linear-algebra

[ruby topaz]

https://en.m.wikipedia.org/wiki/Row_and_column_spaces

[lean aspen]

@dusky epoch can you help me get a better idea of how T-cyclic spaces work

[dusky epoch]

what's a T cyclic space

[lean aspen]

not q sure if I have right idea for what it means if V is a T-cyclic space (where T in End V)

[dusky epoch]

what's the defn

[lean aspen]

Let T ∈ Endo(V), and we'll keep it simply and say V is fdvs. Then we say W is a T-cyclic subspace if there exists v ∈ V such that {T^k(v)} is a basis for W

think that's right

[dusky epoch]

[lean aspen]

i.e., v,Tv,T^2v,...

[dusky epoch]

never seen that before & don't immediately recognize it

[lean aspen]

umm

cyclic/principal module sound familiar?

hmm

[dusky epoch]

cyclic module yes

oh

yeah

[lean aspen]

yeah, it's a bit confusing

[dusky epoch]

K[x]/<f>

K-vector space equipped with an operator <-> K[x] module

[lean aspen]

so you get what a T-cyclic space is?

So um

Can V be T-cyclic when min poly is not full rank?

ie its degree differs from characteristic

sorry, sorta thonking today

@dusky epoch ?

@wintry steppe

[serene stump]

is there any good linear algebra book that you guys can recommend?

[rigid cypress]

linear alg with applications seems alright

explanations aren't too difficult generally

[broken hawk]

There's many linalg books, and many are good, but they go in different directions. What "flavour" of linalg do you want? Do you want the "pure math" approach of full generality and lots of proofs, or do you want to work with matrices and learn about their applications?

[serene stump]

I'd probably prefer the latter, but i think i might be interested on both lol

[dull kettle]

Hi guys, is S1 the only set which is a basis for R^4

[winter reef]

yes

S2 has only 3 vectors and S3 has 4 vectors but two are linearly dependent

[dull kettle]

Thank you! @winter reef what about this question? my answer is that all statements are false due to the inconsistent linear system , not sure about III though

[winter reef]

2nd one should be true I think

because from what I understand spanS is R3?

[dull kettle]

is span(S) R^3?

i thought is would be a subset of R^3

[winter reef]

whaqt subset?

is it defined anywhere?

[dull kettle]

this is the whole question, i'm wrong about that

[winter reef]

I mean vector (-1,-1,10) has to be ins span R^3

not sure what's S

without knowing whats S you cant al;so determine first question I think

because if S is something like (x1,x2,0) then u1,u2,u3 span S

[dull kettle]

S is u1, u2 and u3 in the question

and its values are in the matrix

[winter reef]

ill let sascha answer your questions

[broken hawk]

span(S) will be either all of R^3 (if they’re linearly independent) or a 2-dimensional subspace (if they’re not), I’m too lazy to check if they are

[winter reef]

they gotta be lienarly dependent

[broken hawk]

oh yea cause they are in the row echelon form eh?

[winter reef]

cause you can reduce with 1,0,0

[dull kettle]

yes

[broken hawk]

then it’s a 2-dimensional subspace

[dull kettle]

does that mean v does not belong to span(S)

okay I got it

Is I and III the answer? I is not true because w is a non zero vector and III is not true because there are 4 coordinates when there should be 3 only

[broken hawk]

yea, I agree

[dull kettle]

Thanks! For this question, are II and IV the only statement that is definitely false?

[broken hawk]

the span of a set always includes 0

so II is correct

[dull kettle]

oo I forgot about that, is IV definitely false because we know since the span represent a plane in R^3, that would imply it is linearly dependent, so IV is definitely false

[dull kettle]

if say, span(u1, u2, u3) is linearly dependent, can we say that u2, u3 or u1, u3 etc, is linearly dependent?

[broken hawk]

no

consider the set (1,2,3), (4,5,6), (7,8,9)

each pair is linearly inpendent

but the whole set it dependent

all you know is that there is some finite subset which is dependent, but it need not be a pair

here’s an interesting example:
consider the set
(1, 0, 0,…), (0, 1, 0,…), … together with (1, 1, 1,…) in the space of sequences in ℝ
is this set linearly dependent?

[dull kettle]

so like for eg (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1), (1, 1, 1, 1)?

[broken hawk]

no, infinitely long ones

(countably) infinitely many infinitely long vectors

[dull kettle]

it would be independent?

the last* column would be non-pivot

[broken hawk]

I’m not talking about a matrix

so idk what you mean with “non-pivot”

[dull kettle]

don't you list the vectors down in a matrix

[broken hawk]

I mean you can

but that’s a bit abuse of notation imo

[dull kettle]

to determine if its linearly dependent or independent

[broken hawk]

anyway, infinite matrices are a bit weird

I mean, there can’t be a ”last” column

[dull kettle]

I see, how do you tell if its independent or dependent then? that question you posed

[broken hawk]

and stuff like row echelon form really only makes sense for finite matrices, so forget about all that

they’re linearly dependent if and only if there is a nontrivial linear combination of the vectors resulting in 0

equivalently, they’re linearly independent if the only linear combination resulting in 0 is the trivial one (all coefficients 0)

[dull kettle]

ahh

[broken hawk]

important reminder: linear combinations are finite sums

[dull kettle]

thanks for the help! @broken hawk

[broken hawk]

wait but you haven’t answered the question yet!

are they dependent or not?

[dull kettle]

linearly dependent right?

[broken hawk]

why?

[dull kettle]

you can express it in a non trivial combination but i think i'm thinking about it wrongly because I'm thinking about it in terms of matrix

[broken hawk]

what’s the combination?

