linear-algebra

[winter harbor]

Holy shit

[lavish jewel]

[stable kindle]

first part is trivial

second part i'm thinking:

M^3 = -|a|^2 M iff:

M^3 (b) = (-|a|^2 M) (b) for all b

[lavish jewel]

is the hat a cross prod?

[stable kindle]

yes

but just not really sure how to use the vector triple product at all here

[lavish jewel]

i can't see it either

[stable kindle]

<@&286206848099549185>

ooh

wait no

[glad acorn]

For question (b), is it a not equals to 0 and b not equals to minus 1

[stable kindle]

bruh

[stable kindle]

no, b must not be 0

why can't a be -1

just add u1 once

and you just have bu2

i don't see why a can't be -1

$M^3 = -|a|^2M$ \
iff $M^3b = -|a|^2Mb$ $ \forall b$\
iff $M^2(a\cross b) = -(a\cdot a)(a\cross b)$ $\forall b$

[stoic python]

Kaisheng21

[glad acorn]

Okay I understand now

[stable kindle]

k

wait

holy shit, does cross product distribute over matrix multiplication?

or the reverse idek

[lavish jewel]

don't think so

[stable kindle]

ok well i'm completely out of ideas

[hard drum]

It does for orthogonal matrices w det 1

I believe

[copper locust]

very noob question

$a \times b = |a||b| \sin\theta ^{n}$

[stoic python]

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

[copper locust]

but to get a vector normal to both a and b you just have to do a x b

so why is |a||b|sin(theta) there

[lavish jewel]

the cross product yields a vector perpendicular to the two original vectors and with length equal to the area of a parallelogram defined by the vectors a and b

those terms are there because they define the area of a parallelogram

[copper locust]

that makes a lot of sense

just 1 more question

the distance of a plane from the origin can be found by using d = k/|n| where r.n = k

but how do i derive d = k/|n|

i dont see where it comes from

[lavish jewel]

you can derive this from the equation of a plane

given a normal vector n and a point p0 on the plane, a vector p - p0 is also on the plane if it is perpendicular to the normal

this means $n \cdot (p - p_0) = 0$

[stoic python]

Edd

[lavish jewel]

distributed the dot product to get $n \cdot p = n \cdot p_0$

[stoic python]

Edd

[lavish jewel]

we can divide both sides by the length of n, so that the right hand side yields $$\frac{n \cdot p_0}{\norm{n}}$$

[stoic python]

Edd

[lavish jewel]

after that if a plane does not pass through the origin, p0 cannot be 0, and this quantity is nonzero

and geometrically, what it does is yield the amount of p0 in the direction of n, i.e. the scalar projecto of p0 on n

which is precisely a distance from the origin to the plane

it's also the shortest distance because n is normal to the plane

[copper locust]

ohh n . p_0 is k

i get it

[lavish jewel]

hmm k is n dot p0

so r is p0

in your notation

[copper locust]

wait

[lavish jewel]

oh yeah, you had it right, i misread, sorry

[copper locust]

oh ok

thank you very much

[lavish jewel]

aight

[vital pagoda]

how do i prove that R is a subspace of C?

[lavish jewel]

use the definition of a subspace

consider the set of complex numbers ${z \vert z = a + 0i, a \in \mathbb{R}}$

[stoic python]

Edd

[lavish jewel]

and show they satisfy the def of a subspace

[vital pagoda]

yes that's actually exactly how i did it, i just didn't know if it would've been enough

[lavish jewel]

if you have already shown C is a vector space, then it suffices to show this set is closed under addition and scalar mult

though you have to be careful with the scalar mult

the scalars can only come from R

[vital pagoda]

yes i was going to ask if my field is R or C in this case

oh ok

why?

if the field for the scalar is C then R is not a subspace, if the field is R then it is a subspace

[lavish jewel]

right

[native ore]

I think this is true right

So we say V is a subspace when vectors u, v in V and scalars k

u+v is an element of V

and ku is also an element of v for all scalars k

and the condition for a transformation to be linear is

for a transformation T: Rn -> Rm

let u,v be vectors in Rn and let a be a scalar

T is a linear transformation if

T(u+v) = T(u) = T(v)

T(au) = aT(u)

Or is this not enough information?

I was trying to say that T(V) always being a subspace implies that T must be closed under addition and multiplication but I dont think that implies it distributes over addition and multiplication

<@&286206848099549185> Can anyone give me some insight on this?

[torn stag]

@native ore In $\mathbb{R}$ we have $T(x) = x^3$ as a counterexample.

[stoic python]

IlIIllIIIlllIIIIllll

[native ore]

Oh true.

Ok, that makes a lot of sense actually.

Thank you @torn stag

You cant place powers on vectors though right?

A but you said in R

So I would define it T: R -> R

T(x) = x^3 is valid for a linear transformation?

[torn stag]

I gave a counterexample for $T \colon \mathbb{R} \to \mathbb{R}$.

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

I'm not sure about $\mathbb{R}^2 \to \mathbb{R}^2$ though.

[stoic python]

IlIIllIIIlllIIIIllll

[native ore]

Id imagine R2 -> R2

we could say T: R2->R2 defined by [x^3,y^3]

also achieves the same goal does it not?

[wispy horizon]

from rank nullity theorem

[wispy horizon]

if its linear then T(ax+by) = aT(x) + bT(y) holds.
Im pretty sure yours isnt a linear transform

[teal grotto]

no, x^3 is not linear. but it is a counter example from R to R. it maps all of R to all of R and it maps to trivial vector space to itself. on R^2 its a bit more difficult to come up with a c.e., if there is one

[sour cloud]

is it since its just 5x there is no x^2 and a constant you just put 0 ?

[leaden tide]

yup

they just wrote 5x as a quadratic to make the matching up of the coefficients more explicit

[sour cloud]

thanks

[gray dust]

https://i.gyazo.com/0fd1e03e1ef7b57213ede51f801d1675.png do the products

[sour cloud]

wdym ?

[gray dust]

each side is a matrix product

[sour cloud]

on the left side if i multiply i get [ a -b 0 d ]

[gray dust]

that's not how matrix products are defined

[sour cloud]

im confused :/

how is a negative and c negative ?

c is 0

[gray dust]

you must know how matrix products are defined. see https://www.mathsisfun.com/algebra/matrix-multiplying.html

[sour cloud]

got it thank you sm!

[winter harbor]

I will give you a hint, if $A = [I]^{\gamma}{\beta}$ for some ordered basis $\gamma, \beta$ of $\mathbb{F}^{n}$, then there exists $P,Q \in \text{GL}{n}(\mathbb{F})$ invertible matrices such that:
$$
A = P^{-1} I Q
$$
Now, suppose that $A$ is invertible, try applying elementary row and elementary collumn operations to $A$ to get a basis under which $A = [I]^{\gamma}_{\beta}$.

[stoic python]

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

[granite kraken]

I see the first half

Can you hint what kind of operations

[winter harbor]

You literally just apply gaussian elimination

I don't want to type it down because it would be a pain

https://gyazo.com/18701c74b77c4b1bb0fc3d2ef86006d0

You want to prove a version of this.

[granite kraken]

got it

I think my bottom half is right but not sure about top half

[wintry steppe]

let's make that visible

[granite kraken]

Thanks

[winter harbor]

Your first proof is not correct

We are saying that the map T is invertible

Not that AXB is invertible for some nxm matrix X

it doesn't even make sense because n is not necessarily qual to m

[granite kraken]

I see

Any hint

Is the 2nd half okay?

[winter harbor]

so that it doesn't make sense to ask AXB to be invertible.

[winter harbor]

It is.

[winter harbor]

Idk lol

I was thinking like

[torpid socket]

Hey I have a quick question so when I say modulo 11 how does it work in negative numbers does it go to -11 or 0?

