linear-algebra

[native rampart]

R over R?

[wheat prairie]

vectors having real numbers as their entries

[fringe matrix]

hey guys

i totally forgot stuff..

lets say i have a block matrix

(A I
I A)

and i want to multiply it by a vector (X1,X2)

what is the result?

A - nXn matrix
I - nXn matrix
X1,X2 are nX1 vector, so its combine to a 2*n vector

its

(Ax1+x2,Ax2+x1) right>>

[native rampart]

Should be

[fringe matrix]

What can i say about its eigenvectors? :S

[native rampart]

Say (X1,X2){As per your notation},then $(A-\lambda I)^2 X_1=X_1 and (A-\lambda I)^2 X_2 =X_2$

Where lambda is an eigenvector of the block matrix

I don't think you can say much more

[fringe matrix]

ty

[stoic python]

DrunkenDrake:

[fringe matrix]

lambda is an eigenvalue

and X1 and X2 are both eigenvectors of A, right?

and $(A-\lambda I)^2 is the eigenvalue of the block matrix

[native rampart]

X1 ,X2 may or may not be eigenvalues of A

[fringe matrix]

ok but

lets try to find the eigenvalue of the block matrix. (Block matrix is M)
Mv = av (a = eigenvalue)
meaning:
Ax1+x2 = ax1
Ax2+x1 = ax2

let's choose x1=x2 and move stuff around we will get
Ax1 = (a+1)x1

so we found the eigenvalue of A - a+1

[native rampart]

*(a-1)

[fringe matrix]

yea oops

does it make sense?

[native rampart]

Yea

[vocal isle]

Hi everyone, I have a point in 3D space, point 1, (x1,y1,z1) and it needs to be transformed by a matrix to a point, point 2, given by azimuth and elevation and radius (phi, theta, radius). Fortunately the magnitude of point 1 will always be = r. So the transformation matrix will just be a rotation matrix. Does anyone know what the rotation matrix must be to make this transformation?

[wintry steppe]

yo who ever is in grade 10 or so please help me with my math assignment about Analytical Geometry

[cunning arch]

If I have orthogonal projection onto the xy-plane from R^3 to R^3, how can I tell if it's one-to-one or onto?

And also "Describe the set of vectors that get mapped to 0 under T."

Thank you!

[honest imp]

im confused, would this just be E =
[1, 0, 0, 0]
[0, 0, 1, 0]
[0, 1, 0, 0]
[0, 0, 0, 1]

[gray dust]

@dim venture how do you justify choice 4

[frigid otter]

is this as simple as the coefficient matrix determinant test thing?

[gray dust]

@dim venture it mentions orthogonal complement not just complement

take R^2 w/ dot product. R^2=span{(1,0)} oplus span{(1,1)} but span{(1,1)} isn't the orthogonal complement of span{(1,0)}

[lime laurel]

I’m doing algrabra and I’m stuck on this part...

[ocean remnant]

I’m doing algrabra and I’m stuck on this part...
@lime laurel Cant see the y intercept that great but... if the y intercept is 2 then the equation is y=e^x+1 if y intercept is 3 then its e^x+2

[lime laurel]

Thanks

Also how about this one?

And where did e come from?

[limber sierra]

fyi this isnt linear algebra

[lime laurel]

Oh ok

[limber sierra]

try #prealg-and-algebra , #precalculus , or a #questions-x channel

[lime laurel]

Also is that the frog from Mario rpg?

[limber sierra]

yes.

[wintry steppe]

I'm having trouble understanding how the '1' appears that is circled when finding the general solution.Why does it equal '1' when the third row equals zero

[dusky epoch]

the third row has nothing to do with the third variable

[limber sierra]

the augmented matrix $\begin{pmatrix}1&0&-3&0\0&1&-1&0\0&0&0&0\end{pmatrix}$ actually represents the system [\begin{pmatrix}1&0&-3\0&1&-1\0&0&0\end{pmatrix}\begin{pmatrix}x_1\x_2\x_3\end{pmatrix}=\begin{pmatrix}0\0\0\end{pmatrix}]

[stoic python]

Namington:

[limber sierra]

in other words, solve this system:

\begin{align*}
x_1 - 3x_3 &= 0 \
x_2 - x_3 &= 0 \
0 &= 0\end{align*}

[stoic python]

Namington:

[limber sierra]

note that if you solve it, you get $x_2 = x_3$ (by the second equation) and $x_1 = 3x_3$ (by the first equation)

[stoic python]

Namington:

[limber sierra]

so our solution vectors, in terms of $x_3$, look like $\begin{pmatrix}x_1\x_2\x_3\end{pmatrix} = \begin{pmatrix}3x_3\x_3\x_3\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

factoring x_3 out of the right hand side vector gives their result.

as ann mentioned, the third row has nothing to do with the third variable

in fact, the third column corresponds to the third variable

(assuming your matrix has at least 4 columns, that is)

[wintry steppe]

got it! thanks so much

[cunning arch]

for this transformation (reflected across y=tan\theta x then y=tan\phi x), what's the axis of rotation?

like the angle of rotation would just be phi+theta, i think, but i'm not sure how to answer axis of rotation

i thought it would be the origin but lOL the origin isn't an axis

[odd kite]

@cunning arch axis of rotation really only applies in 3d but sometimes we might consider rotations in the xy plane to be rotations around the z axis even if the axis isn't depicted/mentioned

[cunning arch]

Oh I see, thanks!

[wintry sphinx]

I think it generalizes. In 2D, rotation is around a point. In 3D, rotation is around an axis. In 4D, rotation is around a few planes I think.

[frigid otter]

how do I check this if I don't have n vectors of n length? I only have 3 when they are in R^4?

I think the determinant?

oh okay, it's the linear combination thing, I think I got it

[valid hawk]

Help

[frigid otter]

just from looking at it, this is definitely a subspace right

[green trench]

@frigid otter those are some nasty problems

Certainly quite difficult

[frigid otter]

well that sucks, I just turned it in and I thought it was easy hahaha

[green trench]

Oh

Man 😔

Is that a fun puzzle though ?

[frigid otter]

I could still change it if you have any idea of what I could've done wrong haha

[green trench]

Uhhh u seeeeee

I have not had any experiments with matrices

Nor vectors

[frigid otter]

I just used the subspace test, which is usually relatively easy? and I think it just worked out

although I wasn't really sure how the real numbers condition played into it

[green trench]

I like 3+5 = 8

Do u ever get to do that?

[frigid otter]

that's more my engineering homework 😉

haha

[green trench]

:o

[half ice]

Yes that is a subspace

[frigid otter]

phew, that makes me feel better

[half ice]

But I assume it's under multiplication? I'm not sure

[green trench]

Why is it called a sub space?

[half ice]

No, addition nvm

[frigid otter]

I found it to be closed under addition and multiplication?

that's why it's a subspace right

[half ice]

Vector addition and scalar multiplication

[frigid otter]

scalar* multiplication that is

yes

[half ice]

It also has to have the zero vector, which it does

[frigid otter]

ah crap, I forgot to include proof of that axiom

hopefully he doesn't notice

[half ice]

[green trench]

Omg are u implying HE is not bright?

[frigid otter]

Hopefully he thinks it's soooooooo obvious that the zero vector is in the set that he won't dock me for it

[green trench]

Wait I have 1 question

We’re talkin bout’ the same he rite?

Oh

Predictable

Ofc I knew that

@frigid otter wish me luck ;)

[frigid otter]

I don't know what you're doing, but best of luck!!!

[green trench]

Ty

[lunar viper]

if a system is inconsistent can it still be linearly independent?

[wintry sphinx]

no

[lunar viper]

@wintry sphinx cool thanks

[half ice]

@lunar viper
Yes it can. Consider,
[ 1 | 0]
[0 | 1]
[0 | 0]
No solution there, since the second equation is 0 = 1. But, (1,0,0) alone is a linearly independent set of vectors.

[gritty kelp]

In the statement det(AB), is it the same whether I multiply A and B and then take the determinant of the result versus if I multipy the determinants of A and B?

