linear-algebra

[winter harbor]

T^k(v_i) = λ_i^k v_i

So the eigenvalues of T^k are the k-th powers of the eigenvalues of T

And same eigenvectors

Yeah, that was a mistake, sorry.

[stoic python]

MisterSystem
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[empty hemlock]

What

is the hypothesis that is being proved by induction?

[winter harbor]

Yup, as I said, the idea is proving this result by induction.

[empty hemlock]

Induction on what? On n?

[winter harbor]

Yup, induction on n.

[empty hemlock]

I tried this, but I just don't see how the inductive step is going to work.

[winter harbor]

Ok, I will write down a full proof then using this idea.

[empty hemlock]

For example, in the n + 1 case, what if k = n + 1.

Also, I think we need to make use of the fact that W is complex. For some reason, I feel that this result fails if the field is not algebraically closed.

[winter harbor]

if k = n, then T is an automorphism, and so T^n is also an automorphism and do ker T^n = ker T^(n-n) = ker id = {0} clearly.

So we need only to look for values of k<n

[empty hemlock]

OK.

[winter harbor]

Notice that T is an automorphism because in our hypothesis it has n non zero eigenvalues.

In the case k=n ofc.

So what we can do is make an induction on n

And suppose that k<n.

[winter harbor]

Yeah, we are implicitly assuming existence of eigenvalues we are working over an algebraically closed field yes.

[empty hemlock]

You can have eigenvalues even if the field is not algebraically closed.

Since C is algebraic closed, we can write T in matrix form as D + N where D is the diagonal with the eigevalues and N is some nilpotent matrix.

Maybe we have to use this fact somehow.

[winter harbor]

Yeah, that's a consequence of Jordan decomposition.

In any case

I will write down the proof in full detail

[empty hemlock]

Hmm.. going back to my original argument, we have that <w> is a subset of ker T^k, so then T <w> is a subset of T (ker T^k) = {0}. Now T <w> is basically <w> without the vectors spanned by w. Hmm...

Wait, but then that means w is in ker T, hence in ker T^(n-k).

Hmm... but that doesn't help since I want to show that v is in ker T^(n-k), not w.

[wintry steppe]

does "0" in a 3D vector space over R represent [0, 0, 0]?

like in the equation 0v = 0 (v ∈ V and V is a 3D vector space over R (or C))

does the "0" on the right side of the equation represent the vector [0, 0, 0]?

[wild fulcrum]

yes

[wintry steppe]

damn I kinda don't like how they abuse notation in linear algebra, first it was F^n now it's this

#notationlivesmatter

[wintry steppe]

😋

[lavish jewel]

write v1 and v2 in terms of generic linear combinations of w1 and w2

and show that this boils down to w1 and w2 being lin indep

[wind pasture]

wait what does this mean

[lavish jewel]

do you know what it means for two sets of vectors to have the same spam?

[teal grotto]

alternatively, define the linear map T from span{v1, v2} to span{w1, w2} by T(v1) = w1 and T(v2) = w2

[stoic folio]

will be the augmented matrix?

and what does it mean by geometrical meaning?

like defining what we doing with the rows when doing the reduce echelon form?

[lavish jewel]

i would suggest you just google what augmented matrix means

[stoic folio]

i know what augmented it

im asking will 'B' the augmented matrix of A righht?

[lavish jewel]

what is B

[stoic folio]

this is my first time learning linear algebra so im confirming

[stoic folio]

this B

[wintry steppe]

has anyone here read "advanced linear algebra" by steven roman?

if you have read it, what are the prerequisites for that book?

[finite cosmos]

Can you solve a matrix with x,y,z being exponents ?

[wintry steppe]

help please

[winter harbor]

#prealg-and-algebra

[hollow void]

,rotate

[hollow void]

Is there any shortcut way. It will take so much time

[hollow void]

If the entries in each column of a square matrix M add up to 1. Then eigen value of M is ____?

[torn stag]

@hollow void It's 2x2

[hollow void]

Ans to?

Whoch question

Which

[torn stag]

@hollow void The dimension of that matrix

[hollow void]

Ohh. But i was never concerned about ans. I just wanted to learned approach

Please if you guide me

[torn stag]

@hollow void $P$ is 2 x 3, so the whole matrix inside the transpose is 2 x 2

[stoic python]

IlIIllIIIlllIIIIllll
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[hollow void]

Ok

[torn stag]

@hollow void (). You can think of an $m \times n$ matrix as a map from $\mathbb{R}^n$ to $\mathbb{R}^m$.

[stoic python]

IlIIllIIIlllIIIIllll
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[median ocean]

Does anyone how to do part a on 1

[median ocean]

Would a just be 3+2t+t^2

[teal grotto]

yes

[median ocean]

What about number 2

I’m confused

[dark brook]

First dime doing RREF and its not that straightforward currently. If I've got a matrix of

A = [1 -2 -7 -8 -9; 3 -4 -13 -14 -15]

and wants to find the RREF. I heard that the first thing you need to do is R2 = R2 - 3*R1, but I am confused to why its 3, couldn't it be 5 or another number as well?

[teal grotto]

yes. form the coefficient matrix and pick an element not in the image of that matrix.

[teal grotto]

you need to get every entry under every pivot to be 0

multiplying by anything other than three does not accomplish that

[bold sun]

hey how would part b and c be correct for this question

[dark brook]

Ahhh alright, thats a good point which I didn't observate. Thanks!

[hollow pendant]

lambda = -3, lambda = -2.5

[bold sun]

this is what i did -3k(1 2) and part c -3(1 2 )+1/2(1 2)

[teal grotto]

yea. and lambda = -3k is in R

so ur done

[bold sun]

oh ok thanks then for part d determine 2 different points on the line would i just pick any n.o for lambda like 2 and add the column vectors up???

[teal grotto]

yup. i would pick easy numbers like 0 and 1 for two separate choices of lambda

[bold sun]

ohh ok then thanks did that

but it dont work for the final part?? show if you add them then it aint on the seconf line

i dont know how to go about that

[sonic beacon]

i want to prove hom(V,W) is a vector space, and want to verify the distributively axioms. i have for c in F, cT: V->W, (c * T)(v) = c * T(v)-- so I want to prove (c * (T + T'))(x) = (c * T + c * T)(x). i am kind of stuck early on, i have: (c * (T + T'))(x) and not sure if i can then say c * (T(x) + T'(x)) is that step valid?

[dark brook]

If I have the echelon form of

1     0     1     2     3
0     1     4     5     6 ```
Would I only have 1 pivot-element or do I actually have two? (I could get a R3 and it would just be zeros, so I would assume we've got 2 pivot-elements)?

[sonic beacon]

i think so since (c * T)(v) = c * T(v)

so, (c * (T + T'))(x) = c * (T(x) + T'(x)) = c * T(x) + c * T'(x) = (c * T + c *T')(x)

[bold sun]

hey i am so stuck how do i do 1.7 and 1.8

i dont get it

for r2 woud i just say if they are not parallel then they intersect and if in opposite directions ?? dunno bout 3 dimentional vectors (im not very good at that at alll)

so i dont get that part of 1.7 and i dont get 1.8 either at all

[fiery ibex]

I have an nxn matrix A whose columns have exactly k nonzero entries each equaling 1/k. How can I go about proving that A-Identity has determinant zero? I was given the hint to use linear dependence of the rows, but I'm just not sure why the rows are bound to be lin dependent

Some context that might help is we're kinda describing a random walk

[hard drum]

now if you sum the numbers in each column, you'll get 1

hence if you sum the entries in each column of A-I, you get 0

i.e. determinant 0, as the sum of the rows is the zero vector

[fiery ibex]

oh my god

i appreciate it so much

[vivid field]

I was given a set of polynomials of degree 2 and had to decide whether it is linearly independent in P2. I solved the system and got that the system is NOT linearly independent.

Now how would I determine if this set forms a basis for P2 ?

[hard drum]

well firstly, what is the definition of a basis?

[hard drum]

Np

[vivid field]

Haha Gotcha! I should of found that out first before asking. Thanks. So is it that simple ? If there are a set of unique vectors than there's a basis ?

