming a rectilinear panel
i would like to obtain the rotation matrix at a point p3 on the surface of this panel

[wintry steppe]

Let $P: V \to V$ be a linear map, and suppose that $P^2 = P$. Let $N = \ker P$ and $R = \text{range }P$. Show that $N \cap R = { 0 }$.

[stoic python]

n/c

[wintry steppe]

I have been trying to do this for an hour or more, and I don't know how

[native rampart]

Let x be in N \inter R then Px=0 and there is a u such that Pu=x

That implies P^2 u=0

But Pu=P^2 u

Which implies x=0

[wintry steppe]

How does it imply x = 0?

[native rampart]

Pu and P^2 u are same

Pu=x

[wintry steppe]

Yeah, so?

[native rampart]

P^2u=0

So x and 0 have to be same

[wintry steppe]

Hmm

Oh I see

You're applying P to both sides of Pu = x

Very clever

[hexed spade]

This is better suited to #help-0

[wintry steppe]

lmfao

@hexed spade mine?

I am now trying to show that $N\oplus R = V$

[stoic python]

n/c

[wintry steppe]

I always struggle with showing that V is a subset of the otherset

In this case, $V \subseteq \text{span}(Pv_1,v_2)$

Where v_1 in N and v_2 in R

[stoic python]

n/c

[hexed spade]

oh no, not you

was talking to the other beef guy

[gray dust]

let x in V, write x as a sum of a vector in N & a vector in R

[bitter cape]

Could someone help me with a linear algebra question?

[lavish jewel]

perhaps. what's the question?

[bitter cape]

I'm learning this in a language other than English so lemme know if any term doesn't make sense.

I have that T: R³ -> R² (linear transformation). In which T goes like (x, y + z). [T(1, 1, 1)) = (1, 2). T(1, 1, 0) = (1, 1). Etc]

It's asking whether Nuc(T) [kernel] is a line of R², R³ or R³ in itself, and I can't figure out the reasoning to get there

[wintry steppe]

What is Nuc?

[bitter cape]

Nucleus

[wintry steppe]

So kernel or null space?

[lavish jewel]

hmm could you provide a definition for this? i think it's kernel in english but i wanna make sure

[bitter cape]

I believe it's kernel, yes

[lavish jewel]

is it all the vectors v for which T(v) = 0?

[bitter cape]

Yes

[lavish jewel]

yep, kernel, then

[bitter cape]

Sorry I didn't know the name 😅 it is kernel indeed

[lavish jewel]

no problem

[nocturne jewel]

$T[(x,y,z)]=(x,y+z)=(0,0)$

[stoic python]

Mosh

[lavish jewel]

i would say the null space is all vectors of the form t*(0,-1,1)

[dire thunder]

what

(0, 2,-2) is null space

but not of the form (0,-1,1)

[nocturne jewel]

Edd you mean span of

[lavish jewel]

sure

[dire thunder]

and this basically is line

[lavish jewel]

the span of (0,-1,1)

[dire thunder]

(0, x, y(x)), y(x)=-x

[nocturne jewel]

$Ker(T)=span([0,-1,1])$

[stoic python]

Mosh

[lavish jewel]

so t*(0,-1,1) is the null space

[nocturne jewel]

yes

[dire thunder]

@bitter cape

[lavish jewel]

if you recall the vector definition of a line, it is L = p_0 + t*v

for a scalar t, a vector v, and a point on the line p_0

[bitter cape]

So this is a line in R³, right?

[nocturne jewel]

yes

[dire thunder]

yes

[lavish jewel]

if you set p_0 = 0, you get L = t*v, v = (0,-1,1) which is a line in R^3

[nocturne jewel]

span of a single vector is a line which goes through the origin

[bitter cape]

Yes! That makes sense

Alright, thank you much

[lavish jewel]

no probl

[nocturne jewel]

similarly if you had span of 2 vectors it'd be a plane

[dire thunder]

only if they are independent

[wintry steppe]

Yeah I don't know how.

c1v1' + c2v2' = a1Pv1 + a2v2

I am just so confused..

[gray dust]

this is a load of meaningless letters

we must show V=N+R, ie every vector in V is the sum of a vector in N & a vector in R

use the fact that P^2=P

[bitter cape]

Is there any chance the answer to the question I asked could be a plan in r³? (plane/surface?)

[dire thunder]

what

[bitter cape]

This one.

[dire thunder]

well

wym nullity as surface

and it could be plane yes, if you have nullity of two vectors (spanned by two independent vectors)

[bitter cape]

Got it. Alright, thank you

[dire thunder]

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

then nullity is given by span {0,1,0}, {1,0,0} which is plane

[tame mural]

Is the relationship between cyclic group and cyclic graphs just a coincidence in naming?

Even though finite cyclic groups form a cycle

I ask because the infinite cyclic group doesn't look the same

[dire thunder]

you can define cyclic graph as graph which is a cycle essentially

or a graph that contains cycle

which definition do you use for cyclic graph?

[tame mural]

A 1-generated group

[dire thunder]

[tame mural]

Is that wrong?

[dire thunder]

i am asking abt graph

[tame mural]

oh oops

[dire thunder]

anyway for finite cyclic group of order n we have all elements in group = x^m for 0 <= m <= n

[tame mural]

A cycle is a non-empty trail where only the "first" and the "last" vertices are connected

But I'm wondering about the infinite one

C_∞

[dire thunder]

define G=(V,E) to be graph with V = {1, x, ..., x^n} and (x^m, x^n) being an edge iff x^n = x^{m+1} then it is cyclic graph

i doubt that you can generalize it to infinite case

like consider group generated by x with infinite order

for drawing cyclic graph for it you should have 1 = x^m

but this implies finite order

[tame mural]

I see

mmm

Hmm, this definition of cyclic graph seems slightly different from the one I know

but it does make everything fit again

[rose umbra]

Be invertible matrix A that has eigenvalue of λ , how to proof that 1/λ is the eigenvalue of A^-1 ?

[dusky epoch]

take an eigenvector of A and call it v

it happens that v will also be an eigenvector for A^-1

[rose umbra]

how do you know that?

[dusky epoch]

well take Av = λv and left-multiply both sides by A^-1

[rose umbra]

I cant see that, what next?

[dusky epoch]

what do you get?

[rose umbra]

V= A^-1 * λ * V

[dusky epoch]

wow so many spaces

anyway now divide both sides by lambda

[rose umbra]

ok got it i see now my proof, but I cant see that

wait

[dusky epoch]

A^-1 v = (1/λ)v...

