linear-algebra

[limber sierra]

yes, and you should be able to prove this

linearly independent means theres values $c_1, c_2, c_3$ not all equal to $0$ such that $c_1X + c_2Y + c_3Z = 0$

[stoic python]

Namington:

[limber sierra]

so if we take this same sum for your new set of vectors we have

[cursive narwhal]

dependent

[limber sierra]

$c_1X + c_2(aX + Y) + c_3(bY + Z) = 0$

whoops yeah, linearly dependent*

typo

[stoic python]

Namington:

[limber sierra]

anyway yeah looking at this equation we can just expand

$c_1X + c_2aX + c_2Y + c_3bY + c_3Z = (c_1 + c_2a)X + (c_2 + c_3b)Y + c_3Z = 0$

[stoic python]

Namington:

[limber sierra]

can you see why you can make this equal to 0 with nonzero coefficients?

[cursive narwhal]

@safe garnet

[safe garnet]

hmmm

if c3 is 0 then its good

[limber sierra]

||it's because we already know X, Y, Z are linearly dependent||

[safe garnet]

oh

[limber sierra]

||and we're allowed to choose a, b to be whatever we want||

||so since there's coefficeints such that d_1X + d_2Y + d_3Z = 0, we can just fix the a, b values to force c_3 = d_3, c_2 + c_3b = d_2, c_1 + c_2a = d_1||

||by setting b = (d_2-c_2)/c_3, a = (d_1-c_1)/c_2||

side comment:

if you've seen gaussian elimination before

and used it to check linear independence

this is why that works

[safe garnet]

ooo, ive used gaussian but havent used it to check independence

[limber sierra]

this basically guarantees that "adding rows to each other" will never "change a dependent system to an independent one"

the relation holds the other way; adding rows to each other can't change an independent system to a dependent one

so using gaussian elimination to check linear independence works

[at least adding rows, but it should be obvious that swapping rows won't affect it, and it's not hard to check that scalar multiples don't either]

[safe garnet]

that makes sense

so it holds for any a, b?

[limber sierra]

yep

er

well

yeah

sorry i misstated an argumetn previously, i shouldve said you can pick c_1 and c_2 to make it work

not pick a and b

same idea applies though

[safe garnet]

yeah that makes sense

i was a bit worried, because you were talking about fixing a,b

[limber sierra]

yeah no worries, mentally twisted my words

my bad my bad

[safe garnet]

no problem

thanks for the help. I think it makes sense, ill digest for a bit

I had justified it with determinants, but was having trouble visualizing

[limber sierra]

determinants are pretty bad at visualizing in general haha

[safe garnet]

yeah

they are helpful for arithmetic/algebra

but visualization wise they are pretty abstract imo

thanks again

[polar badger]

maybe stupid question: is every vector linearly dependent with the zero vector (in R^n)?

[pallid rampart]

If a set of vectors contains the 0 vector then the set is not linearly independent

[cursive narwhal]

Let V be a vector space and take any element v of the vector space. If {v,0} was a linearly independent set of vectors, then:

av+b0 = 0 => a = 0 and b = 0

But you can have a = 0 and b = 1. That would still give you the zero vector but not satisfy the required condition. Since our choice of v was arbitrary, we have shown that any set containing the zero vector is linearly dependent.

[polar badger]

yeah ok that was what I was thinking, thanks

[limber sierra]

[and indeed we need this to be the case for "using gaussian elimination to check linear dependence" to make sense]

[half ice]

It's easiest to think in terms of a linear function:
f(a + a', b + b') = f(a,b) + f(a',b')

That is, f "distributes into addition"

Now, f in this case is the determinant where the input is the first row

Linear functions are very important, they are the main object of study in linear algebra. So, it's good to see any examples of them

.
This ends up being an alternative to matrix multiplication, which is the common way to construct a linear function, but there are other ways.

[raven nebula]

can someone explain how to decode an encoded matrix with the inverse of the encoding matrix if the encoding matrix has different number of rows than the encoded matrix?

[pale shell]

How to know if a matrix is surjective or injective or bijective

[dusky epoch]

you don't because it doesn't make sense for a matrix to be any of these things

however, you probably meant "how do i know if a transformation is surjective, injective and/or bijective based on its matrix in some basis?"

to which one might then reply with:

calculate the rank of your matrix

if it matches the column count, your transformation is injective. if it matches the row count, your transformation is surjective.

and of course a transformation is bijective iff it's both injective and surjective at the same time.

[pale shell]

ok

Thank you

[limber sierra]

[it follows that only square matrices can be bijective, which should make intuitive sense as a transformation is bijective iff it is invertible]

[severe magnet]

Need help with a proof since I can't quite come to how to prove it

Let $F:V \to V$ be linear and dim $V = n$ with $r = \rank F > 0$. I need to show equivilence between two statements:

[stoic python]

Nabil:

[severe magnet]

(i) F^2 = F

[stoic python]

Nabil:

[severe magnet]

I assume the representation matrix with respect to the basis is the identity matrix up to r and afterwards 0's on the diagonal for i,j>r

I haven't managed to find an approach to this proof so would appreciate any help

[brittle juniper]

Well first you can easily see that (ii) implies (i), the hard part is to show that (i) implies (ii)

and for that, you may be interested in showing that

$V=\Ker(F)\oplus\im(F)$

[stoic python]

Tuong:

[severe magnet]

for (ii) implies (i), we know that F(a_j) = a_j for some basis A and j = 1,...,r right

would that mean that F o F(a_j) = F(F(a_j)) = a_j?

[brittle juniper]

Yes

[severe magnet]

I can remember having proven that equivilence in the past

lemme think for a sec

Let $V$ be a $K$-vectorspace and $F:V \to V$ be a linear transformation with $F^2=F$. Show that $V=Ker(F) \oplus Im(F)$. $\$ proof: Let $x \in V$. $\$$F(x)=w \neq 0$ $\ \iff F^2(x)=F(w) \ \iff F(x) = F(w) \iff F(w-x) = 0 \implies w-x \in Ker(F \circ F) \supseteq Ker(F)$ (proven in another exercise).

[stoic python]

Nabil:

[severe magnet]

I guess I am missing that Im F n ker F = {0} but that's how I wouldve done it

because that also means that x-w = u is in ker F implying that x = u + w for u in ker F and w in Im F

[brittle juniper]

it's a bit hard to read but the finishing looks alright

and now, from V=Ker(F)\oplus Im(F), are you able to show (ii)?

[severe magnet]

yea we can just choose the basis vectors of Im F and ker F as our basis for V

if I got that right that should give us our representation matrix

is that correct?

[brittle juniper]

Yep

[severe magnet]

ty for the help!

[brittle juniper]

You're welcome

[nocturne arch]

how to find the basis of the null space of a matrix modulo n, n can be a non prime. help please 🙏

[wary terrace]

is it only x3 that's the free variable? or do i need to put it in row echcelon form or something

[nocturne arch]

row echcelon form

[wary terrace]

reduced row echcelon form or

just row echcelon form

ah ok i got it thanks!