[bold sun]

can anyone help me here please- i still dont get it

[hard drum]

Not sure what you mean by a set of unique vectors

[hot willow]

they have to be linearly independent

[gray dust]

if we wanna be thorough we can't skip steps. eg an immediate result of using the definition of cT is (c(T+T'))(x)=c(T+T')(x)

[hot willow]

its either they intersect in a line or point for 1.7

and they intersect in plane or line for wo planes

[dark brook]

If I've got a matrix of

1 -i -i
-i 1 -i
-i -i 1

Would I be able to reduced it to the RREF?

[limber sierra]

yes

every matrix can be put in RREF

add i * the first row to the second row

multiply the second row by the inverse of its second entry

(it will now be 0, 1, something)

repeat this sort of process for the third row

[dark brook]

So something like $R_2 = R_1 \cdot -i $?
What about the second part? I've not sure I understand that.
its my first week with linear algebra 🙂

[nocturne jewel]

Yeah you just perform regular ERO's just with complex numbers and the algebra they come with

likewise for Z_p matrices

[dark brook]

I've no idea what that is

I just try a bunch of things, but I don't really seem to get anything involving 0

[limber sierra]

im gonna be very lazy with my typesetting here

[dark brook]

Would it be easier if the matrix was from the start?
Being:

i 1 1
1 i 1
1 1 i

[limber sierra]

 1 -i -i
-i  1 -i
-i -i  1

row2 = row2 + (i)row1

 1 -i -i
 0  2 1-i
-i -i  1

row2 = (1/2)row2

 1 -i -i
 0  1 (1-i)/2
-i -i  1

row3 = row3 + (i)row1

1 -i -i
0  1 (1-i)/2
0 1-i  2

row3 = row3 + (-1+i)row2

1 -i -i
0  1 (1-i)/2
0  0  2+i

row3 = (1/(2+i))row3

1 -i -i
0  1 (1-i)/2
0  0  1

and from here its simple

1 0 0 
0 1 0
0 0 1

@dark brook

this is a very formulaic process

you can reduce any matrix into RREF by just doing these steps

going down the matrix making each row (0 0 0 ... 1 whatever whatever whatever ...)

(if not possible, put the row on the bottom and 0 it out later)

i mightve made a mistake in my arithmetic somewhere, but the final result is correct in any case.

[dark brook]

Hmm, but how did you come up with it that fast? Is it just fast equations in your head? Because there is quite a lot of equations to be made and wrap around

[limber sierra]

this doesnt really change anything, multiply each row by -i

[limber sierra]

as i said, row reduction is very formulaic; my algorithm is:

for each column:

• swap rows such that the variable in the topmost "unused" row and current column is nonzero; this will be this row/column's pivot variable (if not possible, skip this column)
• 0 out everything to the left of the pivot. this is certainly possible since, if theres something to the left of the pivot, it must have a row with a 1 entry in that position above it; so if that entry is x, add -x * that row to this row. repeat until everything left of the pivot is 0
• now you have a row of the form (0 0 0 ... 0 PIVOT ? ? ? ... ?). multiply the row by 1/PIVOT to get (0 0 0 ... 0 1 ? ? ? ... ?). the contents of ? dont matter
• mark this row as "used" and move on to the next column

once this process is done, you "play cleanup"

you first 0 out all the still-unused rows by using the 1s in the rows above (which wasnt necessary in the above example)

then you go right to left, 0ing out all the elements above the 1s through the same strategy

[dark brook]

I can see it being very formulaic. I really appreciate taking your time to write a way of finding the RREF. I'll get it noted down for later.

[limber sierra]

caveat: if you end up accidentally making PIVOT equal to 0, thats fine, just move it to the bottom and continue on

because obviously you cant multiply by 1/PIVOT if PIVOT is 0

[dark brook]

Ahh yea, thats a good point

[limber sierra]

this only happens if you get "lucky"

i say "lucky" because it saves you some work

[dark brook]

Gotcha! I assume its also a lot of practice that goes into these sort of things?

[limber sierra]

yes, practice is the best way to learn mathematics

especially formulaic processes like these

fyi you dont have to follow that algorithm precisely

row reduction can be very ad hoc if you prefer, and sometimes ad hoc methods will be faster

its just that the algorithm ALWAYS gives you an RREF

so you can code it up in a computer or whatever

with no further logic

[dark brook]

Ah thats actually quite interesting

[limber sierra]

its described more precisely here https://www.math.purdue.edu/~shao92/documents/Algorithm REF.pdf

correctness is easy but tedious to prove with induction.

[dark brook]

Thats a neat document. I see the ERO that Mosh talked about earlier, which I now see what it means (I'm not english)

[limber sierra]

if you prefer actual code, there are various implementations on the internet

https://gist.github.com/sgsfak/77a1c08ac8a9b0af77393b24e44c9547

"elementary row operation" just means a thing you can do when row reducing

this is:

• swap rows
• add scalar multiples of rows to another row
• multiply rows by a scalar

[dark brook]

Yea I noted that MatLab has the easy featureof rref(A) and it outputs it, but I can't use the result for more than just being sure I've got the correct method

[limber sierra]

its good to get into the habit of keeping track of what EROs you're doing

since that matters when using this process to compute determinants

[dark brook]

Ahh I see what ERO's are now, totally different word here 🙂

[limber sierra]

which youll likely see later in your course

(for example, when you swap two rows, it multiplies the determinant by -1)

[dark brook]

It sounded like we would encounter that. The course is not really linear algebra, but linear transformations, but I assume its the same

[nocturne jewel]

linear transformations are a way of justifying things in linear algebra, such as matrices

so yeah, you're doing LinAl

[dark brook]

Ahh thats good to know!

So I've got another problem, I think is alright, but I'm unsure if its what I think it is.
So I need to find all the possible solutions for these equations, where I first find the RREF

Then I've found this (what I assume is the final result)

But they are more formulas than actually solutions, but I am not given any data for x, y, z and w, so this would be the closest I could get to it?

[keen mirage]

what did you get for the rref

[keen mirage]

yeah, your solution is (x1, x2, x3, x4, x5)

and you can rewrite x1 and x2 using what you found

and x3, x4 and x5 don't depend on anything

[dark brook]

Yea, I've not yet fully understanded why they aren't depending on some things, but its my first week with this stuff and we have an upcoming break, so much of time to get a better grasp on it

[keen mirage]

imagine the line y=x: x doesn't depend on anything

so for y=x you'd write (x, x)

@dark brook

[dark brook]

Ahh yea I see

I'll take a look on it in the break we've got. Its also 04:43 in the morning, so I should find the bed soon. Thanks for all the help guys 🙂

[exotic wedge]

Any help? I can give an example of non-diagonalizable matrix but not sure how to make it "generic"

[winter harbor]

There's a nice class of examples we can give

Are you familiar with the fact that a matrix is diagonalizable over C iff the minimal polynomial has no repeated roots?

[teal grotto]

whats your example?

[exotic wedge]

upper tringular matrix of all 1s

[teal grotto]

ones on the diagonal too, or zeros?

[exotic wedge]

ones on the diagonal too

[wintry steppe]

Would anyone be able to help with a numerical linear algebra problem its very math heavy but involves a bit of code too

[winter harbor]

https://www.wolframalpha.com/input/?i=diagonalize+{{0%2C0%2Ca^(2)b}%2C{1%2C0%2C-a^(2)-2ab}%2C{0%2C1%2C2a%2Bb}}

Notice how no matrix of this form is diagonalizable

And this is a pretty generic familiy of matrices in some way

I came up with this as follows

Using the fact that a matrix is diagonalizable iff its minimal polynomial has no repeated roots

I consider f(x) = (x-a)^2(x-b)

This polynomial has repeated root a

And what I did was construct a matrix whose minimal polynomial was f(x)

And I did this by expanding f and then looking at its companion matrix

[winter harbor]

I know such a matrix will have a minimal polynomial with repeated roots and so is not diagonalizable.

[exotic wedge]

is the minimal polynomial is the same thing as characteristic polynomial

[winter harbor]

Np

They are related

the minimal polynomial p of a square matrix A is the monic polynomial of lowest degree such that p(A) = 0.

Notice that by Cayley-Hamilton

we haev that the characteristict polynomial applied to A is also 0

And so the minimal polynomial divides the characteristic polynomial

[exotic wedge]

I am not that advanced into linear algebra xD

[winter harbor]

That's ok

You will see these things at some point I am sure

[exotic wedge]

Yeah, i am trying to solve the problem in terms of the material covered by the text

as i said, i can give an example, i am just not sure to what extent it should be "generic"

[winter harbor]

I don't know exactly what they mean by generic, but I suppose it means like

specifying a family of matrices wich are not diagonalizable

and not only a single example

Oh

There's also this nice example

Are you familiar with nilpotent matrices?