[rose umbra]

yea lmao nvm i was confused for a second

thanks

[wintry steppe]

but what if lambda = 0

[rose umbra]

booom

[dusky epoch]

if 0 is an eigenvalue of A then A^-1 does not exist lol

[gray dust]

@hybrid charm knock it off

feel free to ping helpers

last chance. knock it off

to toge not you

and curb your degeneracy

i told toge to ping helpers, not you

[wintry steppe]

[gray dust]

go ahead

[faint dune]

How do you guys call the lectures where you deal with polynom interpolation and approximation?

We call it numerics, couldnt find the english word for that. It combines linear algebra and calculus, but whenever its LA heavy I ask the question here.

[lean swift]

numerical analysis?

[celest slate]

http://users.math.uoc.gr/~pamfilos/eGallery/problems/HyperbolaGeneration.html

I’m trying to understand the matrix part of this link and I just don’t get it

[lavish jewel]

these are called homogeneous coordinates

[celest slate]

I don’t get why [1 0 0|0 0 r|0 1/r 0] times [x|y|1] gives [x| 1/r | y/r] instead of [x| r | y/r]

I don’t get why he multiplies the vector by r/y and says it’s equal

[lavish jewel]

by convetion in homogeneous coords, you want the last dimension to be 1

[celest slate]

I’m fine with the last dimension being 1

[lavish jewel]

you only care about projections on a 2D plane at z=1

[celest slate]

I just don’t get the actual transformation, not to mention how it equates to a transformation from a circle to a hyperbola

[lavish jewel]

i do agree with this

[celest slate]

wdym you agree with it

like you agree with him that that’s what it gives or you agree with me that it’s confusing

[lavish jewel]

that it should be r instead of 1/r in the result

as you said

[celest slate]

what I want to be able to do

I want to be able to take a hyperbola (ideally of the form xy=c, but if it has to be x^2 - y^2 that’s probably okay too) and transform it into a circle

in such a way that I can choose a point on the hyperbola and see where it lands on the circle

I thought maybe I could use the inverse of this matrix if I could figure it out, and I also tried taking the hyperbola xy=1, doubling it, and subtracting it from the line (x+y)^2 = 3, to get x^2 + y^2 = 1, but I can’t figure out what that looks like geometrically or how to track individual points

I’m not even sure if subtracting equations like that has a meaningful interpretation geometrically

[lavish jewel]

inverting the transformation should work

but you'll have to double-check whether the third column is correct

[celest slate]

I also feel like I need an understanding of why it works

Since I’m going to need to “rotate” it 45 degrees, I’ll need to figure out where I want to do that

and how

but I don’t understand that third column, or why it gets multiplied by r/y, or how it corresponds to a circle-hyperbola transformation

[lavish jewel]

the rotation should be easier, since you could apply a rotation matrix to both sides of the equation

[celest slate]

yeah but

[lavish jewel]

for this, you'll have to go study projective geometries/homogeneous coordinates

[celest slate]

my larger goal is

I take a hyperbola, I transform it into a circle. If I know that there are K integer points on the hyperbola, I want to have a circle with a number of integer points that is dependent on K

[lavish jewel]

by integer did you mean discrete? like discretize the hyperbola and find the corresponding points on the circle?

[celest slate]

yeah

and if I rotate the hyperbola, I think that messes with the discrete points on it, so trying to go directly from a hyperbola with a diagonal transverse axis to a circle might be easier

[lavish jewel]

to be fair, aside from this possible mistake on the third column of the transformation, the website you shared already explains everything

[celest slate]

well it doesn’t because

at one point it goes from [x 1/r y/r] = [rx/y 1/y 1]

and those two things aren’t equal, he multiplied by r/y, which I understand is a valid thing to do but I’d think it would mess with the coordinates, and even if it doesn’t I don’t get why he says it’s equal

[lavish jewel]

you said this, but you didn't understand it, then

you will first have to read what homogeneous coordinates are

and then you can walk through the geometric explanation given on the website from the beginning

all the steps are there

[celest slate]

so

is it supposed to be r or 1/r?

[lavish jewel]

you can work it out yourself following the steps given at the top...

[celest slate]

even if I don’t understand it, I can probably just multiply it by a rotation matrix & then use it

I think it’s supposed to be 1/r

right?

if the goal is to get [rx/y | 1/y] then it has to be 1/r, I think, which means it’s supposed to be 1/r? But why did they put r, was it just a typo they didn’t fix?

[native rampart]

What did you get the basis for Ker A-2I as?

Consider pick a vector v in V/Ker(A-2I)

Try looking at span {v,Tv}

Ok, That's weird

You sure that's correct?

Because Ker(T^2+1) should have atleast 2 elements in basis

Because you can find a vector v such that T^2+1 is the minimal polynomial

So,v Tv will be a LI set

That's kernel of A

Not Ker(A-2I)

one sec

A(1,0,0)=(0,1,1)

So, (A-2I)(1,0,0) is not zero

Should be same thing

[lavish jewel]

,w rank {{0,-2,1},{1,-2,2},{1,-4,4}}

[stoic python]

Results provided by WolframAlpha

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

Requested by Edd#5182

[native rampart]

nvm That's A

I think you should be doing column reduction

{(-2,1,1),(-2,-4,-4),(1,2,2)} span your null space

So basis should be {(-1,0,0),(1,2,2)}?

Weird

Yes

(1,0,0) is definitely not an eigenvector

If your eigenvector of value 2 is (x,y,z),you get -2y+z=0 and x=0

So, it's (0,1,2) only

[lavish jewel]

why does this look so complicated

[native rampart]

Because squirtel forgot about matrix multiplication

I think a good approach to finding a basis of ker(T^2+I) is pretend we are in C and find a basis for Ker(T-iI) and Ker(T+iI)

and then find a basis for Ker(T+iI)+Ker(T-iI) which only has real coefficients

[wintry steppe]

If I wanted to find the matrix for the differentiation map for polynomials D in L(R_3[x],R_2[x]) where the subscript denotes degree of or less than

But I used (1,x,x^2,x^3) as the basis for the domain and (2,2x,2x^2) as a basis for the codomain

Would the matrix then be given by

$\begin{pmatrix} 0 & 1/2 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 3/2 \end{pmatrix}$

[stoic python]

n/c

[native rampart]

Seems correct

[wintry steppe]

Thanks

Now consider the basis (x,x^2) for the codomain

Then the matrix would just be $\begin{pmatrix} 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}$

[stoic python]

n/c

[wintry steppe]

Right?

[native rampart]

What's your operation?

[wintry steppe]

Dp = p'

[native rampart]

You would miss some elements. So that's not the codomain of D

Like 2x gets map to 2 ,but 2 is not in your span

[wintry steppe]

I know

The matrix with (1,x,x^2) as a basis for codomain would be

$\begin{pmatrix} 0 & 1 & 0 & 0 \ 0 & 0 & 2 & 0 \ 0 & 0 & 0 & 3 \end{pmatrix}$

[stoic python]

n/c

[wintry steppe]

So why isn't the one I wrote above correct, if we remove 1 from the basis of the codomain?