[nocturne arch]

B

I think so 🤔

[wary terrace]

its def not A and def not C

[nocturne arch]

The reason i think its B

[wary terrace]

and D is wrong also

so yea i guess B makes sense but why?

[nocturne arch]

non zero determinant

if it had zero determinant

then its columns would be linearly dependent

meaning the null space would have more than simply the trival solution

[wary terrace]

ah ok makes sesne

thanks!

[nocturne arch]

np

i need help please 🙏

how to compute echelon form over ring of integers mod non prime

[wary terrace]

idk how to do that LOL

[nocturne arch]

i have only started learning linear algebra so not 100% sure what the notation means

can u explain the question in words

does it just want the coefficents that result in that linear combination?

D?

[wary terrace]

hmm

idk im a bit stuck on this one too so LOL

[hallow cliff]

what does [x] mean?

[gray dust]

[x]_E denotes the coordinate vector of x wrt the basis E

let $B=\brc{b_1,\dots,b_n}$ be an ordered basis for a vector space $V$, then for any $v\in V$ there exist scalars $c_1,\dots,c_n$ such that
$$[v]_B=(c_1,\dots,c_n)\quad c_1b_1+\dots+c_nb_n=v$$

[stoic python]

RokettoJanpu:

[glass harness]

classify the system by the distinct values of m

someone help me

[quasi vale]

@glass harness Try turning it into a matrix of coefficients, then rref

[glass harness]

got it in rref

what do i do after that

[quasi vale]

what did u get

[quasi vale]

that's not rref since the pivots aren't 1

what's on the other side

[glass harness]

i checked on a matrix calculator and thats what i got for rref

[quasi vale]

@glass harness

this is the rref, got it from a website

you need to take the whole augmented matrix

[glass harness]

oh

[quasi vale]

now during the process of row operations

[glass harness]

ok thats why its wrong

[quasi vale]

we divide by m-1 and (m-1)(m+2)

so m cannot have values 1 and -2

Now if you leave the last column, the rank of the coefficient matrix is 3. If you include the last column, the rank of the augmented matrix(Ab) is also 3, provided that m is not equal to -2 or 1

Since the variables are also 3(x,y,z), we have a unique solution.

[glass harness]

so the system would be one-solution or unique solution?

[quasi vale]

both are same

it's better to say

the system is consistent since there is at least one solution

[glass harness]

ok

i was starting to think that m was a free variable

but then i realised that no it aint

right

[quasi vale]

yeah it can't be a free variable

[glass harness]

i find it funny that linear algebra and calculus are kinda the opposite

the math behind linear algebra is easy while the math behind calculus is hard

but the concepts behind linear algebra are difficult to learn

while calculus the concepts are easy to learn

[ocean sequoia]

Is R^2 a subspace of C^2? my thought is no?

[nocturne arch]

can someone please help me 😭

how to compute echelon form over ring of integers mod non prime

[severe magnet]

@ocean sequoia why do u think it is not?

I mean that is true

[ocean sequoia]

because its not closed under multiplication right? Cant you multiply x * yi and thus are no longer in the set?

honestly im not sure tho

if thats wrong could i get a hint where to go?

[half ice]

There's no yi in R²

[ocean sequoia]

dang

yea that makes sense

duh

[half ice]

Try working with complex numbers of the form x + 0i

You'll find they'll always have a complex part of 0, even when added and multiplied together. Of course 0 + 0i is in there too

This is isomorphic to R², and is a subspace of C²

Whether or not we might call this space R² is tricky

[ocean sequoia]

Ok ill give it a shot I wanted to avoid googling it cause i dont just want the answer

thanks

[severe magnet]

what is the exact question

do u want to check whether these are subspaces?

ah alright I just saw

[half ice]

Here's a question I have. I remember seeing a theorem that says something like:
A vector space over F of dimension n is unique. It's isomorphic to any other vector space over F of dimension n.

I can't find that theorem now though haha. Am I remembering this correctly?

[main drum]

I have an intro linear question

If I have more variables than equations, is it possible for the linear system to have a unique solution

[half ice]

No

[main drum]

Is it because there is a free variable?

So therefore infinite solutions

?

[half ice]

Ya

[main drum]

ok ty

Also, if the null space of n by n contains only the zero vector

Is that matrix invertible?

Sorry for asking so many im doing my hw rn

I was thinking it is bc it has no free variables

[half ice]

Think of it this way,
In order for a system to have a unique solution, its matrix representation should have an inverse. That only can happen if the matrix is square, which means # of variables = # of equations

[main drum]

yea that makes sense

[half ice]

The below are all equivalent:
⇔ Matrix is bijective
⇔ Matrix is invertible
⇔ Matrix has a trivial nullspace

[main drum]

thats the matrix invertible theorem

😄

im doing 1000 true false questions for practice before my final on monday

so i might have a couple more q's here and there 😄 thanks

[slow scroll]

@half ice yea, that is true.

[main drum]

i know that matrices with the same eigenvalues are similar

is the reverse true though? (if they are similar, same eigenvalues)

[limber sierra]

yep

proof idea: say $Av = \lambda v$, then $SBS^{-1}v = Av = \lambda v$

[stoic python]

Namington:

[main drum]

in this textbook it says distninct

does that mean anything?

[limber sierra]

i'd assume it means "not considering algebraic multiplicity"

like for example, if the characteristic polynomial of a matrix is (x-1)(x+3)^2

one might say that -3 is an eigenvalue "twice"

[main drum]

yea

[limber sierra]

but in this case, we don't consider that stuff

[main drum]

ah ok

thanks!

[limber sierra]

we just consider it as the singular eigenvalue -3

that said

[main drum]

right

[limber sierra]

similar matrices have the same eigenvalues with the same geometric multiplicity

[main drum]

that makes sense

but the algebreic is not considered

[limber sierra]

yeah, i'd assume that's what it's trying to say

[main drum]

if matrix A has eigenvalue of 2 with alg multiplicity 2, and matrix B has eigenvalue of 2 alg multiplicity 1, are they similar?

[limber sierra]

similar matrices have the same characteristic polynomial

the characteristic polynomials in those cases vary

so no

[main drum]

oh

ok

So matrices with the same eigenvalues are similar matrices?

or does it depend on the multiplicity

[limber sierra]

with the same eigenvalues factoring in algebraic multiplicity

er wait

not necessarily

consider for example

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

[stoic python]

Namington:

[limber sierra]

these clearly arent similar but they have the same eigenvalues with same multiplicies

[main drum]

same eigenvalues

yea

oh ok

[icy osprey]

can I ask a question? I can wait till y'all are done.

[limber sierra]

so we can only conclude that matrices with teh same eigenvalues are similar

if all of the eigenvalues are distinct

ie if they all have algebraic multiplicity 1

@main drum

[main drum]

yea

that makes sense

thanks

[icy osprey]

How did my prof get matrix A?