[exotic wedge]

aha

I know that non-zero nilpotent matrices are not diagonalizable

[winter harbor]

yup, that's what I was thinking about

[exotic wedge]

yeah that actually could be it

[winter harbor]

yeah

the thing is

[exotic wedge]

but like, how to write that in a generic matrix form ?

[winter harbor]

nilpotent matrices have a ''generic form'' which you will have to find out

[exotic wedge]

We can just find a matric that equals zero when squared

[winter harbor]

\begin{bmatrix}
a_{1} & a_{1} & a_{1} \
a_{2} & a_{2} & a_{2} \
-a_{1} - a_{2} & -a_{1} - a_{2} & -a_{1} - a_{2}
\end{bmatrix}

[stoic python]

MisterSystem
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[winter harbor]

Any matrix of this form is nilpotent

what I was trying to present as an example

was like

the generic form of a nilpotent matrix in a suitable basis

but that won't work

[exotic wedge]

that would work too ?

[winter harbor]

I think this is generic enough for the problem tho

Oh

I get what you mean

[exotic wedge]

like there exist k such that A^k = 0, then we pick k = 2

[winter harbor]

Yeah, yeah

computing A^2 for a 3 x 3 matrix would be a bit painful tho

[exotic wedge]

how is this derived then ?

[winter harbor]

That's a class of examples of nilpotent matrices

notice they square to 0

[exotic wedge]

Yeah, i will try to find its derivation or the derivation of k = 2

[winter harbor]

Np, in any case, looking at families of nilpotent matrices may be the easier way to go.

[exotic wedge]

Thanks a lot !

[wintry steppe]

i want mistersystem to go back in time and teach first year me about dual spaces or some other abstract LA concept i had a hard time grasping

[vivid field]

Ive gotten this far but I am a little confused on what this next part is asking me:

Find the coordinate vector of p relative to S, where p = −1 − 28t + 24t^3

[lavish jewel]

it's similar to what you did with the standard basis B

one way of doing this is with a so-called "change of basis"

the matrix you wrote there, the one you did RREF on, takes vectors in the basis S and changes them into the basis B

you now want the opposite

you could either invert that matrix or solve a system of equations with your favorite method

[vivid field]

@lavish jewel Something like this ? Im not sure if the notation I used is totally correct so apoologies for that.

[hollow void]

Good morning.

Whqt is the best and short method to find out rank

[lavish jewel]

that seems ok, flare

[vivid field]

Row reduce and count the number of pivot columns i think

[lavish jewel]

you can just finally double-check that B_s p_s gives you the original poly as desired

and yeah, i'd say if you do it by hand, (R)REF is about as good as it gets

[limber sierra]

yeah, there isnt a faster way in general than row reduction

unfortunately

[hollow void]

Pivot column? Row reduce?

[limber sierra]

uh

[hollow void]

Echolan form?

[limber sierra]

you might know the term "gaussian elimination"

[lavish jewel]

maybe you know it as gaussian elimination or gauss jordan

[hollow void]

Yes sir

[lavish jewel]

yeah, row echelon form

[limber sierra]

yeah do that

[hollow void]

That method is so long 😭😭

[vivid field]

How could I check again ?

You can use calculator

[limber sierra]

there isnt really a faster way in the general case

[hollow void]

In triangular or scaler?

[limber sierra]

in some specific cases you can try to compute the dimension of the transform's image through another process

but it's rare thatll be faster than just row reducing

(since usually the fastest way to determine the image IS row reducing)

[hollow void]

In exam. No calculator.

[limber sierra]

well, get a lot of practice doing gaussian elimination then

[hollow void]

3 minutes per questions. Negative on wrong ans

[limber sierra]

there isnt really a better way.

[hollow void]

Hmm but is there any special case?

[limber sierra]

if its easy to prove a transformation injective, you can do that instead, and that means it has full (column) rank

[hollow void]

Like in eigen value.

Trace = sum of eigen
Multi Det = multi eigen

[limber sierra]

similar for surjective and row rank

[lavish jewel]

while those are true, they don't help if you have to find each of the individual eigenvalues

[limber sierra]

it usually isnt very easy to do that without just row reducing

[hollow void]

Hmm ok

[limber sierra]

there is the equation rank(T) = dim(V) - nullity(T)

where T is a map V -> W

[hollow void]

They ask 4 × 4 matrix for rank😖

[limber sierra]

but you still need to find nullity(T)

so that likely doesnt help you

[vivid field]

You could try to find the determinant and it could tell you I think

[limber sierra]

yeah, get practice row reducing then.

nothing faster

[limber sierra]

determinant has no relation to rank.

[lavish jewel]

the determinant of a 4x4 by hand will be more difficult than just rref

well, it's 0 when rank deficient

[limber sierra]

the only tangentially connected property is that rank 0 implies det 0 i guess

[lavish jewel]

doesn't tell you "how rank deficient" it is tho

[limber sierra]

oh sure

but thatll never help you compute it faster

outside of really contrived problems

[hollow void]

The maths is interesting. When you study it by good professor

Or ma'am

But in good university. The professors are just term throwing and bookish.

[vivid field]

I have a quick question regarding if I am answering this question correctly. I determined that the set P is not linearly independent in P^2.

I then need to decide if it forms a basis for P^2.

I first decided that the set P does not form a basis for P^2 since it was linearly dependent,

but then I just seen that the vectors 1 and 2 are linearly independent and form a basis for the span of (v1,v2,v3)

So what would the answer be ? That the set does form a basis or does not form a basis in P^2 ?

I am just a little confused on the notations and wordings of problems ?

[lavish jewel]

they form a basis for a subspace of P^2, then

but they are not a basis for P^2

P^2 has dimension 3

[vivid field]

ahh yes, that makes much more sense. Thanks!

[solemn lotus]

Will a non trivial vector space always have infinite subspaces?

[half ice]

No. Consider any 1D vector space, which only has itself, and the trivial space as subspaces

[solemn lotus]

Hmm so can there be any any space with, say, 4/5 subspaces only?

[half ice]

So if you go with finite vector spaces, then you'll find vector spaces with finitely many subspaces

Consider i and j as a basis over Z3. That is,
2i + 2i = i

[dusky epoch]

4/5?

[half ice]

This is a 2D space with finitely many elements, and so finitely many subspaces

I don't know about 4 or 5 specifically

[dusky epoch]

or to put that more formally, kaynex is suggesting $\bZ_3^2$ as a vector space over $\bZ_3$

[stoic python]

Ann

[lucid glacier]

generally once you go above 2D, if your vector space is over an infinite field you will have infinite subspaces

so the only examples you'll find are spaces over finite fields

(And 1D spaces)

[lucid glacier]

This is actually a really nice exercise in basic linear algebra imo

[quartz compass]

that's a good question, how many subspaces does $\mathbb{F}_q^n$ have?

[stoic python]

Meroseous

[deft apex]

how are these 2 different?

<phi| a|psI> = a<phi|psi>. and <psi|a|phi> = a<psi|phi>

how are these different?

only thing i could think of if that a is the same number here

but that seems weird cuz how would one talk about the same number a if this is a general definition

[lavish jewel]

don't forget the complex conjugation

[deft apex]

so are all of these the same vectors?

the psi and the phi here

in both lines

they the same?

[lavish jewel]

not necessarily

just 2 generic vectors

[deft apex]

so why are both different

[lavish jewel]

which two lines did you mean

[deft apex]

i mean 2 and 2'

[lavish jewel]

idk what you mean by 2'

oh what

[deft apex]

the red line and the yellow line

[lavish jewel]

i just saw lol

[deft apex]

ye lol

[lavish jewel]

that's cursed notation

[deft apex]

lmfao

[lavish jewel]

no, they could be different

just 2 generic vectors

they're just giving you 2 definitions

[deft apex]

and the number a could also be different?