[dull kettle]

-1 (1) + 1(1) = 0

[broken hawk]

uh, what?

[dull kettle]

coefficient of -1 and coefficient of 1

wait im confuse

[broken hawk]

for a bit of notation, let’s call e_i the vector with a 1 at the i-th position and 0s everywhere else
and then idk, v is the one with all ones

and remember, these are infinite vectors and there are infinitely many of them

we’re in the space $\mathbb{R}^\mathbb{N}$, if you will

[stoic python]

Sascha Baer:

[dull kettle]

-1( e_1 + e_2 + ... + e_i) + 1(v) = 0 ?

[broken hawk]

so what happens at the i+1th position?

that one’ll still be a 1

[dull kettle]

wouldn't v take care of it

[broken hawk]

the v adds a 1 there

right now, that left sum is 0s for the first i positions

and then 1s afterwards

[dull kettle]

yeah

[broken hawk]

I assume what you actually want is you take v and then subtract all the e_i

[dull kettle]

yeah

[broken hawk]

but you can’t do taht cause that’s an infinite sum

and we only allow for finite ones in linear combinations

[dull kettle]

we only allow for finite ones linear combinations but e_i and v is infinite so

😮

[broken hawk]

yea, vectors can live in infinite-dimensional spaces

[dull kettle]

how am I gonna express it in finite linear combination

when they are infinite

it just keeps going on and on

[broken hawk]

are you happy to say that π+e is a finite sum?

[dull kettle]

π?

[broken hawk]

even though both of them have infinitely many decimal digits?

same idea

[dull kettle]

oh

oh

[broken hawk]

π as in 3.14159etc

[dull kettle]

its not a finite sum right? since the numbers are never ending

same for e

[dire bluff]

is 1/3 finite

[broken hawk]

it is. finite sum just means ”finitely many numbers in the sum”, which numbers they are doesn’t matter

(note number ≠ digit)

(digits don’t matter)

similarly, each of those vectors are just… vectors. you need infinitely many bits of information to write them out, but they’re still just single vectors

[dull kettle]

uh huh

okay but im not seeing how does that help with the question you posed

[broken hawk]

just trying to clear up the confusion

so the thing is this

you cannot find a finite collection of the e_i and v that sum up to 0 in a nontrivial way

you would need all of them

because obviously any collection of just the e_i are linearly independent since they all have a 1 at a different location and you can’t “annihilate” that with the others
so we’d have to add v to it, but then we get infinitely many 1s and can only subtract finitely many

so we cannot get to 0

this means that indeed the set
(1, 0, 0, 0, …)
(0, 1, 0, 0, …)
(0, 0, 1, 0, …)
⋮
together with
(1, 1, 1, 1, …)

is indeed linearly independent

which is counter-intuitive if you ask me

it’s because the vector-space of sequences is actually uncountably-dimensional!

[dull kettle]

"only subtract finitely many" is referring to the condition for calculating linear combination?

[broken hawk]

yes

[dull kettle]

wow

[broken hawk]

there’s a reason why intro linalg tends to just kinda look the other direction when you mention infinite dimensional vector spaces

they’re a lot more counterintuitive and a bunch of important theorems only work in the finite case

for example:
”Let V be a vector space and W a subspace of V. If W is isomorphic to V, then W=V”

this is only true in the finite dimensional case

[dull kettle]

am I right to say that, referring to the example you gave, the set linearly independent because we can't calculate it to be equals to 0 base on the condition of linear combinations

[broken hawk]

could you try and say that again?

[dull kettle]

the set is linearly independent because we can't sum it up to zero

and that's because of the condition

[broken hawk]

I wouldn’t call it a condition as much as a definition

[dull kettle]

yeah so because of the definition we can't find the non-trivial solution

[broken hawk]

well, it doesn’t exist, rather :P

“we can’t find it” makes it sound like there is one, but we’re too dumb to find it

[dull kettle]

oh, I was referring to the idea of summing up e_i and subtracting v which is not allowed by definition

thanks for sharing though learnt something new today

[broken hawk]

oh btw do you want a reason as to why we only allow finite combinations?

like, just an example that would break otherwise

[dull kettle]

didn't you mention it is definition

[broken hawk]

yea but definitions have reasons

and you can always wonder why it was defined that way

and what would change if we altered it

in fact you should always do that

[dusky epoch]

infinite sums are icky in fields that aren't, like, R or C

[broken hawk]

I guess that’s one reason

[dusky epoch]

or generally non-topological fields

[broken hawk]

but I was gonna give a much more pragmatic argument

[dull kettle]

which is? 😃

[broken hawk]

so, polynomials, right?

the polynomials form a vector space

seen that example yet?

[dull kettle]

nope

[broken hawk]

the standard basis is {1, x, x², x³, …}

you can see that every polynomial is some finite sum of these

you can just read off the coefficients, really

3x² + 2x - 1 has coefficients (-1, 2, 3, 0, 0, 0, …)

with me so far?