[limber sierra]

but the converse statement holds in all cases

similar matrices always have the same characteristic poly

[main drum]

yea

makes sense now

[limber sierra]

@icy osprey this is a vandermonde matrix

[icy osprey]

continue 👀

[limber sierra]

each row is of the form 1, x, x^2, x^3

where x is one of the x values

in your dataset table

[icy osprey]

wait, how would that be for the row?

[limber sierra]

our first x value is -2

so our first row is

[icy osprey]

yes

[limber sierra]

1, -2, (-2)^2, (-2)^3

or in other words

1, -2, 4, -8

our second x value is 0

so our second row is 1, 0, 0^2, 0^3

or 1, 0, 0, 0

[icy osprey]

ohhhh I see now !

[limber sierra]

our third x value is 1, so our third row is 1, 1, 1^2, 1^3

etc

[icy osprey]

thanks!

[limber sierra]

if you're wondering why this works

we're basically solving the system

[in this case for n = 3]

[icy osprey]

Gotcha, I see now

[limber sierra]

we're just writing it in matrix form

[icy osprey]

and then you just solve using gaussian elimation

cool

[limber sierra]

if you're doing it by hand, that's the easiest way yeah

there are fancy techniques that let computers compute solutions to vandermonde systems faster

since the matrices involved can often be very large

but they're hard to do by hand

[and unnecessary for small matrices]

[icy osprey]

So

this is the same thing right?

but they need to be orthogonal

[limber sierra]

yeah same gist

[icy osprey]

cool! appreciate the help!

👍

[main drum]

hi ive a question if you guys are done

so I know that a regular stochastic matrix always has a steady state

does a non regular stochastic always have a steady state?

or sometimes?

[main drum]

hey if anyone is on I have another question

nvm figured it out now I have a different question

for anyone online

[azure owl]

just post it and see if anyone replies

[main drum]

Its about least squares solutions.

So I know if A is linearly independent, it has a unique least squares solution\

If A is linearly dependent and has a free variable, does it have infinite number of least square solutions

is anybody online :

Hi

So I know if A is linearly independent, it has a unique least squares solution

If A is linearly dependent and has a free variable, does it have infinite number of least square solutions

anybodddddddddddddddddddddddy

YES I FOUND IT ONLINE AFTER AN ENTIRE HOUR

HWHEWHEHWEHWEHWHEWHEHWEHWHEHW

[odd kite]

and?

[main drum]

its true

where you been timon

past hour has been a headache

doing this one hw problem

lmao

[reef prism]

https://cdn.discordapp.com/attachments/387447715312828416/704802646447620236/unknown.png

which ones are true/false
change of basis matrices T and A from B to S
am i correct saying first two are false last two are true?
i dont have the answer keys with me

[main drum]

hey anyone on for a question

pls

[half ice]

No

[main drum]

So I know if A is linearly independent, it has a unique least squares solution

If A is linearly dependent and has a free variable, does it have infinite number of least square solutions

@half ice if you are still on and want to be kind 😄

[half ice]

I'm not sure what a least squares solution of a matrix is, haha

[main drum]

I got another

Do you know quadratic form

@half ice

[half ice]

What's the question?

[main drum]

So for positive semi definite

That happens when all coefficients are positive

Right?

mr @half ice

[half ice]

I see "positive definite" if Q > 0 for non-zero x

That's not the same thing as "all the coefficients are positive"

[main drum]

Can I show you a diagram from my lecture notes

[half ice]

Sure

[main drum]

Positive semi-definite is when it is greater than or equal to 0

[half ice]

K

[main drum]

And positive-definite is greater

@half ice

[half ice]

What was the question?

[main drum]

Im talking about the coefficients in front of x1 and x2

So if both are positive

Does it mean it is always positive-semi definite

?

x1^2 + x1x2+ x2^2

[half ice]

Consider
Q = x1² + 500x1x2 + x2²
And (x1,x2) = (1,-1)

[main drum]

That would make Q less thhan 0

[half ice]

That form isn't positive definite, but has positive coefficients

[main drum]

is it positive semi definite though?

[half ice]

No since Q < 0

[main drum]

Can I show you this question in my hw

im stumped

[half ice]

Sure

[main drum]

here i write it

True or false:

If every coefficient in a quadratic form is positive, then the form is positive semi-definite

oh

wait why

[half ice]

Oops, I formed the backwards statement when I replied

So given our conversation above, the answer is...

[main drum]

false?

[half ice]

Yaya

[main drum]

i thought it was true

now even bigger confused

[half ice]

Strange, because I gave a counter example haha

[main drum]

is that how quadratic forms work

Where you can insert values and solve for Q

sorry im bad at this lol

Q = x1² + 500x1x2 + x2²

@half ice

i see why its false

You can setup the matrix A and solve for the eigenvalues a

the eigenvalues of ur example are pos and neg even though the coefficients are positive

[half ice]

These are not the same as eigenvalues, one thing to mention

But yeah that's a Q that goes negative

[main drum]

i see

Quick easy question

Not every matrix is a product of elementary matrices right?

Bc it has to be square

@half ice\

[half ice]

Every invertible matrix is a product of elementary matrices

If your matrix isn't invertible, it has a non-elementary factor

[main drum]

Ahh ok

So is my thinking right

Like would that still be true

@half ice

are these 2 matrices similar

or anybody else that knows

should be pretty easy

Hey quick question to anyone tha ti son

If a linear system has infinitely man solutions, does it also have infinitely many least squares solutions?

[cursive narwhal]

Okay, so this really has more to do with the properties of the determinant than anything else. In particular, the matrix on the left is an upper triangular matrix so you can actually prove, by way of induction, that the determinant of an upper triangular matrix is just the product of the entries on the diagonal.

That's why you can disregard (2,3 and 8) in calculating the determinant.

Have you learnt about how the determinant reacts to elementary row operations?

Okay, so imagine you had a 3 x 3 matrix where the diagonal values were just 1,2,3. We will call it A. Imagine how you would get from the identity matrix to A. You'd multiply the second row by 2 and multiply the 3rd row by 3, leaving the first row unchanged.

So, if you know how the determinant reacts to these specific elementary row operations, then it should be clear that det(A) = 2 * 3 * det(E) = 6*1 = 6

Well, the interchange of rows is just one of 3 different types of elementary row transformations.

There are two others that need to be considered and you need to know how the determinant changes when you apply these row transformations to the given matrix.

[cold topaz]

If U is a subspace of V, how do I prove that the orthogonal complement of U is also a subspace of V?

[cursive narwhal]

@cold topaz Do you know how to check if a given subset of a vector space is a subspace or not?

There are 3 things that you need to check

[cold topaz]

yes. It is Subspace if Axioms 1 and 6 pass.

[cursive narwhal]

I don't know what axioms 1 and 6 are

[cold topaz]

and U is not 0.

I mean additions and scalar multiplication.

[cursive narwhal]

You need to prove that it contains the zero vector and is closed under addition & scalar multiplication

Yep. So, use the properties of the inner product to do that

In this case, you'll have to use the bilinearity of the inner product and the definition of the orthogonal complement.

