linear-algebra

[stiff frost]

but what number is it, given T and the vs and the ws

we set k=1 in the equation to answer that question

The equation becomes $Tv_1=A_{1,1}w_1+\cdots+A_{m,1}w_m$

[stoic python]

dirib:

[stiff frost]

Now, here's where our understanding of bases is used without comment, which is a bit confusing

Since $T\in\mathcal{L}(V,W)$, we know $Tv_1$ is some vector in $W$.

[stoic python]

dirib:

[ocean sequoia]

wait quick question

The equation becomes $Tv1=A{1,1}w1+\cdots+A{m,1}w_m$

[stoic python]

brzig:

[ocean sequoia]

so you are saying that the the transformation applied on v1 is equal to the sum of the matrix multiplied by w

so we arent necessarily going from v to w here but using a matrix to make them equal

[stiff frost]

The meaning of that equation was what I was getting to

so just wait one sec and then you can reask your question from a better footing

[ocean sequoia]

ok sorry wasnt sure if i missed something

wanted to make sure

[stiff frost]

Since $T\in\mathcal{L}(V,W)$, we know $Tv_1$ is some vector in $W$. And we know that every vector in $W$ can be written uniquely like $c_1w_1+\cdots+c_mw_m$ for some scalars $c_1,\ldots,c_m$.

[stoic python]

dirib:

[stiff frost]

Does that make sense based on your understanding of the ws being a basis for W?

[ocean sequoia]

yea thats what a linear combination is

and a basis has to span W

[stiff frost]

Well, the uniqueness part is super important

if we just had "equals some linear combination", we wouldn't be able to finish this definition

it equals only one linear combination

[ocean sequoia]

yea

sorry yes

or it would be linearly dependent

if there were multiple ways

[stiff frost]

the ws would be linearly dependent, yes

[ocean sequoia]

some w would be

yea

[stiff frost]

ok

So, there is only one bunch of cs for Tv_1

And we have the equation $Tv_1=A_{1,1}w_1+\cdots+A_{m,1}w_m$

[stoic python]

dirib:

[stiff frost]

So the number $A_{1,1}$ is the coefficient of $w_1$ in the only way to write $Tv_1$ as a linear combination of the $w$s

[stoic python]

dirib:

[ocean sequoia]

so a matrix is a unique represntation?

[stiff frost]

I mean a matrix is unique when you pick T and the vs and the ws

[ocean sequoia]

ok so the matrix is determined by the basis

of V and W

[stiff frost]

but you're picking so much then that I think "a matrix is a unique representation" is something that would do more harm than good to keep in your head

[ocean sequoia]

ok

[stiff frost]

We're building "the matrix for T with respect to the two bases "the vs" and "the ws""

And $A_{2,1}$ is the coefficient of $w_2$ in $Tv_1$, etc.

[stoic python]

dirib:

[ocean sequoia]

that statement lost me

oh wait

no it didnt

[stiff frost]

There is only one bunch of $c$s for $Tv_1$. And we have the equation $Tv_1=A_{1,1}w_1+\cdots+A_{m,1}w_m$

[stoic python]

dirib:

[ocean sequoia]

Tv1 doesnt go to w1 thats your point from earlier

[stiff frost]

Tv1 might be like, 6w_1-178w_5

with coefficients 6, 0, 0 ,0, -178, 0, 0 if W has dim 7

for instance

[ocean sequoia]

ok

[stiff frost]

$A_{4,1}$ is the coefficient of $w_4$ in $Tv_1$, etc. And we can now vary $k$ to get other columns of the matrix of numbers $A$

[stoic python]

dirib:

[stiff frost]

$A_{3,7}$ is the coefficient of $w_3$ in $Tv_7$, for instance.

[stoic python]

dirib:

[stiff frost]

because it has a unique coefficient since the ws are a basis for W

In more words and less symbols, the matrix is the array of numbers where each column is a coefficient list for the ws from a different Tv_k

[ocean sequoia]

ok

[stiff frost]

Since the vs are a basis for V, this is all the information you need to understand what T does to any vector in V

Like, $T(3v_1+2v_2)=3Tv_1+2Tv_2=3(A_{1,1}w_1+\cdots+A_{m,1}w_m)+2(A_{1,2}w_1+\cdots+A_{m,2}w_m)=(3A_{1,1}+2A_{1,2})w_1+\cdots$

[stoic python]

dirib:

[ocean sequoia]

ok hang on let me try to say this in my own way

[stiff frost]

please

[ocean sequoia]

the issue with saying a matrix is a linear transformation T is that T only takes in one vector at a time

[stiff frost]

I would say a matrix is an array of numbers. Which could represent different linear transformations depending on whether you say "this array is for a transformation with respect to this pair of bases" or "this array is for a transformation with respect to that pair of bases", or "this is my report card, it doesn't represent a transformation at all"

[ocean sequoia]

i took a screenshot of that lol i like that

[stiff frost]

The "one vector at a time" business was in response to you writing T(v1,...v,n)

[ocean sequoia]

but we arent applying the matrix on V

[stiff frost]

We aren't applying the matrix on anytjhing

[ocean sequoia]

we are saying that T(v1) = $A{1,1}w1+\cdots+A{m,1}w_m$

is apply the wrong word?

[stiff frost]

Probably

Yeah, that's T(v1)

[stoic python]

brzig:

[ocean sequoia]

in other words we arent moving from V to W but staying in W

[stiff frost]

I'm not sure what you're saying there

T takes a vector in V

and gives you a vector in W

For example, it take v1 and gives you a vector in W "T(v1)"

But A is an array of numbers

it doesn't do anything without the context of "being the matrix for T with respect to the vs and ws"

[ocean sequoia]

yea but the matrix multiplication isnt on V

[stiff frost]

I would recommend not saying the matrix takes a vector at all

[ocean sequoia]

is that probably more accurate?

the edit?

[stiff frost]

matrix multiplication is a different discussion

it turns two arrays into a new array

and that has uses you learned about already

[ocean sequoia]

ok then lets stick with this

[stiff frost]

but we don't have to think about it (matrix multiplication) for this defintion or the problem you were originally working on

[ocean sequoia]

ok so i guess in my own words a matrix is an rectangular array of numbers that is used to represent a transformation of T(v1) to the unique linear combination of W that is given by the coefficenets of the matrix

[stiff frost]

I think you have the exact right idea behind those words, but those words have some bad habits/inaccuracies I worry will confuse you later

[ocean sequoia]

thats kinda an issue i have been experiencing

im self studying so please

rip apart my words

i prefer the rigor when i can get it

[stiff frost]

"a matrix is an rectangular array of numbers that is used to represent a transformation" Yes.
"a transformation of T(v1)" makes it sound like you're taking the single vector T(v1) and then transforming it somehow (under some S from W to a third vector space??)

[ocean sequoia]

ok

Tvk

because a matrix represents the entire change of basis in this case

wow i cant type tonight i have no idea whats going on lmao

[stiff frost]

"represents the entire change of basis" that's...definitely going to confuse you

the matrix represents the transformation T

If T sends every vector in V to the zero vector in W, then T and the matrix are nothing like a "change of basis"

the matrix records the coefficients of the W basis entries for the outputs of T on the V basis entries. But if T sends everything to the zero vector, those coefficients won't be interesting

[ocean sequoia]

ok hm the matrix represents the linear transformation T on all of v1,...,vm

they would all be zeros

[stiff frost]

That's a bad habit to get into because T doesn't care about the basis v1,...,vm

Like imagine T takes ...arrows in 3D to pairs of numbers in some geometric way

T is the same no matter what basis for the space of "arrows in 3D" you use

So it's a bad habit to speak of T as a transformation "on all of v1,...,vm"

[ocean sequoia]

ah ok

that makes sense

[stiff frost]

But if you pick your favorite basis for the input vector space v1,...vn

[ocean sequoia]

standard basis

[stiff frost]

Only the nicest of vector spaces have a standard basis

[ocean sequoia]

can V be nice

do you think its better to make it more complex here?