[winter harbor]

since T is invertible

[torpid socket]

Like example
-12 modulo -1

In -11

[granite kraken]

what can we say

[torpid socket]

Does -12 and -1 have the same value ij modulo 11

[winter harbor]

We can take block matrices:
$$
L_{1}

\begin{bmatrix}
I_{n} & 0 \
0 & 0
\end{bmatrix}
$$
and
$$
L_{2}

\begin{bmatrix}
I_{m} & 0 \
0 & 0
\end{bmatrix}
$$
Which are both $n \times m$ and we can find $X_{1},X_{2} \in \mathcal{M}{n,m}(\mathbb{F})$ such that:
$$
L{1} = AX_{1}B
$$
and
$$
L_{2} = AX_{2} B
$$

[torpid socket]

Can anyone answer ;-;

@winter harbor I’m sorry to bother you but can you answer my question 😅

Its a fast one

[winter harbor]

damn

[torpid socket]

D:

[winter harbor]

Alright

[stoic python]

MISTERSYSTEM

[granite kraken]

No bro me please

Thank ssss

❤️❤️❤️❤️❤️

[winter harbor]

Maybe playing around with this can lead you to something @granite kraken but I didn't think this through.

[torpid socket]

Can you answer me now 😅

[winter harbor]

I need to eat lmao

[torpid socket]

Oh :(

[winter harbor]

Alright

[torpid socket]

Well enjoy

[winter harbor]

-12 mod 11 ?

Is that what you are asking?

[torpid socket]

Yeah yeah

[winter harbor]

ok so

[torpid socket]

Does negative numbers work the same when we say mod 11

[winter harbor]

yup

they do

[torpid socket]

Alright perfect so -12 mod 11 is -1 correct ?

[winter harbor]

and you have to apply euclidean division in the same way

[winter harbor]

It is

[torpid socket]

Alright amazing

[winter harbor]

also

[torpid socket]

thank you!

[winter harbor]

-1 + 11 = - 1 mod 11

so - 1 = 10 mod 11

[torpid socket]

Alright thank you!

[prisma sail]

Is this enf to show T is a linear transformation...

T(e1 + e2) = T(e1) + T(e2)
T(c*e1) = cT(e1)

[golden ermine]

Yes

[wintry steppe]

[empty grotto]

is anyone willing to help me understand what a change of basis matrix is
how to find it

[tight grail]

hey guys, so I get this problem generally

S and T are both just transformations which take some polynomial P

and output a new polynomial

S(P) = P`

T(P) = yP`

Where P is a polynomial in terms of y

for example P = y^2 + y + 1

I'm just a little confused on how I'm meant to write out a standard matrix for a polynomial linear transformation like this?

it says "with respect to the basis {1 to y^3}"

so I can understand the standard matrix for the linear transformation S

will by 4 by 4

but how do I write out these values?

is it that I just plug in?

S(1), S(y), S(y^2) and S(y^3)?

[winter harbor]

You need to find the coordinates of (S+T)(1),(S+T)(y), (S+T)(y^2) and (S+T)(y^3) with respect to {1,y,y^2,y^3}. For example, we have that:
(S+T)(1) = S(1)+T(1) = 0 + 0*y = 0, so
(S+T)(1) = 0*1 + 0*y + 0*y^2 + 0 * y^3

And the coordinates of (S+T)(1) wrt to the standard basis is (0,0,0,0). So that will be the first collumn of the matrix [S+T] wrt the standard basis.

For the second column, we have that
(S+T)(y) = S(y) + P(y) = 1 + y* 1 = 1*1 + 1 * y + 0 *y^2 + 0*y^3. Therefore, the coordinates of (S+T)(y) wrt the standard basis is (1,1,0,0) and that will be your second collumn.

[tight grail]

ah ok that makes sense

we just take the coefficients

[winter harbor]

Yup

[tight grail]

and write it out as the columns

of our matrix

could the question

potentially be rewritten to something like

"with respect to the basis {1, cos(y)}"

or do the y functions have to be

polynomials?

[nocturne jewel]

is cos(y) a polynomial?

[tight grail]

no

[nocturne jewel]

then it cant be a basis vector for a polynomial space

[tight grail]

thanks for that

[nocturne jewel]

(Whatever) spaces has a basis consisting of (Whatever)

[tight grail]

had me dying for a second if cos(y) could be in there

ah wait, so could the question, be written as trigonometric functions spaces

[torpid socket]

How can I determine value of alpha

[tight grail]

@torpid socket You want to write the equations in matrix form

[torpid socket]

I did

Should I do the same until I get alpha on one side only

[tight grail]

kind of

the way I would solve it

instead of row reducing straight away

[torpid socket]

?

[tight grail]

hold up sorry

just drawing it out

[torpid socket]

Oh its fine

Thanks for the help btw

[tight grail]

I would write it out like this first

matrix on the left is all of the coefficients of the equations

the one on the right is the solutions

and then you can multiply both sides

by the inverse of the matrix on the left

[torpid socket]

Oh I don’t think we learned that yet :D

[tight grail]

this is what I would do tbh, it just seems like the easiest way

cause then it means

you need to find out if that matrix can have an inverse, and calculate it

and then multiply the inverse, by the matrix on the right

and it pretty much just directly gives the answer

you can also do stuff like

calculate the determinant

of that matrix

which should tell you if the equation has any solutions

[torpid socket]

I see

[tight grail]

or what "a" has to be for it to have solutions

[torpid socket]

Wait can you explain the x1 x2 x3

What did u do there

[tight grail]

that's just factorizing out x1, x2, x3

in a matrix way

if you take just this part

[torpid socket]

Oh

[tight grail]

and you multiply the 2 matrices on the left together

you'll see that it comes out with the same equation

4 X1 + 5X2 + aX3

[torpid socket]

I see

Yeah we didn’t learn that yet :(

[tight grail]

It would probably be allowed cause augmented matrix form is just another way

of doing the same thing

tho tbh

I suck at row-reducing lol

even tho I shouldn't

[empty ibex]

I did this question like this

[torpid socket]

Oh

[torpid socket]

I will see what get

[tight grail]

that's alg

[empty ibex]

I'm not sure if my method is alright

can someone tell me if i did it correctly?

[tight grail]

sorry I'm not sure if I understand this fully, but what's a+b=0 from?

I'm not too sure how to approach this, but I believe that your final answer is correct

in that if v1, v2, and v3 are all linearly independent

then they must be be able to form a basis for the vector space V

I'm just not sure where you got a+b=0 from

ah nvm you just multiplied

av1 + bv2 + cv3 = 0

tbh the working seems fine, but I would've written a little more

before you started writing a + b = 0

I would've started with av1 + bv2 + cv3 = 0, with no coefficients being non-zero

[empty ibex]

ah sorry. thats from setting scalars for each row and equating them to 0

[tight grail]

yee

that's ok

[empty ibex]

arigatoo!

[tight grail]

the working seems good, just needs to be a tiny bit more clear

but otherwise good job dude

[empty ibex]

thanks1

[tight grail]

does anyone know how to generally find the basis of the orthogonal component to S?

I can get that the orthogonal component of S, will just be a vector space

[winter harbor]

Gram-Schmidt

This is an orthonormal basis for S

[tight grail]

damn so I just apply this algorithm

and it gives me the 2 vectors I need for the basis

[winter harbor]

Yeah

[tight grail]

why is it v2 = u2-v1 tho?

is that just how general formula is done

cause like, what happens

if they gave me 3 different vectors for the original span

S = span{u1, u2, u3}

[winter harbor]

Just search for Gram-Schmidt algorithm

It's a general process

[tight grail]

like this right?