Okay, let me give you a simple example.

Let A be a 3 x 3 matrix such that the diagonal has entries (2,4,6) and everywhere else is 0. Now, there's a simple way of getting this matrix from the 3 x 3 identity matrix E by using elementary row operations. All you have to do is to multiply the first row of E by 2, the second row by 4 and the third row by 6.

Now, what I'm saying is that if we have a matrix A and we turn it into a matrix A' by multiplying a given row by a scalar b, then det(A') = b*det(A). So, in the specific example above, we transformed E into A by multiplying each row by some scalar. In particular, det(A) = 2 * 4 * 6 * det(E) = 48.

In your specific example, what has been done is to reduce the matrix on the left into a diagonal matrix with entries (1,2,8) on the diagonal and 0 elsewhere. That can, then, be further reduced into the identity matrix, though it's not really necessary.

The moral of the story, I guess, is that you probably won't understand much of this if you're not familiar with the properties of the determinant as a map and the consequences that follow from those properties. So, try to cover it and it'll be a bit easy to follow the computations of whichever book you're using.

@cold topaz Were you able to solve your problem?

[cold topaz]

I'm trying to figure out how to use bilinearity to show scalar multiplication.

[cursive narwhal]

Let U be your subspace and $U^{\bot}$ be the orthogonal complement of $U$. Suppose that $\alpha \in \bR$ and $u \in U^{\bot}$. You need to show that $\alpha \cdot u \in U^{\bot}$. Let $v \in U$ be arbitrary. Then:

$\langle \alpha \cdot u,v \rangle = \alpha \cdot \langle u,v \rangle$

[stoic python]

Abhijeet Vats:

[cursive narwhal]

So, what can you do with that?

@cold topaz Assuming you're still stuck, of course

[cold topaz]

@cursive narwhal what you wrote is not bilinear. right?

[cursive narwhal]

What do you mean?

It is an inner product so I did use the fact that it's bilinear

[cold topaz]

so is that it? :/ ur answer seems complete.

[cursive narwhal]

Nope, that's not complete

I mean, if $\alpha \cdot u$ does belong to the orthogonal complement, what is its relation to $v$ with respect to the inner product? What actually happens?

[stoic python]

Abhijeet Vats:

[cold topaz]

v is multiplied by alpha, too.

[cursive narwhal]

Well, no. We know that $v \in U$ and $u \in U^{\bot}$. So, what is $\langle u,v \rangle$?

[stoic python]

Abhijeet Vats:

[cursive narwhal]

Use the definition of the orthogonal complement

[cold topaz]

then v and u are perpendicular to each other.

their product is 0.

[cursive narwhal]

Yes, their inner product is 0. So, what can you say about the inner product of v and au?

[cold topaz]

=0?

[cursive narwhal]

Yeap. So, what does say about au?

[cold topaz]

it says au is also an orthogonal compliemnt.

@cursive narwhal

[cursive narwhal]

Uh, no. A vector is not an orthogonal complement. But that vector does belong to the orthogonal complement

That proves that it is a set that is closed under scalar multiplication

[cold topaz]

....

But since in the answer there is no mention of V, how do we know that it is about V?

[cursive narwhal]

What do you mean?

[cold topaz]

this was my question:
If `U` is a subspace of `V`, how do I prove that the orthogonal complement of `U` is also a subspace of `V`?

[cursive narwhal]

Right, so we already know that V is a vector space over some field and that U is a subspace of V. Once we know that, we just need to show that the 3 given properties for subspaces hold for the orthogonal complement and that will be enough to show that it is a subspace of V

[cold topaz]

so we're basically done?

[cursive narwhal]

Well, you've shown that it's closed under scalar multiplication. Have you shown that it's closed under addition?

Also, I'm rather surprised that you're learning about orthogonal complements but do not know how to check if a subset of a vector space is a subspace

[cold topaz]

let W={x in V | B(x, y) = 0 for all y in U }

Let v,w be in W.
Then B(v+w,y)=B(v,y)+B(w,y)= 0 for all y in U.
So
v+w in W.

Good enough?

[cursive narwhal]

Uh is B(x,y) the notation you use for the inner product?

Yea it works

[cold topaz]

it is for bilinearity.

[cursive narwhal]

Oh

[cold topaz]

and W is the orthogonal component.

[cursive narwhal]

But yes, with that, you're done. I would also show that the zero vector is in there but that's cos i do that mostly out of habit

Complement*

But yes

[cold topaz]

actually I'll just replace W with U upside down T.

[cursive narwhal]

I mean, you can do whatever with the notation lol

(I use phi for the empty set sometimes)

[pallid rampart]

$\varnothing>\emptyset>>>>>>>>>\phi$

[stoic python]

Whoever:

[cursive narwhal]

Don't copswing at me

Otherwise, i'll swing my flesh baton at you

[static bison]

so {i, j, u} is not linearly independent because u = <ai, bj>
{u, v} is linearly because u = <ai, bj> and v = <ci, dj> and they don't each other as components, right?

[cursive narwhal]

Well, I generally think of it as 'I can reduce a matrix to upper or lower triangular form using elementary row operations and, as long as i keep track of what operations i'm using, i can relate the determinant of the final matrix to the determinant of the matrix i started with'

@static bison Your question kind of doesn't make sense as i'm looking at it? Is there a specific problem you're doing?

[cold topaz]

ok, now if W is the subspace of V, how do we prove that the intersection of W and W's orthogonal compliment is the empty set?

[cursive narwhal]

That's impossible

Can you see why?

Both W and the orthogonal complement of W are subspaces of V

So why is it impossible for their intersection to be empty?

[cold topaz]

please note that this is a separate question. it is not related to the other one.

[cursive narwhal]

Yea sure. If W is a subspace of V, you've proved that the orthogonal complement is going to be a subspace

So their intersection can't be empty

Can you see why that's the case?

[half ice]

Saying <ai,bj>
Is like putting a vector into a vector.

ai + bj is the vector you're referring to @static bison

[cursive narwhal]

Looks like an inner product lol

[half ice]

Ooh good point now I'm not sure

[cursive narwhal]

But he asked about linear independence and that, as a concept, doesn't require inner products

He'll probably clarify in a moment

@cold topaz So can you see why it's non-empty or no?

[cold topaz]

so saying This is impossible, because they contain the vector 0 is crroect?

[cursive narwhal]

Yeap

They're both subspaces so they both have to contain the zero vector

So, their intersection contains the zero vector and is non-empty

It's a separate problem to show that the only vector contained in the intersection is the zero vector

[cold topaz]

So, their intersection contains the zero vector and is non-empty```
How do we show that using mathematical language?

[cursive narwhal]

$0 \in W, 0 \in W^{\bot} \implies 0 \in W \cap W^{\bot}$

[stoic python]

Abhijeet Vats:

[cursive narwhal]

But I mean, it's fine to just write it in words lol

I mean, there's nothing wrong with using symbolic logic to write statements. Just don't go overboard with them and make your work entirely unreadable.