[stiff frost]

I think it's useful to have vector spaces like "arrows" or something divorced from numbers, if you've been introduced to such things

not because it's more complex, but because it simplifies this discussion

[ocean sequoia]

i have

[stiff frost]

Because then there are no standard bases, there are no numbers hidden inside vectors

the only numbers come from the coefficients in this matrix, basically

[ocean sequoia]

i honestly think about them like a box with a bunch of random points in them floating around

[stiff frost]

Anyway, once you pick a pair of bases for the input and output space, you could write down all the coefficients in an array and you have the matrix

if you pick a different pair, you get a different matrix for the same geometric transformation

[ocean sequoia]

ok but back to numbers from the coefficients in the matrix if we are talking about the transformation T(v1) the matrix will be the A1,k

[stiff frost]

T(v1) is a vector in W, so I wouldn't call it a transformation

the first column of the matrix for T would be the coefficients of ws for T(v1)

[ocean sequoia]

yea ok

[stiff frost]

This is a pretty abstract concept and was defined in an abbreviated way, so if it's starting to make sense to you, that's really good

[ocean sequoia]

it is a bit

i like the abstraction its kinda nice

[stiff frost]

Well, I need to get some sleep, but I would recommend looking at examples of this matrix construction/definition, and then coming back to the original question

[ocean sequoia]

i will

i appreciate it so much

[stiff frost]

Happy to help. good luck

[ocean sequoia]

you literally jsut spent an hour helping me more or less

[stiff frost]

What can I say, math is fun

[ocean sequoia]

[teal topaz]

<@&286206848099549185> anyone able to help out here?

[dire palm]

I believe you can always find eigenvalues yes. Like the matrix [0 1, 0 0] has an eigenvalue, specifically 0 (with multiplicity two), and a corresponding eigenvector (1,0)

[teal topaz]

I'm more trying to construct a matrix with some desired eigenvectors/eigenvalues, rather than finding the eigenvectors/eigenvalues of a matrix

[dire palm]

hmm. If you are given say 4 eigenvalue and 4 eigenvectors, and you needed to construct a 5x5 matrix, could you not construct a 4x4 with the properties and then pad with zeros?

Just a thought; I don't have proof that works. But unless I'm missing something adding that row and column of zeros shouldn't change the eigenvectors and values of the 4x4

[teal topaz]

Well yeah, but then I get an extra unspecified eigenvector and eigenvalue 0

But I'd like to specify all 5 eigenvalues and all 5 eigenvectors in that 5x5 matrix

Even less -- what if I just wanted to specify all the eigenvectors and not worry about particular eigenvalues? Could I do it then?

[median forum]

@teal topaz https://math.stackexchange.com/questions/472915/what-kind-of-matrices-are-non-diagonalizable

the end should be reasonably useful

[compact basin]

@gleaming adder thanks, i'll try it

[half storm]

I'm trying to solve forthe coordinate function realtive to the basis $\Beta$. I understand how they got to these equations:
For any $ (x,y) \in \mathbb{R}^2 \implies \exists ! a_1, \exists ! a_2 \in \mathbb{R} \text{ s.t.} (x,y) = a_1(2,1) + a_2(3,1)$ Then you apply the coordinate function so that $f_1(x,y) = a_1 f_1(2,1) + a_2 f_1(3,1) \implies 1 = f_1(2,1) \text{ and } 0 = f_1(3,1)$ and then you get to the equations that they have in the example. How do you know that solution always exists when solving for the formulas of the coordinate functions?

[stoic python]

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

[dusky epoch]

{(2,1), (3,1)} is a basis

[half storm]

I know

I know there always exists $a_1$ and $a_2$ so that you can express $(x,y)$ as a linear combination of $(2,1)$ $(3,1)$. I'm asking why is there always a solution to the equations that they have. It may have to do with the fact that $ \Beta $ is a basis, but I'm struggling to see how that relates to the fact that the next system of equations that you come up wtih always has a solution.

[stoic python]

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

[dusky epoch]

i mean tbh

i would look at it this way

f_1( a_1 (2,1) + a_2 (3,1) ) = a_1

so if you can express (1,0) as a linear combination of beta - which you can by virtue of beta being a basis - you know f_1(1,0)

[half storm]

I'm not quite sure I follow. But I just got to thinking, if you look at the equation $a_1 = a_1 f_1(2,1) + a_2 f_2(3,1)$ then technically It's an underdetermined system in $f_1(2,1)$ and $f_2(3,1)$ yea?

[stoic python]

JohntheDon:

[dusky epoch]

...

f_1( a_1 (2,1) + a_2 (3,1) ) = a_1 is true for all a_1 and a_2

[half storm]

Yea, I got you.

I'm just thinking that a_1 is fixed unique scalar as is a_2 say like 1 and 2 and then you're solving for what f_1(2,1) and f_2(3,1) is. So you can express f_1(2,1) as a function of f_2(3,1).

You get a free variable in f_2(3,1).

It's not that big of a deal, I just like to keep track of the logical assumptions that I'm making as I go about solving things.

[dusky epoch]

you're overthinking it.

f_1(2,1) and f_1(3,1) are known

they're 1 and 0 resp by the defn of dual basis

[half storm]

Oooh o.k. I see sorry about that 🤦‍♂️

because the coordinate function $f_1$ gives you the first / scalar relative to the basis $\Beta$ since (2,1) is the first element of the ordered basis $\Beta$, then it has to be that $f_1(2,1) = 1$

[stoic python]

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

[jaunty compass]

I have a question. Why would the last row have to be all zeros in order for the system of equations to have infinitely many solutions

[gray dust]

without any more info, that does NOT follow

it’s entirely possible to have unique/many/no solutions despite the coefficient matrix having a row of 0s

[half storm]

I guess you have to specify whether you're dealing with a coefficient matrix or an augmented matrix

[gray dust]

doesn’t matter, the implication still doesn’t follow

[half storm]

Cool.

I'm just thinking about what the conditions have to be.

It's been a while sense I've dealt with using matrices to solve systems of linear equations and what exactly a row of 0's might mean.

[gray dust]

eg take Ix=0, I=any size identity, obviously has a unique soln x=0. tack on another row of 0s and the soln is still unique

[half storm]

I see.

[gray dust]

recall a linear system can be rewritten as a matrix eqn & vice versa. write out the eqn a row of 0s represents and you see it’s just 0+...+0=0 which gives no info about solutions

[half storm]

I guess you could have a matrix in which you have more rows than columns i.e. more equations than variables. So say that you have a n x n matrix. Would you necessarily have to have a row of zeroes in the RREF for there to exist(s) solution(s)?

The hypothetical matrix in this scenario Is an augmented matrix.

[gray dust]

yes

[half storm]

Can you really say anything about the size of the solution set of a linear system if you have a row of zeroes in the coefficient matrix?

I'm trying hard to think what you can deduce from that.

[jaunty compass]

Can someone check if I did this correct?

This is what happens when I row reduced it

So would the vector u be in Span A?

I said no but not sure on the explanation

[hollow finch]

@jaunty compass Do you understand what the matrix you row reduced represents, or why you row reduce it?

[jaunty compass]

Yeah so we can try to get the value right? @hollow finch

[half storm]

@jaunty compass When you combined those columns together into one matrix, what you created was an augemented matrix. This matrix corresponds to a system of linear equations. You're row reduction gives you a solution i.e. a vector in R^4

So you have your matrix reduced, and assuming that you reduced correctly, then what you have here is telling you that x_4 = -4 x_2 + 1/3x_3 = -26/3 and x_1 + 1/3x_3 = 10/3

Notice that you hav ea free variable in x_3.

Meaning that for any value of $x_3$, you can work back towards a solution, the solution set looks like $ {(x_1,x_2,x_3,x_4) \in \mathbb{R}^4 : (10 - \frac{x_3}{3} , \frac{x_3}{3} -\frac{-26}{3} , x_3, 1)} $

[stoic python]

JohntheDon:

[half storm]

So in short, yup it's in the span of those vectors and there are infinitely many vectors / coefficients that you $(0,-12,12,-2)$

[stoic python]

JohntheDon:

[jaunty compass]

Thank you

[hollow finch]

Right. The system of equations being specifically about finding which (if any) linear combinations of those 4 vectors will give you that new one.
If you were left with a bottom row like {0,0,0,0,1} that's like saying 0c1+0c2+0c3+0c4=1 which means there is no solution.

