linear-algebra

[lavish jewel]

<@&268886789983436800>

thanks

[north anvil]

Assuming this channel is not in use

Is this the simplest way to calculate the cross product of 2 or 3d vectors?

[nocturne jewel]

Cross product is only in R^3

[north anvil]

you can't cross a 2d vector with another 2d vector?

[nocturne jewel]

and I mean those are just the components of the cross vector

no

where's the 3rd dimension in R^2?

[north anvil]

good point

so if a cross product is defined between two vectors P and Q, then both P and Q are 3-dimensional

[nocturne jewel]

Yes

[north anvil]

why can't this apply for higher dimensions (i can't visualize it, anyway)?

[nocturne jewel]

Outer products exist for higher dimensional co-ordinate spaces

[north anvil]

oh okay - i guess different types of products apply for higher dimensions

[nocturne jewel]

Yeah, you create inner product spaces by introducing an inner product like the canonical dot product

outer products just aren't covered in say first year linal

[north anvil]

Does the screenshot I sent above essentially use the formulas for the determinants of matrices that are inside or part of a 3x3 matrix with the components of A, the components of B, and the components of the cross product of A x B

[nocturne jewel]

yes, the 1st row of the matrix is the unit vectors, then components of a and components of b

$a\cross b =det\left(\begin{bmatrix}i&j&k \ a_x&a_y&a_z \ b_x&b_y&b_z\end{bmatrix}\right)$

[stoic python]

Mosh

[north anvil]

right

to get the component of i, find the determinant made by the matrix that has 0 entries that share a row or column with i

[nocturne jewel]

You just compute that det

laplace along the 1st row like normal

[north anvil]

cross out the row + column where i is in, and the matrix formed by the remaining entries is the determinant that becomes the component for i

[nocturne jewel]

yes, the cofactor

[north anvil]

isn't laplace multivar?

idk that ;-;

[nocturne jewel]

laplace / cofactor expansion is how you evaluate determinants

[wintry steppe]

quick question: i've posted a linear algebra problem inside math help question, should I delete from there and eventually post it here so It is more "on topic" or those channels are for every math discipline?

sorry for disturbing though

[nocturne jewel]

doesnt matter

[north anvil]

i think i'm done, but you're good 🙂

ty mosh 👍

[nocturne jewel]

the help channels are for any question, higher level math tends to be put in the corresponding channel though

[wintry steppe]

uh okay, so I'll leave it there. It's in #help-3 in case someone wants to solve a stupid problem I can't understand ahah

[turbid trellis]

Hey how to calculate this, I assume it's with matrices ?

However h and k are not defined, so I'm not sure how to solve

[teal grotto]

multiply the top equation by -3

[turbid trellis]

Yes I figured that out

But I need to find h and k

[teal grotto]

the values of h and k that work are the values h and k such that k = 3h

[turbid trellis]

Ah okay I understand

Since the equations are equal after all

Only we need to divide the final k back to h based upon the left side of the equation and how they're related

[teal grotto]

yea. the image of the linear transformation represented by the given equations is a line, specifically the line y = 3x.

[turbid trellis]

I understand thanks

I just started reading an book on linear algebra

[teal grotto]

cool! not sure what ur second question is asking

[turbid trellis]

I have no second question

I just had to find h and k which make the system consisten

[teal grotto]

oh. was confused on what u meant by this. nvm then

[dusky blaze]

What do you call one "thing" in a matrix?
Like if you have the Matrix

[a b]
[c d]

a is a what of the matrix? Element?

[tidal wharf]

entry

yeah or matrix elements

both are used

[tidal wharf]

There's various ways to generalize the cross product to higher dimensions, depending on what properties you want it to retain. The most similar but also pretty restrictive generalization exists in 7 dimensions: https://en.wikipedia.org/wiki/Seven-dimensional_cross_product

[dusky blaze]

@edgy grail My function that I wrote to determine if lines intersect using linear algebra fails on partially overlapping parallel lines. The lines I tested were [(1, 1), (1, 3)] and [(1, 2), (1, 4)].
The reason it determined that the line segments did not intersect is because the matrix A was determined to be not invertible.
The matrix A being

[x1 - x0  x2 - x3]
[y1 - y0  y2 - y3]

Filling in the values, we get the matrix

[1 - 1  1 - 1]
[3 - 1  2 - 4]

Simplifying, we get the matrix

[0  0]
[2 -2]

Is it true that this matrix is not invertible?

[teal grotto]

yes, it’s determinant is zero.

[dusky blaze]

Interesting. The determinant is zero but the line segments still intersect.

[teal grotto]

well, that’s expected. they’re the same line (the ‘that’ being zero determinant)

[dusky blaze]

But they aren't the same line (segment).

hmm

I either forgot that or I didn't understand the significance.

Alright, thanks for the insight.

[hollow finch]

The line segment is on the line that that matrix describes, however.
Zero determinant means that the system might not have a solution, but it still could. A nonzero determinant means there will always be one unique solution.

You said your function is used to

determine if lines intersect
right? is that in 2 dimensions or 3

Because in two dimensions, unless they are parallel they will always intersect.

Do you maybe mean that the function finds where they intersect? Then the answer is not longer trivial

[dusky blaze]

Zero determinant means that the system might not have a solution, but it still could. A nonzero determinant means there will always be one unique solution.
This is great information. I thought that zero determinant meant that the system will not have a solution. And that a nonzero determinant meant that there will be a solution.
right?
No. It determines if two line segments intersect.
is that in 2 dimensions or 3
That's in two dimensions.
Because in two dimensions, unless they are parallel they will always intersect.
I'm talking about line segments not lines. Non-parallel line segments may not intersect in two dimensions.
Do you maybe mean that the function finds where they intersect?
Currently, the function just finds the values of t and s, the variables for the interpolation/extrapolation functions for the two line segments a(t) and b(s). I use these values to determine if the point of intersection lies on the line segments or on the infinite extension lines (extrapolated). The commented part of the following code sample explains how I use t and s. (The comments start with a hash character)

       # 0 <= t <= 1 describes all points on the first line segment
       # 0 <= s <= 1 describes all points on the second line segment
       # Therefore, if the point of intersection is on both line segments, then the line segments intersect
       return 0 <= t <= 1 and 0 <= s <= 1   # True if intersect, False otherwise

Here is a sample of the code. It's pretty Englishy so if you don't know programming, you might still understand it.

[dusky blaze]

Here's the part that is logically messed up.