[winter harbor]

Yeah lol

[tight grail]

lmao

that's an algs formula

[winter harbor]

Was having trouble typing it lmao.

[tight grail]

that's algs bro

thanks for the help on it

I'll pray I get something easy like least square and polynomial standard matrices

in a couple of months

[tight grail]

does anyone know how to show the coordinate map?

like if p = x

does that mean that [p] = (0, 1)?

and then to calculate the matrix A, I'm assuming

I just need to plug in a = 1, b = x

and then also b = 1, a = x

calculate (a, b) using the integral formula

and then calculate that separate equation at the bottom

and just match them up

to find matrix A?

[zinc timber]

basically you are given be basis {1, x}. Find the matrix of the inner product A

[half ice]

Show the coordinate map? It's just:
[3x + 5]= (5,3)

[zinc timber]

where your $A_{ij} = \int_{-1}^1 e_ie_j \dd x$

[stoic python]

Ryuzaki

[tight grail]

ah ok, just confirming a lot rn, I'm starting to learn that the final numbers

are all just about the coefficients really

and cause it's respect to basis {1, x}

p and q, just have to be some polynomials where
$p = a_1 + a_2x$

[stoic python]

ehff

[tight grail]

there has to be that constant a1

but if the question was different

and just asked for basis {x}

then we would only be looking in the perspective

of polynomials that are just

$p = a_1x$

[stoic python]

ehff

[tight grail]

something like that?

no extra constant

ye I think I get it

[small vigil]

Let $A$ be a positive semi-definite matrix (not be necessarily symmetric).
I would ask that $|(A+A^\top)x|_2^2=0$ implies $x=0$?

[stoic python]

keith_1

[ionic laurel]

Hi there all I have a question

If B is a basis of H consisting of three basis matrices, then if K is a subspace of H, would K share the same basis matrices as H?

[empty ibex]

can anyone confirm the answer for 3. b)?

i got [0,0,5; 1,-1,11, 0,0,2]

[lavish jewel]

looks ok

[teal grotto]

hint : what is the basis of a trivial vector space?

[ionic laurel]

0 vector

I think?

[teal grotto]

the empty set

[wintry steppe]

I have a question.

If A is diagonalizable then what does this imply about the dimensions of the eigenspaces of A?

I find eigenvalues as somewhat related to the nullspace but the connections are still unclear.

[granite kraken]

the book hints to use inner products to solve this but I don't see it immediately. Anyone see ITV

[limber sierra]

cauchy schwarz

[granite kraken]

how?

and what is that

what is Cauchy Schwartz

[lavish jewel]

recognize that the left hand side is of the same form as a dot product

[limber sierra]

oh, you havent seen cauchy schwarz before?

[granite kraken]

Maybe I have maybe we just haven't named it

Is it that famous inequality?

I still am not sure how to use it

[lavish jewel]

you might have seen it as a property of the dot product

have you seen dot products yet?

and how to use them to check if vectors are parallel, orthogonal, or if they make some other angle

[violet pecan]

to show that a set is a vector subspace of a vector space, do we just need to show that it's closed under addition and closed under scalar multiplication? or do we also need to show that the zero vector is in the set too

[lavish jewel]

the zero vector part is related to being closed under addition, since v - v needs to be in the set

[violet pecan]

thank you

[empty ibex]

can someone help me with 6 and 7? been trying to solve these for hours

[zinc timber]

what's assignment 5 question 7

[small vigil]

Let $A$ be a positive semi-definite matrix (not be necessarily symmetric).
I would ask that $|(A+A^\top)x|_2^2=0$ implies $x=0$?

[stoic python]

keith_1

[zinc timber]

I am assuming you have done the previous exercises. Given $\lambda$ is a eigen value of $A$, then for a polynomial $p(x) \ \in \mbb{C}[x]$ we know $p(A)$ eigen value $p(\lambda)$. So, as $A$ is diagonalizable, assume the eigen values are $\lambda_1, \lambda_2, \cdots, \lambda_{10}$. Given $A^2+A = J+2I$, LHS and RHS must have same eigen values. So, $\lambda_i^2+\lambda_i \quad(i=1, 2,\cdots 10)$ is a eigen value of $A^2+A$. But the eigen values are given 12 with mult 1 and 2 with mul 9. So $\lambda_1^2+\lambda_1 = 12$. This gives us $3$. now for other eigen values $\lambda^2+\lambda = 2$ means $\lambda = 1, -2$ now assume $1$ has multiplicity $k$, then $-2$ will have mult $10-k-1$. gien $\tr(A)=0$ we can solve for $k$.

[stoic python]

Ryuzaki

[zinc timber]

@empty ibex

[lavish jewel]

no, since they're only semi-definite

[zinc timber]

No, ex 0 matrix

[lavish jewel]

a simple counter example is to take an identity matrix and replace the last row with all zeros. or a 0 matrix, and ryu says

@small vigil

[zinc timber]

for the 7'th one you can use contradiction. Assume they are disjoint, then their direct sum has dim 10 but it is also subspace, so dim <= 9 is a contradiction.

[wintry steppe]

Is this related to linear independence again?

[zinc timber]

what do you think?

[wintry steppe]

The dim(A) is equal to the number of bases, now I know that the eigenspaces is similar to the null space so can I say that the basis of the kernel of the matrix is the basis for the eigenspace and hence are equal in dim(A)?

I'm not that familiar with diagonalization.

Diagonalization requires linear independence right?

[zinc timber]

diagonalizable means you have enough eigen vectors to construct a basis.

Ok first can you show that eigen vectors corresponding to distinct eigen value are LI?

[ionic laurel]

Not sure what you are implying with this sorry

[wintry steppe]

These eigenvectors must be LI then.

[zinc timber]

NO not necessarily

[wintry steppe]

I see, I see. So they are LI when the EV are distinct and nonzero.

[zinc timber]

what I meant was is if all the eigen values are different, then what can you say about the eigen space ( specifically the dimension of the E space)

There can be cases where A diagonalizable matrix has repeated eigen values, we count them by their multiplicity. example $\mqty[\imat{2}]$ has eigen values $1, 1$ so we say that 1 has multiplicity 2.

[stoic python]

Ryuzaki

[nocturne jewel]

algebraic multiplicity*

[zinc timber]

yeah didn't wanna confuse if further so I omitted it

[wintry steppe]

I guess I'll have to read on this first.

@zinc timber

I've got this idea now thanks.

If there exists a diagonalizable matrix $A_{n*n}$ then the sum of the dimensions of the eigenspaces of A must be equal to $n$?

[stoic python]

Elfenkaiser

[zinc timber]

but I think they asked a more specific answer

What I meant to say is that for diagonalizable matrix, the dim of the eigen space equals to the (algebraic) multiplicity of the eigen value

This is not generally try if A is not diagonalizable. Example $\mqty[1 & 1 \ 0 & 1 ]$

[stoic python]

Ryuzaki

[umbral trellis]

Hiya! I was wondering if anyone knows, if I have two vector solutions to a linear system of equations $\mqty[1 \ 0 \ -1 \ 1 ]$ and $\mqty[2 \ 3 \ 1 \ 0 ]$ how would I start to try to determine a formula to all the solutions in general?

[lavish jewel]

you can't find them all just from that

[stoic python]

foldie

[lavish jewel]

but you can guarantee that the difference of the two solutions is in the null space, so adding a scalar multiple of the difference is always a solution

you need the transformation itself to figure out of the null space has dimension 1 or more, though

[umbral trellis]

hmmm, ok. thanks very much for the help!