[halcyon belfry]

I have A an n x n matrix, and B is the adjugate matrix of A. I need to prove that AB = BA = det(A) * I.

I'm not too sure how to approach this though

my first instinct is to do induction, but that gets quite messy

[hollow finch]

Is the point of the proof to show that the multiplication is commutative, if multiplied by A it gives a scalar multiple (detA) of the identity, or that it works for nxn matrices?

[halcyon belfry]

ultimately it's to show that if A has integer entries, then det(A) = +/- 1

(and that A^-1 exists)

[hollow finch]

Uh but it's not always that

Do you mean if the adjoint has integer entries?

Or adjugate

[halcyon belfry]

err, sorry, "Suppose A has integer entries. Prove that A^-1 exists and has integer entries iff det(A) = +/- 1"

I missed the important part of that 😓

[hollow finch]

Oh okay that makes a lot more sense

[halcyon belfry]

So the first part of that question asks for AB = BA = det(A)*I

[hollow finch]

Are you starting with A^-1 has integer entries implies detA=±1 or the converse

Because I'm pretty sure that property is just a theorem which doesn't seem to be the point of this proof

[halcyon belfry]

Yeah that property is just part a) of this question. I think I know how to do the integer entries question.

Since I'm guessing it has something to do with the only units in Z being +/- 1

[hollow finch]

Hm well can you use the fact that adj(A)=det(A)A^-1?

[halcyon belfry]

I don't think so. This question begins by giving the full definition of adj.

[hollow finch]

Giving or asking you to prove?

[halcyon belfry]

$B$ is a matrix whose $ij$th entry $b_{ij}$ is given by the formula $b_{ij} = (-1)^{i+j}\det (A_{ji})$ where $A_{ji}$ is the matrix obtained by deleting the $j$th row and the $i$th column.

[stoic python]

IamDerek:

[hollow finch]

Aha so the adjoint but you're not told it's the adjoint

[halcyon belfry]

Yeah

[hollow finch]

Alright for that proof I would look at the "entry perspective" of matrix multiplication as my professor would say

[halcyon belfry]

So I think they want me to prove the commutivity and that their product is just the determinant scaling the identity.

So I can do the proof directly without induction?

I hate determinants because of recursion.

[hollow finch]

Yes. If you can show that their product yields a scalar multiple of the identity, there is an inverse relationship going on.
Yes induction wouldn't be appropriate here

[halcyon belfry]

So basically do this directly and show that the diagonal is just det(A) while all other entries are 0?

[hollow finch]

Look at the general entry of the product and see if it looks familiar to anything else you've learned. Start with the diagonal entries.

Exactly

[halcyon belfry]

Okay thank you 🙂

[hollow finch]

Np. Good luck

[halcyon belfry]

I guess ultimately it's to show that if we are working over any ring that A^-1 has entries in R iff det(A) is a unit in R.

[jaunty compass]

I was wondering how you can find if a 3 by 3 matrix is diagonallizable

[hollow finch]

@jaunty compass you need a full set of independent eigenvectors

So you need 3 in that case

The only times you might run into trouble is if the CP has repeated roots. Otherwise you gucci

[halcyon belfry]

If CP has repeated roots it can still have independent eigenvectors right?

[hollow finch]

Indeed it can

[gray dust]

consider the identity

[hollow finch]

But every eigenvalue has at least one eigenvector, so you need a repeated roots for there to even be a chance you're missing one

Well what's the CP?

Put $ on each side

[stoic python]

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

[hollow finch]

Before the start of the code and after the end of the code

[pallid rampart]

You don’t put $ inside the \begin{matrix}

$\begin{bmatrix}
1 & 1 & -4\
2 & 0 & -4\ -1 & 1 & -2\
\end{bmatrix}$

[stoic python]

Whoever:

[jaunty compass]

Haha thank you.

[pallid rampart]

If you can find a basis of eigenvectors then your matrix is diagonalizable. If the eigenvalues are distinct then the corresponding eigenvectors are linearly independent

Therefore find the characteristic polynomial and try to factor it

[jaunty compass]

Okay so I found the lambda values, which is -1,2,-2

[pallid rampart]

Alright

That means that your matrix is diagonalizable

Since you have 3 different eigenvalues

[jaunty compass]

Oh thats it !

[pallid rampart]

Yes

[gray dust]

you can take it for granted now but may want to prove later that distinct eigenvalues have linearly independent eigenvectors

[jaunty compass]

ok so say I have to diagonalize the matrix would that be a differnet problem

[pallid rampart]

It will require more work yeah

[jaunty compass]

ok I think that was my original question, but I didn't know diagonize and diagonalizable were too differnet things

[hollow finch]

A matrix needs to be diagonalizable to diagonalize so it is a necessary part of the other

As in you need to know if it's diagonalizable to diagonalize

[pallid rampart]

But if you just diagonalize it then you know it’s diagonalizable

[jaunty compass]

[hollow finch]

Lol yeah. After you diagonalize for the first time you never really ask yourself that question ever again

You just try to diagonalize it until you maybe don't get an eigenvector

[jaunty compass]

what do I have to do to find the eiegenvectors?

sorry if its dumb. Just learning this stuff

[hollow finch]

Solve
$(A-\lambda I)\vec{v}=\vec{0}$

[stoic python]

nix:

[hollow finch]

That is to say, find a basis for the null space of $A-\lambda I$

[stoic python]

nix:

[half storm]

It's been a while since I've done this but I'm trying to see if I remember this correctly.

You want to find the values of the lambda such that the $det(A - \lambda I) = 0$ yea. Because if so, then it means than $A - \lambda I$ is noninvertible, and so there exists nonzero members of the nullspace of $A - \lambda I$

[hollow finch]

That is true.

[stoic python]

JohntheDon:

[hollow finch]

And how you get the characteristic polynomial

[half storm]

Cool Cool, I've always wanted to really understand why this stuff works the way that it did. I don't ever recall learning that while I was in school but going back and reviewing and getting a deeper intuition and understanding behind the theory behind why this stuff works feels a lot better.

We were always told to resort to this method in order to find eigenvalues but nobody really explained the logic while it yields those values.

[hollow finch]

Since any scalar multiple of an eigenvector is an eigenvector, there need to be infinitely many solutions to Av=lambda v=lambda I v

Yeah I only really understood it after I finished linear algebra

The way you phrased it just about as good as it gets for why we do it this way

[wintry steppe]

The null space is where bad vectors go

But in all seriousness, the explanation @hollow finch gave is also why geometric and algebraic multiplicities are defined the way they are

[jaunty compass]

gotcha

[jaunty compass]

$\begin{bmatrix}
2&1&3/ -4&-2&-4/ 0&0&-1/
\end{bmatrix}$

[half storm]

Why are we evaluating $g_1T(1)$. Why at 1 exactly?

[stoic python]

JohntheDon:

datguy:

[jaunty compass]

Ughh I cant do it a 3 by 3 matrix on here

$\begin{bmatrix}
2 & 1 & 3 \
-4 & -2 & -4 \
0 & 0 & -1
\end{bmatrix}$

[stoic python]

datguy:

[jaunty compass]

Finally lol.

I found the eigenvalues and they are 1, 0 so this wouldn't be diagonable right? Since it expects three vectors right?

[wintry steppe]

@half storm so you can show that a = 1. if you want to discuss this go to a questions channel btw, channel is occupied

NO

slim

i meant the other guy lol

sorry if i gave you the wrong idea

[half storm]

@wintry steppe Is this not a questions channel?

[wintry steppe]

this channel is occupied

[jaunty compass]

me?

[wintry steppe]

@jaunty compass apologies for burying your question, you may continue

[half storm]

Sorry I'm kind of confused and don't really get how discord works.

@wintry steppeSo this was never a channel meant to ask questions in?

@wintry steppe Sorry I didn't know. I joined this from a discord invite without ever reading the readme. But I understand now. I joined another discord and starting asking math questions in there one time not realizing it was meant more for discussing ideas and not really answering questions

[jaunty compass]

Oh i see, @wintry steppe

[half storm]

Oh o.k. I understand.