        # Matrix A must be invertible
        A = Matrix([[x1 - x0, x2 - x3], [y1 - y0, y2 - y3]])
        if not A.invertible:
            print("A is not invertible so lines do not intersect")
            return False   # Lines do not intersect

The reason why I check if matrix A is invertible is because if it isn't invertible, then the determinant will be zero. Zero determinant is a problem because I need to divide by the determinant later on.

        # The inverse of a matrix, A.  is defined as follows
        # [a b]^(-1) = 1 / (determinant(A)) * [d -b]
        # [c d]                               [-c a]

        # Therefore, the inverse of A for our system can be written as
        # 1 / (det(A)) * [y2 - y3    -(x2 - x3)]
        #                [-(y1 - y0)    x1 - x0]

If the determinant is zero and I divide by it, then that's undefined.

[teal grotto]

got a dumb question. the vector space C over R. how do you represent matrix multiplication?
like a matrix A would have real entries, but what dimensions would it have, if i want to do Av for some complex vector v?

[nocturne jewel]

I might be missing something.. but complex numbers arent matrices...?

[thorn robin]

Not with that attitude

@teal grotto If you're just considering C as a 2-dimensional vector space over R and identifying x + iy with (x, y) then I don't think you have a natural way to represent complex multiplication

But if you identify x + iy with a 2x2 matrix (x, -y; y, x) then you can check that complex multiplication corresponds to matrix multiplication

Additionally with this identification you get that |z|^2 corresponds to the determinant

[nocturne jewel]

I mean my prof had mentioned stuff like polar decomposition was like exp form of a complex number, but never understood what he meant lol

[teal grotto]

yea. maybe i didn’t explain that well. just want to have matrix multiplication with vectors in C

like if i want A to represent some linear transformation from C to C

[nocturne jewel]

Oh matrix vector multiplication?

[teal grotto]

i just want to do Av

[nocturne jewel]

ok yeah makes sense now lol

[teal grotto]

but like

v isn’t a two by one matrix like in say, R^2

so…i’m confused

[hollow finch]

I see, line segments. That's interesting. I'd have to think about that before I could help any further.

@teal grotto not sure what you mean. like A is real and v is complex and you want to represent Av with all real matrices?

[teal grotto]

um. so R^2 as a vector space over R. it’s two dimensional and linear transformations from R^2 to R^2 can be represented by 2 x 2 matrices with real entries.

asking for the analogue in the vector space C over R

[hollow finch]

Well you can have v'=(Re(v),Im(v))

C is isomorphic to R2

If that doesn't answer it then I think i still don't understand your question

[teal grotto]

oh wait. did i just not realize

C is a two dimensional R vector space. isomorphic to R^2

bruh

see. dumb question

[hollow finch]

Yeah

[teal grotto]

lol thanks

[hollow finch]

Np

[dire thunder]

@wintry steppe this is not linear algebra

[wintry steppe]

Then what is it?? Where shall I post

[dire thunder]

#prealg-and-algebra or #precalculus

[wintry steppe]

Alright thanks

[drowsy flower]

I know one common rotational matrix is $\begin{bmatrix} -1&0\0&-1 \end{bmatrix}$

[stoic python]

meguuuuu

[drowsy flower]

and its det is 1 as well, would this work?

[dusky epoch]

presenting this as an example of "det(A) = 1 implies A is a rotation matrix" does not constitute proof

just because this particular matrix of det 1 is a rotation doesnt mean they all are

and in fact,

$\bmqty{5 & 7 \ 7 & 10}$

[stoic python]

Ann

[drowsy flower]

hm okay so how do I approach this question then.....

[nocturne jewel]

well is the statement true or not?

[drowsy flower]

My guess is that it might be true ... but not sure

I think such alpha might exist

[nocturne jewel]

well what about the matrix Ann wrote out?

[drowsy flower]

um isee it has determinant as 1

and that its transpose is itself?

im confused what I am supposed to look for there...'

[weak needle]

See what the matrix does to the basis vectors (0,1) and (1,0). Can the matrix possibly be a rotation?

[drowsy flower]

ah no thats not rotational matrix because that matrix * its transpose is not identity matrix

[weak needle]

Correct

[drowsy flower]

Ah so that is one counterexample then

[weak needle]

What I was trying to get at was unit vectors are not mapped to unit vectors

[chilly osprey]

all i need to do is find the derivative of the equation

and then plug in 3 with all the values of x

[lavish jewel]

wrong channel, arvin. try #calculus

[chilly osprey]

right?

oh sry

didnt notice

[dapper scarab]

quick question

if im correct this is independent right for the last row its 000 and there are two pivots remaining

nvm its dependent free variable

oops

[turbid trellis]

Hey how to find this ? Manually I can only figure out one value for h

Nevermind

h != 2

Because they will be parallel at h=2

[wintry steppe]

So, Riesz's representation theorem works for just inner product spaces, right?

Or is it solely for Hilbert spaces?

[glacial mango]

I think it's l.i. iff it reduces to the identity

[turbid trellis]

Okay I have no clue how to do this

Or what they're actually asking

[leaden tide]

"consistent system" just means there's one and only one solution

[turbid trellis]

Oh the book says one or more solutions

However what are they asking ?

The question confuses me

[leaden tide]

Most notably, if you have a row that's like 0 0 0 a with a ≠ 0 it's not gonna be consistent

[turbid trellis]

Okay I understand that part

g and h and k != 0

[tidal wharf]

https://math.stackexchange.com/questions/678656/necessity-of-completeness-of-the-inner-product-space-in-riesz-representation-the

[leaden tide]

nope

[tidal wharf]

also probably better in #advanced-analysis since it involves Hilbert space

[leaden tide]

Do you really think a+b+c=0 has no solutions because the RHS is 0

[turbid trellis]

No I think g,h and k may not be zero ?

[leaden tide]

why couldn't they?

[turbid trellis]

Nevermind I misread your message

I'm new to all of this

[wintry steppe]

So, a nondegenrate inner product doesn't establish an isomorphism between a vector space and its dual?

[leaden tide]

it's aight

[turbid trellis]

But what do they want from me?

I have the answer on my right but I don't know what it means

[wintry steppe]

It shouldn't be problem for real or complex vector spaces, right? Since they are complete?

[leaden tide]

So, in these questions, you want to make a row of the form (0 0 0 something) appear

And it is possible, by doing row operations

If it wasn't, there would be no question, any values of g,h,k would do

[turbid trellis]

Why do they want 0 0 0 if there are only 3 rows of the left size ?

[leaden tide]

what?

