linear-algebra

[torn silo]

then you have to form the hesse matrix and look if positiv or negative definit

[hazy sonnet]

yeah im not sure how to form it

from the equation i mean

[elfin ingot]

why is the dimension of the dual space the same

as the space

[wintry steppe]

iff finite dimensional

[elfin ingot]

yea

[wintry steppe]

Take $(e_1,\hdots, e_n)$ to be a basis of $V$, and define for each $e_i$ a linear functional $\epsilon_i:V\to F$ by $\epsilon_i(e_j)=\delta_{ij}$.

[stoic python]

mart:

[elfin ingot]

whats delta

[wintry steppe]

claim: that gives a basis for V^* and consequently dimV=dimV^*

[elfin ingot]

i see this delta alot but really idk about

in the text

[wintry steppe]

kronecker delta search it up

[elfin ingot]

okay so its like a function that outputs 1 if matrix

sqaure

and 0 if not

square*

?

[wintry steppe]

no

https://en.wikipedia.org/wiki/Kronecker_delta

[elfin ingot]

identity/

[wintry steppe]

no

bruh

[elfin ingot]

the identity of dimension i*j?

yea

i got it

okay

okay i got the proof , for each element in the basis define a linear functional that sends to the corrosponding entry of the identity

thats a basis for the dual

right?

[wintry steppe]

the identity of dimension i*j?

no

please reread the definition

[elfin ingot]

what i read was thtat the delta_(ij) is the ith,jth entry of the identity matrix of dim ij

thats what i meant if i phrased it wrongly

[dusky epoch]

δ_ij is just a shorthand for 1(i=j)

[light brook]

hey,
could someone help me find this permutation`s inverse?

is it just 1, 2, 3, 4, 5, 6

[torn silo]

no?

[light brook]

what is it then?

[torn silo]

can give $\sigma$

[stoic python]

deekaan:

[torn silo]

once you have sigma it's pretty easy

[elfin ingot]

just write it backwards

[light brook]

i dont understand.. wym by having sigma

[torn silo]

🤔

$(1)(2,5) ...$

[stoic python]

deekaan:

[light brook]

yes

[torn silo]

then do like mo2men said and turn that around and you're done

[light brook]

so when i get the inverse does it go like (1), (5,2) and so on?

[torn silo]

yes

[light brook]

how would you write in the form i sent here at the beginning

[torn silo]

1 leads to 1, 5 leads to 2 and 2 leads to 5

how do you write that?

just flip it bro

up becomes down and down becomes up

[light brook]

then in this case nothing changes

so it equals to its inverse right

?

[torn silo]

no you have to flip it

[light brook]

yea but still

[torn silo]

1 5 3 4 2 6

its almost the same

but not quite

🤔

[light brook]

but 5 is mapped to 2 and 2 is mapped to 5 in the inverse.. cant i just write it in ascending order

[torn silo]

I don't think so

you have to go backward

🤔

[light brook]

im confused

i looked it up on the internet its different

[torn silo]

$\sigma (\sigma^{-1}(5))$

[stoic python]

deekaan:

[torn silo]

actually it does work

sorry

ya you can just keep it like it is

because 2 and 5 are the only elements that change

and they map to each other

[light brook]

so it equals to its inverse yeah?

[torn silo]

yes

[light brook]

alright

now i get it

thanks

[gritty frigate]

@torn silo I had my exam yesterday bro

I got 75%

A Parametric Matrix destroyed me lol

Thanks a lot !

[torn silo]

is this a thank you thank you or a fuck you asshole thank you?

[gritty frigate]

Nono it is a thank you true

Lol

You helped me a lot

[torn silo]

oh cool

grats

[gritty frigate]

I could have gotten a 100% if it wasnt for that parameter

Could have gotten is well written ?

[torn silo]

ya it's fine

[gritty frigate]

Oh

[torn silo]

and really well done

[gritty frigate]

Maybe if you want I can send you a pic of the parametr I got

I dont know why it was SO hard

I spent 1/2 the test doing it

[torn silo]

sure

[gritty frigate]

I do not know how to do Latex matrix lol

$\begin{pmatrix}
1 & 3 & \alpha & -2\
\alpha &11 &8 &-3 \
1 & 18 &14 & 1
\end{pmatrix}$

[stoic python]

AfterJack:

[gritty frigate]

Its so simple....

I can not still find the way to solve it

You need to find a value for a that made the system with infinite solutions

Now that I m looking it better I could have switched columns

[torn silo]

hmm that isn't so simple

I mean the matrix is pretty tricky

one trick for the future

if you add the third row twice to the first and three times to the second

then you have the first condition for infinite solutions

now all you have to do is find alpha so that the two upper rows become linearly dependent

it's a bit cheesy but using those kinds of tricks you can avoid doing a lot of arithmetic

@gritty frigate

[gilded night]

Do I need to understand prealg-alg to understand linear-algebra?

[cursive narwhal]

Yes.

You should be comfortable with basic algebra.

[gilded night]

Alright just want to make sure I am doing it right

[cursive narwhal]

Yeap, that's fine. Focus on getting comfortable with basic algebra. The elements of linear algebra are easy to understand with a basic understanding of set theory but it's better to have a strong background since linear algebra is pretty abstract.

[kindred ingot]

I agree with Foundations are important. Make sure you are good with basic arithmetic, linear equations, and functions before you hop in to linear algebra. Having a background in calculus can help, but only in that things like the derivative are secretly linear transformations.

[gilded night]

Yeah I am starting off with basic Arithmetic in KhanAcademy although I have learnt it before its just pretty foggy with so much time going by without using advanced math.

[kindred ingot]

Not to worry. The more you do it, the easier it gets.

[cursive narwhal]

On the other hand, there are a few books out there that focus on teaching linear algebra in 2 & 3 dimensions alongside geometry. If you really do want to get a taste of linear algebra without the full force of the abstraction, you can certainly use those books.

[slender marsh]

A line segment has one end point A at (-3, 5) and a slope of -2/3. State Two locations (x, y) of the other endpoint B.

[gilded night]

Yeah I got the gist of linear algebra as I study them using vectors and matrices

[cursive narwhal]

@slender marsh "Two locations (x,y) of the other endpoint B" ?

I assume that this question is asking you to find two possible points that could be endpoints of the line segment. In that case, what have you tried?

[slender marsh]

idk what to do

[cursive narwhal]

Well, okay, take this to #geometry-and-trigonometry first, it probably belongs to that channel.

[lapis fern]

given a vector space V with vectors w,v, u; is there a more abstract/general name for vector-addition

f: V \rightarrow V
f(u,w) = v \in V

??

(there are vector-additions that don't look like regular old addition, yeah?)

[cursive narwhal]

If you want, you can refer to vector addition in a vector space as $+_V$, where $V$ is the set underlying the vector space.

[stoic python]

Abhijeet Vats:

[cursive narwhal]

That will probably be helpful in distinguishing between vector addition & the addition on real numbers.

Also, no. The map you wrote above is an incorrect representation of vector addition. It is actually:

$+_V: V \times V \to V$

$(u,w) \to v$

[stoic python]

Abhijeet Vats:

[cursive narwhal]

But people don't typically care about such formalisms.