[jaunty compass]

So would I have to find the eigenvectors then to determine that

[hollow finch]

*diagonalizable. sorry to nitpick, but proper terminology is important when learning something imo

np we all make mistakes

[jaunty compass]

how do you take a power of a 2 by 2 matrix

is there a formula?

[gray dust]

yes, involves its jordan normal form @jaunty compass

[wintry steppe]

what exactly does it mean to be a position vector

@dusky epoch

[dusky epoch]

??

oh god. not this again.

[wintry steppe]

what do you mean again

..

[dusky epoch]

sorry, i'm a bit busy. i'll be with you in about an hour.

[wintry steppe]

ok

[dusky epoch]

"position vector" is only a thing you say when you insist on making a distinction between a point in R^n and the vector connecting it to the origin, which might make sense in a context like mechanics when you may need to keep track of multiple coordinate systems with different origins, in which case a single object may have different position vectors in different frames

[wintry steppe]

oh

[dusky epoch]

if you give me more context i can attempt to explain the relevance of the distinction (or lack thereof) for a particular problem of yours

[dusky epoch]

@wintry steppe

[wintry steppe]

@dusky epoch i don't have a problem

i just had a question about the definition

so idk what to tell you

[wintry steppe]

@dusky epoch why do 2 vectors determine a plane, but we need a third to determine where the plane is

[dusky epoch]

????

[wintry steppe]

in R^3

?

[dusky epoch]

two non-parallel vectors determine a plane's orientation but not position

you also need a point on the plane

[wintry steppe]

yeah so what does that mean exactly by the planes position

[dusky epoch]

you can move a plane around w/o rotating it and itll still be spanned by the same two vectors

[wintry steppe]

okay

so if i have

$a\begin{bmatrix} v_1 \ v_2 \ v_3 \end{bmatrix} + b\begin{bmatrix} w_1 \ w_2 \ w_3 \end{bmatrix}$

[stoic python]

polynomial:

[wintry steppe]

this gives me some plane, but not where the plane is? or

[dusky epoch]

this isn't a plane this is a vector

a linear combination of two fixed vectors which i presume you would name v and w

[quartz compass]

well if a,b are free parameters it's a plane through the origin

[wintry steppe]

a, b are not constants

[dusky epoch]

now if you actually MEANT ${ a \bd{v} + b \bd{w} \mid a, b \in \bR }$ then that's the plane parallel to $\bd{v}$ and $\bd{w}$ \underline{and passing through the origin}

[stoic python]

Ann:

[wintry steppe]

ok

so is this different than using a normal vector to describe a plane

i.e. a vector perpendicular to a vector that is parallel to the plane

[dusky epoch]

a normal vector is perpendicular to all vectors parallel to the plane

[wintry steppe]

sure

[dusky epoch]

anyway ${ \bd{x} \in \bR^3 \mid \bd{x} \cdot \bd{n} = c }$

[stoic python]

Ann:

[dusky epoch]

this is the description of a plane in terms of its normal

[wintry steppe]

why would you use this over the 3 vector method

like two vectors that determine a plane

and 1 more than determine the location of the plane?

[dusky epoch]

in this form it's easier to determine whether or not a point lies in the plane ¯_(ツ)_/¯

[wintry steppe]

any examples? 😦

[dusky epoch]

wym

i mean ok like

[wintry steppe]

your last sentence

[dusky epoch]

gimme a moment

the plane P can be described by the equation 3x + 4y - 7z = 11
alternatively it can be described parametrically as (x,y,z) = (1,2,0) + s(1,1,1) + t(0,7,4) where s and t range over the real numbers

if you needed to determine whether or not the point (5, 2, -8) lies in P, which form would you rather use

[wintry steppe]

[dusky epoch]

would you rather check a linear system for consistency or plug some numbers into an equation to see if your point satisfies it?

idk about you but i think one requires way less legwork than the other

[wintry steppe]

@dusky epoch how do you remember the cross product

[dusky epoch]

wym

[wintry steppe]

how do you recall what the cross product is when you are required to use it in linear algebra

[dusky epoch]

you mean how to calculate it?

[wintry steppe]

i mean sure

[dusky epoch]

...

[wintry steppe]

..

[dusky epoch]

if i need to calculate a cross product i usually go with the determinant trick

or the determinant shorthand rather

[wintry steppe]

how do you remember the determinant for 3x3 matrices lol

[dusky epoch]

laplace expansion ¯_(ツ)_/¯

[wintry steppe]

how

[dire thunder]

well it is pretty easy

i mean just row/column entries by signed minors

i.e cofactors

[dusky epoch]

wym how

idk like

do you expect me to give you the equivalent of a magical "never fuck up determinants again" pill

[wintry steppe]

i mean

idk how to remember 3x3 determinant

how do you do it

[dusky epoch]

i did like a few dozen to a few hundred determinants by hand until it stuck lol

[dire thunder]

how do u remember things usually lol, poly

[wintry steppe]

wtf

so there's no algorithm to make it easier

[dire thunder]

all "algorithms" for remembering are crap

[wintry steppe]

uh

no

?

[dire thunder]

i mean for someone it may be easier to remember things by learning while dancing simultaneously

[wintry steppe]

vimes

[dire thunder]

but it does not mean that dancing is good method for remembering things

[wintry steppe]

take your hands off the keyboard

please

[dire thunder]

no u

[dusky epoch]

is this what you were looking for, poly

[wintry steppe]

oh

so now how do you get the cross product

also why does that work lol

[dusky epoch]

some ppl define the determinant to be this way

i don't feel like going all the way through with reducing this to the leibniz formula or to the "unique alternating multilinear form sending the identity to 1" defn

the cross product can be calculated with the determinant shorthand as follows: $$\bd{v} \times \bd{w} = \begin{vmatrix} \bd{e}_1 & v_1 & w_1 \ \bd{e}_2 & v_2 & w_2 \ \bd{e}_3 & v_3 & w_3 \end{vmatrix},$$ if you have the guts to not wince at the matrix having the vectors of the standard basis as entries in its first column and just carry out the computation like you normally would

[stoic python]

Ann:

[wintry steppe]

[dusky epoch]

dunno what else to tell you don't think about it too hard

[wintry steppe]

@dusky epoch ann

if we have a plane consisting of all the points P satisfying $\overrightarrow{OP} = \mathbf{a} + s\mathbf{b}_1 + t\mathbf{b}_2$

[stoic python]

polynomial:

[wintry steppe]

suppose vector $\mathbf{k}$ is a normal vector to this plane

[stoic python]

polynomial:

[wintry steppe]

does that mean we just have to take the cross product of sb_1 and tb_2 and add vector a afterwards to get what vector k is?

[dusky epoch]

no

just b1 cross b2, up to a nonzero scalar multiplier.

[wintry steppe]

why don't we have to add a? it shifts the plane

[dusky epoch]

but the normal is intact.

it is possible for two different planes to have the same normal, poly.

[wintry steppe]

@dusky epoch but if we add a

our solution is still correct right

to the now-crossed product

[dusky epoch]

no it's not

[wintry steppe]

why not

[dusky epoch]

the normal + a will not necessarsily be normal anymore

extreme example: consider the plane z = 5 parameterized as (7, 7, 5) + s(1,0,0) + t(0,1,0)

(1,0,0) cross (0,1,0) gives you (0,0,1) as intended

and then you go and add (7,7,5) into the mix for no reason

and get (7,7,6)

and expect it to still be normal to the plane

so like

[wintry steppe]

ok

thanks

[viscid kernel]

Whats the geometric meaning of innerproducts ?

[dire thunder]

length of projection

[viscid kernel]

@dire thunder Ok so lets say a vector v in 2d undergoes a linear transformation and suddenly leaves the 2d space and gets map to the 3d space. The projection of the vector T(v) is on the 2d plane then ?

[dire thunder]

idk, i just know that one of the interpretation of innerproduct is length of projection

but i am not very good at LA

but if you mean that inner product of (x1, x2, x3) and (x1, x2, 0) is leght of projection then yes

[viscid kernel]

Hmm ok then, but if so isnt that just the same vector but without its z component ( so no hight ) and then the length of that vector without its height?