[turbid trellis]

I learned that in trinomial like that in the question

And have three of them

They need to end up like

1 0 0
0 1 0
0 0 1

[tidal wharf]

It is not a problem for the vector spaces R^n or C^n considered in linear algebra, but it is a problem for general vector spaces like the continuous functions on an interval example (with L^2 inner product) given in the stackexchange link. Note C(I) is complete using the sup norm but not with the L^2 inner product.

[turbid trellis]

So why would we want 000 ?

[leaden tide]

You're mixing up some things

If you're able to get to the identity (the matrix you posted), then great

That means that your system is invertible i.e for any g,h,k, there's one and only one solution to the system

But that's not always the case

[turbid trellis]

I don't know how to get to the identity in this one, because it contains variables

[leaden tide]

variables can be added and multiplied like anything else

but the thing is

you're not going to get the identity

you can try, but at some point you're gonna end up with a row of 0s whether you like it or not

[turbid trellis]

In this scenario ?\

[leaden tide]

yup

Let's go over this again

if you could get to the identity, that would mean the system is invertible

any g,h,k gives you one and only one solution

in particular, you always have at least one solution

[turbid trellis]

I didn't read about invertible systems yet

I am at chapter 1.1 haha

[leaden tide]

Who cares about the jargon, call it "always one solution", call it "invertible", call it "i can get to the identity", it's all the same

Do you really think that they would ask you this question (when does the system have solutions) if you could prove that the system always has solutions?

[turbid trellis]

No

It only has one solution right ?

[leaden tide]

Actually, either it has none, either it has infinitely many

[turbid trellis]

Okay that I understand

[leaden tide]

Usually, you'd expect three equations of three variables to completely determine a point

I mean, you should have enough info, right?

[turbid trellis]

I guess, but it asks me to find an equation or something

[leaden tide]

we'll get there

If you can derive one equations from the other using linear combinations, well then that just means one of your three equations doesn't add any new information

[turbid trellis]

Okay shall I do that now ?

[leaden tide]

Yeah, find which row can be expressed in terms of the others

[turbid trellis]

I'm stuck haha

I can't find anything in here to merge

Is this some other starters have trouble with too ?

[leaden tide]

For example, if you're trying to see if the 2nd row can be expressed in front of the others, you first gotta at least make sure that it works for the first coefficient

The question would be then to find a and b such that a(-1)+b(2)=0

give me some

[turbid trellis]

What do you mean with expressed in front of the others ?

I wish the book actually explained what they're doing instead of just showing the answers

[leaden tide]

Yeah...

If we call your rows R1, R2 and R3

Then we're trying to find, say, a and b such that R3=aR1+bR2

[turbid trellis]

Ah okay like that

Let me try that

[leaden tide]

If this works for the entire row, it has to work for the first coefficient

Then you see if the numbers you found work for the rest

[turbid trellis]

Okay and a and b will be used for the rest too ?

[forest quiver]

Which book are you using sentineL?

[leaden tide]

For the rest of the entries in the row

[forest quiver]

Oh I was using that one for a while

But I switched to another

[turbid trellis]

I just want to learn linear algebra, if it's this book or another, I don't care

[forest quiver]

This one is pretty good for computations I feel like

I came a bit late

So what was ur concern exactly?

Is there somethign specific you don't understand?

[forest quiver]

So just get it into reduced echelon form (remember that g,h,k are numbers just like the other ones)

Syst3ms do you mind if I ask a question really quick?

[leaden tide]

go ahead

[forest quiver]

So

it took me about a day of reflecting

But I finally understand where dot product comes from

this is what my book says

I understand that it works for perpendicular vectors

because of similarity

[turbid trellis]

Btw if this is easy and I'm stupid, please say, so I can switch away from preparing for CS and pick a different career, I feel very dumb

[leaden tide]

nah, it's not trivial

[stoic python]

Tim O'Brien

[forest quiver]

But this similarity is when these vectors x and y are perpendicular

How can a general definition emerge from this?

[leaden tide]

let me get my material on inner products

[turbid trellis]

I found it h = -2g-k

[stoic python]

Tim O'Brien

[forest quiver]

Like understand that the first equation comes because they are perpendicular vectors

[leaden tide]

And that has to be 0 otherwise you get 0=something nonzero

[turbid trellis]

And this just proven that they're equal ?

[forest quiver]

So how do we make this jump with certainty to the general equation for all 2 vectors?

(sorry for the interjection)

[leaden tide]

if you take -2R1+R2-R3, you get a row of 0s and -2g+h-k at the end

So, if you don't want that to be impossible, your RHS has to be 0 too

[turbid trellis]

Okay cool, I'm gonna read back and try to understand what I just did haha

[leaden tide]

Well, I feel like the definition of the dot product isn't so much "=0 for perp vectors" as much as it is "a measure of how much two vectors point in the same direction"

[forest quiver]

hmmm

and that's what is troubling for me

But it's ok

I just need to stop being a perfectionist around this and understand that some things I will understand later

Im just losing time sitting here thinking for hours

[leaden tide]

nah

honestly this doesn't feel like something that requires a lot of prior knowledge to discuss

I'll have you note a couple things

[forest quiver]

ok

[leaden tide]

First of all, there is little actually special about the (canonical) dot product

[turbid trellis]

Last question, why are we trying to find that ?

[leaden tide]

$x_1x_2+2y_1y_2$ is also a valid inner product

[stoic python]

Syst3ms

[leaden tide]

the problem is most easily seen when you have one row of 0s, you can immediately go "oh wait, the RHS has to be 0"

So, if you find something like that, you can just subtract rows to get your row of 0s

[leaden tide]

another thing of note is that given an inner product you can define the norm

as $|x|=\sqrt{(x|x)}$

[forest quiver]

hmm

[stoic python]

Syst3ms

[forest quiver]

x|x /

?

[leaden tide]

general notation for an inner product

[forest quiver]

Idk that

yet

it comes later

[leaden tide]

the dot product is an inner product

[forest quiver]

ohhh

Yeah I get that

[leaden tide]

the dot product has interesting properties, and inner products generalize that

What I showed is also a valid inner product, it just gives rise to a weird norm

[forest quiver]

Yeah yeah that's the pythagorean theorem

the length of the vector is just sqrt of its sides squared

So that's where dot product comes from ?

[leaden tide]

Honestly, I don't know enough about its history to tell you that

[forest quiver]

well I appreciate your help

[leaden tide]

But it is true that there are many possible inner products, but that only one of them gives us distance like we expect

[forest quiver]

hmmm

[leaden tide]

namely the dot product

[forest quiver]

ok

x^2 = ||x|| ^2

[leaden tide]

say what now?