[sick sandal]

can someone explain to me the intuition behind why every hyperspace(dim n-1) of a finite vector space V is the null space of some non-zero linear functional on V
[like i get that f maps to a field of scalars and by rank nullity the null space has dim n-1 but i dont see how to prove each hyperspace is the null space of some f]

[winter harbor]

Let $U \subset V$ be a codimension $1$ subspace of $V$, a finite dimensional vector space of dimension $m$. Let $e_{1}, \cdots, e_{m-1}$ be a basis for $U$ and extend it to a basis $e_{1}, \cdots, e_{m}$ of $V$.
\
\
Let $f : V \rightarrow \mathbb{K}$ be such that $f(e_{i}) = 0, \forall i \in {1, \cdots, m-1}$ and $f(e_{m}) = 1$. Then, by linearity, $f$ extends to a unique rank $1$ linear functional on $V$ such that $\text{ker}(f) = U$.

[stoic python]

MISTERSYSTEM

[winter harbor]

You need to need the linear functional to have full rank btw.

So that there's a one to one correspondence between full rank linear functionals on V and codimension 1 subspaces of V.

[sick sandal]

i dont see how this applies to multiple linear functionals it seems like it proved each hyperspace is the null-space of a unique linear functional but cant we have multiple functionals such that ker(f)=U
maybe im missing something
in the theorem its stated as such
if f is a non-zero linear functional on the space V then the null-space f is a hyperspace in V. conversely ,every hyperspace in V is the null space of a (not unique)non-zero linear functional on V

[winter harbor]

It does not prove that each hyperspace is the null space of a UNIQUE linear functional tho.

I constructed a particular one

What is unique

Is the extension of f

So that f is well defined as a linear functional on the whole of V

[sick sandal]

ye i mean i get that
but i want to generalize that idea

[winter harbor]

For instance

K could be a field of characteristic 0

So that N embeds in V

And if you put f(e_m) = n

For some natural number n

And f(U) = 0

This gives you a new linear functional

That vanishes on U

For each natural number n

So it is definitely not unique

[winter harbor]

You don't need to, you need to prove that there exists at least one.

[sick sandal]

ohhh i see the issue fair enough

[winter harbor]

Yeah

What I am basically using

Is the fact that a linear functional is uniquely specified

By how it acts on a basis of V

And since e_1,...,e_m is a basis of V

If I choose what f(e_i) equals to for i in {1,...,m}

By linearity, this induces a unique linear functional on V

But I could specify how f acts on this basis

In a bunch of different ways

[winter harbor]

And this is where the (generally) non uniqueness of f comes from.

I want f(e_i) = 0 always for e_i in U

But f(e_m) could be anything

As long as it is non zero

[sick sandal]

aight i think i get the catch here ile write it down to fully grasp it
ty ty

[teal grotto]

i’m giving you a counter example. just let K be the zero subspace

[ionic laurel]

Well maybe I should be more specific

I suppose what im asking implies different things

Than the answer i am looking for

[teal grotto]

this is a counter example to your question. let H be the space of 2 by 2 real matrices spanned by B = {e1, e2, e3}. The trivial subspace K = {0} is a subspace of H which shares no basis vectors with H because it’s basis is the empty set

[median plover]

hey can anyone help me out with part (b) of this question

so this is the formula for shortest distance between two lines

given that P is some point on line 1 and Q is the perpendicular point on line 2

this is how far i got but im struggling a little because of the cosθ

i just don't see that going to the answer which is( √ 66 )/33

[median plover]

<@&286206848099549185>

[tired fossil]

Hey guys, does anyone know how to apporach this question?

[lavish jewel]

notice that only basic row operations are taking place

and these affect the determinant in very specific ways

[frail spruce]

I'm given 2 vectors { [cos(theta), sin(theta)] , [-sin(theta), cos(theta)] }. I'm trying to prove these vectors form an orthonormal basis for all values of theta but I don't even know how to find the length.

I know I have to do (cos(theta) * cos(theta)) + (sin(theta) * sin(theta)) = 1 but not sure how to show it for all values of theta.

What I do know are the requirements to show its orthonormal but I'm not sure how to proceed with this specific question.

oh wait im dumb

cos^2(theta) + sin^2(theta) = 1

[lavish jewel]

👍

[frail spruce]

but then -sin^2(theta) + cos^2(theta)

what am i missing here tho

[lavish jewel]

if you have [cos t, sin t] and [-sin t, cos t], the dot prod is -cos t sin t + cos t sin t

[frail spruce]

i dont think i follow

[lavish jewel]

what's the problem?

[frail spruce]

i don't recognize what ur trying to explain

[lavish jewel]

do you know how to take the dot product of 2 vectors?

[frail spruce]

is there a relationship between those trig identities

yes

[lavish jewel]

and do you know what it means when the dot prod of 2 vectors is 0?

[frail spruce]

yes it means it is orthogonal

[lavish jewel]

this adds up to 0

so the 2 vectors are orthogonal

[frail spruce]

i think this is from me not being familiar with how the trig identities cause it to add up to 0

[lavish jewel]

there is no need to use any trig identities

just look at it

it's of the form -x + x

[frail spruce]

ahh i just noticed when u said look at it

but then -sin^2(theta) + cos^2(theta) = 1?

and im not familiar with how its equal to 1

[lavish jewel]

no, that's not equal to 1

i guess you mean when taking the norm of [- sin t, cos t]

the proper way is $\sqrt{(-\sin(\theta))^2 + (\cos(\theta))^2}$

[stoic python]

Edd

[lavish jewel]

the negative goes away

[frail spruce]

this would be different from the 2nd given vector no?

how can u just square root it?

[lavish jewel]

what?

the square root is part of the definition of the length

[frail spruce]

i thought i have to show that (-sin(theta) * -sin(theta)) + (cos(theta) * cos(theta)) = 1

[lavish jewel]

well sure, and the square root of 1 is 1

[frail spruce]

but then why was this?

oh wait

[lavish jewel]

because you had a negative sign that was wrong

[frail spruce]

man im dumb

im so sorry 💀

[lavish jewel]

[frail spruce]

this was super helpful

haha 😢

[median ocean]

1 please

[limber sierra]

this is a definitional question

googling "orthonormal" should suffice

[median ocean]

so i just define what it means and how it was applied?

[thorn yacht]

Can anyone see why the last part implies that null T^{tilde} = 0 ?

[wintry steppe]

Just state what it means for each vector to be orthogonal to each other and normalized.

[median ocean]

ok i got it

goit it figured

whata bout number 2

[thorn yacht]

I've just looked at the book's online errata, and it's supposed to say $null \tilde{T} = {0}$, which makes more sense, but I don't see why $v + null T = 0 + null T$ should imply $v=0$.

[stoic python]

Kaishin

[winter harbor]

Well, they have proved that if $v \in \text{null}(\tilde{T}) \implies v \in \text{null}(T)$. This means that $v$ is equivalent to $0$ mod $\text{null} , T$, but the equivalence class of $0$ mod $\text{null} , T$ is precisely the additive identity (i.e, the zero vector) for the quotient vector space $V / (\text{null} , T)$.

[stoic python]

MISTERSYSTEM

[winter harbor]

It does not imply v = 0

for instance

T could be the zero map 0 : V -> W which sends every vector v to 0.

so that the kernel of T is precisely V

thus

for any non zero v

[nocturne jewel]

how are you defining modular arithmetic wrt a subspace...?

[winter harbor]

we have v + ker(T) = 0 + ker(T)

[winter harbor]

Haven't you seen the quotient of a vector space by a subspace before?