[dull kettle]

yes

[broken hawk]

now the important thing about polynomials is that they’re always finite

like, there’s always a highest power of x

(otherwise they’d be power series)

but now the thing is, if we allowed for infinite linear combinations, suddenly you could get a “polynomial” like 1 + x + x² + x³ + …

e^x would be a polynomial too

[wintry steppe]

lol

[broken hawk]

and we definitely don’t want that

[dusky epoch]

one thing that is to be noted there is that in introductory linalg classes you sometimes encounter findim subspaces of that (i.e. the space of polynomials of degree at most n or sth) in practice problems

[broken hawk]

so if we want the set of all polynomials to be a vector space we really don’t want to allow infinite combinations

[dull kettle]

wow

[broken hawk]

anns argument is pretty important too though

you’re probably only familiar with vector spaces over ℝ and maybe ℂ

which are very “well-behaved”

but there’s e.g. the field 𝔽₂

which is just the numbers 0 and 1, with the rule 1+1=0

(and everything else works as you’d expect)

(e.g. adding 0 does nothing)

[dusky epoch]

take any finite field tbh

[broken hawk]

now, for example 1 + 1 + 1 + 1 + 1 + 1 = 0, which you can verify quickly

shush, examples are easier if you can actually show them

abstract nonsense is nice but not always useful to get a point across

what is 1 + 1 + 1 + 1 + 1 + 1 + 1 + …?

now of course not all infinite sums in ℝ make sense either

but at least some do. but in 𝔽₂ most don’t

[dusky epoch]

in finite fields the only infinite sums that make sense are really just ones with a tail of zeroes

[broken hawk]

also btw the reason I chose 𝔽₂ over all the other finite fields is because it has characteristic 2 and that tends to produce counterexamples like crazy

so it’s a good one to have in the back of your mind

[dull kettle]

hahaha I came here asking a few questions and am leaving with so much more knowledge

[broken hawk]

anyway I should go back to my probability homework

which I’m definitely not just avoiding rn

[dull kettle]

thank you!

[sand hedge]

stats

[muted ore]

https://www.reddit.com/r/learnmath/comments/azjwer/express_vector_of_cash_flows_as_linear/

[primal beacon]

need help w/ this question

i)

nvm

[digital hazel]

U still need help? @primal beacon

[digital hazel]

Just use the midpoint formula

The answer in the pic is correct

Nah it's ok

Just practise more

Depends on the difficulty

Usually more is better but don't overwork ur brain and have some time to take a break

Oh good luck

[hoary osprey]

work smarter not harder

[slim maple]

so i found the matrix for 1.i

[r 2s]
[s r]

but i’m not quite sure what 1.ii is asking for

[slow scroll]

Elementary column operations are rank preserving, right?

[dusky epoch]

yes

[slow scroll]

thanks!

[broken hawk]

yes, but not determinant-preserving!

(they change it in very predictable ways though)

[half ice]

Preserves whether or not the determinant is zero, but that's implied by rank preserving

[winter reef]

Could I get some help with this one?

[brittle juniper]

find 2 linearly independent eigenvectors with eigenvalue 1 and 1 vector in Ker φ, and that'll be your basis

[winter reef]

OOH

I see

thanks

Also a question: if I have transformation R4 into R4 and it has only one eigenvalue but a double one then basis of subspace of these eigenvecotrs are just 2 vecotrs? How would a matrix of this transformation in this basis look like? on first two entries of diagonal 2s and rest are zeroes?

[brittle juniper]

2 vectors can hardly make a basis of R⁴ x')

[winter reef]

Yeah that's what I thought, so if I find eigenvalues and I get only one but a double one (transfromation is in R4) then it's not possible for this to be diagonalizable?

[brittle juniper]

indeed, it's not diagonalizable

[winter reef]

wait maybe im just stupid and did a transpose, the first row in the matrix of this transforamtion should be 2,0,0,0?

Ok I did a transpose I think fuCK

2,0,0,0 should be column right?

[trail valley]

I'm not the one to ask about matrices, though vectors are nice

:\

[winter reef]

I mean how should the matrix look like

of the transformation

I think I did trnaspose instead,

[slow scroll]

@winter reef split the transposed vector up into linear combinations of xk's

[winter reef]

?

I mean how do I build a matrix of transformation phi in standard basis

[slow scroll]

ye

[winter reef]

phi(1,0,0,0) is first column or row?

column right?

[slow scroll]

what i mean is that you can split up that whole vector on the right hand side into:
x1 v1 + x2 v2 + x3 v3 + x4 v4
and use the def of matrix multiplication w/ a vector to create a matrix:
[v1 v2 v3 v4]

[winter reef]

I know

Im just studying for 10 hours

and im just not sure

of anything

so its the columns right?

thats what im asking, if I cant do that I wont do what you said as well lol im tired

[slow scroll]

the transformation here is just transposed. I usually rewrite as column vectors

[winter reef]

could you PLEASE tell me how the matrix of this transformation looks like

or draw in paint

whatver

pLEASE

[slow scroll]

thats how you do it ffr

[winter reef]

Thank you

but I still dont get it

since

it breaks my fucking problem

I mean the question doesnt make sense now I think idk

[slow scroll]

think of the column by coordinate definition of matrix w/ vector multiplcation. This is that in reverse.

[winter reef]

yeah thats what I did, I did it correctly (if you did) but this matirx now is not diagonalizable and my homework doesnt make sense now

so idk

maybe thats a bait

guys I dont udnerstand, one person said to find two eigenvectors with eigenvalue 1, but how do I do taht if its a transformations from R3 to R5? (different dimensions)

<@&286206848099549185>

[slow scroll]

ye

[placid oracle]

@slow scroll why

[slow scroll]

think about the matrix that describes a transformation from R5->R5. It is a square matrix. If the transformation is onto, then its matrix in echelon form has a pivot in every row. If a square matrix has pivot in every row, then it has a pivot in every column (meanings its injective).

If you don't want to introduce matrices, you can think about how you could construct a linear transformation from R5 -> R5 that is surjective. Any surjective transformation you can construct (mappings between basis vectors) would also have to be injective.