[forest quiver]

x^2 = x_1^2 + x_2^2

when n = 2

[leaden tide]

i dislike that notation of the dot product with itself, but yes

[forest quiver]

Wait so

last thing before I go

$x\cdot x=x^2$ ?

[stoic python]

Tim O'Brien

[leaden tide]

idk, that notation sucks

maybe you can define it as such

[forest quiver]

$x\cdot x = x^2 = ||x||^2$

[stoic python]

Tim O'Brien

[leaden tide]

but if you do, x^3 doesn't make sense

[forest quiver]

Ok

[leaden tide]

so i would refrain from writing x²

[forest quiver]

Alright

thank you

[wintry steppe]

In the infinite dimensional case, it depends on the completeness of the space, yes?

[marble lance]

I think what confused you is the order of the material. They didn't "go from" the one equation to the other. They said that we know two vectors are perpendicular iff x1y1 + x2y2 = 0. So this quantity x1y1 + x2y2 is of interest because it can tell us when vectors are perpendicular, so let's call that quantity the dot product. @forest quiver

[forest quiver]

hmmm

Yeah

[leaden tide]

(and on top of that it fits with the common notion of distance)

[marble lance]

So the x1y1 +x2y2 = 0 thing was just motivating why we care enough about this quantity to give it a name

[forest quiver]

But how does that dot product work

when the two triangles don't have 90 degree angle

there

are they still similar?

is the question

I think they are right

no wait

thats dumb

[leaden tide]

A common geometric interpretation of the dot product in the general case is to orthogonally project one of the two vectors onto the other, multiply their norms, and add a negative sign if they point in opposite ways when projected

[forest quiver]

yeah I read about that on stack exchasnge

But whaaaaaaaaaaaaaaaaaaaaaaat

I mean it makes sense that its a projection

because 90 degree will yield 0

but idk

im not gonna worry about that

i cant get stuck on details like this

thank you tho

[leaden tide]

The reason why it's orthogonal projection is because the dot product of two perpendicular vectors is 0 :P

(you can prove this yourself, if you are convinced that for any vector x, you can split any other vector into one component that is aligned with x, and one that is orthogonal to it)

me neither

[forest quiver]

Micheal Penn

I just need to udnerstand where this damn projection comes from

Some cosine stuff?

What's a "signed angle"

[lavish jewel]

it means it's positive if it points in the same direction, negative otherwise

[turbid trellis]

@forest quiver Did you understand the book I am using in the first place ? These questions at the beginning of the book are so damn hard, after only reading 4 pages of linear algebra they ask me to calculate temperatures over a surface using linear systems

[forest quiver]

Ok right

[lavish jewel]

those are the only 2 options you have, since they specify it has to be parallel

[forest quiver]

I understood the book well

it's pretty straightforward

You just learn the computation

and apply it

[turbid trellis]

What things did you do in advance ?

[forest quiver]

Wdym

pre reqs?

[turbid trellis]

I feel like I'm not prepared for this book

[forest quiver]

I just did calc 2

nowim doing this

[turbid trellis]

Ah okay yeah, I only did calc 1, that might explain it

[forest quiver]

No not rly

you dont need calc

to do it

just learn the computation and follow the steps its actualyl a great book. The one im doing now combines linear alg with calc 3 tho

[turbid trellis]

They barely explain anything except how to solve an system, and then just throw a bunch of questions requiring deep understanding

[forest quiver]

Show me a question

Like the one with the temperatures and the nodes

?

[forest quiver]

Yeah this one

I loved this one

it took me like 30 mins to solve

You just follow the instructions

[turbid trellis]

Did you find them hard to do ?

[forest quiver]

The second one you sent isn't hard

let me think about the first one

Consistent for all possible values

[leaden tide]

Yeah not too hard

[forest quiver]

Does that mean its infinite solutions?

no no

[leaden tide]

it has at least one

[turbid trellis]

They mean that it creates consistent system for all values of g and f

And ask what that says about c and d

[forest quiver]

The one with the temperature tho was very nice

It helped me understand it 10x better

[leaden tide]

but since it's for all possible f and g it might as well be asking that the system be invertible

[turbid trellis]

Okay I should first do the temparatures one and then the others ?

[forest quiver]

well if c=1 and d=3 then its not consistent because g neq f

Yeah just create the equations

and line em up

[leaden tide]

it's more general than that

[forest quiver]

then solve

Yeah hmmm

[leaden tide]

for this

[forest quiver]

yeah yeah

hmmm

[leaden tide]

(in general, when can you cancel out an equation by linear combinations)

[forest quiver]

yeah but we dont know the coefs.

[lavish jewel]

no, but you're told the system is consistent

which tells you about the coefs

[tidal wharf]

Yes, otherwise you can take the completion and find linear functionals from there that can't be represented using the non complete space.

[forest quiver]

hmmm

[wintry steppe]

Okay, gotcha! Thanks!

[forest quiver]

not sure for 27

i did some algebra

but I got right back to where I started

[leaden tide]

multiply the first equation, say, by a and see if you can find c and d that make the system not consistent

[turbid trellis]

@forest quiver

I nailed it

[forest quiver]

Yeha it's a great one

it took me like 45 mins of just computation

[turbid trellis]

I now understand it indeed better

Linear algebra is useful when you're calculating stuff which depends on each other

In a circular fashion

Normal calculations would create infinite loop right ?

1 depends on 2 depends on 3 depends on 4 depends on 1

[forest quiver]

It's good when you have multiple linear equations

with multiple variables

because there exist simple algorithms

to isolate variables

and solve with ease

[turbid trellis]

Yeah also that

In this case it fixed the depends on right ?

[forest quiver]

wait

yo

was that ur final solution?

[turbid trellis]

T_4=20

T_3=115/4

[forest quiver]

no

[turbid trellis]

Fuck

I made an arithmetic error somewhere

These matrices are turning me crazy haha

[forest quiver]

This problem was so good that I have my sheet of work for this from like 2 weeks ago

lying on the floor

of my room

yes me too

Try it again

but this time have like a routine

[turbid trellis]

Or I'm just using wolfram alpha, since it was just a arithmetic error

Because this part is correct right ?

Finally it's fun when I grasp it (kinda)

[forest quiver]

@turbid trellis

can ytou explaint his one

this one

[turbid trellis]

Not really

I haven't done it yet, found it too confusing

However I think that they need to be opposites of each other

In order to match at each f and g

-x+y

x-y

Forgive me if I'm wrong

[forest quiver]

udj

idk

theres solutiosn but I wiull check them alter

[zinc lava]

Okay.