[hollow finch]

With some problems, especially conceptual ones, it's often fine for an answer to be all in words.

[cursive narwhal]

Sure, that's fair.

[cursive narwhal]

Yea that's one way of calculating determinants. There's another method that allows you to do it by calculating determinants of smaller matrices. I think what you have above is generally correct.

[river jasper]

I have a quick question regarding eigenvectors, are all eigen vectors linearly independent?

[cursive narwhal]

Uh I actually think the permutation method is more common than laplace expansion. I remember seeing it in some texts a long-ass time ago. Either that or I'm just remembering this wrong

There are some linear algebra texts that do go through that method specifically in a lot of detail, proving theorems etc. For example, Shilov's textbooks and shafarevich's text

Konrad Knopp?

Ohhh idk who tf that is

Hmm idk, the book that i learned from took a bit of an unconventional approach, i think? I mean, I've looked through some other texts and they don't take much of that kind of approach to determinants

Or maybe i just haven't been exposed to enough texts on linear algebra

[cursive narwhal]

I mean, i guess so. The properties that i mentioned earlier are of interest because they're related to the map itself. So, like, we're not just interested in calculating the determinant, cos that might take too long or not even always be feasible. Instead, you want to try and determine how it changes when the matrix changes

It doesn't obviously tell you if the matrix is invertible or not

That's a property you derive from some basic assumptions

The way i learned about it was that the determinant is a map that's linear in all rows of the matrix and if the rank of the matrix is less than n, then the determinant is 0. So, using those two properties, you can fully describe the map and provide a formula for computing it for any given matrix.

Well, any given n x n matrix, i should say

Of course, you also need to include the fact that det(E) = 1, where E is the n x n identity matrix

[limber sierra]

"map" is just synonymous with "transformation" or "function" in mathematical parlance

the determinant is a function from some space of matrices to the scalar field

[cursive narwhal]

So, let $det: M(n \times n,F) \to F$ be a map with the three properties:

1. det is linear in each row of the matrix

2. If rank(A) < n, then det(A) = 0

3. det(E_n) = 1

Fk my life

[limber sierra]

\to \bR

[cursive narwhal]

Oh right mb

[stoic python]

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

[limber sierra]

$\det$ exists in latex btw

[cursive narwhal]

Anyways, that's what i started out with and, from there, you prove every property of determinants

[stoic python]

Namington:

[cursive narwhal]

Oh okay okay

[limber sierra]

anyway

to motivate some of the above properties

the idea behind the determinant is to "transfer" information about the multiplicative structure of matrices to information about the multiplicative structure of the underlying field

for this discussion i'll assume we're working with real matrices, but these ideas generalize

[cursive narwhal]

Anyways, going back to your question, it is exactly a map of the above kind and you show how it is related to matrix invertibility and how it is applicable in solving linear systems

[limber sierra]

the determinant satisfies det(AB) = det(A)det(B), and the above properties are the best way to make a multilinear (i.e. linear in the rows) map that satisfies that equation

there are some cool results of this interpretation; for example

what does it tell us about invertibility?

well, when we consider the multiplicative structure of the reals

the only element without an inverse is 0

so naturally, if a matrix is not invertible, it will have determinant 0

so that's an example of one piece of "multiplicative information" we "transferred" from matrices to real numbers

[and indeed this holds both ways; a matrix is invertible iff it has nonzero det]

this interpretation should also tell you where the characteristic polynomial connection comes from

it's probably worth spending some time thinking about why that's the case

ie why does considering the multiplicative properties of the matrix (A - cI) give us information about the solutions to Ax = 0?

its certainly not the most intuitive notion at first

the determinant is very useful for how it "summarizes" a lot of information about a matrix's multiplicative structure

but of course, this relationship isnt perfect

like we know that $AB \neq BA$ is not true in general for matrices

[stoic python]

Namington:

[limber sierra]

but since determinants come from the underlying field, $\det(A)\det(B) = \det(B)\det(A) = \det(AB) = \det(BA)$

[stoic python]

Namington:

[limber sierra]

so there's clearly some information that is "lost" in the transition

which should make sense, since matrices have a lot of numbers

whereas determinants are only one number

so we naturally lose some info

[quartz compass]

no, he just showed det(AB)=det(BA)

or well maybe he didn't directly but, because it has this multiplicative property, you can commute the real numbers and put it back together

det(AB)=det(A)det(B)=det(B)det(A)=det(BA)

[dusky epoch]

like we know that $AB \neq BA$ is not true in general for matrices
double negative lul

[stoic python]

Ann:

[dusky epoch]

i mean

what you said is true but prob not what you intended to say @limber sierra

[pallid rampart]

[limber sierra]

shit yeah

$AB = BA$ is not true in general

[stoic python]

Namington:

[limber sierra]

alternatively, $AB \neq BA$ sometimes

[stoic python]

Namington:

[static bison]

How do you find the basis of a plane in 3d?

[pallid rampart]

Assuming the plane passes through the origin, then just take any two vectors in the plane that are not linearly dependent

[static bison]

so I did that

how do I check if it's linearly independent?

i know to make sure that they're not parallel

but how do I mathematically show that

do I just do cross product and check if the value is nonzero

[pallid rampart]

Yeah I think you can do that

But

In general, to show two vectors $v_1,v_2$ are linearly independent, you show that the only way $a_1v_1+a_2v_2={\bf 0}$ for some $a_1,a_2$ is $a_1=a_2=0$

[stoic python]

Whoever:

[pallid rampart]

so for example, we can show $\begin{bmatrix}1\1\end{bmatrix}$ and $\begin{bmatrix}1\2\end{bmatrix}$ are linearly independent: so first we write $$a_1\begin{bmatrix}1\1\end{bmatrix}+a_2\begin{bmatrix}1\2\end{bmatrix}=\begin{bmatrix}a_1+a_2\a_1+2a_2\end{bmatrix}={\bf0}=\begin{bmatrix}0\0\end{bmatrix},$$ then it follows that \begin{align*}a_1+a_2&=0\a_1+2a_2&=0\end{align*}

[stoic python]

Whoever:

[pallid rampart]

and then we can solve that system of linear equations

then you see that the only solutions are a_1=0 and a_2=0

and then you proved that the two vectors are linearly independent

This is generally what you do to show two vectors are linearly independent

[static bison]

okay gimme a second to better write my question

@pallid rampart

[pallid rampart]

The basis vectors need not to be like that

and what about these two vectors?

the <-2,1,0> and <3,0,1>

[static bison]

these are two vectors that lie on a subspace plane in 3d

oh wait

am I suppose to get three vector where one vector is dependent on one of the other vectors

[pallid rampart]

No

Remember

A plane is two dimensional

So a basis for the plane will have 2 vectors

Oh

Just fine two vectors in the subspace

Prove that they are linearly independent

And you are good to go

[static bison]

my question is how do I show linear independence of two vectors on a plane in 3d

[pallid rampart]

Well

Same procedure as I wrote above

You show that if a_1<-2,1,0> + a_2<3,0,1> = <0,0,0>, then a_1 = 0 and a_2 = 0

[static bison]

and given by the calculation itself, a_1 = a_2 = 0

so that's it?