Thats what I meant xD

[dire thunder]

well yes if you will like do dot product of 3d vector and vectoor with the same two components and third zero it is projection

[viscid kernel]

Aight

I understand it now xD, but what is actually the purpose of inner products ?

is the inner product also valid for mappings from 3d to 2d ?

[fickle citrus]

Length of a projection is plenty of reason to me

By 3D to 2D, what do you mean

[viscid kernel]

I mean that is it possible if you reverse the proces.

[fickle citrus]

Do you mind drawing

By reversing the process, do you mean recovering the third component?

If so I don't think so unless you can add more information because there would be an infinite amount of possibilities that would project down to the same thing.

Also when you speak of inner product, it strictly produces a real number. You can use this real number in vector projection

[rigid cypress]

mapping an element of a 3d space to a 2d space isnt very specific, and there's not a direct method of mapping like there is for 2d -> 3d

[fickle citrus]

3D -> 2D has some sense of 'best' or 'closest' typically speaking I think

Or at least, you'd want to use those qualities typically

2D -> 3D is addition of information

[rigid cypress]

fair, i was thinking of it as just tacking on a 0, but yeah it can be messy as well

[viscid kernel]

nvm, Its not possible. I just figured it out xD still thanks for yall help.

[hollow finch]

Did someone say 'determinant'?
Just to add to a finished conversation, you can row reduce a determinant before expanding it.This makes determinants bigger than 3x3 not awful.

As for an interpretation, you can think of it as measuring area or volume.
u dot (v cross w) is the same as the volume of the parallelepiped formed by the three vectors (determinant where the three vectors form each column/row). If u is on the plane spanned by v and w, the volume of a plane is just zero. So that is why a determinant can function as a way to define a plane spanned by two vectors.

[latent ledge]

Suppose V and W are finite-dimensional and that U is a subspace of V. Prove that there exists $\mathcal{L}(V,W)$ such that null T = U if and only if $\dim U \geq \dim V - \dim W$.

[stoic python]

Konoha:

[latent ledge]

Suppose that there exists a linear map $T$ such that $\text{null}T=U$ where $U$ is a subspace of $V$, then
\begin{align*}
\dim\text{null}T&=\dim(U)\
&=\dim(V)-\dim\text{range}T\
&\geq\dim(V)-\dim(W)
\end{align*}
since $\dim\text{range}T\leq\dim(W)$. For the other hand, suppose that the subspace $U$ of $V$ satisfies $\dim(U)\geq\dim(V)-\dim(W)$. Let $u_1,u_2,\dots,u_n$ be a basis of $U$ and $w_1,w_2,\dots,w_k$ be a basis of $W$. Extends the basis of $U$ to $(u_1,u_2,\dots,u_n,v_1,v_2,\dots,v_m)$ as a basis of $V$. Let $a_1,\dots,a_n,b_1,\dots,b_m\in F$, and define $T\in\mathcal{L}(V,W)$ such that
$$T(a_1u_1+\cdots+a_nu_n+b_1v_1+\cdots+b_mv_m)=b_1w_1+\cdots+b_mw_m$$
where $m\leq k$.

[stoic python]

Konoha:

[latent ledge]

am I on the right track

[spice storm]

To find the the Orthogonal basis for R^2, Do you do U1 * U1 and U2* U2 or U1*U2?

[slow scroll]

I think that looks fine konoha

and @spice storm I don't know what you mean by

Do you do U1 * U1 and U2* U2 or U1*U2?
You can get an orthogonal basis from a non-orthogonal basis by applying gram schmidt

[spice storm]

Forgot to update. I got the answer

and @spice storm I don't know what you mean by
You can get an orthogonal basis from a non-orthogonal basis by applying gram schmidt
@slow scroll good to know. I haven't learn Gram Schmidt yet

[slow scroll]

ah okay

[neat peak]

how are these 2 definitions equivalent?

the second seems to be more general to me

[limber sierra]

Why does the second seem more general?

[neat peak]

it does not say it has to eat all those entries and send them to the field it is defined on

[slow scroll]

my understanding is that they say the exact same thing

[neat peak]

it does not say where it sends them

[neat peak]

ah ok

thanks

[slow scroll]

np

[weary isle]

In https://towardsdatascience.com/math-and-intuition-behind-affinity-propagation-4ec5feae5b23 , what is the purpose of making the maximum value of a(i,k) = 0 ?

[stoic python]

Added pounce!

This trigger already exists!

Added pounce!

[noble hemlock]

How do you prove that there cannot exist three matrices A, B, C of orders 3x2, 2x3, and 3x3 respectively such that ABC=I?

[half ice]

You're squeezing into a 2D space and expanding back into a 3D space. But you can't get that extra rank out of nowhere

To put it all dumb-like. So, you can't get any full-rank matrix this way, including I

Ah neat I like that proof

[noble hemlock]

@half ice nice proof, thank you!

Alternatively try to show that if ABC = I, then B is invertible
@wintry steppe I don't understand, can you show how this proof works? How do you prove B to be invertible?

[zealous widget]

is it possible to have matrics $A$ and $B$ where $BA=I$ ($B$ is left inverse of A), but $AB \neq I$?

[stoic python]

catfood:

[subtle walrus]

no

the proof depends on what you are willing to believe

$BA = I \implies BAB = B \implies AB = I$, if you are willing to believe that $B$ has a left inverse

[stoic python]

Lochverstärker:

[subtle walrus]

if you are willing to believe that A also has a right inverse, lets say C, the proof is similar

you can also instead consider linear transformations

@zealous widget

[zealous widget]

@subtle walrus OK thanks!

[tame mural]

A very silly question... some linear algebra texts use extra cursive letters for the objects they care most about; do people find it helps with scanning and readability?

[wintry steppe]

do you mean stuff like

mathcal L

or fancy letters like that?

that's super common in mathematics

[tame mural]

yeah

[dire bough]

Mm by 'extra cursive' I assume you mean fraktur font?

[wintry steppe]

as for readability, i think it's nice to have fancy letters here and there

[tame mural]

sometimes they get really curly

[dire bough]

$$\mathfrak T, \mathfrak F, \mathfrak R$$

[wintry steppe]

lets you know what's important and what isnt

[stoic python]

owomorphism:

[wintry steppe]

fraktur letters also show up

good luck reading most of them lol

[dire bough]

Also $$\mathscr{A B C D E F G}$$ etc

[stoic python]

owomorphism:

[dire bough]

haha

[wintry steppe]

a lot of this just boils down to the author's choice, but there are some scenarios in which alternate script letters are standard

e.g. lowercase fraktur for lie algebras

[half storm]

More notation to add to my collection >:)

Didn’t know these script commands in TeX

[wintry steppe]

\mathfrak{letter}

or \mathscr

[dire bough]

Generally speaking, I think it's pretty common to use calligraphic typefaces for 'large' structures like collections of sets, topologies, categories, vector spaces

[wintry steppe]

or \mathcal

[half storm]

Those are the contexts I’ve seen them used.

[wintry steppe]

you can find them all just by looking up "latex math fonts" or something, there are a lot you can choose from

[stoic python]

TTerra:

[tame mural]

oh wow video game font

[dire bough]

There are more established conventions like TTerra said; lower fraktur for lie algebras and ideals

$\mathcal L$ and $\mathcal F$ are more or less the canonical notation for laplace and fourier transforms

[stoic python]

owomorphism:

[wintry steppe]

i've seen \mathcal{L} a lot for the lie derivative

that may or may not be standard (as in, i dont know if it is, but it's all ive seen)

[half storm]

@subtle walrus If you go about proving it that route then how do you conclude that the product of AB is the identity. It seems intuitively true. But would you maybe prove by contradiction?

[dire bough]

Having written papers before, sometimes I'll be using $V$ for a vector space, for example, then realise I need to define a larger class of vector spaces. Then it's like, oh shit, better use $\mathcal V$

[stoic python]

owomorphism:

[half storm]

It’s not really obvious to me.

Like is there a way to prove it directly is what I’m trying to ask

[wintry steppe]

just take the usual letter for your object and let the fancy version denote a set of those objects

ez

[dire bough]

yup haha

[subtle walrus]

@half storm what route?

[half storm]

That it had to be that AB = I. Can you prove that directly?