Linear algebra?

How do you usually solve ax +by = f, ca +dy = g using matrices?

What matrices will give consistent answers?

[limber sierra]

row reduce the augmented matrix [\begin{pmatrix}x & b & f \ c & d & g\end{pmatrix}] and back-substitute

[stoic python]

Namington

[limber sierra]

(the x instead of a may seem weird, but it's because you wrote ca + dy = g instead of cx + dy = g)

if your a, b, c, d, f, g values are actually fixed (so x, y are your variables) then the answer is "you dont solve it using matrices"

you use ac + dy = g to solve for y

which is middle school algebra

and then plug that into ax + by = f and solve for x

you COULD use matrices but why overcomplicate it

you have a linear equation with 1 unknown

solve that

[dreamy iron]

https://cdn.discordapp.com/attachments/678247253822930959/863251910776586250/Screen_Shot_2021-07-09_at_19.41.05.png

I'm trying to write some notes on Dual Spaces. Anything obvious that I missed, notation wise?

(There's been mention in this chat about covariant and contravariant elements, i'd like to work that in but i don't know how )

[wintry steppe]

hi guys quick q

Can someone pls verify this

[dreamy iron]

,rotate

[wintry steppe]

i think it is right

not sure

i have another question

if anyones around

[wraith patio]

https://puu.sh/HWFEL/89f6e6e4c6.png

can someone explain what that actually means? v+W

or what V/W is

[wintry steppe]

this is d and not A right?

result vector i got

matrix*

sorry

[teal grotto]

v+W={v+w : w in W}

[wintry steppe]

coycoy do u by any chance know if the question i linked is D and not A

[teal grotto]

V/W is the quotient space of V mod W and consists of cosets of the the form v+W. the cosets are the vectors in V/W

[wintry steppe]

im being thrown off by the word row

and not pivot in each column

[teal grotto]

let me look

[wintry steppe]

thnk u!

[teal grotto]

yes d looks correct

[wintry steppe]

thank u sir

do u by any chance know fourier series

im struggling w a problem

[teal grotto]

i saw that. i do not know fourier series. sry

[wintry steppe]

coycoy linear algebra god

[teal grotto]

nah nah. thats TTerra

[wintry steppe]

no problem

[teal grotto]

is there more than one correct answer to this? because a is correct too

fourth column is 4(1st) + 1(2nd) - 2(3rd)

[wintry steppe]

no just 1 answer

@teal grotto i had another quick question

[teal grotto]

ok, go ahead

[wintry steppe]

i think the layout is correct

im just not sure where

the cos and -sin goes

like the layout

[teal grotto]

rotations are not really something i’m too comfortable with yet. i’m sure somebody else could help tho

[wintry steppe]

thank u 😦

If someone can verify that would be amazing

This is what I’m thinking so

[wraith patio]

https://puu.sh/HWFWz/51e87d6f54.png

i get that if beta is a basis of V and T(beta) is also a basis of W then that would work for gamma=T(beta) but idk the requirements on beta

[wintry steppe]

how can u solve this fourier series?

plsss someone ❤️

[north anvil]

I wish I knew. I don't know that stuff yet 😦

[wintry steppe]

no problem ;/

[weak needle]

Integrate f(x)e^(-in2pix) to get the nth fourier coefficient

If you want to know how to do that, the hint is to split the integral from 0 to pi and from pi to 2pi

[wintry steppe]

like

im confused

in2pix?

and why e^

like i know the integral of the first part from 0 to pi is

[weak needle]

$in2*\pi*x$

[wintry steppe]

-2pi

and for the second part its

[stoic python]

saketh

[wintry steppe]

its

3pi

[weak needle]

So on the whole the integral is pi, correct. That will give the first coefficient

[weak needle]

Also the way I was talking about fourier series used complex numbers, if you don’t do it that way you can ignore it

But you must have some formula for calculating the coefficients, right?

[wintry steppe]

yeah

its pi

[weak needle]

What about the other coefficients?

[wintry steppe]

wait is it pi

bc i thought for the first coefficient

u divide by 2pi

[weak needle]

You divide all coefficients by 2pi

[wintry steppe]

i thought for the bk terms u divide only by pi

[weak needle]

What you divide by depends on how you define the integral

When you use e^(inpi) stuff you just divide by the length of the interval always

[wintry steppe]

hmm ok

ig my question is

how can i consolidate the two functions into one

if even possible

i think that would make it easier

[weak needle]

Yeah so here is the thing the integral of any function from to 2pi is the sum of the integral from 0 to pi and the integral from pi to 2pi

And inside of those two intervals f is just a constant, it would be -2 in the first case and 3 in the second

[wintry steppe]

so

-2 + 3 is the integral from 0 to 2pi

so -1?

[lavish jewel]

you're supposed to get a function of x

after doing the integral, you take those coefficients and get a sum of sines and cosines

[weak needle]

No -2 and 3 are not the values of the integral, they are the values of the function f. Put those values in f and then do the integral, this should let you do all the integrals for the various coefficients

[wintry steppe]

wait im so confused

wym put them into function f

f(x) is literally defined as f(x) = -2

and f(x) = 3

[weak needle]

I mean instead of f write -2, that is all

[lavish jewel]

right, but only for specific values of x

[wintry steppe]

so integral of -2

[lavish jewel]

so each of the integrals for the fourier coeffs has to be split into two

[wintry steppe]

and integral of 3

ah ok

so

integral of -2 from 0 to pi

and integral of 3 from pi to 2pi

when i do that

i get

-2pi for the first

and 3pi for the second

[lavish jewel]

integral of -2 * cos (...) from 0 to pi

and similarly for 3 * cos(...)

and also sine

[wintry steppe]

what should go in the cos though?

[lavish jewel]

what saketh told you above already

[weak needle]

Yes those are the integrals for the first coefficient, i’m saying you can do the same for the rest

[wintry steppe]

ah ok

so

i know ak is = 0

so now im trying to find bk

there shouldnt be any cos terms in the answer i think

i think i got the answer can i verify w u @weak needle

or @lavish jewel

[lavish jewel]

you can very yourself

plot the sines and cosines and see if it looks like the original thing

[weak needle]

just post your working and i’ll see if i can spot any mistakes, I’m a bit lazy too work out the integrals myself

[wintry steppe]

ok

here

@weak needle

[weak needle]

cos(kpi) is not always -1, it will depend on wether k is odd or even

[wintry steppe]

in which step

[weak needle]

This is in the calculation of bk, sorry I forgot to say that

[wintry steppe]

what should it be instead then?