[lavish jewel]

sure

[deft apex]

but with dirac notation how can we see which vector is multiplied by the scalar a

[lavish jewel]

it's clear in how it was written

they've given you some function apparently called SP with 2 arguments

separated by commas

[deft apex]

the SP is the scalar product

is that a function?

what

[lavish jewel]

yeah

for example when you take vectors in C^n, the scalar product of x and y is y^H x

(or backwards depending on your book, i.e. x^H y)

[deft apex]

can you give me a recource for this

this is not from a book lol

[lavish jewel]

you can look up inner product spaces on wikipedia for a quick read

[deft apex]

ah thanks

[lavish jewel]

but yeah, the inner product is a binary operation that takes two elements from some vector space and yields an element in the underlying field

SP: V x V -> F

and it follows a handful of special properties

[deft apex]

ok so i think i am confused about how this number a is multiplying

does the first line mean <psi|a|phi> ?

in dirac notation

[lavish jewel]

no

[deft apex]

oh ok

what is it then

[lavish jewel]

it means what it says

SP( |x>, a |y>)

a |y> is the scalar multiplication between a, an element of the field, and y, a vector in the vector space

how that is done depends on what vector space this is

[deft apex]

can you maybe write that in dirac notation?

[lavish jewel]

i just did

there is nothing more to write than that without knowing exactly what vector space you're working with

$a \vert y \rangle$

[stoic python]

19eddy4

[lavish jewel]

$\text{SP}\left( \vert x \rangle, a \vert y \rangle \right)$

[stoic python]

19eddy4

[lavish jewel]

it's some operation acting on two vectors

|x> and a |y>

[deft apex]

is SP(|x> , |y>) = <x|y> ?

[lavish jewel]

it doesn't say that anywhere here, but that's usually the case

you can have extra "weighting factors" though

so more generally smth like <x|W|y>

[deft apex]

ok so how do we usually write this

without all this SP stuff

[lavish jewel]

it literally depends on the vector space and the inner product

there isn't a single one

that's why they're teaching it to you this way

the standard scalar product in R^n is x^T y

in C^n, x^H y

for square integrable functions, $\int_{-\infty}^{\infty} x^*(t) y(t) dt$

[stoic python]

19eddy4

[lavish jewel]

and so on

the standard way without dirac notation is simply <x,y>

but still this doesn't tell you what exactly the operation is

[errant mist]

I'm confused in the way the define the transpose here. How do I see that this definition implies (A^T)_ij =a_ji?

[gray dust]

pick x=e_j, y=e_i

[errant mist]

Is e here supposed to be an identity element here? I am not quite sure I understand

[gray dust]

e_i is a vector with 1 in the i'th slot & 0 elsewhere

[errant mist]

right, so when multiplying A e_ij you end up with the same entry A_ij and we apply the same argument for e_j and A^T?

[gray dust]

multiplying A e_ij
that's not the computation going on here

[errant mist]

its not a matrix vector multiplication?

[gray dust]

$(Ae_j,e_i)$

[stoic python]

RokabeJintaro

[gray dust]

work this out step by step

[errant mist]

oh, now I understand. The x,y are just unknowns of our index we are trying to figure out.

[gray dust]

x,y are vectors in R^n,R^m respectively

[errant mist]

ok

[errant mist]

so if ej is vector with 1 in the j slot zero else, then we get another column vector when we multiplying the vector with A right? And ei is another column vector.

[winter harbor]

Yeah, Aej is another column vector.

[errant mist]

so it's like a_i?

[winter harbor]

What do you mean with "so it's like a_i" ?

[errant mist]

every entry could be represented as a_i in the column since i is and index for the rows?

[winter harbor]

Yup

[errant mist]

so, then we have (ai, ej) = (ei, aj)

[winter harbor]

Yeah... This is easier than it looks, really. Notice the following:

[errant mist]

I got it now, just got thrown off a couple of times

thx for all the help)

[torn stag]

@errant mist Really part of showing that the transpose even exists as claimed in the definition is verifying that the matrix $B$ with $b_{ji} = a_{ij}$ satisfies the defining property of the transpose.

[stoic python]

IlIIllIIIlllIIIIllll

[teal grotto]

this is really the adjoint right? it’s the conjugate transpose over complex vector spaces iirc

[winter harbor]

Nah, we are working over real vector spaces. So A^t just denotes the transpose, not the conjugate transpose.

the inner product here is bilinear and not sesquilinear and so on

Denote $A = (\alpha_{i,j}){1 \leq i \leq m, 1 \leq j \leq n }$ the matrix $A$ and $e{i} = (0, \cdots, 1, \cdots, 0)$ the vector that has $0$ in all of its entries except at the i-th entry where it equals to $1$.
\
\
We then have $Ae_{i} = (\alpha_{1,i}, \cdots \alpha_{m,i})$ is the $i$-th column of the matrix $A$.
$$
\langle Ae_{i}, e_{j} \rangle = \sum\limits_{k=1}^{m} \alpha_{k,i} \delta_{k,j}
$$
Where $\delta_{k,j}$ denotes the Kronecker delta, i.e
$$
\delta_{k,j} =
\begin{cases}
0, k \neq j \
1, k =j
\end{cases}
$$
So we have
$$
\langle Ae_{i}, e_{j} \rangle = \alpha_{j,i}
$$
Notice that we also have $A^{t}e_{j} = (\alpha_{j,1}, \cdots, \alpha_{j,n})$. i.e, the $j$-th column of $A^{t}$ is the $j$-th row of $A$.
\
\
This implies
$$
\langle e_{i}, A^{t}e_{j} \rangle = \sum\limits_{k=1}^{n} \delta_{k,i} \alpha_{j,k} = \alpha_{j,i}
$$
\
\
So we have $\langle Ae_{i}, e_{j} \rangle = \alpha_{j,i} = \langle e_{i}, A^{t}e_{j} \rangle$.
\
\
By linearity of both $A$ and $A^{t}$. This means that:
$$
\langle Ax,y \rangle = \langle x, A^{t}y \rangle; \forall x \in \mathbb{R}^{n}, y \in \mathbb{R}^{m}
$$

[stoic python]

MisterSystem

[errant mist]

@winter harbor that's a very nice formulation thank you again)
What is a sesquilinear dot product BTW?

[winter harbor]

You are studying real inner product rn.

They are defined as symmetric, positive definite bilinear forms.

In the case of complex inner product

This changes a little bit

[median ocean]

2 please

[winter harbor]

Because the complex numbers have the complex conjugation.

And we want complex inner products to behave "nicely" with respect to complex conjugation.

In the following sense

Let $V$ be a complex vector space. A map:
$$
f : V \times V \rightarrow \mathbb{C}
$$
is called sesquilinear if $\forall u,v,w \in V$ and $\lambda \in \mathbb{C}$ all of the following properties hold:
\begin{itemize}
\item f(u,v+w) = f(u,v) + f(u,w)
\item f(u+w,v) = f(u,v) + f(w,v)
\item f(\lambda u, v) = \lambda f(u,v)
\
\
\item f(u, \lambda v) = \overline{\lambda} f(u,v)
\end{itemize}

[stoic python]

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

[winter harbor]

The last property is called Anti-linearity

In the second argument to be more specific

So sesquilinear forms are pretty much bilinear forms, except that they are Anti-linear with respect to one of the variables.

A hermitian/complex inner product is then a sesquilinear form that is positive definite and (hermitian symmetric), i.e satisfies:
$$
f(u,v) = \overline{f(v,u)}, \forall u,v \in V
$$

[stoic python]

MisterSystem

[errant mist]

Thanks, I'm heading to sleep, but will read through again tomorrow and give it some thought.

[winter harbor]

Have you guys covered rank-nullity at this point?

[winter harbor]

What have you tried so far?

[nocturne jewel]

They got help

[winter harbor]

Hint, look at the determinant.

Oh, ok then.

[stoic mountain]

can someone help me with this?

[winter harbor]

We have that $S = {(x,y,z) \in \mathbb{R}^{3} , \vert , z = -2y+x}$.
\
\
Notice then if $(x,y,z) \in S$, this implies:
$$
(x,y,z) = (x,y,-2y+x) = x \cdot (1,0,1) + y \cdot (0,1,-2)
$$
So $S = \text{span}{(1,0,1), (0,1,-2)}$.

[cerulean garden]

z=x-2y*

[stoic mountain]

um

i got x-2y = z

i got (1,0,1) and (0,1,-2) but the book's answer is (2, 1, 0) and (1,0,1)

[stoic python]

MisterSystem

[stoic mountain]

how do i reach the books answer?