@lapis fern

[lapis fern]

they don't care because most vector additions act like regular addition?

or

they dont care because it doesn't matter that much?

(i'm reading a pdf where the vector addition and scalar multiplication are defined using symbols that i rarely encounter)

[cursive narwhal]

In many ways, vector addition is very much like regular addition. There are ways in which it is not.

The formalism does matter but it's very rare for it to come up in many problems. Like, when you're doing addition on R, you rarely think of yourself as using a map to associate a real number with a pair of real numbers. You usually just do the addition straight away

So, it's the same story over here

Can you show me the pdf?

[lapis fern]

Axler doesn't make a big deal about it, and just uses the standard notation that I always encounter.

this author wanted to make a point: http://faculty.etsu.edu/joynerm/documents/VectorSpaces.pdf

[cursive narwhal]

He makes a very valid point at that. This isn't just regular addition.

[lapis fern]

goes through the pain of using abstract symbols, but stays in the real field 😆

[sonic osprey]

linear algebra over the real numbers and complex numbers look pretty different from other fields

Even though these axioms stay the same, some of the theorems can differ

[lapis fern]

linear algebra over the real numbers and complex numbers look pretty different from other fields
@sonic osprey i'm sure that's true, but axler said dont worry about it.

[cursive narwhal]

When you begin learning some abstract algebra, authors tend to use $\circ$ to denote binary operations on sets. Once it's clear that we must disassociate the symbols from their typical meanings, some authors start using + or whatever.

[stoic python]

Abhijeet Vats:

[sonic osprey]

I mean, so its still important to look at abstract vector spaces like this, but the trouble of working over general fields is a lot more difficult

[autumn basin]

I think I need to calculate the vector that spans the null space of a given matrix. How do I do this? This is the line in my notes but I'm very confused, they also seem to have transposed a vector?

[lapis fern]

aight, thanks for the discourse folks. I don't know how deep to dive into abstraction, so this sanity check has been helpful.

(im a math noob, i'm more used to concrete things like physics 🤪 )

[cursive narwhal]

It'll stick to you over time. Just work with it and you'll be okay.

[lapis fern]

tyty

[cursive narwhal]

@autumn basin Can you post the entire question?

(battery's about to run out so i might not be around to help)

[half ice]

You can make very odd vector spaces.
But Axler never will.

[autumn basin]

well so this is the entire relevant portion of the lecture notes, I'll also post the matrix I need to consider

and my matrix is
8 0 2
0 8 2sqrt15
2 2sqrt15 8

(Not sure how to format that with latex, sorry 😦 )

For context, this is probability but I'm required to use linear algebra concepts here

[cursive narwhal]

Let $T_{\Sigma}$ be the associated linear map of the given matrix $\Sigma$. A very quick calculation shows that:

$T(x,y,z) = (x+2z,y+3z,2x+3y+13z)$

If we set the above to 0, then we get $x = - 2z$ and $y = -3z$. So, we are essentially looking at a space of vectors of the form $(-2z,-3z,z) = -z(2,3,-1)$. In other words, the space is made up of scalar multiples of $(2,3,-1)$.

[stoic python]

Abhijeet Vats:

[cursive narwhal]

@autumn basin

[autumn basin]

Ah so is your use of T there analogous to the lecturer's use of T that I thought meant transpose?

In that case your explanation makes sense

[cursive narwhal]

No. Your professor is using T to denote the transpose of the given vector. I'm using T to denote the linear map associated with your matrix.

[autumn basin]

hm

I didn't realise transposing a vector is a thing you can do

[cursive narwhal]

Of course you can

A vector is a matrix.

[autumn basin]

yeahh true

[cursive narwhal]

In that case, the vector he wrote is a 1 x 3 matrix & he needed a 3 x 1 matrix because that's the only circumstance where the multiplication would make sense

[autumn basin]

ah right I understand now, the need for the transposition is because of "Theorem 46" which basically just wants you to take that null space vector, and do T* sigma* T transpose

er I used T but I mean (2,3,-1)

[cursive narwhal]

Well, technically, it's $\Sigma \cdot (2,3,-1)^T$ but yea

[stoic python]

Abhijeet Vats:

[cursive narwhal]

$\Sigma$ is the matrix. $T_{\Sigma}$ is the linear map associated with the matrix. They are intimately connected but they're not the same thing.

[stoic python]

Abhijeet Vats:

[autumn basin]

seems like taking a linear algebra module this year would've helped quite a lot with this, huh

thanks for your help mate, think I understand what I need to know 😄

[cursive narwhal]

Indeed, a linear algebra class would've been useful for a module which requires linear algebra

who would've thunk

You're welcome.

[sour jetty]

They're saying the two equations above are both equal to the underlined one, and derive the eigenvector from that assumption. How does one come up with this y = (1-i)x? What is the process to finding this? I'm able to calculate whether they're indeed equal to it, but I want to be able to find it myself

[cursive narwhal]

$(1-i)x-y = 0 \implies y = (1-i)x$

[stoic python]

Abhijeet Vats:

[limber sierra]

the first one, just add y to both sides

[cursive narwhal]

$2x+(-1-i)y = 0 \implies y = \frac{2}{1+i}x = \frac{2}{1+i} \cdot \frac{1-i}{1-i}x = (1-i)x$

[stoic python]

Abhijeet Vats:

[limber sierra]

^^

this one's a bit more involved but

same gist

[cursive narwhal]

@sour jetty

[limber sierra]

alternatievely, 2x- y - iy = 0 implies 2x - y = iy, so divide by i and simplify

[sour jetty]

Oh, I see. Thanks a lot!

[wintry steppe]

Does anyone know how to get from second last line to last line?

Specifically how to calculate Re(B-a|r)

it seems like Im(b-a|r) = 0, but I don't know why

[eager burrow]

Are there any assumptions about alpha and/or beta?

i.e. What's the statement that's being proven?

[wintry steppe]

This is the statement @eager burrow

[eager burrow]

Oh, I think I might understand

Is your scalar product in such a way that you complex conjugate a factor from the right argument? i.e. (x, a y) = comp.conj(a) (x,y)

[wintry steppe]

yes

because you swap it and then pull the constant out

(conj(ay), conj(x)) then conj(a)(x,y)

[eager burrow]

Because then, what's happening is
$\text{Re} (\beta - \alpha | (\beta - \alpha |\alpha - \gamma) (\alpha - \gamma))

\text{Re} \overline{(\beta - \alpha | \alpha - \gamma)} (\beta - \alpha | \alpha - \gamma)$

[stoic python]

Lartomato:

[eager burrow]

pray to god this compiles

Okay, neat! And then, inside the real part, you just have the product of a complex number with its complex conjugate

Hence a real numbah

This is a bit messy, but maybe you get what I mean

(i didn't include the 1/|...|^2 factor)

[wintry steppe]

Wait I'll try to derive this but I don't see how the imaginary part is 0?

[eager burrow]

I'm not sure what you're trying to derive; do you understand what I'm doing in the equation above?