[nocturne jewel]

Nope

[winter harbor]

If you take any vector space $V$ over a field $\mathbb{K}$ and a subspace $W \subset V$, then we can define a vector space $V/W$ over $\mathbb{K}$ as follows. As a set, $V/W$ consists of equivalence classes of vectors in $v$ under the equivalence relation:
$$
v \sim u \iff v - u \in W
$$
We denote the equivalence class of a vector $v$ under this equivalence relation as $v+W$ so that for any $v + W$ and $u + W$ in $V/W$ we may define their sum as:
$$
(v+W)+(u+W) = (v+u) + W
$$
and for any $\alpha \in \mathbb{K}$ we define the scalar multiplication as:
$$
\alpha \cdot (v+W) = (\alpha v) + W
$$
We can verify that these operations are well defined and endow $V/W$ with the structure of a vector space over $\mathbb{K}$.

[nocturne jewel]

k, gonna google the quicker explanation

[winter harbor]

[teal grotto]

this is pretty quick...

there is also the issue of well defined-ness of addition and representatives and blah blah blah, but it works because of group theory stuff

[stoic python]

MISTERSYSTEM

[winter harbor]

Yeah, it's not hard to verify tho.

[nocturne jewel]

See, just saying group theory would have been enough, cause I haven't taken group theory yet.

[winter harbor]

It can be done as an exercise

[nocturne jewel]

Instead of assuming I know (X Y Z)

[winter harbor]

I mean, you don't need any group theory to understand quotient vector spaces

ofc there are related notions

for instance, in group theory we can quotient a group G by a normal subgroup H

and since V under addition is abelian

then any subspace W under addition is a normal subgroup of V

so that V/W under addition has the same group structure as V/W (viewed as a group quotient)

but do you need any of this?

No

The intuitive explanation

is that the quotient V/W

collapses W to to the 0 vector

[nocturne jewel]

so V/W is the 0 map

[winter harbor]

Not really, it's just that we are dealing with V modulo W, in the sense that we are treating two vectors of V as equal unless they differ by an element of W.

[nocturne jewel]

You just said V/W sends everything from W to 0

that's the 0 map

[winter harbor]

That's an intuitive idea

but we can make that precise

[nocturne jewel]

Right, so don't say not really, then yeah

[winter harbor]

we have a canonical projection map

$\pi : V \rightarrow V/W$ which sends every vector in $v$ in $V$ to its equivalence class $v+W$. this is actually a linear map, and turns out that its kernel is precisely the subspace $W$.

[stoic python]

MISTERSYSTEM

[nocturne jewel]

No clue what an equivalence class is.

[winter harbor]

The idea is that we are ''forgetting'' W and treating it as 0 under V/W

[nocturne jewel]

again, you assume I know things when you explain it

[winter harbor]

Oh 😦

Might take a look at a bit of set theory then

[nocturne jewel]

I know set theory

[teal grotto]

bruh how is he supposed to know what you dont know

[nocturne jewel]

ask

[thorn yacht]

Ah, yes, thank you MISTERSYSTEM, I just realized that I hadn't noticed that the definition of the quotient space makes to the affine subset of v and w equal if they give the same affine subset. Just stupid me considering them separate.

[nocturne jewel]

Like if you have a quintesential aspect in an explanation, you would know "ok, you need to have a rough idea of X Y Z in order to understand this explanation of W"

[winter harbor]

I am sorry

I didn't know tho

I hope at least I gave you a brief description of what it is

and what are the prerequisites to understand it

[teal grotto]

sorry they tried to help you but didnt do it in the specific manner you were hoping for?

like...?

[nocturne jewel]

If I explain intervals of increasing/decreasing to someone on here, I need to know if they know calculus or not, so I ask "are you in a calc class"

as an example of knowing your audience

[winter harbor]

It's just that I really have no idea how would I talk about quotient vector spaces without mentioning equivalence relations

They are defined in terms of an equivalence relation

[winter harbor]

Np

[nocturne jewel]

Yeah I know im being a bit irrational, and I apologize however I dont want you to waste your time explaining stuff just for it to be above someone's head

y'know?

[winter harbor]

Aight

In any case, I mentioned set theory because intro to set theory courses usually deal with equivalence classes.

[winter harbor]

You may take a look at Paul Halmos Naive Set Theory book for an explanation of that, or the first chapter of Munkres Topology (it talks about logic and set theory).

[wintry steppe]

[wintry steppe]

[teal grotto]

me when i thought there was a shorter proof of change of variables and didnt pay attention in class

[wintry steppe]

this was a very nice explanation mistersystem

[median ocean]

does anyone know 3

[winter harbor]

❤️

[wintry steppe]

[raven lotus]

All m x n matrix has determinant. true or false?

[teal grotto]

whats the determinant of [0 1]

[wintry steppe]

only square matrices do

[wintry steppe]

zero, obviously. it takes the unit hypercube in R^2 to something with zero two-dimensional volume, so the scaling factor is zero

unit cube?

[teal grotto]

?

[wintry steppe]

[teal grotto]

[wintry steppe]

im pulling your leg

trolling, perhaps

[teal grotto]

bruh i stopped reading after hypercube lol

im so tired

i just want to figure out how hash tables work

[wintry steppe]

if $T$ is a linear operator on $\bR^n$ then $T([0, 1]^n)$ has lebesgue measure $|\det T|$

[stoic python]

TTerra

[wintry steppe]

or something like that

("volume" |det T|)

[teal grotto]

im just starting to learn basics of lebesgue measure theory. i wonder how you show this tho

[wintry steppe]

change of variables theorem

[teal grotto]

aight imma head out

[stoic python]

TTerra

[wintry steppe]

We need a vector v that is orthogonal to every element in W. I am inexperienced but you have the basis for W. Thus let find a vector that is orthogonal to each w_i. So you have 3 equation <w_1, v> = 0, <w_2, v> = 0, and <w_3, v> = 0. Just solve this system. Maybe this helps a bit.

do this with the riemann integral instead and start worrying about orientation and you'll probably get a perfectly good definition of determinant

this also falls out of $$T^*(e^1\wedge\cdots\wedge e^n) = (\det T)e^1\wedge\cdots\wedge e^n,$$ which says the same thing geometrically, and has the benefit of making sense for more general vector spaces

[stoic python]

TTerra

[wintry steppe]

though you need some multilinear algebra for that to make sense

[wintry steppe]

I wish there didn’t say non-zero vector

[winter harbor]

Damn, I remember asking you something related to this almost a year ago.

This

REE

[wintry steppe]

lmao back when i had the gay legosi pfp

yeah

all of what i just wrote makes perfect sense with the riemann integral btw

i just didnt wanna think about orientation so i wrote lebesgue here

francais

[winter harbor]

Now I got to actually learn some exterior algebra memery / differential forms memery and can understand your answer to that old question of mine

[wintry steppe]

[pastel idol]

Wait this isn't his job lol

He's not obligated to help you

Hes just trying his best to help you, and you're retorting...

[nocturne jewel]

Not sure why you're responding to something from 2h ago but ok

[frosty vapor]

can you not be so abrasive

[wintry steppe]

i don't think im understanding what Q is exactly. if someone could explain that would be great

i calculated the distance though

[untold isle]

need help

[dusk drum]

Didn't you draw the Q there already?

To find Q, you can use that vector i-2j+3k is normal to plane T.

[wintry steppe]

yeah lol i drew it but didn't know what to do with it.

sorry you said do what with the normal?

[dusk drum]

(position vector of P_0)+(some scalar)*(vector normal to T) = position vector of Q

You can solve for (some scalar) and hence, find Q

[wintry steppe]

okthanks so much:)

[wintry steppe]

I have three distinct eigenvalues so each of them has AM of 1?

yes

I have found that when the eigenvectors' AM = GM= n , then a matrix $A_{n \times n}$ is diagonalizable.

[stoic python]

Elfenkaiser

[wintry steppe]

Is this correct?

[zinc timber]

for all eigen values, then yes

[wintry steppe]

I have another question.