[native rampart]

You will have to change your domain then

[wintry steppe]

Why?

[native rampart]

For every element in domain,there has to be a element in codomain

If {x,x^2} was your basis 2 has no corresponding element in the codomaim

[wintry steppe]

Oh I see

Our basis has dimension 3

For the codomain

Sorry edited, don't know why I said span

I get it now thank you

I think I understand matrices

When we add two matrices, what kind of operation are we using?

Do we define $+: F^{m,n} \times F^{m,n} \to F^{m,n}$

[stoic python]

n/c

[native rampart]

That's the usual convention

So,you are familiar with linear transforms but not matrices?

[wintry steppe]

That's how the book goes

[native rampart]

You understand given a basis,the isomorphism between linear maps and matrices?

[wintry steppe]

No isomorphisms are the next subchapter

But yes I do understand there's a bijection

Since linear maps are unique

[native rampart]

Matrix addition and multiplication are defined such that the bijection becomes an isomorphism

You want a • operation such that [T]•[U]=[ToU]

Because composition is the "product" wrt linear maps

You would want a + operation such that [T]+[U]=[T+U]

Where RHS is normal addition on functions

You only know the domain and range of +

You also want + to have that property

[wintry steppe]

But to prove that [T] + [U] = [T + U], we can just prove that it holds for a singular column right?

That is

$M_k(S) + M_k(T) = Sv_k + Tv_k = \sum_{j=1}^m A_{j,k}w_j + \sum_{j=1}^m B_{j,k} w_j = \sum_{j=1}^m (A_{j,k} + B_{j,k})w_j = M_k(S + T)$

[stoic python]

n/c

[wintry steppe]

That's how I proved it to be true for a column, and so we can then define the operation to do that without thinking just about a single column

[dusky epoch]

it stops being a basis

[wintry steppe]

I know thank you

[native rampart]

K,I was talking more about motivating the definition of +

You define + pointwise

[dusky epoch]

yes

assuming you are using F^{m,n} to refer to the set of all m by n matrices over F

[wintry steppe]

I am

[wintry steppe]

I see, thank you

Yes the motivation is basically to make it intuitive 😄

[native rampart]

I don't think your book will cover this,but Matrices and linear maps are isomorphic as algebras

[dusky epoch]

i don't think that's worth saying rn

as-is you're glossing over a lot of things

[native rampart]

mb

[wintry steppe]

this is bs right? vector space of functions from K to U Vi?

[limber sierra]

why

[zealous junco]

stupid question but if $(A-\lambda I)v = 0$ then why must v be in the span of $A-\lambda I$

[stoic python]

Anticipation

[zealous junco]

i.e. why must there be a w so that $(A-\lambda I) w = v$

[stoic python]

Anticipation

[zealous junco]

i dont think this is true always, (v nonzero)

[wintry steppe]

How is the set of all functions from K to U Vi a vector space in any natural way?

[wintry steppe]

who told you that? Let A = \lambda I for example then no nonzero v is image (if thats what you mean by span)

[zealous junco]

im dealing with some ODE stuff so maybe its a special case

but if theres like a 2D ODE and I got a repeated root with 1D eigenspace

then apparantly thats true

[wintry steppe]

How do they get the last line

Specifically, how do they move the w_j?

[limber sierra]

note that the Sigmas are interchanged

[wintry steppe]

Yeah exactly

[limber sierra]

[\sum_{r=1}^{n} C_{r, k} \sum_{j=1}^{m} A_{j, r}w_j = \sum_{r=1}^{n} \sum_{j=1}^{m} C_{r, k}A_{j, r}w_j = \sum_{j=1}^{m} \sum_{r=1}^{n} C_{r, k}A_{j, r}w_j][= \sum_{j=1}^{m} w_j \sum_{r=1}^{n} C_{r, k}A_{j, r}]

bleh cut off

one sec

@wintry steppe

its just repeatedly applying distributivity

[wintry steppe]

Where did your k's go?

[limber sierra]

oops

lmao

[stoic python]

Namington

[limber sierra]

they shouldnt have went anywhere

i dont even know how that happened

i just copy-pasted

i didn't do any index fuckery or anything like that

just apply distributivity, swap the sums, apply distributivity again

swapping the sums allows us to factor out w_j since now j is fixed (within the scope of the r=1-to-n sum)

since the sum indexing j is on the outside

[wintry steppe]

Wait I need to write this out lol.

I'm not good enough to tell by inspection

[limber sierra]

first =: distribute C_r, k into each term of the inside sum (note that r, k is fixed since k is a fixed variable and r is the index of the outside sum)
second =: think about it; we can think of this sum as adding together every possible pairing of r, j for 1 ≤ r ≤ n and 1 ≤ j ≤ m, so order shouldnt change anything
third =: factor out w_j from each term of the inside sum (note that j is fixed since it is the index of the outside sum)

but yes, writing this out with some small values (say n = 2, m = 3) can make it easier to follow

[wintry steppe]

thanks for writing it out

I found a mistake in my work (took forever to find)

But now it finally makes sense hahaha

[sly wedge]

hi can someone help me with understanding the simplex method? dm me please thank you !!

[celest slate]

It’s supposed to be r

I can’t prove it but I can verify it experimentally

Okay, now I’m able to transform from a hyperbola with transverse axis y = -x into a circle... let’s see if I can do anything with this or if I’ve been working on hours on nothing lol

not nothing, it’s a fantastic learning experience

I may be able to prove it acrually but I don’t want to bother, that seems like a lot of easy-to-mess up algebra

[wintry steppe]

Suppose V and W are finite-dimensional and T in L(V,W). Show that with respect to each choice of bases of V and W, the matrix of T has at least dim range T nonzero entries.

By definition, the matrix of T has entries in column k A1k,A2k,...,Amk given by sum i=1^m Aik wi

Where we assume that dim W = m and dim V = n

We can now suppose that dim range T = alpha. This implies that Tvi ≠ 0 for i ≥ alpha, as the zero vector is linearly dependent and would not contribute to the dimension of the range of T. Thus, Tvi = ciwi for c in F and wi in W.

As ciwi ≠ 0, ci ≠ 0 by an earlier exercise, and so [T] has at least alpha = dim raneg T nonzero entries.

Changing the basis will not change the dimension of each respective basis, and so it will not change our proof.

Does that argument work?