[wintry sphinx]

I meant that the rows of the un-augmented matrix have to be linearly independent

[soft burrow]

In the statement det(AB), is it the same whether I multiply A and B and then take the determinant of the result versus if I multipy the determinants of A and B?
@gritty kelp yes, that follows from the fact that det(AB)=det(A)det(B)

[gritty kelp]

ok thanks, I knew of that property, I just wanted to make sure that I understood it correctly

I want to state that $det(Q^TQ)=det(I)=det(Q^T)det(Q)=1$ where $Q$ is a square orthogonal matrix, and therefore $det(Q)=1$ or $det(Q)=-1$.

[stoic python]

Blu3_bear:

[gritty kelp]

can I say that is a true statement based on det(AB)=det(A)det(B)?

[soft burrow]

yep that's true for orthogonal matrices

[gritty kelp]

ok thanks

[light ledge]

Can someone help me out with how they got these lines graphed this suppose to be the right answer but I can’t figure out how they got there

[gritty kelp]

for just drawing the graph, its probably easiest to rearrange the equation into slope-intercept form (y=mx+b)

[limber sierra]

also as an FYI, this isnt linear algebra.

see #prealg-and-algebra , #precalculus , or one of the ten #questions-x channels

(talking to cyber dog, not blu3 bear)

[gritty kelp]

I mean, it is a system of equations that are linearly related, but yeah it would fit better in one of those other channels

[light ledge]

Ok thanks

[limber sierra]

it isn't for all matrices, but it is in some specific cases.

for example, the identity matrix commutes with everything (at least, every square matrix of the same size)

the matrix with 1s on the diagonal and 0s everywhere else.

[gritty kelp]

also a matrix of all 0's should commute with everything

[limber sierra]

well youre not allowed to use specific examples for this question

youre just reasoning off the fact that AB = BA and AC = CA

and you want to show A(BC) = (BC)A

if i'm understanding you correctly, yes

when things are equal, that means they're literally the same thing

just alternate "ways" to write it

so you can always "substitute" in equal things

since theyre literally the same

matrix multiplication associates; (AB)C = A(BC)

assuming multiplication makes sense (which it does in this case, since they're square)

and yes, you absolutely need to use that here

i'm not sure what you mean

without seeing the whole question, i dont know what they're asking of you

ah okay

i think it just wants you to give an example then, yeah

of specific values for A, B, and C

so just give an example of matrices A, B, and C such that AB = BA and AC = CA, but it is not true that BC = CB

[cunning arch]

omg i originally posted in the wrong channel. If T is R^4 -> R^3, should my proof even have the u_4, v_4,w_4,r_4?

i'm getting my dimensions mixed up

because the domain is in R^4, but i don't remember if I should be using the codomain R^3

[rose coral]

If your domain is R^4, and you are plugging in the vectors u, v, w, and r, each of them having 4 components seems right

[fringe matrix]

heyy

someone is here for quick question?

[nocturne jewel]

Just post the question

[fringe matrix]

lets say i have n vectors v1,...,vn.
if the dim(span({v1,...,vn}) = n it means that the matrix A with v1,...,vn as its rows doesnt have 0 as eigenvalue.
my question - lets say dim(span({v1,...,vn}) = n-p.
so p is the number of the multply of "0" as eigenvalue of A, am i right?

[native rampart]

Should be

[edgy kraken]

I work out the vectors but how do I compare the result

[fringe matrix]

make v1 and v2 orthogonal

[blissful vault]

if i have a position vector and two points how do i calculate the total distance traveled?

what i did was find t using the two points

and then i derived the velocity vector

and i integrated the velocity from t_0 to t_f

but the answer i'm getting isn't part of the multiple choice

[edgy kraken]

what would the solution to this be

i get [2 -8] + [1 7]

[strange crystal]

Could anyone explain this to me?

[gray dust]

unpack defn of basis & coordinates wrt a basis

given C={v1,v2,v3} is a basis of P_2, there exist unique scalars c1,c2,c3 where p=c1v1+c2v2+c3v3. we say (c1,c2,c3) is the coords of p wrt C

[dapper lance]

this is yes, right?

[warm falcon]

@dapper lance is $(1,0,0) + (1,0,0)$ a unit vector? $(1,0,0)$ is.

[stoic python]

Separable:

[dapper lance]

it'd be 2,0,0, which is just 2(1,0,0)

?

so yes?

[warm falcon]

How is $(2,0,0)$ is a unit vector? What is its length?

[stoic python]

Separable:

[dapper lance]

its sqrt(2)

[warm falcon]

So it's not a unit vector

[dapper lance]

so is S not a subspace of R3?

i dont understand

[warm falcon]

You added two unit vectors in S whose result ended up being a vector not in S.

What do you think?

[half ice]

In order to determine if something is a subspace, we do the subspace test

If you don't know this test, take a second to look in your book haha

[warm falcon]

Hit the book again, @dapper lance

[trail flare]

How to do this?

[wintry steppe]

@trail flare An invertible matrix A has an LU decomposition provided that all its leading sub-matrices have non-zero determinants.

[stoic python]

weilam06:

[grizzled folio]

this is yes, right?
@dapper lance The answer is yes, as $|\vec_{v}|$ is a unit vector in $\mathbb{R}^{3}$ space. This is very simple, I don't understand what this argument is about.

[stoic python]

Joaquin-Revello:
Compile Error! Click the reaction for details. (You may edit your message)

[warm falcon]

@grizzled folio : You sure about that? Try again.

One of the requirements for S to be a subspace is that it's closed under addition. Keep that in mind.

[grizzled folio]

Where does it say that?

[warm falcon]

"Is S a subspace of R^3?" is in the question.

and judging from the context we're clearly talking about linear vector subspace, not any old topological subspace

so yes you need closure of the operation, or you're using the term "subspace" in a rather nonstandard manner.

[grizzled folio]

ok, my confusion then

[warm falcon]

but there are other reasons that S cannot be a subspace (does it contain the 0 vector?)

[coral ferry]

could someone help me understand why basis for kernel of a linear transformation T and the basis for the image of T is linearly independent

[half ice]

The word "basis" implies linearly independent

I assume you're asking why those two spaces are each linearly independent - saying that they're linearly independent from one another doesn't make sense

[coral ferry]

yea sorry

[half ice]

Unless T is a square matrix I suppose

[coral ferry]

if you have a set that is composed of the basis for the kernel and the basis for the image

why is that set linearly independent

i just can't seem to develop the intuition for taht

(trying to understand the rank nullity theorem btw)

[coral ferry]

aaaaaaaaaaaaand i've got it

[gritty kelp]

Can anyone help me with part b on this?

Obviously something about the mutual orthogonality of the rows of Q must give us some reason why this is true

but I cant pinpoint exactly what

[gray dust]

unpack the definition of Q

[gritty kelp]

do you mean the definition that $Q^TQ = I$?

[stoic python]

Blu3_bear:

[gray dust]

yes

[gritty kelp]

ok

so since Q is like one half of I, the effect it has on multiplication is going to be limited, but I dont understand how I can prove that the length of X doesnt change

I guess I could also say that the length of each vector that makes up Q is 1

right?

[gray dust]

try more algebra than reasoning. recall properties of transpose & norm

[gritty kelp]

so looking at the dot product definition of matrix multiplication, the first row of $Q\vec{x} = $ the first row of of $Q$ dotted with $vec{x}$, and since the length of each row of $Q$ is normal then the length of $\vec{x}$ is not going to change.

[stoic python]

Blu3_bear:

[gritty kelp]

when you multiply two vectors together is the length of the product equal to the products of the lengths of our starting vectors?

or is that just for normal vectors?

or is that just not true?