[placid oracle]

got it! the matrix explanation helped

[slow scroll]

np

[placid oracle]

I have a couple more questions for u if thats ok

[slow scroll]

sure

only when b=0

[placid oracle]

ah thats right it need to go tthrough the origin

[slow scroll]

true. also "a" is like the transformation of the basis vector in R. i.e. T(1) = a. Linear combinations of 1 will tell you how T affects the rest of R. b just ruins everything :p

the exception is when 2 of the linear equations are linear combinations of the other i think

if you write the 3x2 matrix that it represents, the two bottom rows should get zeroed out leaving you with only one pivot row/column.

[placid oracle]

ok ill try it one sec

do u not mean a 3x3 matrix?

[slow scroll]

you said 3 linear equations in 2 variables

[placid oracle]

yes but do u not use an augmented matrix?

x1+x2=...

[slow scroll]

oh uh. I wasn't really talking about the augmented matrix. You can do all of that if you want, but it is sufficient to work with Ax = 0, that way the augmented part is just always zero

@placid oracle wait oof that was dumb. yes ur right. its a 3x3 augmented matrix

[placid oracle]

2 zero rows

=> 2 free variables

[slow scroll]

u mean 2 zero columns -> 2 free variables
but yea

wait uh

just one free variable

[placid oracle]

what?

ok im confused

i got 2 zero rows...

[slow scroll]

yea row pivots concern existence. pivot columns concern uniqeness

there are only two columns, so there can only be one free variable here

errr uh two columns in the coefficient matrix lol

[placid oracle]

dont u need to use the third

oh no\

[slow scroll]

ur right that the third column does matter, especially when you are dealing with inconsistent equations where there is a pivot in the last row of the augmented matrix. It just wasn't too important in this case, thats why i keep forgetting about it lol

the free variable is special case. You were asking about the exception. a 3x2 matrix can certainly have a pivot in every column, so there are definitely situations where there are no free variables

Basically, the echelon form of a 3x2 matrix could look something like this:
$ \begin{pmatrix} 1&0\0&1\0&0 \end{pmatrix}$ or this $ \begin{pmatrix} 1&0\0&0\0&0 \end{pmatrix}$ @placid oracle

[stoic python]

bigboy77584:

[placid oracle]

ok, so using an augmented matrix is not necessary

[slow scroll]

It is, but im only paying attention to pivot columns. Pivot columns in the augmented matrix mean basically nothing. If you had something like this $ \begin{pmatrix} 1&0&a\0&1&0\0&0&b \end{pmatrix}$ then we get that the system is inconsistent because 0!=b. i.e. there is a pivot in the last row

[stoic python]

bigboy77584:

[placid oracle]

that there are no free vars

got it

[slow scroll]

other way around xd. this happens $ \begin{pmatrix} 1&0\0&0\0&0 \end{pmatrix}$ whenever the two bottom linear equations are linear combinations of the top

[stoic python]

bigboy77584:

[slow scroll]

no pivot in the second column = free variable

[placid oracle]

yes mb mixed it up

[slow scroll]

{0} is not linearly independent.

[placid oracle]

other than the trivial one mb, it says x is nonzero

[slow scroll]

oh i didn't see non zero. I don't think so. at least not in R^n. There might be examples in C or something tho.

[placid oracle]

what is C

and so the only thing that will determine that x'

[slow scroll]

the complex numbers. I was thinking about something wrong tho. I don't believe there are examples in C. Not sure about other fields tho 🤷

[placid oracle]

and ok, i havent really learnt muich about complex numbers since algebra 2 anyways

[slow scroll]

yea mea either. im def. not qualified to talk about complex numbers

[dusky epoch]

someone say complex numbers?

C as a real vector space is isomorphic to R^2

(and admits a rather obvious basis, too: {1, i})

a set consisting of a single nonzero vector is LI in any vector space.

[frosty pewter]

I have no idea how to get started with this, any help pls?

[dusky epoch]

it suffices to prove the thing in one direction

i.e. if Ac = 0 then Bc = 0

[slow scroll]

suppose the E1E2E3E4...En A = B, where each Ek is an elementary row operation, then
Ac = 0
E1E2E3E4...En Ac = E1E2E3E4...En 0
Bc = 0
If A can be transformed to B by elementary row operations, then the inverse of those operations take you back to A.

so like Ann said, the other direction is kinda trivial.

[frosty pewter]

okay, round 2... I was thinking about using rank(A) in some way, but that way seems kind of flawed now...

[dusky epoch]

why do you even need RREF tho

[slow scroll]

what does the subscript number j on Col_j mean?

[lean aspen]

@wintry steppe so um

linear algebra

T*T

T__T

in the real case T*T is symmetric, in the complex case it's hermitian

yeah idk what else u can do

feel as though you want to apply spectral theorem somehow tho

i don't know

[slow scroll]

@placid oracle are we talking about polynomials here?

[placid oracle]

how do I write a matrix in chat?

i want to show it to you

[slow scroll]

in latex, its \begin{pmatrix} \end{pmatrix}
put &'s between each coordinate and do \\ to start a new row of the matrix

[stoic python]

bigboy77584:
Compile Error! Click the reaction for details. (You may edit your message)

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

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

[slow scroll]

right so you need the transformation $T:\mathbb R^3 \to \mathbb R$ described by $T(\mathbf x) = \frac{a}{3} + 2b + 3c$ assuming that calculus is all right lol.

[stoic python]

bigboy77584:

[placid oracle]

it is, not sure where to go from here though

[slow scroll]

you want to separate the rhs into linear combinations of a, b, and c. and use the definition of a matrix to write [v1 v2 v3][a b c]^T

does that make sense?

[royal star]

did u forgot the word calculations and wrote calculus

[slow scroll]

nah i meant as in "if i am integrating polynomials correctly"

[royal star]

oops

[placid oracle]

@slow scroll not really...