[pallid rampart]

Yes

[static bison]

hmm

thank you

[static bison]

wait so

V = {(a, b) : a, b ∈ R}

(a, b) + (c, d) = (a + c − 1, b + d − 2)
k(a, b) = (ka − k + 1, kb − 2k + 2)

when do u (1,2) + v (3, 4) the sum is (3,4)

isn't that a problem?

[dusky epoch]

no, it's just that (1,2) is now your zero vector as you just saw

[static bison]

woah

I was just going to ask that

so for test 4, There exists an object in v, called the zero vector, that is denoted by 0 and has the property that 0 + u = u + 0 = u for all u in v.

the zero vector changes from (0,0) to (1,2)?

hmm

so the test is successful?

[dusky epoch]

"test 4" why call them tests

the zero vector doesn't "change from" anything to anything

in this space, the zero vector just is (1,2)

the fact that it doesn't look like you'd normally expect the zero vector to look means nothing

[static bison]

ah okay

thanks a ton

[static bison]

i think it fails on axiom 7

yup, just checked four times

[dusky epoch]

did you mean 2x**^**2

the derivative of 1+x+x^2 is not 1+x just sayin'

[static bison]

ah yeah, mistake

running on fumes

so I can't use the method, right?

[dusky epoch]

i mean these didn't arise as a set of solutions of some ODE so i'd be wary of doing that

but also it seems p overkill

[static bison]

hmm

[static bison]

so does an identity matrix have nullspace? no, right?

[ruby locust]

So what is difference between Span of 2 independent vectors in R^2 and 2 independent vector in R^3 even their z exist, don't they make the same plain?

[fathom flax]

Km stuck with differential equations but ill just put it here cuz im not sure whether it goes here or on differential equations cuz it's something to do with matrices.

Where did they come up with all those trig.

[static bison]

is the orthogonal basis of the null space just the row space?

[half ice]

Nah fam, they're not related like that

[gray dust]

this is word salad

[half ice]

Yummy yummy

[pale shell]

If you know how to multiply matrices by just applying one to the other do you need to know how to use the row column rule

[wintry steppe]

they're the same

[dusky epoch]

it's a convenient shorthand and also allows you to write down the product of two matrices more or less explicitly in the general case

[nocturne arch]

which ones are the free variables in this matrix?

[pale shell]

So is it like really necessary to know how to do? The row column rule

[nocturne arch]

??

[half ice]

@pale shell
What's row column rule?

In order to multiply matricies, you have to know how to multiply matricies

[pale shell]

Is the formula

So its like

Uhh

The thing like the first row times the first column and you add them for the first entry

Or something like that

[half ice]

How else do you multiply matricies?

[pale shell]

Yeah that rule

[half ice]

Btw I always just do that to multiply matricies

[pale shell]

I just apply the transformation

[half ice]

Put one above the other and draw the diagonals

[pale shell]

To the other matrix

[half ice]

I don't think I know this method hol up maybe you can teach this to me

I know a matrix can transform a vector - can a matrix transform another matrix?

[pale shell]

Ok so lemme draw it out quicky

[viscid kernel]

Does diagonalizing a matrices mean that you are making sure that the diagonal contains non-zero numbers right ?

[steady fiber]

it means taking a matrix $A$ and representing it as
$A = PDP^{-1}$

[stoic python]

PorosInMyAshe:

[steady fiber]

where P is some invertible matrix, and D is a diagonal matrix

[viscid kernel]

hmm I got it, its about change of basis.

Thanks

Or am I wrong ( correct me if Im wrong ) ?

[dusky epoch]

yes it's a change of basis

[icy osprey]

Can anyone help me with this problem?

[static bison]

so what would I do for this question: "Find an orthogonal basis for the null space of A − 2I." if the null space is span{<-1,-1,1,0>,<-1,1,0,1>}

is it <1/sqrt(3), -1/sqrt(3), 1/sqrt(3), 0> <-1/sqrt(3), 1/sqrt(3), 0 , 1/sqrt(3)>?

[icy osprey]

ok, continue 👀

[static bison]

nah, you're first, I'm just posting

[cursive narwhal]

@static bison I mean, those two vectors are linearly independent and are orthogonal already

So not sure why you normalized them, unless you were looking for an orthonormal basis

[viscid kernel]

@icy osprey I think it has to with crossproduct

[icy osprey]

@viscid kernel HELP ME! I AM SO LOST

lmaooo

[viscid kernel]

Hmm. I don't really know how to solve it 😦

[icy osprey]

how about other problems from my review?

[viscid kernel]

orthogonal projection has to do with dotproduct tho. thats the only thing I know.

@icy osprey Im also not good at solving excercises when it comes to linear algebra. Im now focussed on doing proof based linear algebra.

[icy osprey]

Oof, I am in an applied class

lol

[viscid kernel]

---

If you dont understand the question, I can like explain you, but solving it is a different story

[icy osprey]

Do you know anything about convergence with pde and matrix?

This problem was literally never brought up

[viscid kernel]

what is pde ?

[icy osprey]

partial differential equation

[viscid kernel]

nope, sorry

[icy osprey]

Thanks for the help !

[viscid kernel]

np, sorry for not being able to help you that much

[icy osprey]

don't worry about it, everyone has their specialty!

[viscid kernel]

Hmm

wait a second

I think I found it

https://cdn.discordapp.com/attachments/540211747613704221/705098663705444442/unknown.png

so

[icy osprey]

ok

Im back!

[steady fiber]

find the projection of that vector onto the subspace made by the two column vectors on the side

subtract the projection from the vector

and you get the orthogonal projection left

[icy osprey]

this is the soln my prof has to a similar problem. But my dot product is not 0

problem is my dot product is not 0 for my problem

[steady fiber]

it doesn't matter

[icy osprey]

@steady fiber how do i find find the projection of that vector onto the subspace made by the two column vectors?

[steady fiber]

what I said works for any general subspace

ah actually there's an easier way to do it

find the cross product of the two column vectors

[icy osprey]

I am all ears!

[steady fiber]

that gives a 3rd vector that is orthogonal

[icy osprey]

Yes, I get 4

[steady fiber]

find the projection of the original vector on the cross product

and you're done

[icy osprey]

wait wait wait

So find the cross product of the of the two column vecotrs

I got 4 for that

[steady fiber]

yes

how do you get 4

that's impossible

[icy osprey]

plugged it into my calculator

[steady fiber]

I am talking about cross product

not dot product

[icy osprey]

oh

sorry

ok let me do it and come back real quick

ok i got another vector

it is [6,-4,-10]

[steady fiber]

yes

project original vector onto that

[icy osprey]

now what is the equation for projection again?