[gray dust]

multiply two vectors together is the length of the product equal to the products of the lengths of our starting vectors?
what

[gritty kelp]

wait Im dumb, multiplying two vectors returns a scalar

but if you had a matrix where each row was a vector with a length of two, and you multiplied a vector by it, would the resulting vector have a length twice of the vector you started with?

because I was thinking to answer my homework question that arguing that since the length of each vector in my matrix, Q, is 1 that the length of the vector produced by Qx is equal to ||x||*1

ok I tested it and I think that my claim is true

I only tested the case where I have a matrix with rows and columns that are orthogonal, but not normal. When I multiplied a vector with this matrix, the resulting vector had a length twice that of the original vector

and my matrix was comprised of vectors with lengths of 2

so I think my argument is valid

[gray dust]

if you had a matrix where each row was a vector with a length of two, and you multiplied a vector by it, would the resulting vector have a length twice of the vector you started with?
you should find examples to show this is false

but that's beside my point. this talk of Q's rows is long winded and doesn't show anything

if you want intuition on why Q preserves norms, det(Q)=1 or det(Q)=-1 which corresponds to Q's effect being a rotation or reflection which both preserve norms

but i don't think this is what b) wants. start at https://cdn.discordapp.com/attachments/540211747613704221/771947793576362034/213158691635986432.png & do algebra to get ||Qx||=||x||

[gritty kelp]

So what I have found is that if we multiply $Q^TQ$ by $\vec{x}^T$ and $\vec{x}$ so that we get $\vec{x}^TQ^TQ\vec{x}$ we can use the associative property to go two ways with this equation. We can evaluate $Q^TQ$ first, looking at $\vec{x}^T(Q^TQ)\vec{x}$, so that are are left with $\vec{x}^T\vec{x}$ which equals the length of $\vec{x}^2$, $||\vec{x}||^2$. We can also rearange the equation to get $\vec{x}^TQ^TQ\vec{x} = (Q\vec{x})^T(Q\vec{x})$ and that term is equal to the length of $Q\vec{x}$ squared, or $||Q\vec{x}||^2$. Then since both of terms $||Q\vec{x}||^2$ and $||\vec{x}||^2$ are equal to $\vec{x}^TQ^TQ\vec{x}$ we can say that $||Q\vec{x}||^2 = ||\vec{x}||^2$.

[stoic python]

Blu3_bear:

[gritty kelp]

discord caused me to have to put extra vertical lines to show the length, so it rendered the latex a bit weird

[gray dust]

other than not knowing the difference in expression vs equation this is right

now justify this gives ||Qx||=||x||

[wintry steppe]

Super important question if anyone wants to help me out.

Is it n-by-m matrix? Or is it m-by-n matrix? Clearly only one of these leads us to God's grace. But I'm not sure which it is.

[dusky epoch]

what context

[wintry steppe]

I was to assume the existence of a matrix of arbitrary size, capturing the rows and columns.

[dusky epoch]

the usual convention is m by n so long as neither of those variables is already taken

[fringe matrix]

lets say i have {v1,...,vn} normal vectors. what can i say about the matrix A which
A(i,j) = <vi,vj> ?

so A(i,i) = 1

but what can i know about the relation between dim{v1,..,vn}

to A property

[native rampart]

You mean orthonormal vectors?

[fringe matrix]

no

if they were orthnormal than A(i,j) = 0 for every i!=j and the dimension will be n

[elfin mist]

$E(\lambda,T) = \text{null}(T-\lambda I)$, where $T \in \mathcal{L}(V)$

[stoic python]

the-last-knight:

[elfin mist]

I'm not able to understand the last inequality, where the dimension of the direct sum is shown to be ≤ dimV

It isn't obvious to me

Could someone explain?

[native rampart]

Because the space is a subspace of V

So,Dim(Subspace of V) <= Dim(V)

[tacit sonnet]

Matrix Adjoint is just another name for the Hermitian operation?

[wintry steppe]

@tacit sonnet wdym by "matrix adjoint"?

[humble pumice]

Hey guys, I was having difficulty with this proof question from a practice assignment and I was wondering if any of you could explain it to me.

[tacit sonnet]

The adjoint of a square matrix A = [aij]n x n is defined as the transpose of the matrix [Aij]n x n, where Aij is the cofactor of the element aij. Adjoing of the matrix A is denoted by adj A.

And no the Hermitian operation isn't same as the adjoint... I messed up

[wintry steppe]

oh

@humble pumice

from "Linear Algebra Done Right"

[tacit sonnet]

That's a nice book though on linear algebra

[wintry steppe]

yes

[tacit sonnet]

It helps a lot with the basics though

[humble pumice]

can i find this book online?

[wintry steppe]

@humble pumice i think it isn't allowed making easier how to find this book for free on internet.. 🤔

[tacit sonnet]

can i find this book online?
you can but idts we can help you find it

[viscid kernel]

For question (a) I got this, can yall check

[humble pumice]

@wintry steppe what is the relationship between having a zero kernel and a linear mapping being isomorphic?

[viscid kernel]

@humble pumice
If the dimension of the kernel is 0. It means that only the zero vector gets mapped to the zerovector which is trivial cuz the zero vector stays fixed after a linear transformation. So the definition of the kernel is Ax = 0 If the vector x is the zero vector and if thats only way that the zero vector gets mapped to the zerovector then the mapping itself is isomporphic, bijective. The other way that vector x can be mapped on the zerovector if the determinant of A is 0. If thats the case the dimension of the kernel is at least one. During an isomorphich map A is invertible, but not when determinant is 0

@wintry steppe what if the image was a subspace of W itself ?

Yes I meant that, and I got my answer thanks

[humble pumice]

okay thanks all for the help on a side note :

if U is an element of M nxn (R) and matrix U satisfies the property U transpose = -U

why is it possible to say that U - I is invertible

like U + I is also invertible

[knotty blaze]

I've got 3 kernels (aka 3x3 matrices)
e.g. Kx=[1,0,-1; 2,0,-2; 1,0,-1]
Ky=[1,2,1; 0,0,0; -1,-2,-1]
Ks=[0,-1,0; -1,5,-1; 0,-1,0]
I'm doing convolutions with an image matrix to 'process images'
so [im]*Kx in other words
If I were to apply all the 3 kernels one at a time

[ (([im]*Kx)) * Ky ] * Ks

how can I get an equivalent matrix kernel K_equivalent

such that [ (([im]*Kx)) * Ky ] * Ks = [im]*K_equivalent

[knotty blaze]

~sorted

[green trench]

@knotty blaze hi

[jolly dome]

Is this the right format for the answer?

[limber sierra]

well, there are no variables called $x_1, x_2$; just variables $x, y$

[stoic python]

Namington:

[limber sierra]

also uh, i think something went wrong with your algebra

since those do not solve the given equations

[jolly dome]

oof

well could you help me out a bit?

[austere cedar]

I can only assume the answers are meant to be x = 89/13 and y = -14/13.

[olive atlas]

is it possible to have a 5x5 matrix with only 1 eigenvalue?

[pallid rampart]

The 0 matrix

@olive atlas

[cunning arch]

is it possible for the transformation matrix to not be onto, but have infinite range? i know for sure that my transformation T is not onto, but i need to "describe the range"

so i found my transformation matrix A, put it in RREF, then made a system out of it but that system has infinite solutions so fjdksl;af

[elfin mist]

So,Dim(Subspace of V) <= Dim(V)
@native rampart
Yeah, thank you!

[raven bluff]

Hi Everyone

I have a question in SVM derivation..
Why the distance between the hyperplane $w*x+b=0$ and data point (in vector form) $p$, $d = \frac{w * p + b}{||w||}$ can be simplified to $d = \frac{1}{||w||}$?

[stoic python]

kingslayer12333:

[viscid kernel]

Whats the purpose of transposing a matrix ?

[dusky epoch]

there is no singular "purpose"

but an example of where it comes in is that the dot product between two col vectors $\bd{u}$ and $\bd{v}$ can be written as $\bd{u}^T\bd{v}$

[stoic python]

Ann:

[dusky epoch]

and more generally, $\bd{u}^T$ can be interpreted as the image of a vector $\bd{u} \in \bR^n$ under a particular isomorphism from $\bR^n$ to its own dual space, making $\bd{u}^T$ a linear functional on $\bR^n$

[stoic python]

Ann:

[viscid kernel]

@dusky epoch hast the first thing something to do with duality ?

[dusky epoch]

dual space
yes

[wintry steppe]

Hi, can someone explain me this identity please ?

[native rampart]

Are you familiar with the fact that every matrix is similar to an upper triangular matrix?

Expand e^(tA),change the basis appropriately,(so that you get an upper triangular)and manipulate to get the result

@wintry steppe

[viscid kernel]

What do they mean with the multilinear property of the determinant ?

[wintry steppe]

ok I see the thing, thanks 🙂

What do they mean with the multilinear property of the determinant ?
@viscid kernel maybe that det(a*X) = a^ndet(X) for a real or complex ?