[weak needle]

Well, what is cos(kpi)? Just take a few examples and you’ll see

[wintry steppe]

is it k?

[weak needle]

No

[wintry steppe]

i have no idea :/

should it just be

-4/k

and -6/k

[lavish jewel]

dude just do it

what's cos (0)

what's cos (pi)

what's cos (2 pi)

[wintry steppe]

its either 1 or -1

[weak needle]

Yep, and if you know k, how can you figure out wether cos(kpi) is 1 or -1

[wintry steppe]

well

in the first one

i think its 0 to pi

so

it must be -1

and in the second one

its pi to 2pi

so it must be

1

[weak needle]

Look, the stuff about the integral doesn’t matter here, all I’m asking is, given some integer k, what is cos(kpi). The answer depends on what k is, and what I’m expecting you to tell me is what the relationship is

[wintry steppe]

when k is odd its 1 and when its even its -1

[weak needle]

Exactly

So now do you see how the formulas for bk will change depending on wether k is even or odd

[wintry steppe]

yeah

so then i think for the first one

its 2/pik

and then itll be

either positive or negative depending on

the order

and same w

the second

itll be 3/pik

and then depending on the order itll be positive or negative

does that sound fine?

[weak needle]

No, if cos(kpi)=1 for example, what is cos(kpi)-cos(0)

[wintry steppe]

cos(kpi)-1

so either -1 -1 or 1 -1

[weak needle]

And 1-1=0

[wintry steppe]

yeah

so that eliminates all of that

even order ones i think

[weak needle]

Yes

[wintry steppe]

so then for the odd ones its

-1 - (-1)

wait

it eliminates odd

and keeps even

bc i think its -1 and then the cos(0) is being subtracted

wait no

ur right

[weak needle]

Yeah so the even terms are -4pi/k

[wintry steppe]

ok so

for the first func

i got this

[weak needle]

Sorry -4/pik

[wintry steppe]

just for 𝑓(𝑥)=−2

and so for the second part its

and so final answer i think is

[weak needle]

Why is it 2/3pi and 2/5pi and so on? That seems wrong.

[wintry steppe]

its 4/3pi

and 4/5pi

[weak needle]

First of all, all the odd k coefficients should be 0 right?

[wintry steppe]

oops

[weak needle]

No i’m sorry i think your rightp

[wintry steppe]

other way around

wait no

it eliminates even

[weak needle]

Yes you were right, I got it mixed up

[wintry steppe]

does that look ok

the latest pic is it simplified

[weak needle]

Well now that you got the steps, I think it is a little pointless to verify the addition or whatever, just do what edd said and plot it on a computer or something

(Sorry I’m a bit lazy)

[wintry steppe]

u can plot it?

[weak needle]

There should be graphing calculators available online

[wintry steppe]

thank u saketh for ur help and patience i really appreciate it

[weak needle]

No problem

[wintry steppe]

wait quick q

for f(x) = 3

the second equation

[wintry steppe]

for a0 the first coefficient

should u integrate from pi to 2pi

or 0 to pi

[weak needle]

Pi to 2pi

[wintry steppe]

ok thought so

i think i wrote it wrong bc

i get 3/2 both ways

[weak needle]

Well the integral from pi to 2pi of 3 is the same as the integral from 0 to pi of 3

[wintry steppe]

ah ok

yeah i figured

tys

m

[weak needle]

Np

[umbral birch]

does anyone know how to solve this?

[teal grotto]

try diagonalizing A first

[teal grotto]

just in case, A should be diagonalizable over R (checked because its characteristic polynomial has two real roots, 2 and -2, which are the e-vals of A).

once you diagonalize it, you will have a diagonal matrix X and an invertible matrix P so that A = P X P^-1. then A^6 is P X^6 P^-1 and you’re done, since D = X^6 is diagonal

[half urchin]

Is a co-occurrence matrix by default orthonormal?

Example^^^

[turbid trellis]

Hey, when having this

It is wanted that the pivot is the most right right ? (2, 3)

When it's the last pivot in the augmented system

[radiant yarrow]

What does a sequence mean

I didn't understand

[leaden tide]

well, it's just a sequence of numbers (or something else)

you have a first value, then a second value, then ...

[radiant yarrow]

I didn't get it

What is {a_n} denoting

[leaden tide]

the sequence

For example {3n} is {0,3,6,9,12,...}

[radiant yarrow]

And I don't understand why they say it's a vector space

It just doesn't make sense

[leaden tide]

oh, that

Well, first of all do you know that the set of functions from ℕ to F is a vector space?

[radiant yarrow]

No

Is it because they defined it?

[leaden tide]

Well, no, it's because it meets the definition

You can multiply by scalars

[radiant yarrow]

?

Oh

[leaden tide]

You can add functions

It's a group for +

and generally everything works well

[radiant yarrow]

I don't get what you're saying

[leaden tide]

Do you even know what a vector space is?

[wintry steppe]

a sequence can be considered a vector in a vector space

[radiant yarrow]

I read it a lot of times. I don't get what it means

set of vectors multiplied by scalar?

vectors with inputs from a field F?

[wintry steppe]

its a set of vectors

[leaden tide]

Here's how I understand it

[wintry steppe]

simply put

[leaden tide]

A vector space is a set, along with some given field F. The elements of that set are called "vectors".

They're very much analogous to the vectors you see in geometry/physics

in fact, they're a generalization of that

[radiant yarrow]

Yes I know. But here you can have n tuple array vectors

kinda like vector in n dimensions?

[leaden tide]

yeah?

so long as you define + and scalar multiplication well, why would it be an issue?

[radiant yarrow]

So do I know what vector space is?

[leaden tide]

Well, there are some additional properties that vector spaces need to meet

[radiant yarrow]

Yes

I've read them

commutativity

scalar multiplication

[leaden tide]

basically, adding works well

[radiant yarrow]

associativity

Yeah

[leaden tide]

you can scale things, and that works how you'd expect

[radiant yarrow]

Yeah

[nocturne jewel]

scalar multiplication isnt necessarily a vector space axiom, it's just part of defining the * operation from VxK to V

[leaden tide]

also a vector space has to be a group for +

[radiant yarrow]

group?