[winter harbor]

didn't notice that was a -z

anyways

[winter harbor]

notice that (2,1,0) = 2*(1,0,1)+(0,1,-2)

so really doesn't matter

[cold jasper]

can someone help me with this?

*The given map T:C->C is linear and thinking about it sin^2(x) and cos^2(x) would work but when trying to solve it I cant get anywhere

[cerulean garden]

express x as x(y,z) instead of z=z(x,y)

[stoic mountain]

ok

[winter harbor]

take y_1 = e^(-2x) sin(x) and y_2 = e^(-2x)x sin(x)

notice how both of these are in the kernel of T

so ofc T(y_1) and T(y_2) are linearly dependent

but y_1 and y_2 are linearly independent over C

[cold jasper]

omg ty so much that makes so much sense and i have been working on that question for over an hour

[winter harbor]

👍

The idea would be to try solving an easy ODE that depends on T

and the easiest that came to my mind would be T(y) = 0

[stoic mountain]

I have to determine a spanning set for the nullspace(A) = {(0,0)} so would my spanning set just be (0,0) as well?

[winter harbor]

yeah, (0,0) spans {(0,0)}

You could also take the empty set too

the empty set spans {(0,0)}

Even if that sounds weird

[stoic mountain]

i was never taught what an empty set means can you explain it?

the book answer says null

[winter harbor]

The empty set is the set with no elements.

[stoic mountain]

is there a symbol for that?

[winter harbor]

$\varnothing$ or ${ }$

[stoic python]

MisterSystem

[dull pilot]

why does assuming this not lose generality

[wintry steppe]

it's just poor phrasing

they already said "not all zero" in reference to the c_1, ..., c_k, so they can just say that there's some i such that c_i is non-zero

i isn't something already fixed here

[dull pilot]

ah i see what confused me

i thought it was for any i = 1,...,k and not for some i = 1,...,k

thanks

[pale maple]

Hello, I have a question for finding the rule of a linear relationship. So if i'm trying to find the rule for coordinates set out in a straight line on a Cartesian plane, first I would put the coordinates into a x and y graph right? which I have done but there are so many possibilities of what the rule could be so what would be the easiest and fastest way to do it?

[lavish jewel]

if you know for sure they are on a line, you need only 2 points

then use the so-called slope-intercept form

[half ice]

Sorry they recommended this channel, linear algebra is a very specific study. Consider #prealg-and-algebra in the future

[pale maple]

oh ok sorry, I wasnt really sure with what channel to use but ill use that channel in the future

[prisma kettle]

ah yeah. my mistake

[pale maple]

all good

[half ice]

We're fine here anyway since it's empty

[pale maple]

ok thank you

[pale maple]

Would I use the start and end point of the line

[lavish jewel]

it doesn't matter

a line has only 2 parameters, slope and y-intercept

so you only need 2 points to determine them

[pale maple]

so does it matter if I have more than 2 points?

[lavish jewel]

only if the coordinates are noisy and you need to use statistics to infer the parameters

[pale maple]

oh ok so when there in a straight line no matter how many coordinates there are you only need to use two

[wild fulcrum]

ye

[pale maple]

thank you

[wispy pewter]

unsure how to solve this

[quartz compass]

there are properties of determinants that you can use, do you know what they are?

like for instance, what happens when you exchange two rows or columns, factor out a number, etc

you want to rearrange the one you're given to look like the one you know the answer to

[winter flume]

I should learn linear algebra

[teal grotto]

yeth

[rustic jewel]

Hello, guys! Could anyone help how to solve these two examples? I know it is related to Gram-Schmidt process, but I am not sure how to do that. I solved the first problem by Gram-Schmidt formula, but these two bewildered me

[teal grotto]

it’s the same thing as gram schmidt just with a different inner product

[rustic jewel]

ok, but how to solve this?

[nocturne jewel]

Just go through w/ Gram Schmidt using the canonical basis

[rustic jewel]

ok, I'll try out

[pine lion]

So I'm trying to do this. I'm not too sure how to do it. Intuitively I feel like this vector space has infinite dimensions because I could keep adding variables and it would still be in the space.

So the only thing I can really think of is to just say the span is an infinite sum of some scalar alpha_i times x_i

Am I at all on the right track?

[marble lance]

You are not

Consider the span of some finite set {b1, b2, ... bn} in this space

It is all the vectors of the form
$$a_1 \otimes b_1 \oplus \dots a_n \otimes b_n = b_1^{a_1} \cdot b_n^{a_n}$$

[stoic python]

Lunasong the Supergay

[marble lance]

How many vectors do you need to make the RHS span all the positive real numbers?

[pine lion]

Oh only one no?

[marble lance]

Yeah

[pine lion]

Riiight

So what I gave is a span, but its like super linearly dependent

like incredibly linearly dependent

[marble lance]

And also you need the set to be linearly independent, but any set with one nonzero vector is linearly independent

Yeah lol

[pine lion]

Okay okay I see

Right because I can reach the same real number with one vector than I can with infinite

lol okay thank you so much

[marble lance]

👍

[pine lion]

I just want to make sure I'm expressing my answer properly. I can say that some scalar alpha times x is a basis of V because this x now spans the real numbers and the single vector alpha times x is inherently linearly independent?

[marble lance]

What is x?

You should choose a specific x

[pine lion]

x is a vector in V

Oh okay

[marble lance]

It can't be any vector, so you need to choose one

And it is not the scalar times x that is a basis

Just {x} is the basis

You need to prove {x} is linearly independent, and that span {x} = V

[pine lion]

Oh hmm. Cause I just realized If x is 1 then it wouldnt span V

So I could just fix it at any real number that isn't one?

[marble lance]

Yes

1 is the zero vector in V

That's why it can't be 1

[pine lion]

Whoa that makes a lot of sense

Sorry I'm new to the concept of 1 being zero

[marble lance]

That's okay, remember anything can be 1. Once you have the space, you should always check what the identity is and what inverses look like

[pine lion]

Yeah I'll make sure to keep that in mind

So here since x + 1/x is 1 which is the zero vector 1/x is my inverse correct?

[marble lance]

Yep

[wintry steppe]

What's the geometric interpretation of the transpose of a matrix?

Like you can interpret a matrix as a linear transformation geometrically right? how would that linear transformation be different if you take the transpose of it and then apply it?

[winter harbor]

Are you familiar with dual vector spaces?

[wintry steppe]

no not really, but I have heard of them "a space of linear maps" or something 😅

[winter harbor]

Yeah so

The way I like to think about the transpose geometrically is tied to the dual vector space.

In the following sense

If we have a vector space over a field $\mathbb{K}$, which we will take to be $\mathbb{R}$ or $\mathbb{C}$ for reasons of simplicity, we can consider the space
$$
\text{Hom}_{\mathbb{K}}(V, \mathbb{K}) = {f : V \rightarrow \mathbb{K} , \vert , \text{f is linear} , }
$$
This is the space of all linear functionals on $V$. Notice that this is a vector space with point-wise sum and point-wise scalar multiplication.

[stoic python]

MisterSystem

[wintry steppe]

okay got it

[winter harbor]

We usually denote $\text{Hom}_{\mathbb{K}}(V, \mathbb{K}) = V^{\ast}$ and say it is the dual space of $V$.

[stoic python]

MisterSystem

[winter harbor]

Nice

One very interesting thing with the dual space construction

Is that it is sort of well behaved with respect to linear transformations.

In the sense that is functorial.

Meaning

[wintry steppe]

functorial?

[winter harbor]

I will make that more precise

[wintry steppe]

oh I meant that I got what a dual space is, I don't get how the transpose is tied to the dual space yet

[winter harbor]

You don't need to know what a functor is rn. But only that it is a "nice" kind of construction.

[winter harbor]

You will see now.

Anyways, suppose that we have a linear transformation $T : V \rightarrow W$. The nice thing about the dual, is that it induces a linear transformation between the dual spaces of $V$ and $W$ too! Notice that we have a linear function:
\begin{align*}
T^{\ast} :& W^{\ast} \rightarrow V^{\ast} \
g&\mapsto g \circ T
\end{align*}

[stoic python]

MisterSystem

[winter harbor]

Nice

Anyways, we usually call this map the dual of T, the pullback of T or the transpose of T.