X a n*n matrix

[odd kite]

we'd think of the determinant as a multi linear function of the rows or columns i.e. if the columns of the matrix are $v_n$, then $ \mathrm{det}(a\mathbf{v}_1, ,\mathbf{v}_2,,\mathbf{v}_3,\dots ) = a \mathrm{det}(\mathbf{v}_1, ,\mathbf{v}_2,,\mathbf{v}_3,\dots ) $ and so on

[stoic python]

Timon:

[main moth]

What is the meaning of

[wintry sphinx]

I assume it means all polynomials with degree at most 8

[main moth]

Got it

So P_8 is a vector space then?

[wintry sphinx]

yeah it is

[main moth]

where are the vectors though?

sorry i'm a little new to this concept

[floral thistle]

???

@main moth

[main moth]

nvm

[limber sierra]

the vectors are polynomials of degree at most 8

not all vector spaces have their vectors arranged in nice "column" format

but i mean if you REALLY wanted to, you COULD think of, say, $5x^8 - 5x^6 + 3x^2 + 2x - 1$ as [\begin{bmatrix}5\0\-5\0\0\0\3\2\-1\end{bmatrix}]

[stoic python]

Namington:

[limber sierra]

this isnt really a convenient way to think of them in practice

but you can see how polynomials are kind of "structurally simpler" to the more "generic" vector spaces (like R^n or C^n) youre probably familiar with

@main moth

in practice, it doesnt matter "how we write" the elements of a vector space

just that it behaves the rules we want it to

polynomials behave the rules we want vectors to behave

so P_n is a vector space.

[main moth]

gotcha

thank you! 😄

@limber sierra

[jolly dome]

How would I write the inequalities?

[limber sierra]

this is not linear algebra; try #prealg-and-algebra , #precalculus , or a #questions-_ channel instead

[molten aspen]

DM me if you can solve this please

[limber sierra]

What is linear algebra?
Linear algebra is the field of mathematics that studies vector spaces and linear transformations. Topics included within linear algebra are:

• Vectors, vector spaces, subspaces
• Matrix multiplication, representations of systems, Gaussian elimination, row/column space, invertibility/diagonalization, other topics in matrix algebra
• Eigenstuff and the determinant
• Geometry of vector spaces (e.g. dot/cross products), inner product spaces, similar structures [note: functional analysis also fits in #advanced-analysis ]
• Linear transformations, especially within vector spaces
• Bases, linear combinations, change of basis, and coordinate systems

The above list is not exhaustive, but as a rule of thumb: if you're somewhat familiar with any of the above topics, you probably know whether your question belongs here or somewhere else. If you don't know what most/any of those things mean, perhaps another channel would be more appropriate.

Examples of questions that are NOT linear algebra:

"How do I solve this equation for x? 3x + 1 = 4"
"I have a table of values. How do I draw it on a graph?"
"What is the slope/y-intercept/etc. of this line?"
The above questions are all more suitable for #prealg-and-algebra or for one of the #help-_ channels in the "Available" section. We ask you to please check to make sure the channel is currently unoccupied (hasn't had any recent activity).

"How do I compute the gradient or curl of this multivariable function?"
"I need help with working on integration in higher dimensions."
The above questions are more related to the applications of linear algebra in multivariable calculus. If your question is very "calculus-ey" in scope, #multivariable-calculus is probably a better place. We do acknowledge that the lines can be blurry, however, so if it seems "on the edge", you can ask in either channel.

If you have any questions about where your question best fits, feel free to ask!

[limber sierra]

posted for pin reasons^

[if anyone has any revision suggestions, feel free to suggest - do note that i'm like, right on the character limit, so i'd need to cut some stuff to add more stuff]

[dusky epoch]

thank you so much for this

[wintry steppe]

DM me if you can solve this please
@molten aspen do u need help still or someone already responded to u?

[molten aspen]

I figured it out @wintry steppe thanks

[acoustic path]

Ys is a GOAT.

[ripe steppe]

hey, can someone send me a solution that can be solved by elimination and substitution? like a full solution w the answer thanks ( dm it to me btw, and use the variables as x,y,z

[stone drum]

You don't need to ping mods for that. It happens all the time. The fix is to either ask "What have you tried?" or not provide the solution.

Or both

[wintry steppe]

oh ok srry

[viscid kernel]

Why is a linear map also called a homomorphism ? I mean, doesnt homorphism has to do with groups ?

[subtle walrus]

homomorphisms are "structure preserving"

when it comes to maps between vector spaces, those are exactly the linear maps

[native rampart]

Vectors are groups with additional structure

[half ice]

Lots of things in math are "different but similar" and the names reflect that

A linear transformation is not a group homomorphism, but they're both types of "homomorphisms"

Or "morphisms" if you're short on time

[viscid kernel]

Hmm, its cuz vector mapping preserve the arithmetic and so does combined symmetries of a cube for example?

Am I right ?

[native rampart]

Wdym preserve arithmetic?

[viscid kernel]

f(x+y) = f(x) + f(y)

I meant that it preserves the arithmetic between functions ?

[subtle walrus]

it preserves the structure of a vector space

[viscid kernel]

So is a group a function as well ?

[subtle walrus]

that is vector addition and scalar multiplication

[native rampart]

Well,Groups can be seen as sets with a special function

[viscid kernel]

@subtle walrus I mean like if you have a cube, you rotate it 60 degrees and then 120 degrees, its the same as rotating 180degrees, I meant this

[subtle walrus]

i don't see how this is related

[viscid kernel]

If you see functions as actions, you put something in and it spits something out.

[half ice]

That's not an example of a group homomorphism, at least from what I can tell. Normally a group homomorphism is a function from one group to another.

[subtle walrus]

ok, but how is this related to linear maps or the word homomorphism

[viscid kernel]

@subtle walrus thats exactly my question. A linear map is called a homorphism. But in group theory you use homorphism as well. Why do they use the same term for two different things ?

[subtle walrus]

as i said

we use the word homomorphism a lot

for maps between objects that preserve structure

there are vector space homomorphisms, i.e. linear maps

there are group homomorphisms

there are ring homomorphisms

[viscid kernel]

Whats the connections between those ?

[subtle walrus]

there are homomorphisms of topological spaces

they preserve the structure of the objects we want to study

[viscid kernel]

Ahhhh

[subtle walrus]

when you study vector spaces you only care about vector addition and scalar multiplication

and i.e. not about the names/labels of vectors or scalar

and homomorphisms are the mathematical way to make sense of that

[viscid kernel]

Finally, I understand it thanks 🙏

[tacit girder]

anyone can explain 3b
https://gyazo.com/52e146c527a877d66ebafb877394f46b

[sweet pike]

Can anyone confirm if I'm right in saying this is true:

Let A be an n x n matrix. If Ax = 0 for all x E R^2, then A is the all-zero matrix.

The proof for it is kind of a mess

Well, my proof

[native rampart]

You meant 2 x 2?

Or R^n?

[sweet pike]

r^n

Can be 2 x 2 if you make n = 2

[native rampart]

There's a less messy way of proving this

Just take A (1,0,0,...)(x=(1,0,0,...) ,I mean) that will give the first column of A

Which is zero according to your condition

Repeat for other columns

[sweet pike]

Ah perfect, thank you very much!

It looked a lot more complicated than it was

[native rampart]

What did you do?

[sweet pike]

I basically just made random matrices to see if I can find it is true

No real method

[rugged basalt]

z = [1, 2, x], u = [0, 2, 3], v = [1, 0, 5]

For what value of x can z be written as a linear combination of u and v.

How would I find this? Wouldn't there be infinitely many x-values that satisfy the equation?

[quartz compass]

I'm not really changing anything, but we could put the column vectors (1,0,0,...), (0,1,0,...) like you say @native rampart into a matrix side by side to get the identity matrix then we could write AI = 0 since we know each column in the result is 0, which then shows A=0. I just think it's slightly nicer to see this way @sweet pike

[tacit girder]

anyone can explain 3b
https://gyazo.com/52e146c527a877d66ebafb877394f46b

[quartz compass]

@rugged basalt it might help to write z = su+tv for scalars s and t, and then you can equate all the coefficients

@tacit girder if you want to project onto the span of a vector you should imagine finding the component of a vector in the direction of it, it might be easier to think in terms of the dot product to work it out by some trigonometry

[warm harbor]

hi guys super quick question: how do you read this notation?

is it "v is an element of the span {u, w}"?