[subtle walrus]

ye, but you need at least some lemma before

[half storm]

I guess you would have to prove that the identity is unique in that it is the only matrix such that AI = A and IA = A

[subtle walrus]

that too, but i would consider this well known

[half storm]

What’s the other lemme?

[subtle walrus]

that A has both a left and right inverse

thats probably what i would do

[dusky epoch]

I guess you would have to prove that the identity is unique in that it is the only matrix such that AI = A and IA = A
does this not follow from R^(n times n) being a ring

[subtle walrus]

and then show they are unique and the same

[half storm]

I have no clue.

Does R^(n x n) denote the set of n x n matrixes with entries taken from R?

[tame mural]

I believe Axler uses such notation

and I've seen it around Wikipedia

[subtle walrus]

if you are willing to admit that invertible matrices are just linear transformations, then you can also just argue that GL_n(R) is a group and it follows from general group theory

[tame mural]

Some people also do M(a, b)

[dire bough]

@half storm Yeah that's a pretty standard notation

though it is admittedly ambiguous

[subtle walrus]

but i guess the proof is the same as in the general (group) case

[half storm]

I guess my book uses nonstandard notation.

So you need group theory to prove that?

[subtle walrus]

not really, but you are secretly doing group theory

[dire bough]

covert ops group theory

[subtle walrus]

because invertible matrices form a group and this result is true in groups

[half storm]

So what Ann is suggesting is that it follows that under matrix multiplication the set of n x n matrices is a group yea?

And so you know that the identity is unique

Because for any group the identity under what ever operation your talking about is unique?

Oh nvm she is saying something about rings

I should have paid more attention in Abstract.

[wintry steppe]

It's an K-algebra

[half storm]

I remember this, can it be shown that if a set is a Ring with unity then the identity has to be unique?

Or does this not follow necessarily?

Cool. So you can use this to ascertain that I is unique?

I as in the identity matrix.

Cool I never knew this.

I was gonna say that LA should introduce some elementary group theory but probably not lol.

I was gonna use this as justification

But this is only one such example that it’s useful for proving a fact that I don’t think is trivial

[dusky epoch]

So what Ann is suggesting is that it follows that under matrix multiplication the set of n x n matrices is a group yea?
no.

[half storm]

Yea I figured out this you were talking about rings not groups.

[dusky epoch]

the set of INVERTIBLE n by n matrices is a group under multiplication.

[half storm]

That’s pretty intuitively clear.

[ocean sequoia]

does duality have anything to do with the $T /in L(V,W)$ like is it linked to that specific transformation in any way

[stoic python]

brzig:

[latent ledge]

Suppose $U$ and $V$ are finite-dimensional vector spaces and $S\in\mathcal{L}(V, W)$ and $T\in\mathcal{L}(U, V)$. Prove that $\dim null ST \leq \dim null S + \dim null T$

[stoic python]

Konoha:

[latent ledge]

I suppose the rank nullity theorem is the key, but I don't see how to apply it here so far

[brittle juniper]

Wow someone had this exact same exercise a while ago

[slow scroll]

smells like axler

[pallid rampart]

@ moderators?

Oh right

Yes

I think it is

[brittle juniper]

you can use the rank-nullity theorem on the restriction of T to Ker(ST) (you can check that its kernel is Ker(T) and its image is T(Ker(ST)) which is subspace of Ker(S))

lmao Whoever it was you who had this exercise

in February

[pallid rampart]

[brittle juniper]

20th

[pallid rampart]

Lmao

Oh right

I think I used a basis

I think I just pinged moderators on myself rip

[latent ledge]

suppose $u\in nullST$, then $S(Tu)=0$ implies $Tu\in null S$. this implies $range T\subseteq null S$. In addition, $Tu\in V$. I think I need a different $T$ for mapping from null ST to V, like $T'$. and then use RN theorem, we have the result after gathering all pieces

[stoic python]

Konoha:

[pallid rampart]

This does not imply $\text{range}T\subseteq\text{null}S$, since you did not assume $u$ is a vector in the domain of $T$

[stoic python]

Whoever:

[pallid rampart]

In fact if $\text{range}T\subseteq\text{null}S$ then $ST=0$

[stoic python]

Whoever:

[latent ledge]

Sorry about the confused notation, there were two different T. I should say T':null ST -> V, Tu\in null S implies range T' \subset null S

I believe that is what Tuong said

use this restriction map and then use RN theorem

[wintry steppe]

Can someone link me a resource to how to solve this?

[pallid rampart]

What is B

If B is a given matrix then you can just set each entry of D (or E) to be a variable, write out matrix multiplication, and solve for those entries

It will just be a system of linear equations in terms of the entries of D (or E)

[wintry steppe]

That makes sense, thanks!

[cobalt tartan]

Given that $Ax \geq b, Dx = f$, and that all entries of $x$ are less than or equal to 0, how do I find necessary and sufficient conditions for the existence of a solution?

[stoic python]

Liria ^(;,;)^:

[cobalt tartan]

Where A and D are matrices, b, f are vectors

[wintry steppe]

Well $Ax - b \geq 0$ and $Dx = f \implies Dx - f = 0$

[stoic python]

robo™:

[wintry steppe]

So $Ax-b \geq Dx - f$

[stoic python]

robo™:

[wintry steppe]

And u know the rest

@cobalt tartan

I assume A and D have the same dimensions

And that dim(b) = dim(f)

[cobalt tartan]

Oh I see

Thank you!

[half storm]

Need a little bit of help on this one. Having trouble even starting it to be honest.

[dusky epoch]

start by proving {f1, f2, f3} is linearly independent

this, combined with the knowledge that dim(V*) = 3, will give you that it is a basis

[half storm]

Got you

It kind of sucks not being able to see how it spans though 😦

But I do know that the linear combination depends on how the coordinate functions act on a basis.

i.e. the scalars that allow any linear functional to be represented as a linear combination of the coordiante functions relative to $ \beta $

[stoic python]

JohntheDon:

[half storm]

It might be easier then.

[dusky epoch]

well

you could show that the forms (x,y,z) |-> x, (x,y,z) |-> y and (x,y,z) |-> z can each be expressed as a linear combination of your f_i

[half storm]

Yea I tried that lol

But couldn't really make any progress

[median forum]

choose your factor and make the coefficients in a way they can cancel

there are already a few cues you can notice
for instance
f2 will have coeff zero for both x and y

oh mb

didnt notice the z in f3

brightness too high

youll linear systems anyway

[half storm]

You're talking about showing their linear independent?

[median forum]

not really

to show they span x,y and z

say you want a,b,c st af1+bf2+cf3=x

[half storm]

Yup yup

[median forum]

see what happens when you factor out x, y and z

[half storm]

You mean factoring them out of the coordinate functions?

[median forum]

not really

[half storm]

you mean for some arbitrary linear functional g

[median forum]

well, yeah

mb

basically factoring them out

x(....)+y(...)+z(...)=x

can you see where Im going from there?

[half storm]

Yea I think so.

[median forum]

what should ...., and ... equal

[half storm]

... is something like $f_1(e_1), f_1(e_2), f_1(e_3) $ ?

[median forum]

okay so

lets say you want af1+bf2+cf3=x

[stoic python]

JohntheDon:

[median forum]

when you factor x, y and z
youll have x(expression 1)+y(expression 2) + z(expression 3)=x

what are the expressions equal to, respectively?

[half storm]

$f_1(1,0,0) + f_2(1,0,0) + f_3(0,0,1), f_1(1,0,0) + f_2(0,1,0) + f_3(0,1,0), f_1(0,0,1) + f_2(0,0,1) + f_3(0,0,1)$

[stoic python]

JohntheDon:

[half storm]

Oh nvm there are a's b' and c's

I see

[median forum]

I dont think youre understanding me

lemme get to the pc

we want $\begin{pmatrix}a & b & c \ d & e & f \ g & h & i \end{pmatrix} \begin{pmatrix}f1(x,y,z) \ f2(x,y,z)\ f3(x,y,z)\end{pmatrix} = \begin{pmatrix}a & b & c \ d & e & f \ g & h & i \end{pmatrix}\begin{pmatrix}x-2y \x+y+z\ y-3z\end{pmatrix} = \begin{pmatrix}x \ y\ z\end{pmatrix}$

right?