[slow scroll]

,$ \frac{1}{3} a + 2 b + 3 c = \begin{pmatrix} \frac{1}{3} & 2 & 3 \end{pmatrix} \begin{pmatrix} a \ b\c \end{pmatrix}

[stoic python]

bigboy77584:

[slow scroll]

the standard basis for polynomials, yea

its like x^2 = (1, 0, 0), x = (0,1,0), 1 = (0, 0, 1)

[placid oracle]

how do uk thats what it is?

how did u fidn tht

[slow scroll]

its just the "standard basis" just like how we use linear combinations of (1, 0, 0) (0, 1, 0) (0, 0, 1) to describe 3 tuples.

@placid oracle err i should say the standard basis is linear combinations of x^2, x, and a constant.

[placid oracle]

oh i knew it was the standard basis, i just thouight it was diff for polynomials

ok, so what next

[slow scroll]

different how?

[stoic python]

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

[slow scroll]

\begin{pmatrix} 1&0&0\0&1&0\0&0&1 \end{pmatrix}\begin{pmatrix} a\b\c \end{pmatrix}

[stoic python]

bigboy77584:
Compile Error! Click the reaction for details. (You may edit your message)

[slow scroll]

this? what?

well thats a map from R3 to R3. An integral maps to R.

its what I had before:
https://cdn.discordapp.com/attachments/540211747613704221/554917938847678465/195237017049759744.png
a 1x3 matrix

1/3, 2, 3 are the columns i.e. the transformations of the basis vectors T(x^2) = 1/3, T(x) = 2, T(1) = 3

then I just did the reverse of the column by coordinate definition of a matrix in order to figure out the right hand side

[placid oracle]

so thats the standard matrix

do u think T is one to one or onto here?

[slow scroll]

its onto, but not one to one

[placid oracle]

why, what matrix u using to check

[slow scroll]

take a look at the picture I have above. We have
1/3 a + 2b + 3c where a, b, c can be anything we want. Is there a number we can't represent as a linear combination of 1/3, 2, and 3? Nope. Therefore the columns span all of R, and this equation is onto because it always has a solution for input (a, b, c).

Is it one to one? Well for it to be one to one, the columns of our matrix have to be linearly independent so that the solutions are unique. For what entries of (a,b,c) does 1/3 a + 2b + 3c = 0. An infinite number of entries. Therefore the columns are linearly dependent and solutions to our transformation are not unique, so its not one to one.

You can also use the usual column/row pivot argument too, just thought it seemed a bit overkill for this particular setup.

[placid oracle]

thank u mate, i get it completely !

u help me out almost everyday btw, ur amazing 😄

[slow scroll]

yea np. I'm a beginner in LA too. Answering people's questions just helps me get more comfortable, so I'm happy to help

[placid oracle]

👌

[ionic island]

is the < > notation consistent with the definition of "a cyclic group generated by"?

[dusky epoch]

<> is a notation used for different things in different contexts

[ionic island]

Take <M, v> for example

[empty copper]

Not really linear algebra I don't think

[ionic island]

I've seen the notation in linear algebra, so wondered if it's consistent with the abstract algebra definition

[empty copper]

Wouldn't say so

Angle brackets in linear algebra often denote scalar products lmao

[dusky epoch]

span

[wintry steppe]

It's C. Because that's the one with a slope of 3 that contains (3,0). Also this isn't linear algebra.

[lean aspen]

reality check pls: if $T \in \End V$ induces an eigenbasis ${l_k:1 \leq k \leq dim V}$ we have by def that
$$
T(x) = T(\sum \alpha_k l_k ) = \sum \alpha_k T(l_k) = \sum (\alpha_k \lambda_k)l_k
$$
for $\lambda_k$ the eigenvalue associated with $l_k$.

[stoic python]

flimflam:

[dusky epoch]

ye

[lean aspen]

whew

[lean aspen]

quick question

For $V=U⊕W$ does $T ∈Endo(U⊕W)$ generally refer to like
$$T = T_U + T_W∈ Endo V: T_U|{U} ∈ Endo U, T_W|{W} ∈ Endo W$$
or can it still be anything in $T ∈ Endo(V)$?

feel as though I might want like
$$T = T_U +T_W \in Endo(U) \oplus Endo(W) \subset Endo(U \oplus W) = Endo(V)$$

[stoic python]

flimflam:

[lean aspen]

@dusky epoch r u there

@jagged pendant, @lean aspen

@honest swift does this make any sense

[jagged pendant]

you do not necessarily have such a nice decompostion

[lean aspen]

well

I know that

I meant more like notation

Endo(U+W) not really meaningful then on its own I imagine?

since that's just EndoV?

[jagged pendant]

ya like the way i view this in my head is

like

(for niceness purposes V is finite dimensional)

you split the matrix of T into blocks

the nice decomposition is when the scribbles are 0

[lean aspen]

Mentally I imagine it works something like the following:
Say, $V= V_1 \oplus V_2$, we have
$$\End(V,W) \cong \End(V_1 \oplus V_2,W) \cong
({V_1 \oplus V_2})^\ast \otimes W \overset{?}\cong
(V_1^\ast \otimes W) \oplus (V_2^\ast \otimes W)$$a

then if you let V=W the matrix u posted will have a map like
$$\End(V_1) \oplus \End(V_2) \cong
(V_1^\ast \otimes V_1) \oplus (V_2^\ast \otimes V_2)$$

[stoic python]

flimflam:

[lean aspen]

tbh not even sure why (U⊕W)* would be U*⊕W*
e: (U⊕W)* = Hom(U⊕W, k) = Hom(U, k)⊕Hom(W, k)?