I have a characteristic polynomial $\lambda^3+3\lambda^2+4\lambda-12$. What does this have to do with the inverse?

[stoic python]

Elfenkaiser

[lavish jewel]

the determinant is the product of the eigenvalues, so if you find this poly has a root that is 0, the inverse does not exist

[zinc timber]

with the information Edd already mentioned, can you determine if 0 is root of you polynomial just by looking at it? hint || constant term ||

[lavish jewel]

this seems like a better fit for #prealg-and-algebra

[true jacinth]

i'm gonna put that in a help room

[vital shoal]

hi, can anybody help me with this one?

[winter harbor]

First, find the matrix (with respect to the standard basis) corresponding to the linear transfomration which projects a vector the the y-axis.

Then, find the matrix corresponding to the reflection with respect to the line -2x.

Since we are dealing with matrices wrt to the standard basis of R^2

We basically just need to know how both of these transformations act on the vectors (1,0) and (0,1)

Then, you take the corresponding product of the the projection matrix with the reflection matrix.

[mystic dagger]

how can i get spanning vector for $U=[(x, y, z) \in \mathbb{R}^3 | 2x-y=0]?$

[stoic python]

leonardomoura

[lavish jewel]

do you recognize the geometric shape?

[mystic dagger]

and is there only 3 spanning vectors? or could be infinite?

it's a plane right?

[winter harbor]

We have that $y = 2x$, thus, if $(x,y,z) \in U$, then:
$$
(x,y,z) = (x,2x,z) = (1,2,0) \cdot x + (0,0,1) \cdot z
$$

[stoic python]

MISTERSYSTEM

[winter harbor]

Can you continue from here?

[winter harbor]

Yup, it is.

[mystic dagger]

what about y?

[winter harbor]

y is completely determined by x. So in some sense it restricts the degree of freedom for U.

y = 2x

[winter harbor]

Well, the minimal number of vector vectors that span U is 2.

but we could have more

[lavish jewel]

if this makes it look more familiar to you, what you are given is equivalent to $\begin{bmatrix} 2 && -1 && 0 \\ 0 && 0 && 0 \\ 0 && 0 && 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

[stoic python]

Edd

[lavish jewel]

you can just find the null space of this as usual

rref and look for free parameters

[winter harbor]

For instance, U itself is a spanning set for U.

So yeah, we could have an infinite amount of vectors that span U.

But

[mystic dagger]

ohh

[winter harbor]

It is useful

[mystic dagger]

i get it

[winter harbor]

To try finding a minimal spanning set for U

this is called a basis

[mystic dagger]

ohh

i haven't learned basis yet

but thank you!

i got it

[gaunt marsh]

Quick question: what's the third condition for a something to be defined as a vector subspace. I know two of them are closed under addition and closded under muliplication

[sick sandal]

scalar multiplication*

[lavish jewel]

that's enough if you know it's a subset of a vector space

[gaunt marsh]

These were the lecture notes I got, does the highlighted portion mean that the 0 vector is in subspace?

[lavish jewel]

that's a consequence of me being dumb

but you do find it as a separate condition in some cases, sure

[marble lance]

It is not 👀

[lavish jewel]

hmm i suppose not

[marble lance]

It's a consequence of being closed under scalar multiplication assuming nonempty

So you usually either have W is nonempty or 0 in W

And yes, the highlighted part means the zero vector is in W

[lavish jewel]

ah, that's the right explanation

[gaunt marsh]

👍 Thanks

[vital shoal]

how can I find the first matrix

[half ice]

See what that matrix does to (0,1) and (1,0)

Then have it be an arbitrary matrix
a b
c d
And solve for a,b,c,d

Or, recognize this as a matrix multiplication of other matricies

[wintry steppe]

show that not (definition of nilpotent) holds

[marble lance]

An arbitrary vector space does not have a standard basis, there can be many different bases for the same vector space. And sometimes you know what the basis looks like, and sometimes you just know there is one but don't know what it looks like. That's context dependent.

If {e1, e2} is a basis then span{e1, e2} = V. That is correct.

[wintry steppe]

there must be simpler ways. do you have a specific example in mind?

one way is to show that it has a non zero eigenvalue

that might be the easiest

cause if A^n = 0 and Av = cv, then A^nv = c^nv = 0, and since v isn't zero, c^n = 0 giving c = 0

or if you like element-free proofs, some monomial x^n annihilates A, so the minimal polynomial divides x^n, hence has only the root 0

[marble lance]

I'm totally unable to read what you wrote

It looks like computer code more than math

Why is your function's name a whole phrase just call it f or T

[wintry steppe]

it's fine if you write like this for yourself but this isn't how people communicate math lol

feynman was a crank and a misogynist

[marble lance]

So what I think you have is that V is a 2 dimensional real vector space with some fixed basis {b1, b2}. That means every v in V can be written as a1b1 + a2b2 for real numbers a1, a2, and the column vector [a1, a2] is called the coordinate vector. Now you are considering the functions which maps v to [a1, a2]?

Note that you need to fix a basis for V before considering this function, because the coordinate vector depends on the basis

That is the same as the function I described

What's your question though?

There is no standard basis in V

There are many different bases

And each different basis defines a different phi

Phi depends on the basis

Yes

Only K^n has a standard basis

Yes

K^n also has many bases, but the standard basis refers to a specific one

[teal grotto]

i would argue that polynomial rings have a standard/canonical basis as well (but thats kinda besides the point)

[marble lance]

Yes

I don't know what you mean by "to see the vectors"

Well, now you don't, because V is an arbitrary vector space. But for a specific vector space, you can know what its elements look like

Sure

You can have a vector space of all polynomials of degree 5 or less

Then the elements of the vector space looks like ax^5 + bx^4 + cx^3 + dx^2 + ex + f

So I know what the elements look like

That's one basis

Yes

If you use the basis you defined up there

Then phi(ax^5 + bx^4 + cx^3 + dx^2 + ex + f) = (a,b,c,d,e,f)

Or (f,e,d,c,b,a) since you reversed the order

The polynomials are the vectors

A vector is an element of the vector space. If your vector space consists of polynomials, then the polynomials are the vectors.

I don't know what you mean, sorry

[half ice]

Vectors are things you can add/subtract. So, at the very least, you have to be able to do that. How exactly you do this depends on the vectors.

You can't really have a vector space where the vectors are completely intangible because of this, haha

But you could have one where "assigning them to a coordinate" is difficult

Yes all finite dim vector spaces have a map to F^n

Where n is the dimension

I should say an isomorphic map, as you can also map back from F^n

There's a layer of abstraction here for sure. When you say v, w in the context of "any vector space", you don't need to declare which vector space u and v belong to.

Obvious pro: You can prove things about all vector spaces this way

Con: It can be hard to think about u and v since they aren't "real things" until you come up with a vector space to attach them to

I can say "They're 3rd degree polynomials" and suddenly you know what the elements look like, and how they add

But the process of "adding vectors" always exists. You're guaranteed of that, at least

Just, you don't always know how I guess, haha

Yes, an isomorphic map is invertible and respects addition (Or, I like to say the function splits over addition). I should have said linear transformation though, which is more common in linear algebra

Haha so high school uses a different idea for vectors. They can move and be placed anywhere, and be line segments.

This is useful for solving physics problems, but not useful for talking about algebraic structures

Linear algebra has vectors always start at the origin, and prefers an algebraic interpretation

[wintry steppe]

what does it mean to "start at the origin"?