[stoic python]

Fractal:

[half storm]

Yes.

[median forum]

ok so

lets do the first expression

$a(x-2y) + b(x+y+z) + c(y-3z) = x(a+b) + y(-2a+b+c) + z(a - 3c) = x$

[stoic python]

Fractal:

[median forum]

can you see now where we going?

[half storm]

Yea

you want a + b = 1 and (-2a + b + c) = 0 and a - 3c = 0

I'm a little confused why showing this show that this spans the set of linear functionals though.

Like how does this show that for $ g \in \mathscr{L}(V,F) g(x,y,z) = af_1 + bf_2 + cf_3$

[stoic python]

JohntheDon:

[median forum]

if these are solvable, you are showing f1,f2,f3 span the base functions

since any linear functional is a combination of x,y,z
you are showing they span any functional

[half storm]

Oh o.k.

[median forum]

I should say basis, but Im way too tired

[half storm]

any linear functional is a linear combination of x y and z since $ g(x,y,z) = x \cdot g(1,0,0) + y \cdot g(0,1,0) + z \cdot g(0,0,1)$ ?

[stoic python]

JohntheDon:

[median forum]

I guess so

seems tautological to me anyway

is there any other way to express them?

I mean, they are linear

but yes, your reasoning is right

[half storm]

o.k. I now see how solving the above system would tell you whether or not they span.

[median forum]

you can think of it as a change of basis

that matrix is the change of basis matrix

@half storm also, good to see you here

[half storm]

lol yea I spend a lot of time in here.

[median forum]

I cant stop answering questions

really messing my studying

[half storm]

Lol i like to when I can.

It actually is fun

[median forum]

ye

[half storm]

You just got of class today, is that why you're tired

[wintry steppe]

if we have the graph of x = 4-2t, y = 2+t, z = 2-3t

we can write this as

$\begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 4 \ 2 \ 2 \end{bmatrix} + t \begin{bmatrix} -2 \ 1 \ -3 \end{bmatrix}$

[stoic python]

polynomial:

[wintry steppe]

if we want to find a vector that's parallel to this graph

is it any vector in the form of

$\begin{bmatrix} x \ y \ z \end{bmatrix} = t\begin{bmatrix} -2 \ 1 \ -3\end{bmatrix}$

[stoic python]

polynomial:

[wintry steppe]

@dusky epoch

[dusky epoch]

oh, you have a non-default pfp now

[wintry steppe]

yeah i found it on google

do you like it

[dusky epoch]

yes

[wintry steppe]

thanks

[dusky epoch]

anyway yes a vector is parallel to a line iff it's parallel to its direction vector

[wintry steppe]

but what if we add something to said vector

not parametrized

then it's automatically not parallel if it's not a scalar multiple of the direction vector nor the zero vector?

[dusky epoch]

wym

as in if you take [-2; 1; -3] and add to it something that isn't a multiple of [-2; 1; -3]?

[wintry steppe]

$\begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 123 \ 2 \ 6 \end{bmatrix} + t\begin{bmatrix} -2 \ 1 \ -3\end{bmatrix}$

ugh

[stoic python]

polynomial:

[wintry steppe]

is this still parallel

i don't think so

[dusky epoch]

as in if you take [-2; 1; -3] and add to it something that isn't a multiple of [-2; 1; -3]?
if so then yeah what you'll get is another non-multiple of [-2; 1; -3]

okay so like

this LINE is parallel to the original line.

but the position vector of a point on it won't be.

[wintry steppe]

where can i graph these

[dusky epoch]

geogebra, i guess

[wintry steppe]

in 3d?

[dusky epoch]

yes

[wintry steppe]

@dusky epoch what does it mean that the position vector of a point on it won't be

i'm asking if that (last posted) point for some value of t will be parallel to my original graph?

[dusky epoch]

and i answered that negatively

[wintry steppe]

i don't understand what i am doing wrong

i have the points

(-1,2,4), (3,-1,2), (5,1,6)

i am trying to find the point (x,7,10)

so i have

$\begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} -1 \ 2 \ 4 \end{pmatrix} + t\begin{pmatrix} 3 \ - 1 \ 2\end{pmatrix} + s\begin{pmatrix} 2 \ 2 \ 4\end{pmatrix}$

[stoic python]

polynomial:

[wintry steppe]

then this gives t = -1, s = 2

but that's not right

why?

[dusky epoch]

uh

can i have the exact problem statement?

[wintry steppe]

find x if the the point (x,7,10) is in the plane through the points (above)

[dusky epoch]

well then your parameterization is wrong

[wintry steppe]

how

if the points above are A, B, C respectively

then my parametrization is

[dusky epoch]

then you have OA + tOB + sBC

maybe you meant AB and not OB

[wintry steppe]

thanks

sigh

isn't

OA + tAB + sAC correct...

[dusky epoch]

it is but what you wrote up there isn't OA + tAB + sAC

[wintry steppe]

i know

but i mean

$\begin{pmatrix} -1 \2 \4 \end{pmatrix} + s\begin{pmatrix} 4 \ -3 \ -2\end{pmatrix} + t\begin{pmatrix} 6 \ -1 \ 2\end{pmatrix}$

[stoic python]

polynomial:

[wintry steppe]

is not correct either

[dusky epoch]

wym

this does parameterize your plane now

[wintry steppe]

this gives (s,t) = (-2,-1)

[dusky epoch]

ok, and?

wait, does it now?

no it doesn't. the y-coordinate ends up as 9, not 7.

[wintry steppe]

nevermind

found my arithmetic mistake

finally

[ivory trout]

I get this

[wintry steppe]

how is that linear algebra

[ivory trout]

Euh idk i learn this on linear algebra

[wintry steppe]

no

[ivory trout]

Introduction to limear algebra

Ok...

In my country we do...

Where should i ask then

Thank you

[wintry steppe]

@dusky epoch

if a normal vector to a plane is [a;b;c], then the equation of the plane will be ax + by + cz = D

where a, b, c are the components of that normal vector

[dusky epoch]

you said the same thing twice

[wintry steppe]

what

[dusky epoch]

normal vector to a plane is [a;b;c]

where a, b, c are the components of that normal vector

i got it the first time round no need to repeat yourself

[wintry steppe]

i'm talking about the equation

of the plane.

[dusky epoch]

you introduce the variables a, b and c here

if a normal vector to a plane is [a;b;c],
and then use them here
then the equation of the plane will be ax + by + cz = D
you do not need to restate that a, b and c are the components of your normal bc youve already established them as such

anyway what were you actually gonna ask me lmao

[wintry steppe]

why is this the case

[dusky epoch]

ax + by + cz = [a;b;c] · [x;y;z]

so the equation will be $\bd{n} \cdot \bd{x} = D$ (where $\bd{n}$ is your normal)

[stoic python]

Ann:

[wintry steppe]

and what is x

[dusky epoch]

$\bd{x} = (x,y,z)$

[stoic python]

Ann:

[wintry steppe]

just an arbitrary point..?

[dusky epoch]

yes

[wintry steppe]

@dusky epoch can you help me understand some things about planes and lines

[dusky epoch]

maybe

[wintry steppe]

if we have like

x + y + z = 1

and then some line AB = v + tw

then what can we say about the intersections of these two from the coefs?

[dusky epoch]

the intersection exists iff the equation (v + tw) · [1;1;1] = 1 has a solution for t

[wintry steppe]

why = 1?

[dusky epoch]

x + y + z = 1

this is [x;y;z] · [1;1;1] = 1

[wintry steppe]

lol

[cobalt tartan]

Wait @wintry steppe I dont' see how to get necessary and sufficient conditions from that?

[pallid rampart]

For an operator on a finite dimensional complex vector space how do you find a basis such that the matrix of the operator with respect to the basis is in Jordan form?

[pallid rampart]

Of course

[pallid rampart]

So you find all the distinct eigenvectors $\lambda_1,\dots,\lambda_n$ and find a vector in $\ker((A-\lambda_iI)^d)$ for every $d$ and $i$?