[nocturne jewel]

$\cdot :V\cross\mathbb{K}\to V$ is scalar multiplication

[leaden tide]

I'm confused as to how you got to vector spaces without knowing what a sequence or a group is

[stoic python]

Mosh

[nocturne jewel]

group theory comes after linal usually

[leaden tide]

well that's dumb and stupid

[radiant yarrow]

I know what sequence is. Didn't understand what it meant in LA

[nocturne jewel]

same thing in like calc / other settings

[wintry steppe]

can confirm

[radiant yarrow]

I was advised to take LA after calc 1

and have calc 2 in parallel

[nocturne jewel]

${a_n}=a_1,a_2,a_3,...$

[stoic python]

Mosh

[leaden tide]

anyway, group for + just means that there is something like 0, and every element has an additive inverse

(which you'd expect vectors to)

[radiant yarrow]

Can't you just say existence of inverses?

[nocturne jewel]

I mean existence of inverses and every element has an additive inverse technically mean different things

since you can have operations which define only certain elements having inverses

[radiant yarrow]

Okay

Oh yeah, "there exists" is different from "for all"

[leaden tide]

Some of the vector space axioms are really stupid things, like $1\cdot x = x$

[radiant yarrow]

makes sense

[stoic python]

Syst3ms

[leaden tide]

(if you multiply by 1 it doesn't change anything, duh)

[nocturne jewel]

yes, identities are stupid

[leaden tide]

by stupid i mean stupidly obvious

[radiant yarrow]

trivial

got it

[wintry steppe]

depends on the identity and element

[leaden tide]

as long as you call 1 the multiplicative identity like any reasonable human being, this must work

[nocturne jewel]

Commutativity of +
Associativity of + and *
Existence of additive and multiplicative identities
All elements have an additive inverse

• and * distribute

those are the 8 axioms in few word basically

[leaden tide]

Now bear in mind, as long as all these axioms work, things can be n-tuples, but they can very well not be

In fact, that's the point : going beyond tuples, and looking at the concept in its full generality

So now, back to the main point

Why is the set of functions from ℕ (or any set) to F a vector space?

Well, you can define function addition by (f+g)(n)=f(n)+g(n)

You can define scalar multiplication by (λ·f)(n)=λf(n) (recall, λ∈F)

And it isn't hard to see that all properties mentioned before hold

By definition, this set is a vector space (over F, at least)

[radiant yarrow]

But isn't a vector space about representation of n tuples? I don't get when a function is a vector space

a set is a vector space right?

How is a function one?

[leaden tide]

a function isn't a vector space

a set of functions can be

[radiant yarrow]

?

[leaden tide]

Vector spaces are about generalizing the operations and concepts that notably apply to n-tuples

In this case we're replacing the n-tuples with functions

[wintry steppe]

just a side note on identities maybe im stating the obvious but in case you miss the point
identities are operations that keep the element of the set as is
so even something as trivial as 1* x=x 1 being the identity for multiplication
the identity of lets say a matrix A with the multiplication identity would be A*(the identity)=A that you would know as In the identity matrix
and the identity of a set M with the union would be M U (identity)=M thus being phi

ik your topic is vector spaces but had to point it out

[leaden tide]

formatting

[wintry steppe]

ye discord has this thing for *

[radiant yarrow]

How can you give functions as inputs. I don't get what you're saying here

And why doesn't this book explain what's going on

[leaden tide]

Inputs of what?

[radiant yarrow]

You said you're replacing the array of n tuples with functions

why isn't there anything explained in the book?

[leaden tide]

I wouldn't know

[leaden tide]

i don't quite get what you think i am implying by that

[radiant yarrow]

let 'f','g','h' be a function. Instead of a vector x =(a_1,a_2,.....) didn't you say a vector x can also be =(f,g,h)?

[leaden tide]

No, that's not my point

[radiant yarrow]

Okay

Oh wait got it

[leaden tide]

My point is that you know how we can add and scale vectors?

[radiant yarrow]

x=(N,F)

[leaden tide]

No..?

[radiant yarrow]

yes we can add vectors

[leaden tide]

Well we're defining an analogous way to add and scale functions

Which works in a very similar way to vectors

[radiant yarrow]

Okay

got it

[leaden tide]

But the two kinds of objects have nothing to do with one another

Just so we're clear

[radiant yarrow]

The book sucks

[leaden tide]

anyway, addition and multiplication of sequences is the same thing

just with different notation

Though, one last tidbit that was in the original definition

[wintry steppe]

you can always use another book for reference, helps

[leaden tide]

It specifically considers sequences that are nonzero for a finite number of terms only

[radiant yarrow]

Bruh then what's the point of the original book

I sincerely thought I can be done with LA in 6 months without hassle if I stick to one book

[leaden tide]

yeah but some books aren't good teachers

heck, we don't have textbooks here

i mean, in highschool we kinda do but we mostly use them for exercises

i digress

[radiant yarrow]

I'm using Friedberg,Insel,Spencer BTW. his definitions are marked as "Example:n"

[wintry steppe]

well you can
1-complain about the book and sit in a corner
2-think reasonably and get another book that will solve the lack of competency from the book

[radiant yarrow]

you digress with?

[leaden tide]

digress, not disagree

[radiant yarrow]

my bad

[leaden tide]

anyhow

[radiant yarrow]

Yeah

[leaden tide]

this

[radiant yarrow]

Yes. On the search now

[leaden tide]

we still haven't proved that this was a vector space

we proved that the set of functions from ℕ to F is

But not with that restriction

When you have a vector space that is a subset of another vector space (with the same operations), you call that a vector subspace (how creative)

Checking if something is a vector space is tiresome

You have like 8 axioms and it sucks

It's much faster to see if something is a vector subspace of a vector space you already know

I chose to start by explaining how the set of functions from ℕ to F was a vector space because it's nice to keep in mind, since it's really general

Checking if a set is a vector subspace is easy on the other hand

You just have to check :

• the set is included in the original vector space (duh)
• 0 is in the set
• the set is stable by scalar multiplication
• the set is stable by addition

i think you won't have trouble checking that the first two hold

the set of sequences that are only nonzero for a finite number of terms also includes the sequence which is always 0

it's stable by multiplication because you're not going to change any of the zero terms by multiplying them

And it's stable by addition, because you're still going to end up with only a finite (but perhaps larger) amount of nonzero terms

So there you have it, the set of sequences with a finite amount of nonzero terms is a vector space

[wintry steppe]

[radiant yarrow]

I..... umm

[leaden tide]

take your time

[radiant yarrow]

I don't understand when you say "a set of functions" is a vector space

How are functions depicted as vectors?