I mean for fdvs everything seems to check out at least

[subtle wharf]

Would these classify as proper rotation matrices?

These are rotations for 4D but I multiplied the orthogonal planes together which is why XY and ZW are in the same matrix, XZ and YW are, etc

Does this work, or am I screwing up the math since matrices aren't commutative?

[lean aspen]

first one looks okay since it's the direct sum of rotations in 2D

@jagged pendant how simultanuous diag work

[jagged pendant]

there exists a basis for which they both diagonal

consequently they will commute

[lean aspen]

why does there exist such a basis

also umm

just thonks but

if you assume you have commuting maps, their rank is equal (and thus kernel). If they're diagonalasible , then

how do you show that there will exist a basis diagonalising them simultaneously

@jagged pendant

you said you didn't get to weight spaces right

[royal fiber]

is there a "trick" to this question or is just expanding out P*P and solving the only method

[wintry steppe]

diagonalize it

and take the square root of the diagonal matrix

[wanton zenith]

I'm trying to find a linear transformation to send functions f which are continuous over the interval [a,b] to g. Is there some property I'm overlooking? Some way I can manipulate the integral?

[shadow hinge]

Do you already know it’s possible because it doesn’t seem doable to me

[broken hawk]

wait, L sends f (a function) to g(x) (a value)?

something is not defined right here

[lean aspen]

should probably be f →g, and seems like convolution

[wanton zenith]

it's a homework assignment, so im presuming it's possible, and yes, f is a function put in the transformation L sent to g(x)

[random wasp]

g(x) = Stuff( g(t-x) )

seems a bit, circular

BUT, assuming stuff is well defined

wouldn't L{ f(t) } = ∫ f(t)g(t-x)dt = g(x) be linear anyway?

[wanton spoke]

There's a problem in the definition sinxe it says it sends f from C[a,b] to C[a,b]

But g(x) is a value

So it should be g, not g(x), or it should be R instead of C[a,b] on the right

[dull kettle]

Hello guys, the two conditions for a set to be a basis for V is when the set is linearly independent and it spans across V

when checking if a set is linearly independent, assuming it is linearly independent, wouldn't the results of gaussian elimination tells us also that the set spans across V? since the linear system would be consistent

[slow scroll]

yep

[dusky epoch]

no

you can have, say, a set of two vectors in R^7 be linearly independent

but those won't span all of R^7

also it's "spans V", not "spans across V"

[wanton spoke]

If the number of vector linearly independant is equal to the dimension of the space, then linear independance is enough to say it's a basis

[slow scroll]

ok but a spanning set of vectors is consistent for all v in V

[dull kettle]

I was about to say that @wanton spoke , thanks for confirming

thank you @dusky epoch as well !

[dusky epoch]

what do you mean by "a spanning set is consistent"

[slow scroll]

if you wrote the the set of vectors in a matrix, there is a solution x that solves Ax = b for all b in the space

a set of vectors spans a space iff Ax=b is consistent for all b im pretty sure

and u can use elimination to figure out whether thats the case

[dull kettle]

When checking if a set of vectors spans a vector space, if it is verified (using gaussian elimination) that one of the vectors in the vector space is a linear combination of the set of vectors, is it safe to assume that the set of vectors spans the vector space?

since the numbers in the linear system would always be the same when doing gaussian elimination except for the results which is dependent on the vectors in the vector space

actually no, scrap that, we can't assume

[lean aspen]

how abusive notation is it to use A for a matrix and linear map if V = F^{n x m}

[broken hawk]

idk I prefer using $L_A$ for the map, or alternatively $[T]_\mathcal{B}$ for the matrix

[stoic python]

Sascha Baer:

[broken hawk]

wait, hang on, the space is the space of matrices

[lean aspen]

wait

[broken hawk]

then a linear map would be a matrix of matrices

[lean aspen]

did i mean that or n x 1

[broken hawk]

idk you said that V was the space of n×m matrices over F

I assumed you actually meant that A: Fⁿ → Fᵐ

slash A in F^{n×m}

[lean aspen]

Clarification: A in M(n,F), also A in Endo F^n

[broken hawk]

imo, it’s fine as long as it’s absolutely clear what the basis is

[lean aspen]

s t a n d a r d b a s i s

[broken hawk]

but a matrix representation always fixes a basis

I don’t really believe in the standard basis tbh

like in its existence

every basis is the standard basis if we label it that way

[lean aspen]

we typically distinguish 1_F from other units (nonzero elements)

[broken hawk]

you can embed F into any 1D subspace though

it’s all too symmetrical

[lean aspen]

such is linear algebra

does this make sense:
Write
$$A = [T]{\mathcal J}= \bigoplus{k} [T|{W_k}]{\mathcal J_k} =
\bigoplus_k J_{n_k}(\lambda_k)$$
for the jordan block decomposition $T$ with respect to jordan base $\mathcal J = \bigcup \mathcal J_k$ formed from the jordan bases of the irreducible blocks.

[stoic python]

flimflam:

[lean aspen]

@wintry steppe

@tranquil ermine

@jagged pendant

[tranquil ermine]

what

[lean aspen]

[tranquil ermine]

the fuck is this saying

[brittle juniper]

What are A and W_k and J_k?

[tranquil ermine]

yeah this is correct

you are just saying that jordan blocks are a thing

and yeah they are

[lean aspen]

yeah I haven't proven anything yet

[tranquil ermine]

looks fine

[lean aspen]

thanks

[peak cloud]

hey does anyone know good programs for linear algebra/multivariable calculus functions?