[wintry steppe]

No, I don't

I can't see what you are doing 😦

[eager burrow]

Awright; so, I'm inserting this $\tau$ (which you called r) into the inequality

[stoic python]

Lartomato:

[eager burrow]

I'm focusing on the first summand, so this part

[wintry steppe]

Oh, I see, hold on

Yea, I got the idea

[eager burrow]

Oh, so does it make sense to you now?

[wintry steppe]

yea

[eager burrow]

neat

[wintry steppe]

thanks

[eager burrow]

np

[wintry steppe]

kunze pisses me off

my poor brain cannot understand his "Of course cj = f(aj)" in the last part

and why f(a) =c1x1+...+cnxn

[elfin ingot]

same lmao

[wintry steppe]

<@&286206848099549185>

[elfin ingot]

quickie

is Hom(V,W) the same as L(V,W)

im looking for psets and i keep seeing this Hom

(L(V,W) is the set of all linear transformations between V and W vec spaces)

[wintry steppe]

whats your definition of Hom, all ring homomorphism?

[elfin ingot]

idk i never seen Hom b4

but id guess its the set of all homomorphisms ig

[wintry steppe]

I believe they are the same but I could be wrong

you should ask experts, but a ring homomorphism satisfies

1. f(a+b) = f(a)+f(b) 2)f(ab) = f(a)f(b)

so essentially I think L(V,W) is a subset

of Hom(V,W)

[elfin ingot]

yea

cool ty

[wintry steppe]

though I don't see how homomorphism satisfy cf(x) = f(cx)

so I'm probably wrong

[subtle walrus]

linear maps are the homomorphisms of vector spaces

[modern palm]

If an mxn homogeneous system has non trivial solutions then m < n. Is this True or False?

I'm thinking that it is True, since rank = min(m, n) = m since m < n. Then m = r < n which means infinitely many solutions which means it has non trivial solutions. I don't see anything wrong with this statement...

[modern palm]

wait nvm, its False, there are counter examples

[cursive narwhal]

@elfin ingot L(V,W) and Hom(V,W) are two notations for the same set. Both represent the set of linear maps between V & W

[elfin ingot]

got it

tysm

[wintry steppe]

can someone explain to me question 5c?

I do not understand why T(b_1) = T(1) = (t+5)(1)

Since if you take T(b_1) = T(1) = 0 (the derivative of it)

How is it still possible that (t+5) is in the answer?

[cursive narwhal]

? Why take the derivative of it?

[wintry steppe]

because the example in the book does it

also a person on yt did the same

I just dont know what I have to do ... so I follow the steps from example and a vid I found on yt but the things dont add up

[cursive narwhal]

Maybe you should do a bit of reading up from an actual book? It will be easier to approach such problems if you have a good understanding of the theory

[wintry steppe]

the book is good

I just dont get this point

there is not much more to read lmao

do you know what I am doing wrong?

[cursive narwhal]

Okay let me try and show you how you might think about this in an easier way.

Let $p(t) = at^2+bt+c$. Then we know that:

$(t+5)p(t) = at^3+(b+5a)t^2 + (c+5b)t+ 5c$

So, a good way of thinking about what $T$ does is to think of it as transforming the coefficients of the polynomial. In other words:

$T(a,b,c) = (a, b+5a,c+5b, 5c)$

Now, finding the matrix relative to the given bases shouldn’t be too difficult

[stoic python]

Abhijeet Vats:

[wintry steppe]

now I know how xD

[cursive narwhal]

What they’ve done in their working is to consider the image of the polynomial p= 1. That’s the same thing as setting a=0 and b=0 and c =1. In other words, the image is going to be the vector:

(0, 0,1,5) —-> t+5

[wintry steppe]

but what does the image of the polynomial p = 1 mean?

[cursive narwhal]

It refers to the polynomial that p =1 is mapped to under the transformation. A linear map is just a function of a particular kind. It’s no different from, say, f(x) = x^3, for example. So when we say that f(1) = 1^3 = 1, then we can say that 1 is the image of 1 under the map f.

(Of course, they are very different in many ways. The basic principle still applies in both cases.)

[wintry steppe]

I understand the gist of what you are telling me

and there are more ways to the same answer

this is what my book does

[cursive narwhal]

It’s exactly the same thing. The rule for how T transforms a general 2nd degree polynomial gives you the images of the basis vectors too

[deep flicker]

In a survey of 200 students of a school, it was found that 120 study Mathematics, 90 study
Physics and 70 study Chemistry, 40 study Mathematics and Physics, 30 study Physics and
Chemistry, 50 study Chemistry and Mathematics and 20 study none of these subjects. Find
the number of students who study all three subjects.

Does this question fall under linear algebra?

[cursive narwhal]

No.

Take this to #discrete-math or #proofs-and-logic

whichever is unoccupied

[deep flicker]

Okay thanks

[wintry steppe]

i will try to understand

[cursive narwhal]

For the example in the book, think of it as a map:

$T(a_0,a_1,a_2) = (a_1,2a_2,0)$

[stoic python]

Abhijeet Vats:

[cursive narwhal]

So mapping coefficients of one polynomial to the coefficients of another polynomial

[wintry steppe]

YES!

I finally understand 🙂

thanks Abhijeet Vats!

[cursive narwhal]

You're welcome.

[vocal breach]

Shouldn't the determinent of this be 0

E_2 - 4E_3

Then you have the same equation at the top an middle

[limber sierra]

do you?

what's 2 - 4*1

[vocal breach]

-2

you get
2 1 2
-2 -1 -2
1-log(8) 3 1+log(8)

[limber sierra]

er

what's -1 - 4 * 3

i'm not sure how the -1 was magically untouched

[vocal breach]

oh,

ok

lol

Was too focused on getting rid of the log(8)

[cunning sonnet]

sorry if this is the wrong channel, but is there a variant of https://en.wikipedia.org/wiki/Bresenham's_line_algorithm, that is specifically for "discrete" grids that only use the 8 basic directions (NESW, NE SE SW NW)?

Basically my goal is to find which grid cells are between two other cells, in a simpler grid cell. Such as

X1 = 5, Y1 = 5
X2 = 10, Y2 = 5

I want to get a result of 6, 5, 7, 5, 8, 5, 9, 5

or is there a simpler way to determine that I am stupidly overlooking?

[torn silo]

maybe check out redblob games

https://www.redblobgames.com/grids/hexagons/

that could give you a starting point

[cunning sonnet]

ty will give a read

[split heart]

How can I define eigenvalue?

To understand it myself

[limber sierra]

do you want a formal definition or an intuitive one?