[steady fiber]

$\frac{\mathbf{u}\cdot\mathbf{v}}{\mathbf{v}\cdot\mathbf{v}}\mathbf{v}$

[stoic python]

PorosInMyAshe:

[steady fiber]

for projection u onto v

[icy osprey]

more dot products!!! YESS

lol

thanks, I think I got right answer

[pale shell]

Well

Do you know the definition of linear independance?

[steady fiber]

it's because there is a non-trivial linear combination of the vectors so that you can get 0

[pale shell]

^

Also

Wait

[steady fiber]

$c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 + c_4 \mathbf{v}_4 = \mathbf{0}$

[pale shell]

The set

[stoic python]

PorosInMyAshe:

[steady fiber]

if there is a way to do this where not all the c are 0

[pale shell]

The set contains the zero vector

[steady fiber]

then it is linearly independant

[pale shell]

Therefore

[steady fiber]

ya, the zero vector makes it very easy to see

[pale shell]

It will always be linearly dependant

[steady fiber]

also the fact that there are more then 2 vectors

[pale shell]

^

Because think about it this way

If you take zero times all the other vectors in the set

Then you take some scalar k times the zero vector

No matter what k is you will always have zero vector

So you have more than the trivial solution of just all zeroes

[flat depot]

Can somebody give me an idea how to prove that every linear map on a subspace of V can be extended to a linear map on V?

I am a little bit puzzled how to approach this...

Not talking about trivial subspace and subspace which equals the space, obviously

[sharp merlin]

How do i answer 2

this is rref of A btw

I just said x3 is a free variable

[flat depot]

x_1 = x_2 = x_3

[sharp merlin]

huh

[flat depot]

I mean, that's probably what they want for the result, because it's pretty

Obviously that's the same as stating that x_1 - x_3 = 0 and x_2 - x_3 = 0.

[sharp merlin]

so it has infinite solutions righ

@flat depot

[flat depot]

Yeah. But solution space is one-dimensional, I'd say

Because, well, you have one free variable. Other variables are tied by this equation

[sharp merlin]

ok

And for this one do i just show W(0) is not equal to 0

[dusky epoch]

W(0) doesn't make any sense

W isn't a function

[sharp merlin]

oh yeah oops

Thats for 3.2

How do I do 3.1

[flat depot]

You can just point out that W doesn't contain additive identity, I think

[dusky epoch]

3.1 can be answered in the negative by showing 0 ∉ W

[flat depot]

By proving that you can't set a,b, such that you get [0,0,0]

[cursive narwhal]

@flat depot About your question, if you have a map $S:U \to W$ and you want to extend it to a map $T:V \to W$, where $U$ is a subspace of $V$, then there's a nice way to do that. Of course, I will assume that $V$ is finite-dimensional.

Let $(u_1,\ldots,u_m)$ be a basis for $U$. Take any basis of $V$ and you can extend the basis of $U$ so that it is, now, a basis of $V$. Call that $(u_1,\ldots,u_m,u_{m+1},\ldots,u_n)$.

Now, I'll make the following definition:

$T(u_i) = S(u_i) if 1 \leq i \leq m$

$T(u_i) = 0 if m+1 \leq i \leq n$

Once you do that, it's easy to check that this is linear and well-defined.

[stoic python]

Abhijeet Vats:

[sharp merlin]

So i just explain whatever i set a and b, I will never be able to get 0,0,0 right

@dusky epoch @flat depot

[cursive narwhal]

Ah fk bad latex but you should be able to parse through it

[sharp merlin]

is there a way to prove it or do i just explain that

[cursive narwhal]

Yes, hydra, you can prove it. Let -a+8 = 0. So, a = 8. But that means that 8-3b = 0 so b = 8/3. However, 4b+a = 32/3+8 = 56/3, which is not 0.

[flat depot]

@cursive narwhal That's interesting idea, and I had similiar in mind, but thing is that just before this question I had this one:

Which seems to be in odds with your line of reasoning...

Maybe I get this wrong somehow?

[dusky epoch]

T(0) ≠ 0 is a sufficient but not necessary condition for lack of linearity.

also, abhi didn't say your extension was 0 everywhere outside U

[icy osprey]

can anyone help with an applied la problem?

[flat depot]

It is 0 for all base vectors outside U, and so for all their linear combinations? Isn't that the same as saying "for all vectors in V, but not in U"?

[icy osprey]

@real plaza question alpha

[sharp merlin]

@dusky epoch For 3.2

Can I just show T(0) is not equal to 0, and therefore it is not linear

[dusky epoch]

yes

[sharp merlin]

oh sorry didnt see u wrote above

[dusky epoch]

It is 0 for all base vectors outside U, and so for all their linear combinations? Isn't that the same as saying "for all vectors in V, but not in U"?
no

[sharp merlin]

@dusky epoch This explanation works right

[dusky epoch]

,rccw

[dusky epoch]

extraneous

T(0) ≠ 0 is enough

[ocean sequoia]

wouldnt the integers work?

[cursive narwhal]

No

$\bZ \subset \bR^2$ is false.

[stoic python]

Abhijeet Vats:

[ocean sequoia]

what

how?

I think it wasnt an example of the subset

hence why i figured Z?

[cursive narwhal]

$\bZ$ is a set of integers, $\bR^2$ is a set of ordered pairs. Not the same thing.

[stoic python]

Abhijeet Vats:

[limber sierra]

elements of R^2 are of the form (a, b)

elements of Z are singular entities

[ocean sequoia]

oh

ah

could i use Z^2?

[limber sierra]

you could say that the set {(a, 0) | a is an integer} is a subset of R^2

but thats a different set

[cursive narwhal]

^That, in fact, would be your required set. It has the properties that the question asks for but isn't a subspace

[ocean sequoia]

its not a subspace cause you could multiply by 1/2 and thus be out of the set right?

[cursive narwhal]

Yeap

[ocean sequoia]

yea ok

[cursive narwhal]

In other words, it's not closed under scalar multiplication

[ocean sequoia]

yep

thats why i wanted to use Z but thats too general got it thanks guys!

[cursive narwhal]

?

It's not about being general at all

Z is not a subset of R^2 and that already fails what the first thing that the question is asking you to find

[ocean sequoia]

ahhh

ok

[limber sierra]

Z^2 totally works

what

am i an idiot

why doesnt it answer the question

[cursive narwhal]

No it does

It does lmao

[limber sierra]

...

[cursive narwhal]

Disc, you're not helping

[limber sierra]

anyway yeah, Z^2 works for the same reason my example worked

ie it's not closed under scalar multiplication

but everything else works

[cursive narwhal]

Z is the thing that failed. That's what they suggested previously. It fails cos it's not a subset of R^2

Anyways, the question has been resolved

[ocean sequoia]

that makes sense

thanks guys

all good! I appreciate the help

[icy osprey]

can i ask a question? Or should I wait?

wanna make sure your question is fully answered

[ocean sequoia]

nope your good

[icy osprey]

So I found the eigenvalues and eigenvectors

[ocean sequoia]

Did that video help with the regression problem?

sorry if it didnt

[icy osprey]

How do I go about the next part

I have yet to watch it but I have 2 tabs open, lool

coming back to that problem after this one

[fluid wyvern]

Hi everyone! So i need help on the following proof

Given two vectors OC and OD. If P is a random point of segment CD, prove that

OP = (1-r) * OC + r * (OD) where 0 ≤ r ≤ 1 and r = (distance between C and P)/(distance between C and D).