[quartz compass]

v in the span of u and w

yeah what you said is good

[warm harbor]

v in the span of u and w
@quartz compass as in v is in the span that is made by the 2 vectors u and w right?

oh ok ok

[quartz compass]

yeah

[warm harbor]

ty @quartz compass !!

[quartz compass]

you're welcome

[rugged basalt]

Gotcha

Is this vector in reduced-row echelon form?

[spice storm]

yea

[rugged basalt]

ok thx

[spice storm]

I mean you can try making the 3, 1 or zero

but that's good

[rich dawn]

How exactly do I go about doing this?

[quartz compass]

I'd expand along the first row, then expand along the second row

[rich dawn]

if i just expand along the first row

wouldnt that end up giving me the determinant anyway

[quartz compass]

sure

but you want to keep an eye on the dets of A,B, and C

[rich dawn]

i end up getting a1a4(detC)-a2a3(detC)

[quartz compass]

you're almost done

factor out detC

[rich dawn]

right

makes perfect sense

thanks

[quartz compass]

yup

you're welcome

there might be some other way to do this

like for putting general nxn matrices in a 2n x 2n matrix

at least, if you're interested in trying to generalize this

I think there's a generalization that's something like, if you have a lower triangular matrix but made up of square matrix blocks, then the determinant is the product of the determinants of the matrices on the diagonal

[rich dawn]

hmm

but we dont exaclty have a lower triangular in this one right

so would it work?

[quartz compass]

it's lower triangular in the sense that the blocks of matrices are

[rich dawn]

aho kay

okay

[quartz compass]

$\begin{pmatrix} A & 0 \ B & C\end{pmatrix}$

yeah like this

[stoic python]

Merosity:

[rich dawn]

good to know

[quartz compass]

woah I just was guessing

I don't know if in general that's true, but I think I've seen it before lol

[rich dawn]

oh haha

[wintry sphinx]

@dim venture A is a projection, so you could conceivably search up the result online, but if you can show that the intersection of Null(A) and Col(A) is just the zero vector, you're done

since you can just use the rank-nullity theorem after that

[hidden ember]

why is it that a n*n matrix with linearly dependent columns, cannot have an inverse

[limber sierra]

theres a lot of different ways to explain this

[hidden ember]

try all

i am quite confused

[limber sierra]

are you familiar with the determinant? thats the fastest way to justify this

[hidden ember]

just uh, i have not studied ranks

yeah i am familiar with determinants

[limber sierra]

if so, then note that row operations don't affect whether a determinant is zero or nonzero

and if a square matrix has linearly dependent columns, when you row reduce it, you will "zero out" one of those columns

[hidden ember]

yes they dont, unless you change the position of two rows

[limber sierra]

and we know the determinant of a triangular matrix is the product of its diagonal

[hidden ember]

yes

[limber sierra]

which will be 0 since one of its columns is 0

hence the determinant of the matrix will be 0

and so it wont be invertible

if you want a more "direct" and less determinant-focused approach

[hidden ember]

oh so that means multiplying the matrix with a 0 column if it had an inverse would result in a 0, instead of an I?

[limber sierra]

it's just a fact of determinants that, if your determinant is 0, your matrix cannot be invertible

this is because $\det(AB) = \det(A)\det(B)$

[quartz compass]

way I think of it is if it has linearly dependent columns then it can transform more than one vector to the same point, and there's no way to undo this

[stoic python]

Namington:

[limber sierra]

so if $\det(A) = 0$, then $\det(AB) = 0 \cdot \det(B) = 0$, so there cannot be a $B$ such that $AB = I$ the identity (since the identity has determinant $1$ but $\det(AB)$ is always $0$)

here's another approach which i think is more direct/fundamental:

[stoic python]

Namington:

[hidden ember]

linearly dependent means that, a column can be expressed as a scalar multiple or sum of two other columns i.e a linear combination right?

[limber sierra]

let's say a matrix $A$ has linearly dependent columns; that means that, if $c_1, c_2, \dots, c_n$ are the column vectors of $A$, there exist not-all-zero scalars such that $a_1c_1 + a_2c_2 + \dots + a_nc_n = 0$.

then we can write $Av = 0$ for $v = \begin{bmatrix}a_1\a_2\\vdots\a_n\end{bmatrix}$.

But then multiplying by a hypothetical inverse $A^{-1}$ on the left, we get $A^{-1}Av = A^{-1}0$, and of course $A^{-1}A = I$, so $Iv = A^{-1}0$, hence $v = 0$. But this is a contradiction, since we said that not all $a_k$ are zero, yet $v$ is made up of only $a_1, a_2, \dots a_n$.

[stoic python]

Namington:

[limber sierra]

(if you're not sure why we can write Av = 0, just recall how matrix multiplication works; the first column of A gets assigned to the variable a_1, the second column of A gets assigned to the variable a_2, and so on... so we eventually have the equation a_1c_1 + ... + a_nc_n which we said is 0)

[hidden ember]

give me a sec

let me just go through what you wrote

wow okay that makes so much sense

thank you!!

[ionic lynx]

A matrix is linearly independent if it's non singular, right?

[limber sierra]

a square matrix with linearly independent rows/columns is nonsingular, yes.

[ionic lynx]

Yes thanks. I indeed forgot to add the part that it should be n*n

[limber sierra]

oh i read your question backwards but

same idea holds

a nonsingular matrix is linearly independent (i.e. has lin. ind. rows or columns) and square

[ionic lynx]

I knew that there are four items defining a (non) singular matrix, I just forgot about the independence

[queen gate]

Hi, little question about series.
One of my friend say that 0 is important in a serie... but I don't understand what ? Can you explain that to me plz ?

[limber sierra]

that is incredibly vague

maybe they mean that $\sum_{n=1}^{\infty}s_n$ converging to a finite value implies $s_n$ converges to $0$?

[stoic python]

Namington:

[limber sierra]

but without more information, it's impossible to tell what is meant by that statement.

[round coral]

Can I ask a question?

[queen gate]

If you want the context, he say that 1 + 2 + 3 + 4 + 5 + ... can be proof by the grandi, and alternance of entiers series... but I say a lot of arguments like linearity, regularity, and stability cant be used in the same serie... and finally I say : if we use 0 + 1 + 2 + 3 who is the same sum, it didn't work with these two series

and he say that 0 + 1 + 2 + 3 + ... + inf different of 1 + 2 + 3 + ... + inf

but he don't know why

@limber sierra

[round coral]

Could anyone please give me an example of a linear operator Q from a vector space V to V, such that Im(Q) = Ker (Q) and Q^2 =0 ?

[quartz compass]

map the first half of the components to 0, map the second half to the first half

[limber sierra]

uh

okay so its worth noting that series like 1+2+3+... dont actually converge

when we introduce alternate summation methods

its more in the vein of "if these converged, what would they converge to?"

there are a lot of fairly sophisticated analytical reasons we choose a given representation but

in general, adding 0 shouldnt affect these alternate series

[queen gate]

ok

so in this super sum, there is no big difference between 0 + 1 + 2 ... and 1 + 2 + ... ?

[round coral]

@quartz compass You mean like for eg. I map (x,y,z,w) -> (z,w,0,0) ? am I right

[quartz compass]

yeah

derivative on basis {1,x} also works as a kind of silly example lol

[round coral]

but the kernel is not equal to image in that case

[limber sierra]

what's the kernel? what's the image?

of this map

[round coral]

@limber sierra you have to give an example for that, to give an example of Q which satisfy the given conditions

[limber sierra]

what

im asking you

take the map merosity defined

(x, y, z, w) -> (z, w, 0, 0)

whats its kernel? whats its image?

you'll notice that ||they're the same.||

[round coral]

Ah!! I am sorry I was so dumb

I get it

@quartz compass Thank you very much

@limber sierra Thanks a lot

[thorn lichen]

so this is just 4 and 8 right

[gray dust]

yes

[thorn lichen]

p=4 and q=8?

[thorn lichen]

bump

[rich dawn]

Is this possible? https://i.imgur.com/EE2cQar.png

[wintry steppe]

yes

[rich dawn]

how so?

[wintry steppe]

do you know what it means for (2,4) to have B-coordinates (1,1) ?