[lean aspen]

does eigenvector mean generalized eigenvector or eigenvector

😰

like $$A := \begin{pmatrix} \lambda & 1 \ 0 & \lambda\end{pmatrix}$$
has eigenvector $(1,0)^T$, and generalized eigenvector $(1,1)^T$right?
e: also eigenvalue λ & generalized eigenvalues λ

but the latter isn't an "eigenvector"since
$$(\lambda I - A)(1,1)^T \neq (0,0)^T$$

[stoic python]

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

[lean aspen]

@wintry steppe

[winter sage]

is there any special properties about a transpose of a matrix multiplied by a matrix

then taking the inverse of that matrix

like whats the relationship of the inverse of that matrix and the inverse of the original matrix

[half ice]

Both the inverse and the transpose are order reversing.
(AB)' = B'A'
(AB)¯¹ = B¯¹A¯¹

[winter sage]

my question pretty garbled it might be easier if u have the questions for context

[half ice]

K¯¹
= (A'A)¯¹
= A¯¹(A')¯¹
= A¯¹(A¯¹)'

So that's ez if you already know A¯¹

[winter sage]

what does ' mean

'

[half ice]

' being transpose

[winter sage]

ah i c

tysm!

[half ice]

Np! Feel free to ask if you have anything else!

[winter sage]

ok sry for asking again so soon

but the solution has an 85 in the bottom right and i have no idea where that comes from

im not getting it from A-1(A-1)'

[half ice]

Wat

85 is a number, K¯¹ is a matrix

[winter sage]

this is what the inverse of K looks like

but that is nowhere near A^-1(A^-1)'

oh crap

i solved A-1 wrong lmao

nvm nvm yikes

[winter sage]

how does B span all of R2?

idk how 3 coordinates can fit in 2 dimensions

[slow scroll]

you only need two linearly independent vectors to span R2 so....

[winter sage]

but for B there's only 1 and 2 in XY

so thats only covers a line?

[slow scroll]

the third vector has a y coordinate, so it works out. In fact, you only need 1 of the first 2 columns and the third column to get a spanning set

[winter sage]

dam im actually a bumbling tard i was looking at it like (1,2,3) and (0,0,4)

😬

[slow scroll]

ah alright.

[winter sage]

ty

[slow scroll]

np

[flat pasture]

can someone help me on 37

why is the anwwer clockwise rotation of theta = pi/2

<@&286206848099549185>

[dusky epoch]

surely you mean counterclockwise rotation by π/2

[flat pasture]

yes why tho

i did a times x1 and x2

[dusky epoch]

try considering what happens to the standard basis vectors

[1;0] and [0;1]

[flat pasture]

I did Ax and got [-x^2; x1]

but i dont know anything past that

[dusky epoch]

try considering what happens to the standard basis vectors

what happens to [1;0]? where does it get sent?

[flat pasture]

sorry i dont gt wat that means

[wintry steppe]

the unit vectors in the x axis and y axis

[dusky epoch]

what is A*[1;0]?

[flat pasture]

0 1

[wintry steppe]

and what is A * [0;1]

[flat pasture]

[-1;0]

[wintry steppe]

right

[dusky epoch]

so do you see now

[wintry steppe]

so would you agree that the basis vectors are rotated 90 deg counterclockwise

[dusky epoch]

they both got rotated a quarter turn ccw

[flat pasture]

ohhhh

[dusky epoch]

in general
the matrix [cos(θ) -sin(θ); sin(θ) cos(θ)] rotates ccw by an angle of θ

[wintry steppe]

owo

[flat pasture]

what is this called? Linear map rotations?

[wintry steppe]

rotational matrix

[flat pasture]

ok

[wintry steppe]

i mean the matrix that ann posted is called that

[flat pasture]

sorry for asking dumb questions this is a quick intro to linear algebra unit

[wintry steppe]

it wasnt a dumb q

also

you can read off what happens to the basis vectors by looking at the columns of that matrix

the 1st column is [0;1]

so i gets sent to 0,1

[dusky epoch]

i mean

that's the definition of the matrix of a linear map lol

[wintry steppe]

lol

[flat pasture]

is it okay if you walk through 39

[wintry steppe]

multiply through the 1/2

like take the 1/2 inside the matrix

[flat pasture]

ok

i saw on the internet that [0 -1; 1 0] is 90 ccw, [-1 0; 0 1] is 180 ccw and [0 1; -1 0] is 270 ccw

[wintry steppe]

that's right

[flat pasture]

are we just applying this info

[wintry steppe]

...kinda? not really those specific matrices but did you see the matrix that ann posted

[cos(θ) -sin(θ); sin(θ) cos(θ)]

[flat pasture]

yea and when u half it then it looks like root3 over 2

which is pi/6 right?

[wintry steppe]

exactly

rt3/2 is sin(pi/6), yes

but we are looking for something else

...cos, since the first entry of that matrix is cos theta

[flat pasture]

[cos pi/6 -sinpi/6;

sinpi/6 cos pi/6}

[wintry steppe]

oop

yeah you're right lol

i thought sin(pi/6) = rt3/2 for a sec

[flat pasture]

i mix up the unit circle all the time haha

[wintry steppe]

so the linear map would just be a rotation by pi/6 rad

[flat pasture]

these seem to always be ccw

do i need to worry ab direction?