[wintry steppe]

I guess I need to change it a little bit

Since I am assuming it's equal to some vector wi in W but actually wi is a linear combination of vectors in W

Though it definitely doesn't seem like I am using it in that way

[arctic stone]

So, quadratic forms can be represented with the usual x^T M x, I was wondering, is it possible to represent cubic forms in a similar way by using rank 3 tensors?

[arctic stone]

Nvm, I found a way to make it actually work, not sure how useful it'd be but discovering it was quite fun

[celest slate]

yeah it didn’t work for what I hoped it would but it was still an enlightening exercise

I was hoping I could map a hyperbola xy=c onto a circle, in such a way that lattice points on the hyperbola translate to something interesting on the circle that could give insight into factoring problems

the issue is that the y coordinate a point maps onto has a “c” term in it, and the x term has a “square root of c” term in it, and they’re both divided by “x-y”, which really makes just about everything useless

[wind pasture]

if i have lin. dep. vectors, does one of them have to be a scalar multiple of another one?

no right? one vector just has to be a linear combination of the other 2 vectors

[celest slate]

correct

[wind pasture]

if a linear system Ax = b is consistent, at least one solution must be in row(A) right?

@celest slate

[celest slate]

I am not familiar with the notation row(A) sorry

[wind pasture]

actually i think what i said is wrong

if 2 functions are linearly independent, are their derivatives also lin. indep.?

function in the form e^x

[dreamy iron]

i wrote this a year ago. and i don't know what i did, but im almost sure i flubbed up the last half by flipping my \bigcaps, and turning them upsidedown into \bigcups

https://docs.google.com/document/d/160hxN3Os8AkIPCQR-mDATwKKrgA-kOJaBUibkW4JFi4/edit

which is to say, this is all wrong, right? i need to flip the cup into a cap?

[dusky epoch]

yes

[rose umbra]

why does geometic multiplicity of eagevalue<= algebraic multiplicity of eagenvalue ?

i could get 2 eagenvectors of one eagenvalue

[dusky epoch]

it's spelled eigen, just for reference

anyway, the geometric multiplicity is the dimension of the eigenspace, so you would need two linearly independent eigenvectors

something you may not have

loosely speaking, you can pick a basis for the eigenspace (consisting of as many vectors as the geo mult, by defn) and extend it into a nice enough basis of the whole space wrt which your operator will be block-triangular, with the top left block diagonal - thus showing the determinant has (x - λ)^(geo mult) as a factor

[rose umbra]

got it thx

[waxen flume]

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

https://i.imgur.com/j4fCi2s.png

Is the dimension of S just three?

how would i find this

[nocturne jewel]

set of solutions to $2x_1+x_2-x_3=0$ form a plane

[stoic python]

Mosh

[nocturne jewel]

and a plane is defined by the span of 2 vectors

[waxen flume]

so its R2?

@nocturne jewel

[nocturne jewel]

dim(S)=2

yes

[waxen flume]

thank you!

https://i.imgur.com/2KaXirh.png

https://i.imgur.com/Hmx7Xux.png

how would I do this?

I know I have to use a projection vector but not sure for my setup

[dusky epoch]

is there any reason why you are screenshotting one line at a time

anyway you want to project (2,1,1) onto S

[waxen flume]

thanks I already figured it out

and nah

i just thought it was neater

https://i.imgur.com/eovMk5s.png

thats where its from if you wanted to know

[short coral]

How to think of adjoint of a matrix geometrically?

[mossy lodge]

does anyone know the basis for quadratic polynomials

I don't know if its just {1,x,x^2} or an infinite set

[wind pasture]

if i have kerT = span(

1 0 0 0
0 0 0 1

)

those are 2 matrices

whats dim(kerT)?

[nocturne jewel]

span consists of 2 vectors

so dim(span(u,v))=2

[wind pasture]

@nocturne jewel but im interested in the span of two matrices not vectors

[nocturne jewel]

yeah, matrices can be vectors

[wind pasture]

ok

[nocturne jewel]

$\mathbb{R}^{m\times n}$ is the R-vector space of m by n matrices

[stoic python]

Mosh

[nocturne jewel]

so the kernel of T is defined as the span of 2 vectors, so dim(ker(T)) = 2

[wind pasture]

aight

thanks

[wintry steppe]

anybody care to explain?

please tag me, as i will be searching for the answer elsewhere

[nocturne jewel]

which part dont you get?

[wintry steppe]

7,9,-2,4; why?

[nocturne jewel]

yeah shouldnt be that vector

[wintry steppe]

well, the answer sheet tells me that's the answer

and im very sure its not an error

[nocturne jewel]

yes cause a piece of paper will always be right. . .

[wintry steppe]

well, you solve it then, does the point (1,0,2,1) lie on the line between (-2,1,1,0) and (3,1,0,4)?

or if you can explain to me how to apply rref to 3 dimensional systems, that would be fine as well, and i suspect this is quite a bit easier

[nocturne jewel]

it's not on the line.

[wintry steppe]

yeah but why

[nocturne jewel]

find the equation of the line, then show that the parameter has to be different per component

[wintry steppe]

what is the equation of the line

[nocturne jewel]

Find it out yourself

[wintry steppe]

i dont know how, i would not be here

[nocturne jewel]

Check the answer sheet

[wintry steppe]

answer sheet is what i have in the picture

[wintry steppe]

answer sheet says its ${\left(\begin{matrix}-2\1\1\0\end{matrix}\right)+\left(\begin{matrix}7\9\-2\4\end{matrix}\right)*t|t\in\mathbb R}$, which you disagree with

[stoic python]

Betelguse

[nocturne jewel]

Yes, and you say it's right and were firm on that

so go with what the answer sheet says since it's apparently right

[wintry steppe]

i do not understand it

[wintry steppe]

i am also open to the possibility that it is wrong. if you come up with a solution that would solve the problem and contradicts with the answer, i obviously will be willing to accept that

i am not prioritising you or the answer sheet, i just need an answer that i understand, or an explanation that would make me understand

[nocturne jewel]

You did prioritize the answer sheet.

and the way you came off was disrespectful, so you got the amount of help I am willing to give.

[wintry steppe]

well, answer sheet has an answer and you dont.
you should refrain from engaging conversations on the internet, as the slightest implication that someone else other than you might be right offended you.

[nocturne jewel]

You asked for help, I gave you help, you wrote it off as "I doubt the answer sheet is wrong"

[wintry steppe]

i will not suck you off so that you give me an answer. i am guilty of nothing that requires apologising. if you do not have an answer, do not come in and challenge the only answer i have.