[rich dawn]

Yes

the x vector (2,4) has the B-coordinates (1,1) relative to the basis B

I am trying to find the Basis B

my question is, can I find the basis given only x and the b-coordinate of x

[calm hamlet]

Yes

You have the canonical basis, x in the canonical basis, x in basis B and you search basis B

[hidden ember]

how can i say if this statement is true or false?
If det(A) = det(B) then det(A − B) = 0.

how do i go about justifying my answer

[slow scroll]

you would find a counter example: matrices A and B such that det(A) = det(B) but det(A-B) != 0

[hidden ember]

oh and how do i get that example

should i start by considering any arbitrary 2*2 matrix?

[floral thistle]

Thinking about it....

Well, the first think is that those matrices must have equal size, right?

[hidden ember]

yeah we are talking about n*n matrices

[floral thistle]

Do it with 2x2 matrices

@hidden ember Do it with 2x2matrices. A has elements [a b, c d]

B has elements [d c, b a]

Work from there

[hidden ember]

oh so is it the best practice to work around with variables for such questions?

or should i take some real numbers

[floral thistle]

Variables, always

Variables are general and do not rely on particular cases

[hidden ember]

that makes sense

lemme try

I am getting something like d^2 + c^2 - (a^2 + b^2) as the determinant @floral thistle

[floral thistle]

Yup

[hidden ember]

so what does this mean

how do i infer these numbers add up to 0 or not

[floral thistle]

Let a= 2, b= 1, c=4, d=5

Replace that in your result

[hidden ember]

well that is not equal to 0

[floral thistle]

Then the statement is not true in general

[hidden ember]

yeah that makes sense

[floral thistle]

It may be true in certain cases but not for every case

[hidden ember]

If A and B are symmetric, then the matrix AB is also symmetric.
Just wondering if i am thinking it correct, so AB = A'B' = (BA)'hence it is false?

It may be true in certain cases but not for every case
totally got it, thanks

[floral thistle]

If A and B are symmetric, then the matrix AB is also symmetric.
Just wondering if i am thinking it correct, so AB = A'B' = (BA)'hence it is false?
@hidden ember Can you phrase your proof differently?

[hidden ember]

symmetric means thats A' = A

so I took the transpose of AB

which is B'A'

OH i see i made a mistake there

but then A and B are symmetric too, so uh B'A' = BA

idk how to do it lol

[floral thistle]

You did it

If AB is symmetric, what is its transpose?

[hidden ember]

B'A'

[floral thistle]

No

It's AB

[hidden ember]

hm

[stoic python]

Max Hetfield:

[hidden ember]

(AB)' = B'A'?

ohhh

sorry

[floral thistle]

Dude or dudette

Calm down, please.

[hidden ember]

i mixed up two things lol

[floral thistle]

You're trying to arrive at the answers with desperation

[hidden ember]

i feel so

youre right

[floral thistle]

If $AB$ si symmetric, then $(AB)^T = AB$

[stoic python]

Max Hetfield:

[floral thistle]

Now try to see if that equality holds

[hidden ember]

it does not hold

because I got BA when I took the transpose and replace the transposes with their equivalents

[floral thistle]

Algebraically...

You can do it, you have intuition.

[hidden ember]

no just wondering now, if what i said is correct

[floral thistle]

Once again, you're going for the answer

Let's do it together

[hidden ember]

okay 🥺

[floral thistle]

What's $(AB)^T$ equal to?

[stoic python]

Max Hetfield:

[hidden ember]

It is B'A'

[floral thistle]

But B is symmetric

Then $B^T = B$

[stoic python]

Max Hetfield:

[hidden ember]

and $(A)^T = A$

[floral thistle]

What do we end up with?

[stoic python]

Nobita:

[hidden ember]

we end up with BA

[floral thistle]

Look at the hyphotesis we had at the beginning, the equality

Does it hold?

[hidden ember]

If $AB$ is symmetric, then $(AB)^T = AB$

[floral thistle]

That one

Does it hold?

[stoic python]

Nobita:

[hidden ember]

this one right?^

it does not hold

[floral thistle]

it does not hold
@hidden ember Exactly. Because $AB \neq BA$

[hidden ember]

ohhh

thanks a lot!!

[floral thistle]

You're welcome!

[stoic python]

Max Hetfield:

[hidden ember]

can you check if i am thinking it right for this one?
If A and B are skew-symmetric, then the matrix AT + B is also skew-symmetric.
for A' + B to be skew symmetric (A' + B)' = -(A' + B) and we get -A' - B which is -(A' + B)

[floral thistle]

What's T?

[hidden ember]

T is transpose

[floral thistle]

Oh

Yes, that's a hypothesis

Now you have to verify it

[hidden ember]

i just proved it right

for A' + B to be skew symmetric (A' + B)' = -(A' + B) and we get -A' - B which is -(A' + B)

[floral thistle]

No you haven't

You said what should happen if the statement is true

[hidden ember]

No I took the transpose of (A' + B)

I just said if it were true

my bad, i should have phrased this in a better way

[floral thistle]

This is your hypothesis

for A' + B to be skew symmetric (A' + B)' = -(A' + B)

Now work the equality from left to right, and see if it holds

[hidden ember]

it holds

[floral thistle]

Algebraically...

[hidden ember]

just the way we did it before

it holds

[floral thistle]

Dude/Dudette, do it algebraically

[hidden ember]

$((A)^T + B)^T = A + B^T = -A^T -B$ since A and B are skew symmetric

[floral thistle]

Then it holds

[stoic python]

Nobita:

[hidden ember]

yeah

[rugged basalt]

could someone help me out in vc?

[floral thistle]

What's the problem?

@dim venture For a and b, read this https://math.stackexchange.com/questions/3018066/prove-that-if-a-is-a-real-skew-symmetric-matrix-then-i-a-is-invertible-where

[hidden ember]

Question:Suppose that det(A) = −2. Let B be another 3 × 3 matrix (not given here) with det(B) = 3. Find the determinant of each of the following matrices.

To find the determinant of D = 4(A^−1)(B') i did something like det(D) = 4* det(A^-1)det(B') since determinant of the transpose is the same as the determinant of the original matrix det(B') = 3 and det(A^-1) = 1/|A| = 1/-2; where n is the number of rows/columns
so 4 * 1/-2 * 3 = -6

but i am getting my answer wrong for some reason

[floral thistle]

I don't think you can distribute a determinant like that

[hidden ember]

det(AB) = det(A)*det(B)?

[floral thistle]

Oh, didn't know that property

[hidden ember]

so where did i go wrong

[latent ledge]

det(cA)=c^ndet(A)

@hidden ember

[latent ledge]

I suppose n here is 3

[humble pumice]

could someone please explain how i could solve this question

[hidden ember]

but there is no c anywhere @latent ledge

[latent ledge]

c is a scaler

which is 4 in your D

[hidden ember]

yeah i know but i cant see it any scalar coming out?

oh

hm

i dont think i get it

there is no det(cA)=c^ndet(A) form anywhere?

[latent ledge]

https://en.wikipedia.org/wiki/Determinant#Properties_of_the_determinant

[hidden ember]

oh

i see

so it should 4^3 * 1/-2 * 3

[wintry steppe]

how do i do this

[latent ledge]

https://en.wikipedia.org/wiki/Rotations_and_reflections_in_two_dimensions this is the hint I can give you for p5, the first one, you need to read your note

[native rampart]

Memes belong in #chill

[wintry sphinx]

who knows he might be asking about the validity of the proof; it would go in #real-complex-analysis then

[simple hornet]

hello i have a question

how do we even know what f(x) + g(x) is here?

[dusky epoch]

F is a field

for every x in S, f(x) and g(x) are elements of F

[simple hornet]

Do we use whatever definition of addition we have on F?

[dusky epoch]

yes exactly

[simple hornet]

ah i see thank you

[dusky epoch]

• here denotes the addition F comes with

[simple hornet]

yus thank you :)

also nice trans flag

im genderfluid lol

nice to see other lgbtq+ people in math

also the product $\lambda f(x)$ is defined by the multiplication on F, right?