And the name transpose makes sense

Because this induced map

Is the transpose of T

When we see it as a matrix

[wintry steppe]

so the transpose of a matrix (that transforms something from V to W) is the matrix that transforms an element from V* to W*?

[winter harbor]

W* to V*

[wintry steppe]

ahh

[winter harbor]

Yeah

I recommend you doing this calculation for yourself

[wintry steppe]

what's the g -> g o T?

[winter harbor]

What this transformation T* does

Is that it takes a linear functional g : W -> R

And maps it to a linear functional on V

And this linear functional is g ° T

Notice that T : V -> W and g : W -> R

So g ° T : V -> R is a linear functional

[wintry steppe]

I think it makes sense

[winter harbor]

Try doing this calculation by yourself

[wintry steppe]

yeah that might help

[winter harbor]

Let V and W be two real vector spaces that are finite dimensional

Take a basis for V

And a basis for W

When you view T* as a matrix

With respect to the dual basis of V

And the dual basis of W

This matrix is the transpose of T

A few weeks ago someone asked the same thing btw

[wintry steppe]

alright I'm gonna go try doing that

thank you very much!

[winter harbor]

.

[nocturne jewel]

Putting this here cause can't decide where it fits more, MultiVar or LinAl.

More just confused on what's actually being said than anything.

cause in my head for say part a, it's because (A+tb)f(A+tb)=Af(A)=I_n at t=0. so it's a "constant matrix" which differentiates to the 0 matrix

Or actually it's I_n for all t

[winter harbor]

That's correct, yeah.

[nocturne jewel]

Ok.. but why?

cause you can "distribute" the operator to the entries?

cause Im good with the fact it's $\dv{t}I_n$ but from there it only makes intuitive sense that it becomes 0, not proper understanding

[stoic python]

Mosh

[winter harbor]

$\dfrac{d}{dt} I_{n} = \lim\limits_{h \to 0} \dfrac{I_{n}(t+h) - I_{n}(t)}{h} = \lim\limits_{h \to 0} \dfrac{I_{n} - I_{n} }{h} = 0$

[stoic python]

MisterSystem

[winter harbor]

Where we are viewing I_n as the constant map t -> I_n

[nocturne jewel]

right but doesnt that mean t is a R^n vector?

not a scalar?

[winter harbor]

Why would t be a vector in R^n? that's just notation. I said before I am using I_n to mean the same thing as the map t -> I_n

So when I write I_n(t) this should be I_n

[nocturne jewel]

Oh I thought you meant I_n as in the identity map

[winter harbor]

No, we are viewing I_n as the constant map t -> I_n here

[nocturne jewel]

you mean $L:\mathbb{R}\to {I_n}$

[stoic python]

Mosh

[nocturne jewel]

with L(t)=I_n

[winter harbor]

Yup, that's what we are doing here.

Lemme take a screenshot from Tu's book.

[nocturne jewel]

wait what I thought would've still made 0, oh well

no, that's 1

yeah I get why it's 0

[native cipher]

Quick question if you have det(A+B) is that equal to Det(a)+Det(b)

[winter harbor]

No, it's not.

For instance

A and B can both be invertible matrices

And their sum not be an invertible matrix

[native cipher]

ok

Coolio

thanks

[fallen quest]

How do i solve a out of these vectors that make up a pyramid?

[wintry steppe]

What exactly is meant by the last paragraph?

[winter harbor]

What exactly? The bit where he discusses the vector space F^S?

[wintry steppe]

Yes

[winter harbor]

We usually define $\mathbb{F}^{n}$ as the vector space consisting of all n-uples of elements in a field $\mathbb{F}$ with point-wise addition and multiplication by scalar.
\
\
What this paragraph is point out to us, is that we can instead also think about these n-uples as certain functions.
\
\
More specifically, we can think of an n-uple $(a_{1}, \cdots, a_{n}) \in \mathbb{F}^{n}$ as a function $f : {1, \cdots, n } \rightarrow \mathbb{F}$ where $f(i) = a_{i}$ is the i-th component of the n-uple $(a_{1}, \cdots, a_{n})$.
\
\
And the nice thing the paragraph is pointing out is that the vector space consisting of those n-uples and the vector space consisting of functions $f : {1, \cdots, n} \rightarrow \mathbb{F}$ are in fact "the same". More formally, they are isomorphic, so we can think of these as the same.

[stoic python]

MisterSystem

[winter harbor]

The same line of reasoning applies to sequences elements $(a_{n})_{n \in \mathbb{N}} \in \mathbb{F}^{\infty}$, which are sequences of elements in $\mathbb{F}$ and functions $f : \mathbb{N} \rightarrow \mathbb{F}$

[stoic python]

MisterSystem

[sonic beacon]

@gray dust thanks, then would the complete proof be: (c * (T + T'))(x) =c(T+T')(x) = c * (T(x) + T'(x)) = c * T(x) + c * T'(x) = (c * T + c *T')(x)

[gray dust]

there's still a 'jump'

[sonic beacon]

i am kind of confused, how do we go from (c * (T + T'))(x) to c(T+T')(x) -- i am confused by scoping here

[gray dust]

recall the definition of cT

[sonic beacon]

(cT)(x) = c * T(x)

[gray dust]

apply it to c(T+T')

[sonic beacon]

thats the confusing part, i want to say cT + cT'

[gray dust]

think of the structure of the definition not the exact letters used

the definition of constant*map

(const*map)(x)=const*map(x)

[sonic beacon]

ok i see consider (T+T') = "T" then (c*"T")(x) = (c * "T")(x) which is the same thing as c * (T+T')(x)

wait

[gray dust]

now think of it for const*(map1+map2)

[sonic beacon]

let me rethink

[gray dust]

replace map by map1+map2 in the definition

[sonic beacon]

((c*"T"))(x) = (c * "T")(x)

(const * map1+map 2)(x)= const *(map1+map2)(x)

[gray dust]

careful brackets

(const*(map1+map2))(x)

[low pecan]

anyone can explain by using orthogonal decomposition to solve for the matrix to find the reflection of an arbitrary vector where the reflection line passes through the origin but does not lie on the axis lines? i hope what i typed makes sense

[sonic beacon]

yeah im still confused.

(c*(T+T'))(x) = c *( (T+T')(x) ) is what i think step 1 and 2 should be

[gray dust]

the brackets are fine here

() around (T+T')(x) is extra but still ok

[sonic beacon]

(c*(T+T'))(x) = c *(T+T')(x) = c * ( T(x) + T'(x) ) = c * T(x) + c * T'(x) = (c * T + c * T')(x) ?

[gray dust]

it's ok up to

c * T(x) + c * T'(x) = (c * T + c * T')(x)

this is the jump i mentioned

[sonic beacon]

i dont see what the jump is

[gray dust]

(c * T + c * T')(x) doesn't 'immediately' equal cT(x)+cT'(x)

you used the definition of multiple & sum at the same time

which is jarring compared to each other step

[sonic beacon]

i agree, i dont know how to remedy it

[gray dust]

do it 1 step at a time

it helps to start at (c * T + c * T')(x)

then reverse whatever you write to complete the proof

[sonic beacon]

c * T(x) + c* T'(x) = (c * T)(x) +T'(x) = c * T(x) + (c*T')(x) = (c * T + c * T')(x) ?

[civic nacelle]

B 1 and 2

please

[gray dust]

many typos

pls don't rush

[sonic beacon]

ok

[winter harbor]

ce n'est pas de l'algèbre linéaire...

[wintry steppe]

francais...

[sonic beacon]

@gray dust you are suggesting i prove (c * T + c * T')(x) = c * T(x) + c * T'(x) ?

and then reverse the proof for the steps to complete my proof

[gray dust]

saying 'prove .. separately' implies much more effort than required

you already did it above. it's just littered with typos

[sonic beacon]

ok i will take a look at it again, might be best to stop and work on tomorrow used my brain on other things today and cant think well now

[gray dust]

no it's done

[gray dust]

just fix the typos

[sonic beacon]

ok

going to try now

c * T(x) + c * T'(x) = (c * T)(x) + (c * T')(x) - are these steps correct

[gray dust]

yes that's just by definition of multiple

[sonic beacon]

now how do i justify (c * T)(x) + (c * T')(x) = (c * T + c * T')(x)

which is the last step i wanted to show

its by def of scalar mult?