[split heart]

an intuitive one

Kind to see more clearly what it is

[limber sierra]

an eigenvector is a vector $v$ such that $Av = \lambda v$, where $\lambda$ is an eigenvalue

[stoic python]

Namington:

[limber sierra]

to entangle this more "intuitively"

this means that, when we multiply some vector v by the matrix A

this is the same thing as just scaling the vector

by a scalar factor we call the "eigenvalue"

[split heart]

So each eigenvalue is associated to it's eigenvector

Alright, thank you 🙂

[limber sierra]

yes, eigenvalues are defined by the eigenvectors we scale

so if you think of the matrix A as a linear transformation

it's basically saying "there's some vector v for which A is just multiplying v by a constant lambda"

"then, v is an eigenvector and lambda, the constant, is its eigenvalue"

an example of this might be if we have a linear reflection which "reflects" the coordinate plane about the line y = x

in that case, a vector on the line y = x would be untouched

so Av = v

therefore this would be an eigenvector with eigenvalue 1

meanwhile, a vector along the line y = -x would be reflected "strictly"

but the reflection would correspond to multiplication by -1

[split heart]

Those eigenvalues should be obtained by doing det(A - λI) = 0

[limber sierra]

like if the white line is the line of reflection, the red line is the original vector v, then the reflected vector is the blue line

(if everything is centred at the origin)

then you can see thsi corresponds to just multiplying by -1

so this red vector is also an eigenvector

but it has eigenvalue -1

Ax = -x

the blue vector is the red vector times -1

[split heart]

mhm, I start to see it's purpose

Can I think it as a way to simplify a matrix for determined vectors?

the matrix job*

[limber sierra]

sure, although it's perhaps better thought of as "special behaviour" at a certain vector

[split heart]

Right

I'll take note of all this, ty

[quartz compass]

I think of the eigenvectors as being the nicest basis you could wish for

[gritty frigate]

Sorry for the quesiton I will do but

Can I change columns ?

For example x = y

Specially for parametric equations

[foggy lodge]

are you talking about column switches as a linear transformation?

yeah, you can

[wintry steppe]

I am breaking my head over this proof question

Any help would be greatly appreciated

If V⊆W, then dim(V)≤dim(W)

Where: V and W are subspaces of Rn

[limber sierra]

suppose B is a subset of V that forms a basis of V. clearly, then B is also a subset of W. what must B be in the context of W?

suppose for contradiction that dim(V) > dim(W). what does the above tell you, then?

you should be able to derive a contradiction

[wintry steppe]

@limber sierra Question

I did it this way

I said that if V is a subset of W, this implies that the basis of V is: (v1,...,vn) and the basis of W is: (w1,.....wk) where both sets of vectors are linearly independent. This implies that n ≤ k and so dim(V) ≤ dim(W).

Is this the wrong approach?

[limber sierra]

"This implies that n ≤ k"

how?

[wintry steppe]

Since V is a subset

It has to have at most k vectors

But could have less than that

Right?

[limber sierra]

why?

[wintry steppe]

Since it's a subset right

[limber sierra]

you dont sound very confident of this fact you're asserting without justification

sure, it's a subset; what does that mean?

[wintry steppe]

That every element in V is also in W

[limber sierra]

so...

[wintry steppe]

So at most V would be equal to W

[limber sierra]

so...

[wintry steppe]

So this means that the basis of V would be at most equal to basis of W

[limber sierra]

i think i get what you mean

but youre not really explaining the full chain of reasoning

[wintry steppe]

Yeah I'm not sure how else to explain it, trying to think

[limber sierra]

let $B_V$ be the basis of $V$ and $B_W$ the basis of $W$. Suppose for contradiction that $\abs{B_V} > \abs{B_W}$. But, since $B_V$ is a linearly independent subset of $V$, this means it must also be linearly independent in $W$. So, $B_V$ is a linearly independent subset of $W$; but it is a theorem that a subset of a vector space is a basis iff it is a maximal linearly independent subset, so $B_W$ can't be a basis, since it is smaller than some linearly independent subset (i.e. $B_V$) and thus is not maximal.

[stoic python]

Namington:

[limber sierra]

contradiction, QED

i was really explicit, you dont have to be that wordy

but that's the chain of reasoning

[wintry steppe]

why is this wrong?

So for contradiction, are we saying P->Q is the same as P and not Q?

I am always confused by contradiction

[limber sierra]

um

we're saying $\neg \neg P \implies P$

[stoic python]

Namington:

[limber sierra]

if it's not true that (not P)

then P is true

so we're assuming not P

ie we're assuming dim(V) > dim(W)

[wintry steppe]

Got it

[limber sierra]

and proving that this is false by deriving a contradiction

so not(not P)

i.e. P

[wintry steppe]

Thank you!!

can someone respond to my question

idk why the answer is wrong

[limber sierra]

p(t) = -4 - 2t

what is p(-2)? what is p(0)? what is p(1)?

here p is a function

you're not multiplying it by anything, you're evaluating a function at certain values.

[wintry steppe]

oh

i see

wow

So I wanted to clarify what I said earlier - since V is a subset of you don't we know the entire basis of V is in W?

[limber sierra]

right

[wintry steppe]

So then the basis of W should have at least the number of vectors in V

[limber sierra]

since V subset W we know B_V is a subset of W

[wintry steppe]

Yes so that's why I was saying n<=k

[limber sierra]

you skipped all the intermediate reasoning, though

the logic is "B_V must be linearly independent, but then W would have a linearly independent subset bigger than its basis, contradiction"

this proof doesnt have to be phrased by contradiction, fwiw

it could be rephrased another way:

[wintry steppe]

Is it possible to do a direct proof?

As I had attempted?

[limber sierra]

"B_V is linearly independent since it is a basis, and is contained in W since V is a subset of W. the dimension of W is the size of a maximal linearly independent subset of W. so, |B_W| is an upper bound on the size of a linearly independent subset of W. Since B_V is a linearly independent subset of W, this means that |B_V| <= |B_W|"

[wintry steppe]

So you are saying I should explain all of that instead of just stating n<=k?

Just want to make sure

[limber sierra]

you probably don't need to be quite as explicit as I was.

the point is just that you can't randomly say "therefore n <= k", you have to give some sort of justification

of why these two facts connect

[wintry steppe]

Got it

Much appreciated

What do you mean by maximal linearly independent?

I know what linearly independent is

Not maximal

Okay I just found the definition online

[limber sierra]

sorry, i just mean "the largest possible linearly independent subset"

maximal means "largest"

[wintry steppe]

No problem, thank you!

[limber sierra]

you might not have proved that "the dimension of W is the size of a maximal linearly independent subset of W"

but the proof isnt too hard

at least in the finite dimensional case

[wintry steppe]

I have another question I'm working on so if I'm facing an issue I'll post it here

I don't think we have

[limber sierra]

well here's a quick proof for you:

let B = {b_1, ... b_k} be a basis for W

so dim(W) = k

B must be linearly independent, since it's a basis

[wintry steppe]

Yes

[limber sierra]

now suppose we "added" an element, b_{k+1}, to B

to make a new set B'

[wintry steppe]

Yeah

[limber sierra]

since B was a basis, we can write any element of W as a lin. comb. of b_1, b_2, ... b_k

including b_k+1

but this means B' is not linearly independent

since b_{k+1} can be written as a linear combination of the rest of B'

hence B' is not a basis

so, any subset larger than B can't be a basis

therefore, if B is a basis, then it is a maximal linearly independent set

hence dim(W) = |B| = size of a maximal linearly independent subset of W

QED

[wintry steppe]

Oh I see what you mean

That makes a lot sense

[limber sierra]

one notes that this only works "nicely" in the finite dimensional case

but fortunately, subspaces of R^n are necessarily finite dimensional

so you dont have to worry about that

[wintry steppe]

So you are saying because b_k+1 is in span of B that B' is linearly dependent?