[nocturne jewel]

Ok then gl

[wintry steppe]

thanks

anybody care to explain?
answer sheet might be wrong, but please come with a proof.

i specifically do not understand why the free element is (7,9,-2,4). how does one get there?

alright, found it

(7,9,-2,4)*t is parallel to the line given in the question, and you just need to translate it by (-2,1,1,0) to get there

[odd kite]

is it though?

notice the points have the same y value

adding 9t to y is going to change y

for any t besides 0

[wintry steppe]

see the explanation here.
a free vector of a line from point a to point b is $\begin{pmatrix}b_x - a_x\b_y - a_y \ \vdots \ b_{dimension} - a_{dimension} \end{pmatrix}$, and translating it to the right place gives you the exact same vector.

[stoic python]

Betelguse

[wintry steppe]

so free vector starts from origin and is parallel to the line

thats why you need to translate it

[wintry steppe]

what do you mean? can you rephrase

[odd kite]

$$\begin{pmatrix}3\1\0\4\end{pmatrix} -\begin{pmatrix} -2\1\1\0\end{pmatrix} =\begin{pmatrix}5\0\-1\4\end{pmatrix}$$

[stoic python]

Timon

[wintry steppe]

your right

how did i not see that

[odd kite]

The answer is wrong, maybe they changed the problem and forgot to update the answer or something

[wintry steppe]

thanks a lot man, i am embarrassed to not have noticed it before

@nocturne jewel you were right

[odd kite]

@untold pier wrong channel, use one of the pre-university ones like #prealg-and-algebra

[wind pasture]

$$\begin{bmatrix}
1 & 1 \
-1 & 1 \
\sqrt2 & \sqrt2
\end{bmatrix}$$

[stoic python]

ABCDE

[wind pasture]

how can I calculate the standard matrix of the orthogonal projection onto Row(A)?

matrix above is A

[gray dust]

pick an orthonormal basis $(e_1,\ldots,e_n)$ of Row$(A)$. writing vectors as column vectors, the projection of any $x$ onto Row$(A)$ is
$$\sum_{i=1}^n(x\cdot e_i)e_i=\sum_{i=1}^n(e_i\cdot x)e_i=\sum_{i=1}^ne_ie_i^Tx=\br{\sum_{i=1}^ne_ie_i^T}x$$
so $\sum_{i=1}^ne_ie_i^T$ is a matrix representation of the projection

[stoic python]

RokabeJintaro

[wind pasture]

if a linear system Ax = b is consistent, does it have to have a solution in row(A)?

[dusky epoch]

does that even make sense?

if A is m × n, then elements of its rowspace have size m while the solutions of Ax=b have size n

[lavish jewel]

you sure about that one?

[wind pasture]

good point

[lavish jewel]

i'd say the elements of the column space have size m, while those of the row space are of size n

[dusky epoch]

fuck shit

lmao

my bad

[lavish jewel]

i think as long as by solution they meant the vector x, it should be right

[dusky epoch]

still i dont see why x should be in row(A)

like, an element of row(A) is a vector of the form uA for u a suitably sized row vector

while x is a column

so why would x be A^T times something

[lavish jewel]

i'm pretty sure by row space they mean to address the orthogonal complement of the null space

if Ax = b is consistent, it has at least one solution, and it corresponds to the projection of x onto the row space

it might have infinitely many solutions, in which case the solution is the projection of x onto the row space + anything from the null space

[wind pasture]

so is my statement true?

[lavish jewel]

it should be

consistent means b is lin dep with the columns of A, yeah?

and A maps elements from the row space to the column space

[dusky epoch]

im not convinced

[lavish jewel]

it should be a consequence of the fundamental theorem of linalg

(i might be wrong, so always take it with a heap of salt)

[dusky epoch]

you seem to be mangling terminology here because like

row space and (nullspcae)^perp may be isomorphic but they arent the same space?

[lavish jewel]

hmmmmm

are they not?

when you take an SVD, for example, the right singular vectors form a basis for R^n; the first r vectors span the row space, and the remaining n-r span the null space, and they're all orthonormal

this should be a consequence of being able to diagonalize symmetric/hermitian matrices with orthonormal eigenvectors

[wind pasture]

@dusky epoch when i say solutions i mean x not b

[dusky epoch]

of course i understand that you mean x

we are given a matrix A ∈ R^(m × n) and a vector b ∈ R^m, and we are told that there exists x ∈ R^n such that Ax = b

must x necessarily be expressible as A^T u for some u ∈ R^m?

[lavish jewel]

not necessarily all of x, but at least some component of it

because the part that isnt just yields a 0 vector

[dusky epoch]

??

so what you're saying is that there must exist a solution to Ax = b in the form A^T u?

i.e. that $AA^Tu = b$ must be consistent?

[stoic python]

Ann

[lavish jewel]

im saying thw direct sum of the row and null spaces is R^n

yeah

this is the same as saying that you can pseudo invert A

as long as b is in the image of A

and the pseudo inv is the map from image to row space

so if x = xrow + xnull, then xrow must be nonzero. though looking at it again, if b is 0, then what i said doesnt hold. is the vector b nonzero?

if nothing is said regarding that, i just wasted a whole bunch of everyone's time

[thorn robin]

Your argument is fine whether b is zero or not

What's wrong with xrow = 0?

Just decompose x = xrow + xnull and compute that b = Ax = A(xrow + xnull) = A(xrow) boom solution in the row space

[thorn robin]

This corresponds exactly to what you said way above

[lavish jewel]

yea but if b is 0, then you could just take any xnull

[thorn robin]

Yeah but the problem only wants one solution to be in the row space, in the case b = 0 that's x = 0

Like you also said before you can get more solutions by adding xnulls that don't have to be in the row space, but the one solution is

[lavish jewel]

i would argue that xrow = 0 and xnull =/= 0 means x is not in the row space, since it had an orthogonal projection of 0 onto the row space. but this is only the case for b=0. otherwise, what i said still holds, yeah

[thorn robin]

One of us is very confused about something silly

The row space contains the zero vector surely?

[lavish jewel]

it does, otherwise it wouldn't be a space, sure

[thorn robin]

Right so the solution to Ax = 0 that lies in the row space, is just x = xrow = 0

[lavish jewel]

fair enough. then the assertion that x is in the row space is correct, but one needs to be careful depending what is known about b.

but since the question only asks for existence of a solution in row(A), then yeah

yes

yeah sorry, i had to go back and read that bit again

[lavish jewel]

both ann and abcde disappeared

[dawn fractal]

"basis is a maximal linearly independent set"

does this mean that there exists an s not in the basis such that the union of the basis and the set containing s is linearly dependent?