[gray dust]

no

c*T+c*T' is short for (c*T)+(c*T'). think pemdas

what's the definition of c*T+c*T'

[sonic beacon]

(c * T + c * T') = (c* (T + T') ) ?

er

[gray dust]

cT & cT' are maps

the sum cT+cT' is a map defined by (cT+cT')(x)=(cT)(x)+(cT')(x)

[sonic beacon]

(c * T)(x)+ (c * T')(x) = c * T(x) + c * T'(x)

[gray dust]

so thinking 'backwards' we can write (cT)(x)+(cT')(x)=(cT+cT')(x)

[sonic beacon]

this is my confusion i guess maybe a brain fart, how does (cT+cT')(x) = (cT)(x)+(cT')(x)

ah

wait a minute

its because (T+T')(v) = T(v) +T'(v) - def of addtion on my linear maops

is that right

[gray dust]

the sum cT+cT' is a map defined by (cT+cT')(x)=(cT)(x)+(cT')(x)
that's exactly what i cited here

[sonic beacon]

ok i see

[gray dust]

so i think it would've helped you to write out things to help you write the proof like

T & T' are maps, so cT, cT', cT+cT', c(T+T'), etc are maps, and write the definition of each

so that you don't get caught up on what can be rewritten as what

[sonic beacon]

i think i have all the pieces, i need food/mental break. i'll try to write up a complete proof tomorrow

[gray dust]

no the proof is complete

you just need to cut all the typos & deadends, string up what you wrote

let me do it

[sonic beacon]

ok

[gray dust]

$(c(T+T'))(x)=c(T+T')(x)=c(T(x)+T'(x))\=cT(x)+cT'(x)=(cT)(x)+(cT')(x)=(cT+cT')(x)$

[stoic python]

RokabeJintaro

[sonic beacon]

ok that helps a lot

ty very much for helping me understand this

[gray dust]

you're very welcome

[pine lion]

Ok I think I have something for this

Where w1 is in E+F and w2 is in E+G

right and I defined w1 = u1 + v1 where u1 is in E, v1 in F, then w2 = u2 + v2 where u2 is in E, v2 in F

This seems right to me but I'm just not sure how legal it is to say if span(w1)=spawn(w2) then w1=w2

Like generally that wouldn't be true but because there's only a unique way to form w since its a direct sum it would be...

that's what I'm thinking

[winter harbor]

You have defined w_1 as an element in E direct sum F. That's ok, but it may not be the case that span{w_1} = E direct sum F

So the argument doesn't hold.

Same goes for w_2, it is not always the case that E direct sum G is the span of only one vector w_2.

Hint. Notice that if $v \in F$, then either $v = 0$ or $v \neq 0$. If $v=0$, then $v \in G$ since $G \subset V$ is a subspace of $V$. If $v \neq 0$, then since $E \cap F = {0}$, by definition of direct sum, and $ v \in E \oplus F = E \oplus G$, it must be the case that $v \in G$. The converse is proved analogously.

[stoic python]

MisterSystem

[pine lion]

Oh I see. Thank you!

And now that I know that my v_1 and v_2 are both in F and in G I can use the property that w_i is unique to complete the proof, right?

[winter harbor]

Yup

[pine lion]

Okay I'm attempting to prove this

This is my attempt

Does this make sense?

[nocturne jewel]

1. You dont need span when defining the basis (Basis is a lin indep set of vectors which span the space)
2. 3rd line of the actual working feels random/out of thin air (dim(U+W)=<dim(V))
3. You already knew UnW was nonempty since U and W are both subspaces

[wintry steppe]

the conclusion shouldn't be that U cap W is non-empty, but that it's non-trivial

[winter harbor]

Hint. Prove this by contradiction.

[pine lion]

Hmm okay yeah I see your points. And thank you!

I’ll look into this contradiction 👀

[winter harbor]

You will readily see that dim(U+W) > dim(V), which can't be the case since U+W is a subspace of V.

So the intersection has to be non trivial.

[pine lion]

Oooh yeah I see you

Thanks!

[wintry steppe]

mistersystem been farming this channel recently

love to see it

[winter harbor]

[wintry steppe]

[winter harbor]

I would like to farm #point-set-topology tbh

But I am not so confident

Yet...

Even tho I think I could answer one some questions here and there.

But not very specific stuff.

[nocturne jewel]

Mister's replacing Terra

[winter harbor]

Nah, TTerra's the GOAT I can tell you that.

[wintry steppe]

i couldn't have chosen a better successor

[nocturne jewel]

could've been me, cause I'm shit so good at LinAl

[wispy pewter]

can anyone help me solve this

[limber sierra]

do you know how elementary row operations affect the determinant?

[wispy pewter]

Yeah I think so

[robust owl]

If a book defines span(A) as the set of all linear combos of elts of A, but forgets to define span({})={0} is it reasonable to claim span({})={0} by considering any y in span({}) and claiming it is an empty sum hence 0?

Or am I making a mistake/assuming too much foundational junk in doing something along those lines?

Actually, now that I think about it I can only really see how that guarantees span({}) is a subset of {0} but not the reverse inclusion.

[torn stag]

@robust owl $\text{span}(A)$ can be seen as the smallest vector space containing $A$. Thus $\text{span}(\emptyset) = {0}$.

[stoic python]

IlIIllIIIlllIIIIllll

[robust owl]

Yeah, but in this case the book didn't define it that way.

[torn stag]

@robust owl Your logic is also right. The only linear combination of elements in $\emptyset$ is the empty sum, which is $0$.

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

@robust owl The book definition is equivalent to the smallest vector space definition, so that's why I mentioned it.

[robust owl]

Doesn't that only give $span\emptyset\subseteq {0}$ though?

[stoic python]

DootDooter

[robust owl]

What stops span{} from being the emptyset?

[torn stag]

$\sum_{v \in \emptyset}3v = 0 \in \text{span}(\emptyset)$.

[stoic python]

IlIIllIIIlllIIIIllll

[robust owl]

So just the fact that we can take empty sums at all?

[torn stag]

I guess.

[robust owl]

That's really trippy lmao

[torn stag]

Yeah, thats why the smallest vector space definition is somehow satisfying.

Or at least justifies the conclusion

[robust owl]

It seems like a nice succinct way to put it without risking forgetting stuff like defining span{} that would cause a bunch of extra work.

[torn stag]

Well $\text{span}(\emptyset)$ is defined, as long as you know what an empty sum is.

[stoic python]

IlIIllIIIlllIIIIllll

[robust owl]

Yah true.

[exotic wedge]

Are there general properties for a 3 x 3 matrix to be non-diagonalizable ?

[odd kite]

https://en.wikipedia.org/wiki/Diagonalizable_matrix#Characterization

it can depend on what field is being used - for example C vs R

[exotic wedge]

The book assumes C for eigenvalues related matrices

[wintry steppe]

one characterization of diagonalizability is the minimal polynomial being a product of distinct linear factors

[exotic wedge]

I am not sure how to make it "generic" using that

[wintry steppe]

me neither

im just telling you a cool condition

im not even sure what it means by generic

[exotic wedge]

. Ideally you should give an equivalent condition (i.e. “A is non diagonalizable if and only if it has these properties…”).
This is what the TA said

[winter harbor]

.

😭 😭

[wintry steppe]

lmao

[exotic wedge]

The nilpotent solution we worked out before is not "generic" enough

[hollow finch]

what does it mean "no special structure of the matrix could be seen"?

[exotic wedge]

probably it want some properties for any 3x3 matrix to be non-diagonalizable

[hollow finch]

maybe that it is the sum A=D+N where D is diagonalizable, N is nilpotent (and nonzero), and D and N commute.

[runic oar]

ngl the late night LA homework sessions are actually becoming kinda a vibe

[exotic wedge]

I've done all the problems, and i am stuck with this for the last two days

[winter harbor]

Yeah, I talked before about the minimal polynomial thing tho.

But I didn't know if like

You have covered this in class yet or not

[exotic wedge]

the text didn't go over that

but maybe i can figure out, what about the minimal polynomials that makes the matrix non diagonalizable ?

[winter harbor]

Yeah, the problem here to me seems more about what the exercise means by "generic".

[exotic wedge]

The TA said ". Ideally you should give an equivalent condition (i.e. “A is non diagonalizable if and only if it has these properties…”)."