[stoic python]

Whoever:

[pallid rampart]

@warm briar what do I do after this

[half storm]

So is this asking me that for any plane in R^3 passing the through the origin, I need to show that there exists a linear functional on R^3 such that every point lying on the plane gets mapped to 0? And if that is the case, is it asking me to find an explicit formula for that linear functional?

[pallid rampart]

I want a more systematic way

[wintry steppe]

@half storm a bit more precisely, you have to show that the null space of your linear functional is exactly that plane, not just that your functional sends all vectors in the plane to zero. this channel is occupied by other people, so if you have any questions about this, @ me in a questions channel

[half storm]

👍

[pallid rampart]

Can you be more specific

I'm not sure what to do if you just tell me to find the generalized eigenvectors

How do I find the size of the blocks

and if I find these vectors, it is very possible that I find more than n vectors so I will just need to randomly select n vectors to see if they're linearly independent?

Alright I'm still not sure how to do it

I'm just gonna try to find something online again

[void slate]

Hi, “If <x,y>=<x,z> for all x in V, then y=z” to prove this as I supposed to say “Supposed <x,y>=<x,z> then...” or how would I show this? This problem is working with the nxn matrices in the real numbers. I’ve proved the first four components of an inner product space but I’m slumped on how to start this one off

To prove this am* I... sorry for the typo

[limber sierra]

are you familiar with proof by contrapositive?

that's how i'd want to approach this

wow for some reason that didnt even cross my mind

but yeah it works too

(and is more directly constructive!)

[void slate]

Ah okay, let me try it out! Brb

Is this what you were getting at?

I would type it out with the texit but I haven’t learned the language yet, it’s on my lists once my summer classes are over sorry😅

[void slate]

Yess <A,B>=tr(AB)

Should I not write iff?

Ah okay, we might learn that tomorrow in class, I’m getting a bit ahead so we haven’t touched the topic at all. And thank you for the correction and help! I really appreciate it

[void slate]

I saw an example in the book with it going to C but I’m curious as to how it would change now that it’s going to Complex? I’ve proved <x,y+z> but now that I gotta price <x,cy>= conj(c) <x,y>. Is this possible? Normally for the field being R taking the conjugate would just flip the functions for me to manipulate it how I’d need to (since a in R is also a+0i in complex so it’s conjugate would be a-0i=a) am I overthinking it?

Prove* not price

Ugh another typo, an example in the book with it going to R* man it’s late

[dusky epoch]

the formula doesn't give a complex inner product

you need to integrate f * conj(g)

[wintry steppe]

how does $\vert| \mathbf{u} \times \mathbf{v} \vert| = \vert| \mathbf{u} \vert| \vert| \mathbf{v} \vert| \sin \theta$ fail if $\theta > \pi$

?

[stoic python]

polynomial:

[half storm]

Why would it fail?

[wintry steppe]

it should be |u x v| = |u| |v| |sin theta|, and if sin theta > 0, you can drop the absolute value sign and get the thing polynomial posted

if 0 < theta < pi, then the formula polynomial posted works

[half storm]

Does it have something to do with the vectors forming a parallelogram

[wintry steppe]

for some values of theta > pi, the formula polynomial posted fails, because of the missing absolute value signs

[half storm]

Oh ok

But slapping the absolute value signs on there should do it yea?

It should always work then ?

[wintry steppe]

yeah, you can see it if you take this to be your cross product

[wintry steppe]

what is n

@wintry steppe

n is a unit vector perpendicular to the plane containing a and b in the direction given by the right-hand rule

from wikipedia

my keys are fucked up right now so im copypasting

never understood the right hand rule

😩

[wintry steppe]

@dusky epoch please help me 😦

[dusky epoch]

idk how i can help you

[wintry steppe]

i am trying to understand the scalar triple product

why is that the area of the parallelpiped

[dusky epoch]

the volume you mean?

[wintry steppe]

sure

[dusky epoch]

can i use without explanation the following fact

the magnitude of v×w is the area of the parallelogram with v and w as sides

[wintry steppe]

well i think i can explain that myself. consider a parallelogram with sides v, w. suppose t is an angle of the parallelogram. then |v| * sin t is the height of the parallelogram, and |v| * sin t * |w| is the area.

this is equal to v x w from the thing i posted above

though not really sure how the get that but w/e i guess..

[dusky epoch]

i didn't ask you to explain that yourself

i asked whether or not i can use it without further explanation

as i was going to rely on it

[wintry steppe]

well, i wanted to explain it so that you can see if it's necessary or not

[dusky epoch]

...

[wintry steppe]

what

[dusky epoch]

aight im out

[wintry steppe]

???

what did i do wrong

i am so confused..

@dusky epoch can you explain without using the fact?

[dusky epoch]

guess not

[wintry steppe]

why not??

sorry

i meant

you don't have to explain

the derivation of v x w

being the area

[dusky epoch]

you said you wanted me to explain why u · (v × w) is the volume of the fucking parallelepiped without using the parallelogram thing

[wintry steppe]

i think we're miscommunicating, if you want me to accept that v x w = |v||w| sin t

then i can do that

(and hence area of a paralellogram with sides v, w)

[dusky epoch]

hrgh ok lemme try to cook up a visual

[wintry steppe]

how do you get that first equivalence

lol..

[dusky epoch]

the plane spanned by u and v×w is perpendicular to that spanned by v and w

[wintry steppe]

no i mean why is that thing equal to cos t * |u| * |v x w|

where did the cos come from

[dusky epoch]

dot product

[wintry steppe]

of what

[dusky epoch]

u and (v×w)...

[wintry steppe]

oh

that's just the fucking

lol

i get it now

jesus

in that form i didn't see it at all

honestly it's crazy that this stuff gets discovered

anyway

do you know

$\begin{pmatrix} + & - & + \ - & + & - \ + & - & +\end{pmatrix}$

[stoic python]

polynomial:

[wintry steppe]

this for remembering the 3x3 det?

do you know if everything besides the first column is irrelevant? because i feel like it is

[dusky epoch]

wym

i mean sometimes you can expand in terms of a row instead of a column

[wintry steppe]

so if you want to compute 3x3 det

[dusky epoch]

and maybe a column other than the first

[wintry steppe]

do you remember the

[dusky epoch]

as a shortcut

[wintry steppe]

lines you did

the red lines?

ugh

1 sec.

[dusky epoch]

the strike-out

yeah that's still a thing

[wintry steppe]

$\mathbf A=\begin{pmatrix}
{\color{red}a}&{\color{blue}p}&{\color{black}x}\
{\color{red}b}&{\color{blue}q}&{\color{black}y}\
{\color{red}c}&{\color{blue}r}&{\color{black}z}
\end{pmatrix},$

[stoic python]

polynomial:

[dusky epoch]

you still compute the minors corresponding to each entry in the row or col you're expanding over

[wintry steppe]

$\text{det}(\mathbf A)={\color{red}a}{\color{blue}q}{z}
-{\color{red}a}{\color{blue}r}{y}
+{\color{red}b}{\color{blue}r}{x}
-{\color{red}b}{\color{blue}p}{z}
+{\color{red}c}{\color{blue}p}{y}
-{\color{red}c}{\color{blue}q}{x}.$

[stoic python]

polynomial:

[dusky epoch]

ok that's basically sarrus' rule

[wintry steppe]

$+\begin{pmatrix}
{\color{red}a}&{-}&{-}\
|&{\color{blue}q}&{\color{black}y}\
{|}&{\color{blue}r}&{\color{black}z}
\end{pmatrix} -\begin{pmatrix} {|}&{\color{blue}p}&{\color{black}x}\ {\color{red}b}&{-}&{-}\ {|}&{\color{blue}r}&{\color{black}z} \end{pmatrix} +\begin{pmatrix}
{|}&{\color{blue}p}&{\color{black}x}\
{|}&{\color{blue}q}&{\color{black}y}\
{\color{red}c}&{-}&{-}
\end{pmatrix}.$

[stoic python]

polynomial:

[wintry steppe]

$\begin{pmatrix} + & - & + \ - & + & - \ + & - & + \end{pmatrix}.$

[stoic python]

polynomial:

[dusky epoch]

ok sure

[wintry steppe]

notice how only the pluses/minuses

for the leading terms matter?