[dire thunder]

yes

[dawn fractal]

thx

[dire thunder]

let {v_1, ..., v_n} be basis

let s be not in basis

then {v_1,..., v_n, s} is linearly dependent

[dawn fractal]

lmao that name

[lavish jewel]

(for v_i and s in the same subspace)

[dire thunder]

bold of you to assume there exist spaces different from R^{pi}

[lavish jewel]

i don't know enough math to know if that was a joke or a real thing 😛

[wintry steppe]

Let $V,W$ be vector spaces over a field $\mathbb{F}$. Let $T \in \hom(V,W)$.

1. Prove that if $\mathbb{F} = Q$, then the homogeneity property of linear maps is redundant.

Let $n,m \in \mathbb{Z} \setminus { 0 }$. It follows that $n^{-1}$ exists as it must have a multiplicative inverse in the field $\mathbb{Q}$. Thus, $n^{-1} \in \mathbb{Q}$ by definition of $\mathbb{Q}$.

Using the additive property of linear maps, we have that [ T(u) = T\left( \sum_{i=1}^n n^{-1}u \right) = \sum_{i=1}^n T(n^{-1}u) = nT(n^{-1}u). ] This implies that $n^{-1}T(u) = T(n^{-1}u)$.

Then observe that $T(mn^{-1}u) = T\left( \sum_{i=1}^m n^{-1}u \right) = \sum_{i=1}^m T(n^{-1}u) = mT(n^{-1}u)$. But we know that $T(n^{-1}u) = n^{-1}T(u)$, so $T(mn^{-1}u) = mn^{-1}T(u)$. By definition of, $mn^{-1} \in \mathbb{Q}$ as it is a ratio of two nonzero integers, and so we have shown the property the scaling property to be redundant for that case. If $m = 0$, then $T(0u) = 0$ and $0T(u) = 0$, so $0T(u) = T(0u)$. If $n = 0$, then a multiplicative inverse does not exist for it in the field (by definition), and so we are done.

[stoic python]

n/c

[wintry steppe]

Does this proof work?

[dusky epoch]

appears so

[wintry steppe]

Thanks

[wintry steppe]

How do I prove the converse is not true?

If F = Q, then homogeneity does not imply the additive linear property

[dire thunder]

you may just construct counterexample

[wintry steppe]

I know

I'm having a hard time thinking of one lol

I need vectors v,u such that T(v + u) ≠ T(v) + T(u)

[dire thunder]

maybe this function

f(m/n)=m+n

ah no

not homogeneous

but f(m/n)=m

looks to be homogeneous

[wintry steppe]

1/n f(m) = m, and f(m/n) = m, but f(a/b) + f(c/d) = a + b, however 1/b f(a) + 1/d f(c) = a/b + c/d = (ad + bc)/(bd). Then f(a/b + c/d) = f((ad + bc)/(bd)) = 1/(bd) f(ad + bc) = ad + bc

[dire thunder]

here problem is possible tho

oh

it is not homogeneous

hmm

there is an example

f(x,y)=x if xy > 0

and 0 otherwise

[wintry steppe]

So c f(x,y) = f(cx,cy) = cx if xy > 0 and 0 otherwise

[dusky epoch]

not well defined

[dire thunder]

anyway not homogeneous

so we do not care abt it

[dusky epoch]

you can have a function on Q^2 which is homogeneous but not linear

[dire thunder]

well f(cx, cy) would be cf(x,y) since sign of c^2xy is the same as xy

[dusky epoch]

say, define f by f(1,q) = q^2 and f(0,q) = q and let the rest follow by homogeneity

[dire thunder]

and on Q^n with n=1 looks like homogenuity implies additiveness

[dusky epoch]

so f(p,q) = q^2/p for p != 0

f(1,2) = 4, f(1,1) = 1, f(2,3) = 9/4 != 5, boom, not additive

[wintry steppe]

That's kind of a crazy counterexample

But I suppose it does work

[dusky epoch]

crazy how?

[wintry steppe]

Not obvious

[robust pelican]

uhm can somene teach me how to do age linear problems?

[dusky epoch]

wrong channel, go to #prealg-and-algebra.

[robust pelican]

whoops sorry

[wintry steppe]

Suppose V, W are finite-dimensional and T in L(V,W). Show that with respect to each choice of base of V, and W, the matrix of T has at least dim range T nonzero entries.

Suppose that dim range T > 0, otherwise if dim range T = 0, then null T = V and so the matrix of T will be the zero matrix and so it is true.

Then assume that M(T) has at least p nonzero entries, where p = dim range T. for p ≥ 0.

Without loss of generality, assume that T(vi) ≠ 0 for i = 1,2,...,p.

Then T(v1) = c1,1 w1 + ... + cm,1 wm ≠ 0, where ci,1 are not all 0

Assume that, without loss of generality again, that c1,1 is nonzero, and the rest can be zero.

This tells us that in the worst case, T(v1) = c1,1 w1, which implies that T has one nonzero entry for v1.

We can repeat this process for all of vi for i = 1,...,p to see that each of the vi have at least one c_(i,j) such that it is nonzero

[dusky epoch]

this feels verbose.

[wintry steppe]

Because there are p such v_i, it is implied that there are at least p = dim range T nonzero entries in the matrix of T

I know

But I don't know how to make it less... handwavy?

[native rampart]

There's a much better proof

columns of matrix of T span range(T)

[dusky epoch]

Then assume that M(T) has at least p nonzero entries
i mean

assume your goal

[native rampart]

and there are atleast rank(T) nonzero elements in a set that spans range(T)

[lavish jewel]

proof by assumption

[wintry steppe]

Sorry, change the wording to T(v_i) is nonzero for i = 1,...,p (wlog)

[dusky epoch]

and v_i is what?

[wintry steppe]

It's a basis vector of V

[dusky epoch]

a basis of V you pull out of thin air

ok

[wintry steppe]

Where the basis of V is v1,...,vn

[dusky epoch]

i mean ok like

you say there are p columns in the matrix of T which span the range of T, and in each col there is a nonzero entry

that's it

no need to fuck around with which rows they're in cause you don't care

[wintry steppe]

Yes

That proof is much better I guess

Suppose V, W are fin dim and T in L(V,W). Prove that there exist a basis of V and a basis of W such that with respect to these basis, all entries of M(T) are 0 except that the entries in row j, column j, equal 1 for 1 ≤ j ≤ dim range T

Are they asking about a single entry at M(T)_(j,j)?

[lavish jewel]

they're asking for a sorta diagonalization

say, M(T) = U I V^T, with I as an identity-like rectangular matrix with 1's along the diagonal

[wintry steppe]

What is U?

[lavish jewel]

presumable a change of basis matrix

same with V

[wintry steppe]

Have not gone over that yet.

[lavish jewel]

i see. maybe someone has a different approach, then

[wintry steppe]

I think I get it, but I just want to make sure

They want 1 at (1,1),(2,2),(3,3),...,(dim range T, dim range T)

And zeros everywhere else?