I tried making a diagram showing the different vectors and plugging in real numbers, but i get confused on how to prove the r part of the equation...

[cursive narwhal]

Let $C$ and $D$ be two points and let $O$ be our origin. So, we have the vectors $OC$ and $OD$. Then, $CD = OD - OC$.

Pick a point $P$ on the segment $CD$ and notice that $CP = OP-OC$ is a vector parallel to $CD$. In other words, there is a real number $r$ so that $CP = r\cdot CD$, where $0 \leq r \leq 1$.

Now, $CP = OP-OC = r \cdot (OD-OC) = r \cdot OD - r \cdot OC$. Solving for $OP$, we get:

$OP = (1-r)\cdot OC + r \cdot OD$

That was your desired result.

[stoic python]

Abhijeet Vats:

[cursive narwhal]

@fluid wyvern

If you didn't understand anything I said above, then ask questions. If you want to visualize it, start by drawing a diagram of the vectors OC, OD and OP. You should be able to see why the above is true.

[ocean sequoia]

{(a,b): a+b=1} would work?

x = (a,b), cf(x) = c = f(cx) = f(c(a+b) = c

but it doesnt contain the 0 vecotr?

im not sure if my scalar multplication is wrong

[cursive narwhal]

If it's closed under scalar multiplication, it has to contain the zero vector

[pale shell]

Hi abhijet

[wintry steppe]

why does first chapter of Hoffman Kunze Linear algebra introduce fields (and use them throughout the book)

[cursive narwhal]

Hallo earlten

[wintry steppe]

but no other linear algebra book does

[cursive narwhal]

I will wait for the day when you can spell my name properly

What do you mean 'introduce fields'?

[pale shell]

Abhijeet vat

[cursive narwhal]

There are other linear algebra books that do so as well

[wintry steppe]

@cursive narwhal First chapter of the book it defines fields etc

Yeah but most books I've seen don't

[pale shell]

Guys how do you think gilbert strang ocw course is

[cursive narwhal]

Klaus Janich's text does it, i believe anne schilling's text does it

[wintry steppe]

It's pretty much only H&K

Okay then

[ocean sequoia]

I liked it

[pale shell]

Is it like more computations or theory/proofs

[wintry steppe]

What's the point of excluding/including them? are they not fundamental?

[cursive narwhal]

Indeed, they are fundamental but their properties are explored in greater detail in abstract algebra

[wintry steppe]

I see

[cursive narwhal]

You just need them to define your vector space, which is the object of discussion in linear algebra

[wintry steppe]

So they skip them in LA to introduce them in AA

H&K is basically just a teaser

for fields

[ocean sequoia]

computational i believe it the consenous

/inutitive

[cursive narwhal]

They don't really skip them, do they? You do get all of the axioms and you do prove some properties depending on your textbook

[pale shell]

Is he good at explaining

[ocean sequoia]

yes

very

[pale shell]

Also

[cursive narwhal]

I'm not sure about hoffman and kunze's book because i've never used it

[wintry steppe]

They do skip them

[pale shell]

Do they have problem sets

On the website

[ocean sequoia]

I believe so

yes

[pale shell]

Or do you have to use txtbook

[cursive narwhal]

Janich's text definitely does it but, once again, the details are left to an abstract algebra course

[wintry steppe]

yeah

thats what i suspected

thanks

[pale shell]

Abhijeet dont vector spaces have like ten rules

[cursive narwhal]

In fact, Janich's text goes into groups as well. So, you prove some things but just enough to organize whatever it is you're doing

[pale shell]

Unlike subspace

[ocean sequoia]

on website

[pale shell]

ok

[cursive narwhal]

Why're you asking me? I don't know what vector spaces are.

[pale shell]

o

Are you a linear algzbra

[elfin ingot]

hoffman kunze is more rougyh

he also does rings with polynomials iirc

[gray dust]

I don't know what vector spaces are.
lies

[cursive narwhal]

Yes, i am, in fact, the general linear algebra

[pale shell]

Hi rokabe

[gray dust]

iH

[pale shell]

when you transpose a matrix

[fluid wyvern]

If you didn't understand anything I said above, then ask questions. If you want to visualize it, start by drawing a diagram of the vectors OC, OD and OP. You should be able to see why the above is true.
@cursive narwhal Thank you so much for your explanation, it helped a lot!:)

[pale shell]

Does it matter if you take columns to rows or rows to columns

[cursive narwhal]

You're welcome.

[gray dust]

idk wym

[elfin ingot]

this is the transpose of that

[pale shell]

When you take the transpose of a matrix

[elfin ingot]

column/row is the transpose of row/column

[ocean sequoia]

If it's closed under scalar multiplication, it has to contain the zero vector
@cursive narwhal wait so is {(a,b): a+b=1}
x = (a,b), cf(x) = c = f(cx) = f(c(a+b) = c wrong?

[pale shell]

Well yeah but like

[ocean sequoia]

im guessing that f(c(a+b) = c is wrong?

[cursive narwhal]

What is f(x)

[pale shell]

Does it matter if you move rows to columns or vica versa

[ocean sequoia]

x is just (a,b)

[cursive narwhal]

What you have to ask yourself is

How have you defined the set

[ocean sequoia]

a + b must be one

[cursive narwhal]

So cx = (ca,cb). Does cx belong to the set?

[gray dust]

i still don't know what you're asking

[cursive narwhal]

Well, ca+cb = 1 in order for that to happen

[pale shell]

Ok

[cursive narwhal]

But is that necessarily the case if a+b = 1?

[pale shell]

Do you know what you do when you take the transpose of a matrix

[ocean sequoia]

oh yea i guess thats not closed under multiplication

[gray dust]

tell me whatcha do

[ocean sequoia]

shoot

thanks

[pale shell]

Well ok I am not sure because

[cursive narwhal]

Sure

[pale shell]

You either swap rows/columns or vica versa but im not sure which one

Or eityzr works

[cursive narwhal]

??

Are you having, like, a stroke at random moments while typing?

[gray dust]

it's ok, he doesn't spellcheck himself

[pale shell]

o

me?

[gray dust]

you

[pale shell]

ok it says or either way works

I am not sure

[gray dust]

when you transpose, the entry in the (i,j) spot is sent to the (j,i) spot

[pale shell]

Yes but in english that means you just swap the positions

So I am asking

Does it matter if you swap the rows to the columbs

Or vice versa

[ocean sequoia]

when you transpose, the entry in the (1,3) spot is sent to the (3,1) spot

maybe with numbers it helps?

[cursive narwhal]

They're thinking in terms of rows and columns

[gray dust]

"swap positions" is a butchering of what i said