[half ice]

Yes. f can be written as a matrix, but that depends on the coordinate basis of V and W

Mind you, any two matricies for f will be similar matricies

[wintry steppe]

equivalent not similar

[half ice]

That is, if A and B are matrices for f, then
A = PBP^(-1) for some invertible matrix P

[wintry steppe]

no

they said f is from V to W

V not necessarily same as W

A=PBQ^(-1) for invertible P and Q

[half ice]

Yes I think I'm mixing change of basis into this hol up

No you cannot

[winter harbor]

Not really. You see, a matrix associated to a linear map between vector spaces is not an ''intrinsic notion'', in the sense that in general there's no notion of canonical/natural basis for both V and W such that we can write f wrt to these bases.

The choice of a basis for a vector space is pretty much arbitrary in nature.

Are you asking if these diagrams commute ?

Well

I kind of see what you are going for, I guess your question is about if a map f : V - > W between finite dimensional vector spaces and a choice of isomorphisms phi_1 : V - > F^n and phi_2 : V -> F^m induces a map f' : F^n - > F^m which is pretty much ''the representative of f wrt to phi_1 and phi_2''.

Is that so?

You can do phi_2 ° f ° phi_1^(-1) : F^n -> F^m

This is basically what the map f ''looks like''

wrt to the basis induced by the isomorphisms phi_1 : V -> F^n and phi_2 : W -> F^m

Yeah, but any choice of isomorphism, say phi : V -> F^n gives us a basis for V if we take the i-th basis vector b_i of V to be b_i = phi^(-1)(e_i). Where, e_i is the vector consisting of zeroes everywhere except for the i-th entry.

and this is what we are gonna do now

wdym

No

n is fixed

and I am not choosing any map

but an isomorphism

so dim(V) = n

Remember that to construct a matrix for a linear map f :V -> W between finite dimensional vector spaces V and W is basically to make a choice of basis for V and to W and see how f acts wrt to both of these basis.

and choosing a basis for a finite dimensional vector space E (over a field F) is basically to make a choice of isomorphism phi : E -> F^n

and this is what we are using now

Yeah, phi is an isomorphism

and to get a matrix represenation for f : V -> W wrt to these isomorphisms

Exactly

The idea is that a coordinate representation (wrt to given bais) for a map f : V -> W is basically an induced map f' : F^n -> F^m where dim(V) = n and dim(W) = m.

There's a less abstract way to define the matrix of a linear map

And I will give you a description of that

but now

I want to try to make sense of your intuition

[wintry steppe]

mistersystem literally carrying this channel recently

[winter harbor]

Now we have an induced map phi_2 ° f ° phi_1^(-1) : F^n -> F^m. And we define the i-th column of matrix f wrt to phi_1, phi_2 to be the image of e_i = (0,...,1,...,0) under phi_2 ° f ° phi_1^(-1). Where e_i is the vector where all entries are equal to 0 and the i-th entry is equal to 1.

This is kind of abstract lol

So I will give an example

[mystic dagger]

if $A=[v_1, v_2]$ is a spanning set of $v_1=(1, -2, -1)$ and $v_2=(2, 1, 1)$ what does the det of the matrix being zero implies? $\begin{pmatrix}
1&2&x\
-2&1&y \
-1&1&z
\end{pmatrix}$

[stoic python]

leonardomoura

[winter harbor]

No

Not really

Or at least idk how to phrase it in a better way

[mystic dagger]

hmmm

[winter harbor]

Let $V = \mathcal{P}{1}(\mathbb{R}) = {f \in \mathbb{R}[x] , \vert , \text{deg}(f) \leq 1 }$, i.e $V$ is the real vector space consisting of all real polynomials of degree less than or equal to 1. We have that $\text{dim}(V) = 2$, since $\mathcal{B} = {1,x}$ and $\mathcal{B}' = {1-2x,2+3x}$ are both basis for $V$. As such they induce isomorphisms $\varphi{1}, \varphi_{2} : V \rightarrow \mathbb{R}^{2}$ given by:
\begin{align*}
\varphi_{1}& : V \rightarrow \mathbb{R}^{2} \
1 \mapsto& (1,0) \
x \mapsto & (0,1)
\end{align*}
and
\begin{align*}
\varphi_{2}& : V \rightarrow \mathbb{R}^{2} \
1-2x \mapsto & (1,0) \
2+3x \mapsto & (0,1)
\end{align*}

[stoic python]

MISTERSYSTEM

[winter harbor]

Is this good so far?

To make myself clearer

phi_1 is the map associated to the basis B

And I am only defining how it acts on B

because we can extend it by linearity

so phi_1 is well defined

similarly, phi_2 is the isomorphism associated to B'

Now

Vector space of real polynomials with degree less than or equal to 1.

R[x] is the set of all real polynomials of one variable

that's a notation

deg(f) is a notation for the degree of f

yup

now

let's take the map f : V -> V

which takes a polynomial p(x) and sends it to its derivative p'(x)

Alright

So now let's calculate phi_2 ° f ° phi_1^(-1)(1,0)

this will be the first coordinate of the matrix of f

wrt to the basis B and B'

so we have
$$
\varphi_{2} \circ f \circ \varphi_{1}^{-1}(1,0) = \varphi_{2}(f(1))
$$
But f(1) is the derivative of 1, which is 0, so:
$$
\varphi_{2}(f(1)) = \varphi_{2}(0) = (0,0)
$$

[stoic python]

MISTERSYSTEM

[winter harbor]

so the first column of the matrix f wrt to B and B'

will be (0,0)

now let's calculate the second column

1 is a polynomial too

and f is the derivative of this polynomial

the derivative of 1 is 0

$$
\varphi_{2} \circ f \circ \varphi{1}^{-1}(0,1) = \varphi_{2}(f(x))
$$
Now, $f(x)$ is the derivative of $x$, which is $1$, so:
$$
\varphi_{2}(f(x)) = \varphi_{2}(1)
$$
Now, we have to try to find the coordinates of $1$ wrt to the basis $\mathcal{B}'$, we then see that we in fact have:
$$
1 = \dfrac{3}{7} \cdot (1-2x) + \dfrac{2}{7} \cdot (2+3x)
$$
and then:
$$
\varphi_{2}(1) = \varphi_{2}\left(\dfrac{3}{7} \cdot (1-2x) + \dfrac{2}{7} \cdot (2+3x) \right)
$$
Since $\varphi_{2}$ is linear, this is in fact equal to:
$$
\dfrac{3}{7} \varphi_{2}(1-2x) + \dfrac{2}{7} \varphi_{2}(2+3x)
$$
since $\varphi_{2}(1-2x) = (1,0)$ and $\varphi_{2}(2+3x) = (0,1)$, then this simplies down to
$$
\dfrac{3}{7} (1,0) + \dfrac{2}{7} \varphi_{2} (0,1) = (3/7, 2/7)
$$

[stoic python]

MISTERSYSTEM

[winter harbor]

So the second collumn of $f$ is given by
\begin{bmatrix}
3/7 \
2/7
\end{bmatrix}

[stoic python]

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

[winter harbor]

and the matrix of f wrt to $\mathcal{B}$ and $\mathcal{B}'$ is given by
\
\
\begin{bmatrix}
0 & 3/7 \
0 & 2/7
\end{bmatrix}

[stoic python]

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

[winter harbor]

That was some work

Is this so far so good?

Np

I'd recommend you coming up with more examples

And doing this all by yourself

like

take a map f : V -> W between finite dimensional vector spaces

take a basis for V and a basis for W

construct the induced isomorphism phi_1 : V -> F^n and phi_2 : V -> F^m

and then calculate the values for phi_2 ° f ° phi_{1} : F^n -> F^m on the canonical basis of F^n

I will tell you

This is a big ammount of work

And you don't necessarily need to do this

to find a matrix for V

But I will give you some time to play around with this idea

And later I will show you how to find a matrix for f (wrt to given choice of basis)

without doing this

[steep stratus]

I have a question that asks me to solve least squares straight line to fit some data points and then the minimum square error. What is the minimum square error? Is it the same as the least squares error?