[lavish jewel]

yes

[wintry steppe]

Okay

Thanks, I just waned to clarify that

[wintry steppe]

How do I make entries of a matrix zero, relative to basis?

i.e. suppose for a basis v1,...,vn and basis w1,...,wm

The entry at j,k of the matrix is A

For nonzero A

How can I adjust the basis so that A = 0?

It's not possible, correct?

[dire thunder]

if matrix wrt to some basis is zero it would be zero for any basis

since if A is matrix wrt basis and B is change of basis matrix then T=BAB^(-1)=B0B^(-1)=0

[wintry steppe]

No A is an entry

At (j,k)

But yeah I don't see how we could make it zero.

[dire thunder]

i doubt that there is procedure for this

unless certain assumptions are made

otherwise it implies every matrix is diagonalizable

[lavish jewel]

SVD moment

not really "diagonalizing" tho

at any rate, you can use givens rotations to do this

[wintry steppe]

I understand the proof that two vector spaces are isomorphic iff they have the same dimension in the direction where you assume that they're isomorphic, but I don't understand the other direction.

In other words, could someone help me prove that two vector spaces are isomorphic if they have the same dimension?

[wintry steppe]

@wintry steppe what's the most obvious way to try to construct an isomorphism between them?

We map v1 to w1, v2 to w2, ..., vn to wn

did you try that

But is that enough?

To show there exists some function which "links" them like that?

ofc not

do you know what an isomorphism is?

Yes

It's an invertible linear map

so

maybe such a map is invertible

Okay

I see, thanks

[dawn fractal]

do i need to prove that a set orthonormalized from a linearly independent subset of an inner product space V through the Gram-Schmidt orthogonalization process and vector normalization is a basis for V?

it was given that the original LI subset of V was a spanning set of V

and that that subset is also LI

[lavish jewel]

don't think so

[dawn fractal]

why is that

[lavish jewel]

it should be easy to do it, but the definition of the gram schmidt alg. is a procedure that takes a basis for a subspace V and produces an orthonormal basis for the same subspace V

each step of the algorithm takes one vector and keeps only the component in the orthogonal complement of all the previous vectors

in other words, you can relate the gram schmidt basis, let's call it G, to the original basis, let's call it B, through a full rank transformation T, i.e. B = GT

and so B and GT have the same image, meaning their columns span the same space

but anyway, the short answer is you don't need to, but it should be pretty easy to do so

[dawn fractal]

ty

[dawn fractal]

real subspace vs. complex subspace of C^{nxn}

im guessing real means real scalars and complex means complex scalars?

both of which are multiplied to vectors in said respective subspaces

[lavish jewel]

yea

[short coral]

Spanning set is a set containing "all" the basis?

[lavish jewel]

what do you mean

[short coral]

Whether is it correct?

Is the spanning set a set containing all the basis?

[lavish jewel]

what do you mean by "all the basis"

that expression alone doesn't make sense

because there usually isn't a unique basis

[short coral]

What's the difference between basis and spanning set?

[lavish jewel]

a basis is a linearly independent spanning set

[short coral]

So, every basis is a spanning set but not all spanning set is a basis?

[lavish jewel]

indeed

you can have "extra vectors" in a spanning set

[short coral]

What's the significance of spanning set?

[lavish jewel]

pretty similar to that of a basis, but there may be more than one way of representing vectors in the subspace

with a basis, the representations are unique

[short coral]

Like for example?

[lavish jewel]

the set [1,0], [0,1], [1,1] is linearly dependent

it spans R2, but it is not a basis

[short coral]

Thanks

Basis will be just [1,0] and [0,1] right?

[lavish jewel]

well

that one works

but so does [1,0] and [1,1]

or [0,1] and [1,1]

those 3 are all valid bases

[short coral]

Is it necessary for one of the vectors to have a single one and remaining all zeroes?

*atleast one of the vectors

[lavish jewel]

not at all

[-1,1] and [1,1] is another valid basis

the only thing necessary is that the vectors are linearly independent and span the subspace

[wintry steppe]

how do u prove a linear function is continous?

[native rampart]

Context?

I don't think a linear function has to be continuous in general

[wintry steppe]

Mmmm

i am asked to prove that given a diagonal matrix, if a_n is bounded, then the matrix has a bounded operator (?)

[native rampart]

https://en.m.wikipedia.org/wiki/Discontinuous_linear_map

[wintry steppe]

idk if this is the terminology

but we have to do 2 things: first prove that if a_n is bounded, then the linear application is continous

https://gyazo.com/6ecfb57d4cae926bb4e88892667c0fae

If a_n is bounded, then

https://gyazo.com/f3d9d4fc0cf05b23aa51b04c04b1dfd8

T is continous

[dusky epoch]

$|a_n| \leq M \implies \nrm{Tx} \leq M \nrm{x}$

[stoic python]

Ann

[dusky epoch]

@wintry steppe

[short coral]

How to prove that Transpose (Adjoint (A)) = Adjoint (Transpose (A))?

[lavish jewel]

what are you calling adjoint? and is A a matrix?

[wintry steppe]

adjunta

[wintry steppe]

i found this too. But on my case i need to prove the imply on both sides

this one is left way or right?

and what is x there?

did u mean T(x)?

[short coral]

Yes, A is a matrix. Adjoint = Adjugate = transpose of its cofactor matrix

[dusky epoch]

x is a sequence

@wintry steppe wrong channel, see pins.

[wintry steppe]

and how does that prove a_n is continous?

[dusky epoch]

what?

a_n is a sequence

[wintry steppe]

i mean, the linear application*

sorry

T, in this case

[wintry steppe]

does this just mean it is continous?

okey yes. And how do i do the other way? if T is continous, how can i prove a_n is bounded?

[short coral]

How come the Eigen vectors of a matrix and its inverse are same?

[native rampart]

Let Av=cv then A^-1(cv)=v

i.e., an eigenvector of A is an eigenvector of A^-1

Similarly repeat for the other direction

[short coral]

Similarly, how to prove for Transpose of a matrix?

[wintry steppe]

it's false, a matrix and it's transpose need not have the same eigenvectors

they will have the same eigenvalues, though

[whole solstice]

i am

a dumbass so be patient with me

Suppose that U and W are both five dimensional subspaces of R&9 prove that U intersection W != 0

[native rampart]

Suppose U inter W is 0

You can pick five basis vectors for U and 5 basis vectors for W

These 10 vectors will be LI in U+W

But U+W is a subspace of R^9

And hence a set can only have a max of 9 LI vectors

[whole solstice]

what are LI vectors

[native rampart]

Linearly independent

[whole solstice]

o

[native rampart]

This is not direct

Try to see why that's true

[tardy cypress]

i do math

[frosty vapor]

Prove it.