[leaden tide]

a vector is no longer necessarily just an n-tuple

it's an element of a vector space

that sounds circular i know (and it kind of is)

but really, a vector in the most general sense is only defined by how it transforms with regards to addition and scalar multiplication

does it behave well and expectedly with regards to those two?

If so, it can be considered a vector

[radiant yarrow]

maybe I'll try tomorrow

after some headbanging

I heard this subject is supposed to be easy

[leaden tide]

it isn't all that bad, but there are some initial subtleties

main take away from this is that a vector is no longer just an n-tuple, it's way more general now

[wintry steppe]

well it is once you get the hang of it

its not as hard to grasp you'll be fine

[radiant yarrow]

Okayy thanks

I'll just take a break

[wintry steppe]

[fallen scaffold]

can someone help me with these

[vital plume]

I have a question. If I have an augmented matrix and almost all the numbers are filled in, but there is an unknown coefficient in the matrix how do I solve for it.

[wintry steppe]

subtract row 1 - row 2 and then solve for h?

@vital plume

[dusky epoch]

depends on what you're asked for

are you asked to find h such that the system has no solutions?

or are you asked to find h such that the system has (420, 69) as a solution?

or what

[vital plume]

@dusky epoch has infinite solutions

[dusky epoch]

okay

so when a system has infinite solutions, row-reducing it will eventually lead to a row of all zeros

act accordingly

[wintry steppe]

An augmented matrix is inconsistent when a row of zeros leads to a number right?

if two rows are exactly the same

then the system can't be inconsistent

[dusky epoch]

theres no such thing as an augmented system

[wintry steppe]

augmented matrix*

sorry

[limber sierra]

the system can still be inconsistent if two rows are the same

consider the augmented matrix $\begin{pmatrix}1&1&0\1&1&0\1&1&1\end{pmatrix}$

[stoic python]

Namington

[wintry steppe]

Sorry i should've specified more

[limber sierra]

this row reduces to $\begin{pmatrix}1&1&0\0&0&1\0&0&0\end{pmatrix}$ which translates into the system $x + y = 0, 0 = 1, 0 = 0$

[stoic python]

Namington

[limber sierra]

and obviously 0 = 1 is a nonstarter

[stoic python]

jswatj

[limber sierra]

not necessarily, consider $\begin{pmatrix}0&0&1\0&0&1\end{pmatrix}$

[stoic python]

Namington

[limber sierra]

that said, "most of the time" yes

in that if a or b are not equal to 0

OR if the rightmost column is equal to 0

then yes, it will be consistent

that is to say, the ONLY inconsistent matrices of that form look like $\begin{pmatrix}0&0&k\0&0&k\end{pmatrix}$ for some $k \neq 0$

[stoic python]

Namington

[wintry steppe]

hmm ok

[limber sierra]

("inconsistent matrix" here meaning augmented matrix representing an inconsistent system)

right

unless a = b = c = 0 or e = f =g = 0

then the system isnt inconsistent

so?

[wintry steppe]

😂

[limber sierra]

"if these are inconsistent, then theyll be parallel" doesnt mean "if these are parallel, then theyll be inconsistent"

all dogs are animals, but not all animals are dogs

the contrapositive would be "if these are not parallel, then they are consistent" - which is actually true!

(assuming they form planes, at least)

if your planes arent parallel they'll cross at some point

(more appropriately, some line)

the line on which they cross is your solutions

right

[wintry steppe]

so its true

[limber sierra]

to form a plane you need all your coefficients to be nonzero

yeah

i added that but it was redundant in hindsight

so forget that bit

basically, if we have two planes, then there are these possibilities:

• The planes are not parallel, meaning they cross at some point [CONSISTENT]
• The planes are parallel, meaning either:
  --- They are "the same", so they intersect at every point [CONSISTENT]
  --- They are "stacked over each other" but never meet, so they never intersect [INCONSISTENT]

not necessarily that the coefficients are the SAME

[wintry steppe]

Ohhh

multiples of eachother

[limber sierra]

right!

one equation is a scalar multiple of the other (except the constant term/rightmost column of the augmented matrix)

sure

[wintry steppe]

Does Riesz's representation theorem work for inner products that are not positive definite?

[leaden tide]

that's a contradiction

[twilit anvil]

can you get an orthonormal basis for such a space

[leaden tide]

"inner product" literally implies positive definite

but it wouldn't work if it wasn't positive because you need cauchy-schwarz

[wintry steppe]

LOL, pseudo inner product.

[leaden tide]

and at one point the proof uses the fact that <a,z>=0 for all z implies a=0

[wintry steppe]

It could, but there would be the signature involved.

[leaden tide]

which uses the definiteness of the inner product

and that's not even mentioning the more advanced theorems which may implicitly come from either of those two properties

[twilit anvil]

what are the sort of theorems you are talking about?

im thinking if basic ones like triangle inequality, pythagoras, and gram schmidt hold

(and i dont rlly think so)

[leaden tide]

idk i just skimmed over the proof

[wintry steppe]

Does results like Riesz's representation theorem exist for spaces with a pseudo inner product?

Otherwise, how can one use the "metric" to "lower" or "raise" the indices in an expression as they do in General Relativity where the Lorentzian inner product on the tangent space is a pseudo inner product?

[wintry steppe]

all you need to raise and lower indices is non-degeneracy of the "metric tensor," no? and that's built into the definition

[fast ice]

if you have a free variable after reducing a matrix into row echelon form, does that mean that the vectors that make up the matrix are linearly dependent?

[lavish jewel]

yes

[quartz compass]

yeah, it's like saying positive definite matrices are the only invertible matrices, that's just not the case

[wintry steppe]

Isn't this non-degeneracy of the inner product that is the isomorphism from Riesz's representation theorem?

[wintry steppe]

Can I establish an isomorphism between a vector space and its dual with a pseudo inner product?

[marble lance]

How do you define a topology using pseudo inner products to talk about continuity? Genuine question because I know nothing about pseudo inner products.

[wintry steppe]

If there's some way to make a norm out of a pseudo inner product, I think you can; I don't think you can make a norm out of a pseudo inner product though.

[marble lance]

So then how do you construct the dual space?

[wintry steppe]

You don't need an inner product to construct the dual space.

[marble lance]

You need to know what the continuous linear functionals on your space are

For which you need a topology to talk about continuity?

[wintry steppe]

Isn't the dual space just bounded linear functionals?

[marble lance]

How do you define bounded then?

[wintry steppe]

And linearity implies continuity, IIRC.