[winter harbor]

Let $f : V \rightarrow W$ be a linear transformation between finite dimensional vector spaces over a field $\mathbb{F}$ and $\mathcal{B} = {e_{1}, \cdots, e_{n}} \subset V$ be an ordered basis for $V$ and $\mathcal{B}' = {b_{1}, \cdots, b_{m} } \subset W$ be an ordered basis for $W$.
\
\
Thus $\forall j \in {1, \cdots, n}$ we have that $f(e_{j}) \in W$ is a vector in $W$, thus, since $\mathcal{B}'$ is a basis for $W$, there exists unique constants $\alpha_{1,j}, \cdots, \alpha_{m,j} \in \mathbb{F}$ for which $T(e_{j})$ is given by the linear combination:
$$
f(e_{j}) = \sum\limits_{i=1}^{m} \alpha_{i,j} b_{i}
$$
We define the matrix of $f$ wrt to the ordered basis basis $\mathcal{B}$ and $\mathcal{B}'$ to be the matrix $[f]^{\mathcal{B}}{\mathcal{B}'}$ whose $(i,j)$-th entry is $\alpha{i,j}$. Thus, the $j$-th column of $[f]^{\mathcal{B}}{\mathcal{B}'}$ is precisely given by the coordinates of $f(e{j})$ wrt to the ordered basis $\mathcal{B}$ and $\mathcal{B}'$, i.e the $j$-th column of $[f]^{\mathcal{B}}{\mathcal{B}'}$ is
$$
(\alpha{1,j}, \cdots, \alpha_{m,j})
$$

@wintry steppe here's another way to find the matrix of a linear transformation between finite dimensional vector spaces

wrt to ordered basis

[stoic python]

MISTERSYSTEM

[winter harbor]

phi_2 maps to the 0 vector because phi_2 is a linear transformation.

phi_1 and phi_2 are both defined on a basis

and by linearity we can extend it to the whole of V

to a linear function

which is unique and well defined

and that satisfies phi_2(1-2x) = (1,0) and phi_2(2+3x) = (0,1)

idk if you have seen this before

This is usually how we define the matrix of a linear transformation btw.

yup

1 = ...

because basically

since 1-2x and 2+3x are a basis for V

I want to find constants a and b

such that 1 = a(1-2x) + b(2+3x)

this is good

because varphi_2 is linear

and we know how it acts on 1-2x and 2+3x

and to find a and b we basically have to solve a system of linear equations (can you see how?) with unknowns a and b.

Nice

[winter harbor]

Try to make sure you understand why both these descriptions of a matrix of a linear map are equivalent.

Did you come up with the phi_1, phi_2 stuff on your own tho?

That's actually really good

This comes up later

In some sense

good luck

Nah, I'm good

I should be studying something else now tho

[random axle]

Can you apply linear transformation to functions? Could the rotation matrix be used to rotate $\ln(x)$ by any angle for example? $$ \begin{bmatrix}
\cos(\theta) -\sin(\theta) \\sin(\theta) \cos(\theta))
\end{bmatrix}
\begin{bmatrix}
x
\
\ln(x)

\end{bmatrix}$$

[stoic python]

azeem321

[winter harbor]

You can do that

https://www.youtube.com/watch?v=BPgq2AudoEo

Basically what they do in this video

[random axle]

thanks just what i was looking for

[winter harbor]

❤️

[golden ermine]

On the left side, you have < f + g, f + g > = < f, f + g > + <g, f + g >
= < f, f > + < f, g > + < g, f > + < g, g >
= < f, f > + < f, g > + conj(< f, g >) + < g, g >
= < f, f > + 2 * Re(< f, g >) + < g, g >

Note that < g, f > is the conjugate of < f, g > and they are just complex numbers

no, the field you’re working in is C so < f, g > is an element in C — call it x + iy. Then Re < f, g > = x while Im < f, g > = y

But < g, f > is the conjugate of < f, g > so if < f, g > = x + iy, then < g, f > = conjugate of x + iy = x - iy

So adding the two inner products together gives you 2x = 2 * Re < f, g >

[dusk kayak]

could someone help me out with a matrix question

[nocturne jewel]

https://dontasktoask.com/

[dusk kayak]

does anyone know how i could approach this

[teal grotto]

apply gaussian elimination with an augmented identity matrix

[dusk kayak]

how would i do that

[teal grotto]

do you know how to do gaussian elimination?

[dusk kayak]

yes

[teal grotto]

well, then you know how to do this problem

when your doing gaussian elimination here, tack on a four by four identity matrix to the right of the original matrix

proceed with gaussian elimination normally with the identity matrix tacked on

the rightmost four by four matrix you get afterwards will be the inverse of the matrix

[wintry steppe]

[teal grotto]

fuck it lol

dont know why it wont render right

[wintry steppe]

post it again

[teal grotto]

perform gaussian elimination on this matrix

\begin{pmatrix}
1 && 0 && 0 && 0 && 1 && 0 && 0 && 0\
0 && 1 && 0 && 0 && 0 && 1 && 0 && 0\
0 && m_1 && 1 && 0 && 0 && 0 && 1 && 0\
0 && m_2 && 0 && 1 && 0 && 0 && 0 && 1
\end{pmatrix}

[stoic python]

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

[teal grotto]

wot

[dusk kayak]

something like this

[teal grotto]

yes

that

[dusk kayak]

okay lol

let me try this

[teal grotto]

god why does the bot hate me

[wintry steppe]

$$
\begin{pmatrix}
1 & 0 & 0 & 0 & && 1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 & && 0 & 1 & 0 & 0\
0 & m_1 & 1 & 0 & && 0 & 0 & 1 & 0\
0 & m_2 & 0 & 1 & && 0 & 0 & 0 & 1
\end{pmatrix}
$$

[teal grotto]

okeh

thanks man lol

[stoic python]

TTerra

[dusk kayak]

thank you guys

[wintry steppe]

yeah idk bot was being funky

[dusk kayak]

ill lyk if i get it

[wintry steppe]

$$
\begin{pmatrix} 
1 & 0 & 0 & 0 &  & 1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 &  & 0 & 1 & 0 & 0\\
0 & m_1 & 1 & 0 &  & 0 & 0 & 1 & 0\\
0 & m_2 & 0 & 1 & & 0 & 0 & 0 & 1
\end{pmatrix}
$$

i think it just doesnt like double dollar signs in the matrix?

double and sign

not dollar

latex is stupid

[teal grotto]

yea only one ampersand ig

[dusk kayak]

i get it guys

thank you so much appreciate yall

[dusty zodiac]

im new to linear algebra, was just wondering if this was right?

[half ice]

They should have arrow heads to indicate direction

But yes, it would be right

A+B starts at the beginning of A and ends at the end of B

[dusty zodiac]

thanks

ill fix that

[winter harbor]

That's a really nice work,

Tho there's some mistakes

specifically on page 5/6

This can't be a basis for P_3

because it is not linearly independent

no subset of a vector space that contains the zero vector is linearly independent

thus, can't be a basis.

In this particular case

this isn't even a spanning set for P_3

there's no way we could write x^3 as a linear combination of {0,1,2x,3x^2}

You didn't particularly have to write down this map also

You could have worked through all your calculations using this map you defined in the first page

if you wanted to find the basis of the differentiation map wrt to the basis {1,x,x^2,x^3}.

But you pretty much got the whole idea correctly

what you needed to do was to calculate phi ° f ° phi^(-1) : R^4 -> R^4 applied on the canonical basis vectors e_1,e_2,e_3 and e_4.

[winter harbor]

Where phi is the map defined on the first page.