[limber sierra]

yep, exactly

and we know it's in the span since B is a basis

[wintry steppe]

Got it, makes sense!

If you don't mind - could I post one more question and have you check my solution?

If V⊆W and dim(V) = dim(W), thenV=W. Suppose that dim(V) = dim(W) and V ⊆ W means that the basis of V: (v1,....,vn) of linearly independent vectors spans V, and also, because V is a subset of W, this means that basis of V lives within W and spans W

So span(V) = span(W)

V = W

[limber sierra]

"that basis of V lives within W and spans W"

you should probably say "and spans W because it is a linearly independent set of the same size as a basis of W" or something similar

also uh, dont say span(V) or span(W), say

span(basis of V)

but yeah since

V = span(basis of V) = W

V = W

[wintry steppe]

"and spans W because it is a linearly independent set" so what exactly does it being a linearly independent have an affect on this?

I get the same size part of your explanation

[limber sierra]

well, something like $\left{\begin{pmatrix}1\1\end{pmatrix}, \begin{pmatrix}2\2\end{pmatrix}\right}$ does not span $\bR^2$

[stoic python]

Namington:

[limber sierra]

despite the fact that $\dim(\bR^2) = 2$ and this set has two vectors

[stoic python]

Namington:

[limber sierra]

because it's not linearly independent

for example, you cant write $\begin{pmatrix}1\0\end{pmatrix}$ using these vectors

[stoic python]

Namington:

[wintry steppe]

But since it was a basis in V it was linearly independent right

[limber sierra]

right, in the case of your problem

we know its lin. ind. because it was a basis of V

[wintry steppe]

Should I still restate it?

[limber sierra]

so this issue doesnt arise

i mean, I would

personally

just to make it clear that that's the reasoning

[wintry steppe]

Got it

Also, could $\begin{pmatrix}1\0\end{pmatrix}$ and $\begin{pmatrix}0\1\end{pmatrix}$

[stoic python]

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

[wintry steppe]

Idk Latex - um could the vectors (1,0) and (0,1) be a basis of R2?

I don't think it spans right?

Google says: "In fact, any collection containing exactly two linearly independent vectors from R 2 is a basis for R 2"

I'm an idiot - it spans

My bad

[limber sierra]

those span yeah

[dreamy iron]

Quick QQ about polynomials according to Axler.

[stoic python]

ninnymonger:

[dreamy iron]

Or maybe constant functions is a better word here?

[quartz compass]

yep

[dreamy iron]

Quick, indeed. TYVM

[quartz compass]

you're welcome

[shy atlas]

so $\mathcal{P}_0 (\mathbb{F}) = \mathbb{F}$ ?

[stoic python]

soαρ:

[sonic osprey]

ignoring some set theoretic stuff, yes

[wintry steppe]

Does anyone know why Pa is that given value?

nvm thats just the definition of orthogonal projection onto compl(W) since compl(W) has degree 1 right

[wintry steppe]

I have a question:

Larger picture in #help-3

[eager burrow]

From the equation $f(\alpha) = (\alpha|\gamma) \cdot \frac{f(\gamma)}{|\gamma|^2}$, you can see that you can express $f(\alpha)$ as a scalar product $(\alpha | \beta)$ if you define $\beta$ the way that they do

[stoic python]

Lartomato:

[eager burrow]

You're just pulling the factor $\frac{f(\gamma)}{|\gamma|^2}$ into the right side of the scalar product, which means that have you have to complex-conjugate it on the way

[stoic python]

Lartomato:

[eager burrow]

@wintry steppe does that make sense

[wintry steppe]

No,..

I'm so confused what is going on..

[sonic osprey]

What about what he said is confusing to you?

[hollow oracle]

i have no clue

whatsoever

how to do this red question

it makes 0 sense

[wintry steppe]

You can express f(a) as a scalar product (a|B) if you define B the way they do

[sonic osprey]

yeah

Go look at the properties of the dot product

especially the one about scalars

f(y)/||y||^2 is just a scalar

[wintry steppe]

Yea I know all the properties its just that I don't know where everything comes together

[sonic osprey]

Okay, well use the property

what do you get when you use that property on the scalar

[wintry steppe]

on which scalar?

Like where do you begin with B? what is B equal to ?

[sonic osprey]

f(y)/||y||^2 like I just said

You have $ (\alpha|\gamma) \cdot \frac{f(\gamma)}{|\gamma|^2}$

[stoic python]

Zopherus:

[sonic osprey]

use the property of the dot product

[wintry steppe]

isn't • just the ordinary multiplication

[sonic osprey]

ordinary multiplication of scalars yes

You said you knew your properties of dot product? There's a property that looks a lot like this

[wintry steppe]

(a, y compl(f(y)/|y|^2))

[sonic osprey]

yeah

And that value on the right is how they get B

[wintry steppe]

why does that give B..

like how

[sonic osprey]

B is equal to that value right?

And you should think about what B even represents in the theorem

[wintry steppe]

How is B equal to the RHS value?

[sonic osprey]

Look back at your theorem, what property do you want B to satisfy?

[wintry steppe]

Actually I'm not even sure of the meaning of a linear functional

[sick robin]

A linear functional is a mapping from a vector space to its underlying field

[eager burrow]

Beta is being defined as that thing

[sick robin]

Typically its a map from a vector back to R or C

[sonic osprey]

a linear functional is a linear function from V to the field that V is defined voer

[sick robin]

Linear functional != linear function

[sonic osprey]

I.e., R^n to R

[wintry steppe]

ok, I see, so its saying for every linear functional there is a b such that the linear functional is represented by f(a) = (a|b)?

[sick robin]

For a linear function you are correct

[sonic osprey]

linear functionals are required to be linear

Yes that's true LHC

[eager burrow]

Yes, exactly

[sonic osprey]

(its important that V is finite dimensional here too)

[eager burrow]

And if they define beta the way that they do, then it fulfils that property

[wintry steppe]

so here B is defined as the vector such that <a,b> = f(a), and if b = y*compl(f(y))/|f(y)|^2 this holds?

[sonic osprey]

kind of

The whole point of the proof it to show such a vector b exists with that property

We don't know if it exists, we're trying to prove that it does

And taking the value of b = that makes b have that property, so we know such a value of b exists

[wintry steppe]

oh god

im so stupid

I realized

thanks

do you plug B into the RHS in y to show this?

[sonic osprey]

Ah, they actually do show that b is unique, but not really clearly

I'm not sure what you mean by that

[wintry steppe]

like f(a) = (a|y)f(y)/|y|^2

and do they just let B = y

since they defined f(a) already

Sorry this wasn't clear, I think what I'm trying to say that they are trying to show that f(a) = (a|b) by defining b to be compl(f(y))y/|y|^2 and then replacing y with b

I don't know if this is wrong

[sonic osprey]

you don't really replace y with b?

b is replaced with that thing

[wintry steppe]

Oh right

yea

makes sense

thanks, and they did prove uniqueness using first proof it seems

This is confusing me. Why do they define f to be both the thing that is integrated as defined by the inner product <f,g> AND also the linear functional?