[stoic python]

funnynumberatyahoo.com:

[dusky epoch]

yes

[simple hornet]

ty

[ember hatch]

I'm kinda fucking retarded

[simple hornet]

Remember that average speed is $\frac{\Delta d}{\Delta t}$

[stoic python]

mirzathecutiepie:

[simple hornet]

so we need the total distance divided by the total time

since dick makes the trip twice, and the distance is 120 miles, we know that the total distance must therefore be 240 miles

Now we get to the tricky part, we need to know the total time he spent going to detroit and back to lansing

$t_{total} = t_{going there} + t_{coming back}$

[stoic python]

mirzathecutiepie:

[simple hornet]

So now we need to remember that $v = \frac{d}{t}$, and therefore we get that $t = \frac{d}{v}$

[stoic python]

mirzathecutiepie:

[simple hornet]

Now we calculate $t_{going there}$ by saying $t_{going there} = \frac{120 \text{miles}}{65 \text{mph}}$

[stoic python]

mirzathecutiepie:

[simple hornet]

we do a similar thing to calculate $t_{coming back}$, then all we have to do is add them both together to get the total time. The answer follows from there :) @ember hatch

[stoic python]

mirzathecutiepie:

[viscid kernel]

How can I show that my output vectors are linearly independend ?

[old flame]

Here is my attempt. Suppose V is a complex vector space and $T \in L(V)$. By theorem, $T$ has an upper triangular matrix with respect to a basis of $V$, let the basis be $(v_1,...,v_{\dim V})$. Then by proposition, $\span(v_1,...,v_{\dim V})$ is invariant under $T$ for each $k=1,...,n$. So $T_{v_k}$ is an operator for each $k=1,...,\dim V$. Hence, each subspace of dimension $k$, where $1 \leq k \leq \dim V$ is invariant under $T$. Thus, $T$ has an invariant subspace for each dimension $k$, for $k=1,...,\dim V$.

[stoic python]

Otoro:
Compile Error! Click the reaction for details. (You may edit your message)

[old flame]

<@&286206848099549185>

[ripe steppe]

<@&286206848099549185> Does anyone have a solution from like ur school book, that has the answers? the equation is that it should be solved by elimination and substitution? btw 2 variables

[torn zealot]

<@&286206848099549185>

I have this exercise for my linear class, i'm completely lost, the resistors of the circuit are all 10 Ohm, we are asked to find all the Voltages of the circuit.
I dont know how to turn this problem in a linear system.

[native rampart]

Do you have the kirchoff equations?

[torn zealot]

yup

[native rampart]

Just convert that system of linear equations into a matrix form

[old flame]

<@&286206848099549185>

[trail dirge]

take three mesh and use kirchoof's voltage and current law

[tropic garnet]

not sure if this question is for here or else where, "The final value of the impulse response of the system described by the transfer function", would this be when you make the s tend to 0 or infinite?

[wintry sphinx]

the shorter the impulse, the higher frequencies it contains

[tropic garnet]

ah so it would be when it tends to infinite?

[hollow finch]

Is that linear or diff eq?

[tropic garnet]

well the transfer function given is like

in this example, s+1/s^2+3s

i assume it would be infinite since that would make sense actually

cuz u would get 0, if it was 0 it wouldnt make sense

[torn zealot]

@native rampart i cant, i suck at physics

im so confused

been looking this for an hour

[native rampart]

Are you an engineer?

Because why else would you get this in a la class?

[torn zealot]

yes

[native rampart]

Ask in the physics server

[torn zealot]

wheres that

[gray dust]

#old-network

[native rampart]

I don't know why you do engineering if you suck at physics

[torn zealot]

i wasnt being serious, im just confused but eh

[native rampart]

mb

[warm niche]

what does <.,.> notation mean, I swear I've seen it before

or is it just saying any vector

[dusky epoch]

< , > is the notation for inner product

[native rampart]

Any pair of vectors are allowed

[warm niche]

@dusky epoch right but the little dots

I remember there was notation describing stuff

ught

I forgot

oh

(V, +, .)

Oh wait, but dats a definition of a vector space

[dusky epoch]

the dots are just argument placeholders

[warm niche]

oh ok ye that makes more sense

I was trying to relate (V, +, .) to the above definition

Now I realize

ty

[wintry sphinx]

@tropic garnet idk what your question is; for realistic impulses, you basically lose information about the transfer function at high values of s. For an idealized impulse, the filtered value is just the inverse Laplace transform of the transfer function

[distant matrix]

<@&286206848099549185>

I need help on question 2 and 3

[wintry sphinx]

what part of it

also bruh wrong channel

[distant matrix]

Oh im new in this server

Waht channel am i supposed to be in

[wintry sphinx]

one of the questions channels or the #prealg-and-algebra #precalculus something like that

linear algebra is different from what you're doing

[muted holly]

LOL

[nocturne jewel]

This is why Ys made the pin

[knotty blaze]

how do I get the phase of the green vector

would it be 2*pi-arctan(0.9803/3.65)

[warm harbor]

Can someone give me some examples of applications of knowing vector spaces?

I am learning about them right now and know what it is, but I just can't picture the use vector space in real-life applications

[knotty blaze]

imagine you're trying to build a building

and it's windows are curved cause you want a cool building

[warm harbor]

yrd

yes*

[knotty blaze]

you won't be able to build this if vector spaces didn't exist

I mean the applications are endless to be honest

engineers would be doomed

[warm harbor]

hmmmm could you elaborate more on your example..?

of curved windows

[knotty blaze]

right, so vector spaces (along with all the good stuff they come with namely matrices, eigenvectors etc) allow architects and engineers to account for all of the forces that will be acting on the building so that it doesn't collapse

I'm sure if you look up vector space applications on google you'll find numerous examples with explanations

[warm harbor]

right, so vector spaces (along with all the good stuff they come with namely matrices, eigenvectors etc) allow architects and engineers to account for all of the forces that will be acting on the building so that it doesn't collapse
@knotty blaze i see ok makes sens now...

well yeah i did try to search it online but most of the examples are not really concrete

ty @knotty blaze !

[knotty blaze]

send help ;-; I'm looking at the phase of the green vector in rads and I'm unsure so I need confirmation

would it be

$ 2*\pi-\arctan(0.9803/3.65) $

[stoic python]

Blackout:

[knotty blaze]

where 0.9803 comes from 5.19-4.2097

[thorny hemlock]

i cant calculate the eigen value here? im not sure

[wintry sphinx]

@warm harbor signal processing is another; if you want to make a voice changer or if you want to filter out noise, you will frequently take things like the Fourier transform, and understanding the Fourier transform requires a knowledge of a vector space of functions

[stuck stratus]

Compute a matrix A with eigenvalues 1 and 4 and eigenvectors (3, 1)^T (corresponding to 1) and (2, 1)^T (corresponding to 4).

How do I do this?

[wintry sphinx]

A is 2x2

so you can write out the diagonalization of A

[stuck stratus]

thank you :)

This is for k =/= 2 right?

[stray granite]

anyone got a sec ?

Determine whether the following sets are bases of the indicated vector spaces.

[thorny hemlock]

found the eigen value

2

how to find eigen vector?, idk

unsure what to do after this

[vernal pebble]

@thorny hemlock plug the eigenvalue you found to find the set of eigenvectors

namely, you want to find the null set of (A - ℷI)

[stuck stratus]

can someone explain this?

[slow scroll]

b is a unit vector means that
<b,b> = ||b||^2 = 1.
so you can compute <b,b> and set it equal to 1 to find all possible values of c

[main moth]

Could somebody help explain to me the concept behind this question?

If a subspace is spanned by the given vectors, then would it only be the linearly independent columns? (like with the free variables removed)

[robust pond]

im in this class too, but my understanding is you pick up a dimension in span for every linearly independent column

so you have a good idea of the upper and lower bounds here

@main moth i dont really get what you mean concept

i dont think id say only the linearly independent columns

its more a question of directions of freedom

[main moth]

i went ahead and rrefed it

octave:2> rats(rref([[1,-1,-7,5];[0,1,4,3];[2,-1,-10,7]]))
ans =

          1          0         -3          0
          0          1          4          0
          0          0          0          1

[robust pond]

sorry i was gonna react rats

but too much work

[main moth]

haha

[robust pond]

i just like that word

[main moth]

i did run rats

😛

[robust pond]

yea so that pretty much answers the question?

[main moth]

hmm

[robust pond]

glad i was right

i think

[main moth]

well looks like we have three pivot columns

column 3 is a free variable

wait i'm blanking out rn

is this set linearly independent?