[hollow finch]

more specifically you can do

$A=P(D+N)P^{-1}$

where $P$, is invertible, $D$ is diagonal, and $N$ has all nonzero entries except for ones on the super diagonal corresponding to repeat entries in $D$

[stoic python]

nix (@ me for the love of euler)

[hollow finch]

so basically A has a jordan block of size 2 or 3

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

[exotic wedge]

The text didn't cover jordan block neither

[hollow finch]

idk it seems like everyones given you pretty much all your options

how do you know it isnt "generic" enough

[exotic wedge]

I don't know, I might investigate that minimal polynomial thing

[winter harbor]

Maybe you can even explicitly prove things in the 3×3 case by yourself if you are clever enough I guess.

So maybe that's the way to go.

[hollow finch]

i still dont know whos telling you what is or isnt generic enough

is it a teacher?

or are people giving you valid answers and you just dont like them?

help us out here

[exotic wedge]

The teaching assistant.

[hollow finch]

oh great. okay yeah now i get it

[exotic wedge]

I thought any family of nilpotent matrices would be generic, but she said this isn't generic enough

[hollow finch]

the question doesnt seem to be asking "characterize all nondiagonalizable 3x3 matrices" though

[exotic wedge]

probably this is what the answer should be

[hollow finch]

nilpotent matrices are all nondiagonalizable. that sounds generic enough for me tbh

[winter harbor]

Yeah, I would interpret it maybe as like "Give an example of an infinite family of non diagonalizable matrices over C" or something.

[hollow finch]

probably they have their own proof in mind which may not be entirely correct since theyre still just a student right? not a professor with a degree?

honestly idk much about TAs

[exotic wedge]

tell me again about the minimal polynomial thing ?

[winter harbor]

Are you familiar with the Cayley Hamilton theorem?

[exotic wedge]

No . Okay i think i can't use that solution

[hollow finch]

lol okay what does your book actually cover

[winter harbor]

The minimal polynomial of an endomorphism T : V -> V on a finite dimensional vector space V is the monic polynomial P of lowest degree such that P(T) = 0.

What the Cayley Hamilton theorem tells us

[exotic wedge]

I mean, the book is called linear algebra done wrong. maybe this is why xD

[winter harbor]

Is that the characteristic polynomial χ of T satisfies χ(T) = 0. So the minimal polynomial divides the characteristic polynomial.

Form that we readily see that a matrix is diagonalizable over C iff the minimal polynomial splits into distinct linear factors.

At least that's the proof I know

But it is highly non trivial to a student.

[hollow finch]

but a matrix can still be diagonalizable if it doesnt split into distinct linear factors
(EDIT: shut up nix)

[winter harbor]

The minimal polynomial

Not the characteristic polynomial

A matrix can have repeated eigenvalues

But still be diagonalizable yeah

But its minimal polynomial has to have no repeated roots/split into distinct linear factors in order for a matrix to be diagonalizable.

[hollow finch]

right

@exotic wedge idk id probably ask your prof exactly whats wrong with using nilpotent matrices as your answer

tell them the TA has a problem with it but you dont know why

[winter harbor]

Yeah, I don't think students can rediscover Cayley Hamilton by themselves

Or Jordan Canonical Form

Which is the bad way to do it 🤮

[torn hornet]

no jcf is bad way to prove it

the right way is topologically

[hollow finch]

i mean i think C-H is fairly straightforward and intuitive in the diagonalizable case (using diagonalization since its pretty obvious for diagonal matrices)

but yeah i get what you mean

[winter harbor]

Btw, your post on the Cayley-Hamilton theorem is pretty good

Tbh, I particularly don't like that proof which uses some adjugate memery.

It's pretty much all algebraic manipulation.

And I am not that good at remembering this kind of stuff.

But with the Zariski Topology it's way more intuitive to remember the proof.

[torn hornet]

shameless plug :https://johndsmathblog.wordpress.com/2021/08/05/proof-of-cayley-hamilton-using-zariski-topology/

[winter harbor]

And is way more elegant.

[torn hornet]

yeah i never like the LA stuff where u have to do some random factorings that are super tedious etc

[exotic wedge]

The TA gave us a hint and it was "A is non-diagonalizable if we can find basis such that A looks like ?, then it's non-diagonalizable". And I still can't figure it out with that hint

[hollow finch]

oh

i think i might know what theyre saying

[winter harbor]

I like Schur decomposition tho, because you can rephrase it in terms of like, how endomorphism of finite dimensional complex vector spaces stabilize complete flags of ortonormal basis or something.

But most factorizations in linear algebra like SVD or LU decomposition or whatever idk how to phrase without having to explicitly make mention of matrices.

[hollow finch]

maybe theyre referring to something like a Shear matrix?

[torn hornet]

yeah. diagonalization is nice too cause we are finding a basis in the eigenvectors basically.

[exotic wedge]

elaborate ?

[hollow finch]

or actually more generally i guess an upper triangular matrix that isnt diagonal. idk im tired lol

[exotic wedge]

that makes since but not sure if it's generic enough

anyway

thanks a lot

[low pecan]

need help with this qn….

[wintry steppe]

Is -1 an element in every field corresponding to a vector space?

[low pecan]

i dont know/ dont think so….? i was given a clue that the way to solve this was doing projection of an arbitrary vector to the reflection line

[wintry steppe]

I was asking a completely unrelated question

[low pecan]

oh mb

[gray dust]

the question is just about fields. nothing to do with vector spaces

each field has a multiplicative identity in most cases we call 1. additive inverses exist for each scalar, so in particular each field has what we call -1, the additive inverse of 1

[wintry steppe]

Ok, that makes sense

What is (a)?

[turbid trellis]

How to find the determant of any matrix ?

Any square matrix *

I can only find methods for 2x2, 3x3 and 4x4

However I want to know how they work, what the general procedure is

[teal grotto]

use either laplace expansion or the Leibniz product formula

[teal grotto]

wdym?

[wintry steppe]

Example (a)

[teal grotto]

but what is ur issue with it?

[wintry steppe]

How is it a sub space?

[lavish jewel]

just use the definition of a subspace

check what is needed for it to be closed under addition and scalar multplication

[tall rock]

hey i want to ask a question, but its more like a rant, really.

I'm a cs student learning lin alg.

i'm really frustrated with my linear algebra course lately. I basically understand how to calculate numbers and do stuff with it, but I always do badly in proving things. I just still see no benefits of me doing proving. should i just like fail this semester class and retake onto another class which is more suited to programming?

[lavish jewel]

that's up to you, but not understanding the meaning behind linalg stuff will severely limit the things you can do with coding

for example if you wanna do machine learning or physical simulations/engines stuff

you simply won't be able to without copy-pasting what others did before

if you wanna do stuff like web development, then sure, you won't need it

[tall rock]

i do want to learn this, but damn the difficulty is just so high for proving stuffs...

[lavish jewel]

you'll have to learn, then

no way around it

[tall rock]

is there any course i should take to practice my proofing?

[lavish jewel]

this is right about the first one for many people, so

it's common to struggle with it at first

[tall rock]

dang ok

i hope i can pass with c- atleast 😓

[crystal hound]

When working with column vectors, so you’re matrix multiplying on the left, if you have 2 transformations followed by one another, so say transformation A followed by B, would the matrix representing this transformation be BA? Since you do A, then B?

[lavish jewel]

yes

since matrix mult is associative, you can see this from B(Ax)

Ax yields a new, transformed vector. then B transforms this vector once more

[crystal hound]

Perfect thank you

[wintry steppe]

How do you show that the set is not a subspace of $F^{4}$ iff $b \neq 0$?

[stoic python]

justini

[lavish jewel]

show that it's not closed under addition

[wintry steppe]

Why is it not closed under addition though?

How does adding "b" make it not closed?

[lavish jewel]

so, x3 must be 5 x4 + b, yeah?

let's take 2 vectors

(x1,x2, 5x4+b, x4) and (y1,y2, 5y4+b, y4)

this yields (x1+y1, x2+y2, 5(x4 + y4) + 2b, x4+y4)

and you see that the third element is not 5 times the 4th element + b unless b = 0

because you need 2b = b for this to be true

if b is nonzero, adding two vectors in this set yields an element outside of the set

[wintry steppe]

b in the first vector has to equal b in the second vector, right? Or no?

[lavish jewel]

yes

[wintry steppe]

Oh, so b is like an arbitrary number? Say 2, and then you get x4 + y4 + 4 which doesn't work