L(f) = f(z) is a notation I don't understand also

and finally I am clueless to why (hf)(z) = 0

if h = x-z

the other parts are ok

[sonic osprey]

L is a function from V to C, where V is the vector space of polynomials with complex coefficients, and C is the complex numbers

so L takes in polynomials, and outputs complex numbers

so if f is a polynomial, then L is defined to be the function such that L(f) = f(z)

[wintry steppe]

Ok, I understand that, now I don't know why (hf)(z) = 0?

[sonic osprey]

do you understand what hf means

[wintry steppe]

not quite, is h h(x) = x-z or h(z) = x-z

[sonic osprey]

h(x) = x - z

since z is already a fixed complex number

[wintry steppe]

Ok, so L(f) = f(z) because a linear operator on a polynomial is still a polynomial?

[sonic osprey]

No

Not even close

[wintry steppe]

then why?

[sonic osprey]

Think about it again, read it again

Read what I said again

[wintry steppe]

So they defined L(f) as f(z)? its not something required?

[sonic osprey]

That is how they defined L yes

[wintry steppe]

I see, and this definition makes L linear

Still I don't know why h(f(z)) = that integral - z = 0?

[sonic osprey]

its not h(f(z))

its (hf)(z)

you multiply the polynomials h and f together

[wintry steppe]

ok, but it becomes x•(integral)-z•(integral), which I don't see is 0

or I don't know how to multiply them together

[sonic osprey]

f isn't an integral?

f is a polynomial?

[wintry steppe]

it says for suppose we have f(z) = from 0 to 1 of f(t)compl(g(t))

[sonic osprey]

Okay sure

[wintry steppe]

even if it is polynomial, I cannot see why its 0..

[sonic osprey]

(x-z)(integral)

(hf)(x) = (x-z)(integral)

what is (hf)(z)?

[wintry steppe]

oh.

but why does hf become a function of x..

wait i see

[sonic osprey]

it doesn't matter what the letter is

I can write (hf)(y) = (y-z)(integral)

[wintry steppe]

No like, I mean

I was confused why z was not a variable

but then realized its fixed

So I was misinterpreting the proof by thinking that L is all linear functionals, instead they just chose 1 of them and showed with that particular L they can reach a contradiction

[sonic osprey]

L is not an arbitrary linear functional correct

[wintry steppe]

thanks again

[wintry steppe]

How do you show the basis used here is orthonormal? and what is the special case of the corollary?

[slow scroll]

the standard basis for C^{nx1} is orthonormal. The special case of the corollary is the line where they show that the conjugate transpose of A is the adjoint operator. @wintry steppe

[wintry steppe]

oh ok, just a further question to clarify: given any ordered orthonormal basis, and some matrix A corresponding to transformation T, is conjugate transpose of A always the adjoint?

I think thats true right? directly follows from the corollary

[slow scroll]

isn't that exactly what it says?

[wintry steppe]

right

ok thanks

[slow scroll]

npnp

[wintry steppe]

the words here confused me a bit because it says its a special case of the corollary

so I thought maybe there was cases within the corollary to explore

[slow scroll]

ah yea, ig by special case, they just meant the case when V = C^nx1

[wintry steppe]

Also do you know how they can rearrange like this?

like from 1st line to second line, and subsequently to 3rd line.

is trace of product commutative or what?

[slow scroll]

yep it is, as long as the multiplication is defined ofc.

[wintry steppe]

so trace of standard matrix multiplication?

is commutative?, but not necessarily self-defined matrix multiplication?

nvm i got it I think, I think I skipped over early sections of the book

[slow scroll]

idk what u mean by "self-defined" but basically y'know if u have two non-square matrices where the multiplication works one way, it doesn't necessarily make sense the other way.

[wintry steppe]

yea, my question was not well thought

[wintry steppe]

Does anyone know why U1 = (U1)* and U2= (U2)*. Is it because of the property T** = T?

[sonic osprey]

Not really

[wintry steppe]

then why

[sonic osprey]

You should note that this isn't true for all U1 and U2

such that T = U1 + iU2

It's saying that there exists U1 and U2 such that T = U1 + iU2 and U1 = (U1)* and U2 = (U2)*

[wintry steppe]

One thing I'm confused is it talks about self-adjoint T=T*, then suddenly it talks about representing any T linear on complex inner product space as T = U1+iU2

like I cannot see the connection of that to self-adjoint

[sonic osprey]

Given any complex number z, you can represent it as a + bi right

[wintry steppe]

Yea, though a,b real is the whole purpose of doing that

[sonic osprey]

And that's basically what they're saying

self adjoint operators are like the real numbers

T here is a complex operator, its an operator on a complex vector space

[wintry steppe]

Ohhh

ok

[sonic osprey]

So you can write it like T = U1 + iU2

just like you can write z = a + bi

and the property of real numbers is that conjugate of a real number is itself

just like for operators

[wintry steppe]

yea, thanks

[wintry steppe]

Im lost on how this shows B is in W

oh nevermind I get it

its proving that B is also an eigenvector

[neat halo]

What are quantifiers in the context of systems of equations?

How does this relate to the concept of free and bound variables

[neat halo]

What is the difference between a quantified variable and a bound variable

[wintry steppe]

Hey all

Just checking that x^2+y^2+z^2<=3 describes all points contained in a sphere ?

yes

[neat halo]

Seconded

[neat halo]

Let's say I have 3 independent linear equations with 8 variables in total. Not 8 variables in each mind you, but 8 unique variables total.

What can be expressed in terms of what else? There are rules clearly. For example, if each variable only appears in one equation, the variables in each equation cannot be related to those in the others at all.

[swift frost]

Anyone interested in working through a linear algebra course as a group?

Or know where I can find a group if not

[solid latch]

Hi

[wintry steppe]

Prove that dim(U+V) = dim(U) + dim(V) if and only if U∩V={0}

I am trying to prove: if U∩V={0} then dim(U+V) = dim(U) + dim(V)

I am sort of just not sure where to even start

[elfin ingot]

U+V = { u + v | u in u v in V }

find the general case first

let b be a basis of u and b' be a basis of v

how can u write any element in U+V

dimension wise

@wintry steppe

[wintry steppe]

Okay, so just a rough outline of what I am thinking - if we let b be a basis of U and b' be a basis of V, then b + b' - (b∩b') = U + V

Right?

And since (b∩b') = 0

Something like that? @elfin ingot

[elfin ingot]

an element in u would have to be written as n linear combinations and in v as n' linear combinations

n and n' dimension of U and V

and the chance of an element beingf in both

but yea ig ur right

[wintry steppe]

One second, let me think about this

Okay I'll write it out and let you know if I have any questions

[elfin ingot]

sure

[wintry steppe]

@elfin ingot I started off by saying: let (u1,....un) be a basis for U and (v1,....vn) be a basis for V

Not sure what to do next - would I then pick an element u in U and v in V and write them as a linear combination of (u1,....un) and (v1,.....vn)?