[robust pond]

how would you feel say

if you just saw columns 1 2 and 4

a 3x3 matrix with those columns

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

[main moth]

it's linearly independent

it's invertible

blah blah blah

[robust pond]

yea

[main moth]

inv matrix theorem

[robust pond]

whats the span

[main moth]

columns spams R3

[stoic python]

jan Niku:

[main moth]

right?

[robust pond]

so these columns span R^3

then look at the 3rd column

i mean this might not be the most intuitive way to see it

but to me, with 1 2 4 you have R^3

so whatever the 3rd column is, if it lives in R^3, you have it covered

[main moth]

yeah exactly

makes sense

[robust pond]

so yea, i think your intuition is right

sorry you were already there

lol

[main moth]

haha np

column 3 can be represented as a linear combination of the three pivot columns

so in conclusion though, the subspace spanned by the vectors is R3 though right?

[robust pond]

I believe so, yea

[main moth]

and the dimension of that subspace is 3

beecause duh we just showed it

but also because dim R^n = n right?

[robust pond]

I_3 gets you to R^3, and your span can't be larger than R^n

[main moth]

just making sure i have this down

lol

[robust pond]

yea

[main moth]

talking it through is helpful 😆

cool! thanks 😄

[robust pond]

[main moth]

🙂 is so unenthused

[robust pond]

yea

[main moth]

😄 is better

[robust pond]

its one of my favorite ones though

bc it looks uhh

[main moth]

😛

[robust pond]

idk

[main moth]

haha

true

[robust pond]

disingenuous

or something

like if someone gave you a cup of water and then went 🙂

you'd be like hmm sus water

[main moth]

🙂

😂

💯

[hollow finch]

would it be accurate to say that similar matrices perform the same transformation just on a different basis?

[native rampart]

Yes

[hollow finch]

thats bodacious

[plush lava]

i was wondering if anyone could see if i calculated something wrong with a cross product

https://www.symbolab.com/solver/vector-cross-product-calculator/\begin{pmatrix}\frac{400}{\sqrt{2}}%26\frac{-400}{2}%26\frac{-400}{2}\end{pmatrix}\times\begin{pmatrix}\frac{\left(2\cdot10^{-6}\right)}{\sqrt{3}}%26\frac{\left(2\cdot 10^{-6}\right)}{\sqrt{3}}%26\frac{\left(2\cdot 10^{-6}\right)}{\sqrt{3}}\end{pmatrix}

https://www.emathhelp.net/calculators/linear-algebra/cross-product-calculator/?ux=(400)%2F(+sqrt+(2))&uy=(-400)%2F(2)&uz=(-400)%2F(2)&vx=((2*+10^(-6)))%2F(+sqrt+(3))&vy=((2*10^(-6)))%2F(+sqrt+(3))&vz=((2*10^(-6)))%2F(+sqrt+(3))&steps=on

these two calculations should produce the same answer, but they dont

i was wondering if there was anything i did wrong?

the numbers should be the same

200\sqrt[2] = 400/sqrt{2}

and \sqrt{3}/1500000 = (2x10^-6)/\sqrt{3}

[hidden ember]

Can someone explain this problem to me

I dont get what they mean by p(-1) etc..

[dusky epoch]

P_3 is the space of polynomials of degree at most 3

p(-1) is the value of the polynomial p at x=-1

is there anything else in particular that you need explained @hidden ember

[sleek veldt]

really really simple question i think but whats B' here?

is it transpose maybe?

[dusky epoch]

yes

[sleek veldt]

thank you ^^

[hidden ember]

Thank you @dusky epoch

[acoustic path]

guys i need help with a really advanced linear equation

3x+7=4

[hidden ember]

is a span of vectors always linearly independent?

[native rampart]

No

Take Span{(1,0),(2,0)}

The span set is not linearly independent

[hidden ember]

but by taking a span we mean to say a set of vectors that can represent every element in a space?

Take Span{(1,0),(2,0)}
but this span would just represent a straight line?

[cunning arch]

I'm a bit lost on how to write the transformation matrix across a plane

he gave us a few hints but i'm not sure what to do with them LOL

[wintry steppe]

Why does a square homogeneous system only have a non trivial solution if the rank is smaller than n?

Oh wait

I think I know

For the system to have a non trivial solution we need to get a free variable, so the column space will be less than n?

[viscid kernel]

@wintry steppe the rank null theory says dim(im(A)) + dim(ker(A) = dim(V)
Consider a 3x3 matrix A. We now have to form an equation like this Ax = 0
The solution to this is the ker(A) if ker(A)=0 the solution is trivial, if not that means that the dim(ker(A)) is at least 1. If its 1 it means you have a whole line of solutions, whole line of vectors that get mapped to 0 and that is only possible if the columnvectors are linearly dependent, ....

If the dim(ker(A) = 1 then dim(Im(A)) = 2
So 1 + 2 = 3

1 free variable means thag dim(ker(A) = 1

[wintry steppe]

Thanks , I'll go and digest that :) @viscid kernel

[acoustic path]

Hey if we have an odd by even or even by odd matrix and use gaussian elimination are we guaranteed to get a free variable?

Assuming it has a solution and has a trivial result

Cause i cant simplify it any further

[tall thunder]

Can anyone lead me in the direction of how to solve this

[viscid kernel]

@acoustic path if you for example have a 2x3 matrix there is guaranteed a free variable because that means that ur 3 column vectors for example the standard basis vectors in 3D got mapped on the 2D dimensional plane, which means that ur original space got crushed into a lower dimensional space, so there is at least a full line of vectors that got mapped to the 0 vector, this means that the dim( ker(A)) = 1. Also it means that there are 3 vectors in 2d, but in 2d you can have max 2 vectors being linearly depended. If those three vectors after being mapped to 2d gets mapped on the same line then you can see that there happened another dimension crush, that means that there is a whole plane that got mapped onto the 0 vector. So dim(ker(A))

For 3x2 you dont always have a guaranteed free variable. Because you have 2 columnvectors in 3d which are linearly dependend which means there was no dimension reduction. If they happen to fall on the same line, then dim(ker(A)) = 1

[gray dust]

@tall thunder you got x in terms of the b's & the b's in terms of the c's so you can get x in terms of the c's

[cunning arch]

is it ok if I bump my question down?

[viscid kernel]

I mean yes

Sorry

I read that wrong

@cunning arch go ahead

[cunning arch]

haha it's ok, i got scared for a moment sorry:

I'm a bit lost on how to write the transformation matrix across a plane

[viscid kernel]

I dont really understand the questionw. Are the “ three a’s ” each a coordinate of one unitvector. Or does each of them represent a different unitvector ?

[acoustic path]

@viscid kernel is it correct to say then that an n-1xn matrix has at least one free variable and that we have n vectors in n-1 dimensions. And what do u mean that a whole plane got mapped to the 0 vector

[cunning arch]

i think each of them is a component of one unit vector, but i'm not 100% sure either

[viscid kernel]

@acoustic path

1. yes, indeed

2. when dealing with a matrix vector multiplication equation. Such as Ax = 0, you always think of as the columns being the vectors and the rows the coordinates, because thats how matrix multiplication is defined. There is a theorem that says dim(ker(A) + dim(Im(A)) = dim(V)

Suppose Ax =0.
Dim(V) = coordinates of x.
The kernel is the vectorspace that gets mapped to the zerovector ( also the solution ) , and thats the case when there happened a dimension crush. The image is the vectorspace the columvectors span. If you have your 3 standard basis vectors in 3D suppose they get mapped onto 2d that means that atleast one of them should be in the span of other two vectors. In the case where dim(Im(A) = 2 this means that that columnvectors span a 2d plane so one of them shouldnt not be included in the dimension, it also means that there is a whole line of vectors that got mapped to the 0 vector. If there happened to be a double dimensional crush then the three columnvectors should land on a line this means that there is a whole plane of vector that got mapped to the vector.

@cunning arch is that the whole question, cuz I have the feeling that there is missing something

[cunning arch]

here's the whole question, i finished everything else

[viscid kernel]

Ahhhh

[cunning arch]

but like, i know how to write the inverse matrix and multiply to verify i'm correct, so all i need to do is figure out how to get the transformation matrix

[viscid kernel]

Lemme see If I can figure it out