[wind pasture]

is it possible for me to find an orthogonal matrix thats row equivalent to

-1 2
2 -4

I don't think so, but how do i prove this?

[lavish jewel]

the rows are linearly dependent, it does not have an inverse

so the inverse can't be its transpose

maybe like that?

[dusky epoch]

row-equivalence preserves rank, an orthogonal matrix has full rank, yours does not

[unkempt shale]

We have a test for screening cancer that is 90% sensitive and 90% specific. Assuming that 1% of the population has cancer, what is the fraction of positive tests that correctly detects cancer?

[limber sierra]

this isnt a linear algebra question

try #probability-statistics or a #questions-_ channel

[zealous junco]

dumb q but theres no difference between "cone" here and matrices right, wonder why they write it as cone

(not exam btw)

[limber sierra]

i dont understand the question

that set is called a cone

a matrix is not a set

and certainly not a cone

[zealous junco]

i mean isnt the set S^n+ just the set of matrices that are symmetric positive definite

based on the definition of all X in R^nxn so that X = X^T and X ≥ 0

idk why they call it a cone

[zealous junco]

wait im being dumb

thanks

i thought they were calling the matrices cones

[limber sierra]

they could just say "set" instead of "cone" sure

[zealous junco]

lol

[autumn wing]

Does someone know how to solve a?

[woven fossil]

If ~v4 ∈ span(~v1, ~v2, ~v3), then replacing ~v3 with ~v4 does not alter the span. That is,
span(~v1, ~v2, ~v3) = span(~v1, ~v2, ~v4).
please help it is urgent

<@&286206848099549185>

[gray dust]

@woven fossil please wait 15min before pinging helpers

[woven fossil]

sorry I will remember next time

[marble lance]

@woven fossil this is false

Consider R^3 with the standard basis e1, e2, e3. Then x = 2e1 is in span(e1, e2, e3) but span(e1, e2, x) ≠ span(e1, e2, e3)

[glad acorn]

what is the meaning of symbol "~" on a basis vector

it appears in coordinate system transformations but what is the actual meaning of the symbol ~

[wintry steppe]

post image

[lavish jewel]

it's just to distinguish it from the old e_i

[marble lance]

This is gross

Just replace the arrow

[viral flint]

\tildearrow?

[wintry steppe]

$\tilde{\vec{x}}$

[stoic python]

TTerra

[wintry steppe]

[crude ridge]

Would it be true to say that for A to be diagonizable, the multiplicity of each eigenvalue needs to be equal to the dimension of the eigenspace corresponding to that eigenvalue?

[lavish jewel]

sounds about right

[crude ridge]

Great, thank you

And that comes from the fact that the P matrix with the vectors needs to be invertible?

[lavish jewel]

yeah, otherwise you can't get the similarity

[crude ridge]

Ah that makes sense, thanks

[azure bough]

can't you just do $\vec{x'}$

[stoic python]

Conan

[wintry steppe]

cf

[azure bough]

yeah x' is definitely better than x~

disgusting

[stoic python]

TTerra

[wintry steppe]

ew

this is why i don't use \vec lol

[zinc wasp]

Can anyone help me understand how they get b and or how to better understand this markov chains stuff? I wouldnt mind doing it via voice either to really understand better.

[glad acorn]

covector is called covector because it acts as a function on vectors?

[odd kite]

well, that's what it means, it's a linear function that takes a vector as input

the co- prefix sort of means "accompanies" or something

[main thistle]

Hi, I would like to confirm something.
If vectors a & b are linearly dependent, then a & b are scalar multiples of one another?

[viral flint]

If they're both non-zero, yes.
If say a = 0, b non-zero then a is a multiple of b but not true other way round.
Basically yes with the caveat that non-zero vectors can't be scalar multiples of the zero vector.

[main thistle]

thanks!

[woeful fable]

hi, i'm having trouble understanding what the answer key (on the right) means. the question is asking for the coordinate vector of p(t), which should be [x]ß. but why is the key.. finding [p]ß :_ D..? or does [p]ß actually mean the coordinate vector (the book says the opposite thing tho)?

[odd kite]

@woeful fable [ ....]ß just means coordinate vector of whatever vector is in the square brackets. It can have different names.

[woeful fable]

ohhhhh thank you!!! that clears it up

[dusky epoch]

wrong channel, go to #prealg-and-algebra or #precalculus

[dreamy iron]

Confession: I don’t think I’ve ever proved that the zero vector is a vector space in its own right.....

.....I guess I’ll just have learn to be okay with that.....

(No one ever needs to know, right?)

[gray dust]

proving {0} is a subspace of the ambient space proves it's a vector space in its own right

[wintry steppe]

[heavy crown]

hmm any idea?

[dusky epoch]

Z linearly independent => E invertible, huh...

[heavy crown]

yea I don't get how it relates haha I think it's disprove?

[dusky epoch]

this looks true to me, but i'm only like 80% sure

[heavy crown]

hmm

[dusky epoch]

let's see

let's consider the subspace $U_E$ of $M_2(\bR)$ defined by $U_E = { XE \mid X \in M_2(\bR)}$

[stoic python]

Ann

[dusky epoch]

i have a hunch that U_E is a proper subspace of M_2(R) iff E is non-invertible

ah actually

$\dim(U_E) = \begin{cases} 4 & \rank(E) = 2 \ 2 & \rank(E) = 1 \ 0 & \rank(E) = 0 \end{cases}$

[stoic python]

Ann

[dusky epoch]

this could be written as dim(U_E) = 2rank(E) but this would have been a little unintuitive as there are different arguments to be made for each case

@heavy crown i can explain why this is true to you if you'd like, or you could take my word for it for now and i could tie it to the original problem

what would you like me to do first?

[heavy crown]

unforunately they didn't teach us rank yet, the last subject was about R(A) and C(A) Row and column spaces, and before that Grassmann's formula
maybe if you could explain this to me intuitively I could understand how to make it?

thank you for your kindness btw 🙂

[dusky epoch]

hrm

i mean ok these are two by two matrices

so you can replace rank(E)=2 with "E is invertible", rank(E)=1 with "E is singular but not the zero matrix" and rank(E)=0 with "E is the zero matrix"

[heavy crown]

how would you know how much is dim depends on if E is invertible ? I mean why is it 4

[dusky epoch]

well the dimension of M_2(R) itself is 4, is it not?

[heavy crown]

oh, right

[dusky epoch]

• if E is invertible then U_E = M_2(R)
• if E = 0 then U_E = {0}
• if E is neither invertible nor 0 then dim(U_E) = 2

i think the third point is the trickiest to prove