[subtle walrus]

this assumes both spaces have the same dimension

[elfin ingot]

yea

[wintry steppe]

Sorry, I mean v1.....vk

[elfin ingot]

yea

now find dim(U+V)

[subtle walrus]

do you see why the dimension of U+V can be at most the sum of the dimensions?

[wintry steppe]

Yes, I do

[subtle walrus]

then it suffices to find a set of n+k linearily independent elements in U+V

(this will then be a basis)

[wintry steppe]

Okay, one sec... So now I did u+v = m1u1+...mnun+n1v1+...nkvk -- so this should be linearly independent in U+V?

[subtle walrus]

what is u and v?

[wintry steppe]

I let u be an element of U and v be an element of V -- u1...un is a basis for U and v1...vk is a basis for V

[subtle walrus]

ok, what do you mean by "this" when talk about linear independence?

[wintry steppe]

u+v should be linearly independent?

[subtle walrus]

that statement makes no sense

[wintry steppe]

Lol

Um

[subtle walrus]

linear independence is defined for sets of vectors

and a singleton is always linearily independent in that sense

like ${u_1, u_2, \dots, u_n}$ is a linearily independent set for example

[stoic python]

Lochverstärker:

[subtle walrus]

(because it is a basis)

you now need to find such a set of size n+k in U+V

||there is an obvious choice||

[wintry steppe]

I see

Okay one sec, I'm kind of losing sight of this problem - so we are given U intersect V is 0 and we want to show that equality

So what exactly am I supposed to do to show this again?

[subtle walrus]

the dimension of a vector space is defined as the size of a basis

[wintry steppe]

Yes

[subtle walrus]

so ultimately you need to find a basis of U+V

[wintry steppe]

That has the same size as U + size of V

[subtle walrus]

a basis is a set of vectors that is both linearily independent and spans the space

yeah

[wintry steppe]

Yes

[subtle walrus]

you are given a basis of U and a basis of V

so it is natural to construct a basis of U+V from that

[wintry steppe]

So just basis of u + basis of v

[subtle walrus]

if + is set union, then yes

[wintry steppe]

Yes that is what I mean

So dimension is n+k

[subtle walrus]

well, you still have to prove that this actually is a basis

so that it a) spans the space and b) that it is linearily independent in U+V

[wintry steppe]

Okay, so after I prove this

What should I think about?

[subtle walrus]

huh?

[wintry steppe]

I meant, after I prove the fact that it is a basis

How would I use the fact that U intersect V = {0} satisfies the equation

dim(U+V) = dim(U) + dim(V)

[subtle walrus]

oh, so the other direction?

[wintry steppe]

Wait, I'm proving right now: if U intersect V = {0} then dim(U+V) = dim(U) + dim(V)

[subtle walrus]

yes

[wintry steppe]

Yes

[subtle walrus]

if you find a basis of U+V of size dim(U) + dim(V), you are done

[wintry steppe]

Okay got it, I'll do that

And I need to prove that basis spans and is linearly independent

[subtle walrus]

otherwise it wouldnt be a basis

[wintry steppe]

Got it

Thank you

[sick dragon]

how do i do this?

[slow scroll]

the determinant is 0 whenever you have linearly dependent rows/columns

[pallid rampart]

Calculate the determinant of M in terms of k, then set that expression to 0, then solve for k

[quartz compass]

by imagining expanding along the first row or last column (cause there's a 0 there it's convenient) you can tell it's a quadratic. That means it has at most 2 distinct roots for k. Doing what kxrider says by looking at the first two columns and last two rows makes it obvious what they are.

[sick dragon]

i got k=1,-5

is it saying to add those ks

[eternal finch]

Yes.

[quartz compass]

does sum mean add?

[sick dragon]

yes

[quartz compass]

[sick dragon]

ok so i add them, then put them back into the original, which SHOULD equal 0?

[quartz compass]

no way

they just want you to add them, has nothing to do with plugging them back in

it's only 0 when you plug in 1 or -5

[sick dragon]

i mean to prove it

if it is 0, then it is lin ind

[quartz compass]

no if it's 0 it's linearly dependent

[sick dragon]

k

[quartz compass]

plug in k and tell me what the linearly dependent vectors are

[sick dragon]

x1=-x2
x2=x2
x3=0

[quartz compass]

no no

https://prnt.sc/sp0qw0

if you plug in k=-5 then the first two columns are the same vector

if you plug in k=1 the last two rows are the same vector

that's what's making the determinant spit out 0 by being linearly dependent

[sick dragon]

ahhh

[sick dragon]

How about htis one?

[pallid rampart]

Think about how each type of row reduction will effect the determinant of a matrix

And think about how you can row reduce M into A, since the determinant of A is given

[wintry steppe]

Hey everyone

What's this called in english

In french it's Composante

the formula is

(dotP(a, b))/norm(a)

[eternal finch]

@wintry steppe A scalar projection of b onto a, I think.

[wintry steppe]

thanks

[hushed igloo]

any help for mine? @eternal finch

I've done this before but it's been a while

so all I need is something to jot my memory

[gray dust]

that isn't linalg

it goes in #prealg-and-algebra

[hushed igloo]

oh sorry

[swift frost]

Most people here are in uni classes I'd imagine right

[gray dust]

on the left you can see this channel is in the "early uni" category

[swift frost]

That's the level of math I thought, not necessarily related to the people here

[wintry steppe]

engineering school

so uni yes

[wintry steppe]

Could someone explain back substitution to me over voice? Specifically this problem

[slow scroll]

its where you put the system in row echelon form, and then you start from the bottom, z = [something] and plug it into the equation above, solve for y, and then plug in y and z into the top equation and solve for x

[wintry steppe]

so always start from z?

and then go to y

my priority is to find z first?

[quasi vale]

@wintry steppe When you have a matrix in row echolon form/reduced row echolon form, you will know which variable's value you can find first.

Then others will follow.

[haughty lagoon]

so the spectral theorem says that in the complex case, all normal matrices can be unitarily diagonalized, but are there any matrices that can be diagonalized, just not unitarily so?

because complex numbers are algebraically closed there'll definitely be "enough" eigenvalues

but is it possible that the eigenspaces will have high enough dimension, but there is no orthogonal set of eigen vectors that span the whole space?

[placid oracle]

Consider an operator defined by the rule
L1(P(t)) = P(t).
How can I find its eigenvectors and eigenvalues and prove that it is not diagonalizable?

[slow scroll]

umm is it just me or does that look like the identity @placid oracle

[placid oracle]

Hmm which one

[slow scroll]

L(p) = p

[wintry steppe]

Prove that dim(U+V) = dim(U) + dim(V) if and only if U∩V={0} - I have hit yet another snag with this question

After picking a basis for U and V and adding them together

I am not sure how to show that it is linearly independent

[sonic osprey]

Well, you still haven't used the fact that U intersect V is {0} yet, so try to use that

[wintry steppe]

How would I use it? if U intersect V is 0 then I know the basis of U and basis of V - their linear combinations - none would be the same

Right?

So let's say basis U is u1.....un and basis V is v1....vn. Then, c1u1+....+cnun+a1v1+...akvk = 0...

I know c1u1 != c2u2 != .... akvk, right?

[radiant jasper]

You can't assume c1u1 != c2u2 here. A counter example would be when c1 = c2 = 0.

[wintry steppe]

@sonic osprey

Hmmm, so any idea how I would use the fact U intersect V is 0?

[radiant jasper]

I think you are on the right track with the bases

[wintry steppe]

I am just slightly confused how to show c1......ak = 0

To show linearly independent

I've been stuck on this for a while, it's frustrating me :/

[radiant jasper]

but you do know how to show linear independency of vectors, right?

[wintry steppe]

Yes

[sick dragon]

what's the solution to this?

[foggy lodge]

how do you add vector spaces

is it the same as union

[wintry steppe]

Show that the relation c1v1+...+ckvk = 0 can only be done when c1...ck = 0

@radiant jasper I am pretty sure that's how to show linear independence

[radiant jasper]

ye that's correct

so your just one step away

[wintry steppe]

So the issue is I don't know how to show those constants are 0 lol

I don't understand how if U intersect V = {0} should help with this

Since I am proving if U intersect V = {0} then.....

[radiant jasper]

If you don't have that condition then how could you assume u1..un and v1..vl are linear independent

@foggy lodge This is the definition

[wintry steppe]

Oh yes, you're right

Well then if they are linearly independent then we know using that fact the constants are 0

So that's it?

[foggy lodge]

Oh ok

I forgot

its been a hot minute

[wintry steppe]

Given that we know U intersect V is {0} we know that basis of U and basis of V are linearly independent

[radiant jasper]

But you have to prove it first right?

[foggy lodge]

||yeah, the basis of the sum will become the basis elements of U and V, because then any element of a basis will not be expressable as a linear combination of the other terms.||

[wintry steppe]

Hmm, yeah but I still don't get how I show the constants are 0

I don't see what information they provided can help me arrive at that conclusion

By they I mean the question

[foggy lodge]

U and V are subspaces of some vector space W, right?
Then that means that U + V is all possible linear combination of all the vectors in U and V. However, we can use a basis of U and a basis of V to find all possible sums, which creates a new subspace
You create a new basis, and at most it can have all the elements in both of the original bases, however, if we can express one or more of the vectors in the bases as a linear combination of the other, i.e. U intersect V being nonempty, then the dimension will be smaller

This is my understanding so far

[wintry steppe]

Yes

Exactly

That’s exactly correct, whatever you have stated

[foggy lodge]

then what are you stuck on

it seems like you're done with the problem

[sick dragon]

https://cdn.discordapp.com/attachments/540211747613704221/715423432711471104/unknown.png anyone know how to solve this?

[foggy lodge]

ill give it a look @sick dragon

[wintry steppe]

U intersect V is empty though so this means that dimension will be the exact same

[foggy lodge]

not quite, this just means that there isnt an element in both of those subspaces

so the basis elements are unique across both of the bases

[wintry steppe]

I see

[foggy lodge]

so if they are unique, all of the elements in both of those bases form a new basis for this new subspace

which means that dim(U+V) = dim(U)+dim(V)

where dim(U) = |some Basis of U|

Ok @sick dragon, what have you tried so far

[sick dragon]

i made them into two separate vectors

then i took their cross product 🤷

[foggy lodge]

Vectors?

You're dealing with a matrix transformation though

[sick dragon]

yeah this isnt in my class slides

[foggy lodge]

I see

[sick dragon]

so im just trying to remember stuff from calc

[foggy lodge]

Are you familiar with applying matrices to other matrices? I can give a quick rundown

[sick dragon]

sure

[foggy lodge]

When we have Ax, that represents a linear transformation A being applied on a vector x.

[sick dragon]

Yeah

[foggy lodge]

Now, if we have BAx, we first apply A to x, giving us x', then we apply B to x'

However, we can represent BA as one composite transformation

When we multiply the matrix B with A, B acts on the ROWS of A. Similarly, A acts on the COLUMNS of B

Therefore, in your problem, we can introduce a new matrix B that does row operations in order to get the modified matrix

The first two rows are the same, so in this newly introduced matrix, we will take 1 of row 1, 0 of row 2, and 0 of row 3

So our first row is 1 0 0

Our second row is unchanged, so we want 0 of row 1, 1 of row 2, and 0 of row 3, so the second row is 0 1 0

Try the third row yourself

[sick dragon]

0 of row1, 0 of row 2, 1 of row 3. 0 0 1

[foggy lodge]

Not quite

That would just give us the original row

We want 3g+a, 3h+b, 3i+c

How would we get that

[sick dragon]

1 0 1?

sorry 1 0 0

[foggy lodge]

Try again

We want x quantity of row 1, y quantity of y, and z quantity of row 3

When we do that, we get x y z

We would scale row 1 by x, row 2 by y, and row 3 by z

Then add them all up

[sick dragon]

you lost me sir

well i figured it out but im interested in where you were going

[foggy lodge]

Ok great

So you see, 3g+a, 3h+b, 3i+c is a sum of 3 times row 3 + 1 times row 1

So our bottom row is going to be 1 0 3

We then have

1 0 0
0 1 0
1 0 3

We apply this matrix to matrix A

Since det(AB) = det(A)det(B), we get det(this new matrix)*10

The determinant of this new matrix is 3, so the answer is 30

@sick dragon

[sick dragon]

interesting

heres what i did haha

[foggy lodge]

Lmfao

This is wrong

Somehow it gives the right answer though lol

My method is the intended one

[sick dragon]

😄

[foggy lodge]

Honestly if I was the professor I wouldn't even be mad

But just to clarify, you can only pull out scalars from a matrix if you divide ALL of the elements by that scalar

[sick dragon]

i see

[foggy lodge]

Also that middle step is illegal lol

[sick dragon]

can i just plot these and take the area

[foggy lodge]

It's a good idea to plot it yeah

[quartz compass]

since this is linear algebra and you know it's a parallelogram

you should probably use a determinant to compute it

[foggy lodge]

I'm assuming it's a type of triple product

Idk I never did that in my linalg class

[quartz compass]

well, triple product gets you volume

[foggy lodge]

Yeah

[quartz compass]

you want cross product effectively

knowing it's a parallelogram, just subtract one vector from the other three to put it at the origin

one vector should be obviously the sum of the other two

[sick dragon]

i think im supposed to move it to the origin (dunno why)

[quartz compass]

so you just take the cross product of those two to get the area

[sick dragon]

then use determinant 🤷

[quartz compass]

yep that's what I just said

[sick dragon]

cool

[quartz compass]

subtracting one vector from the other 3 is translating it

[foggy lodge]

I think of it as finding the side vectors of a parallelogram because visually that's what subtraction gives you

Though, what's the intuition for taking the cross product

[sick dragon]

this is what i was talking about earlier btw

this is how i did it

😄

[foggy lodge]

Huh ok

I think it's certainly an interesting thought you have

Because the determinant is a measure of how much you're scaling the volume/area

I never thought about that

However, I think it's incredibly difficult to do that problem that way, and I think the point is to use the multiplicative nature of the determinant function to find the answer

[cold topaz]

dimension of a matrix M_mn, is the result of m times n. right?