[wintry steppe]

not really

the rotational matrix is usually defined for rotation by theta in ccw direction

[flat pasture]

ok thx

[dusky epoch]

rt3/2 is sin(pi/6), yes

cos(π/6)

[wintry steppe]

i corrected myself afterwards

[dusky epoch]

@flat pasture cw is just ccw but negative

[flat pasture]

thank u so much

[sour garden]

If dot product is -1

Does that tell you anything about linear Independence

Example:

(1,-1,0)•(0,1,-1)=-1

But they are linearly independent

[dusky epoch]

dot product being -1 tells you absolutely nothing

for any unit vector u, the dot product of u with -u is -1

[wanton spoke]

Dot product tells you if two vectors are orthonormal or if two vectors are in the same sense

To test colinearity there's cross product instead

[neat jolt]

https://i.imgur.com/xhyn2tk.png

how do i know if that transformation im supposed to find is in the span

[wintry steppe]

I have to use cramers rule for this, but I can't use an inverse matrix so I don't know how to approach this question.

I think you'd find the adjacency matrix then use that to find the determinant and then use that to find the inverse, but you can't use an inverse matrix wtf

[half ice]

Do you know the statement for cramer's rule? You only need to calculate determinants @wintry steppe

[wintry steppe]

Yeah I figured it out after looking back at the rule. Thanks tho @half ice

[hazy fiber]

Fast question...

A subspace basis can be formed with just one vector?

[dusky epoch]

if the subspace is one-dimensional, why not

[hazy fiber]

Okay 😉

[broken hawk]

(in fact, a subspace being one-dimensional is defined as having a basis formed by one vector)

[sand hedge]

Cramers rule smh

[lean aspen]

my favorite subspace are 1D

[rocky hill]

quick question

let H, K be subspaces of V

show H intersect K is a subspace too

is it enough to simply say that since both H and K satisfy the properties needed for being subspaces of V, then any vector both in H and K also satisfy those properties, or do I very explicitly need to show all three properties?

[dusky epoch]

those properties are not of the form "for all v in H, v satisfies <property not involving H>", so no, you cannot do that

[rocky hill]

ugh

lol

Wait, why? H and K are subspaces, so all vectors in H and K satisfy the properties. Why is that not enough to imply that all vectors in both H and K satisfy the properties too? I don't understand.

[half ice]

Note that the properties of a subspace don't belong to each vector of that subspace, but instead belongs to the space as a whole @rocky hill

If you chop a vector out of a vector space, it will usually (pretty much always) cease to be a space. So it's pretty surprising that the intersection of two spaces is also a space, as that's usually not the case with other structures

[rocky hill]

mmm

[sand hedge]

[rocky hill]

Note that the properties of a subspace don't belong to each vector of that subspace, isn't this contradicting the definition of a subspace though? The definition is for all vectors in the subspace, we are closed under addition, scalar multiplication, and we have the zero vector

...right?

[broken hawk]

basically you just have to spell it out more explicitly

[rocky hill]

so then all vectors in the subspace have those properties...?

[half ice]

It's a property that these vectors have because the space is the way it is

[broken hawk]

like, you just have to argue rigorously why $v,w \in H \cap K$ implies $v+w \in H \cap K$, and you’re absolutely right that it is not hard

[stoic python]

Sascha Baer:

[broken hawk]

a few properties you can indeed take as induced

e.g. addition being associative and commutative

[rocky hill]

yeah it's just tedious and inelegant lol

[broken hawk]

but the closedness you have to show

[rocky hill]

(in my opinion)

[broken hawk]

anything on the operations is induced. closedness you have to show

which is like, one line each, but you must actually do that

[half ice]

It is tedious and ugly lol. That's common in "prove this has this structure" questions. Thing is, the result is beautiful and widely applicable

[rocky hill]

so something like, Let $u, v \in H \cap K$ then $u, v \in H$ and $u, v \in K$, then show closedness?

[stoic python]

blorgon:

[broken hawk]

first of all

do you hvae the statement on which things you need to prove that a subset is a subspace?

there are three things you have to check

there’s a general theorem about this

[rocky hill]

0 vector, closed under addition, closed under scalar multiplication

[broken hawk]

yep

each should take you no more than one or two lines

if you want I can spoil it and show one of the arguments

the other runs exactly the same way

[rocky hill]

so if $u+v \in K$ and $u+v \in H$, that implies $u+v \in H \cap K$?

[stoic python]

blorgon:

[broken hawk]

yes

[rocky hill]

that seems tenuous

[broken hawk]

I mean that’s the whole proof, really

it’s one line each

and then it’s done and you never ahve to do it again

[rocky hill]

ye, it just feels.... I dunno. I'm still not used to proving stuff. It all feels hand wavey to me to a degree

lol

[broken hawk]

this is exactly why you hvae to do it

[rocky hill]

fair enough

[broken hawk]

these are easy statements that you can do rigorously on your own

so it prepares you for much harder ones

what was handwavey is what you did before

the full, rigorous argument would run like this

[rocky hill]

fair enough

[broken hawk]

Let $v, w \in H \cap K$ be arbitrary. Since $v, w \in H$, $v + w \in H$ because $H$ is closed under addition. analogously, $v, w \in K$ and thus $v + w \in K$.
Thus $v+w$ is both in $H$ and $K$ and thus in $H \cap K$. Therefore $H \cap K$ is closed under addition.

[stoic python]

Sascha Baer:

[rocky hill]

👍

[broken hawk]

note that I’m implicitly using the definition $H \cap K = {v \in V: v \in H \text{ and } v \in K}$

[stoic python]

Sascha Baer:

[broken hawk]

I did not feel the need to spell that out

but of course it’s important that that’s the definition

[rocky hill]

since you're here, can I get a little hint on how to get started on detAB = detA detB

[broken hawk]

wow how did we suddenly get from elementary subspace stuff to determinants

which are way more complicated

[rocky hill]

lol

well the subspace one is for HW, but the determinant one was left as an extra thing my prof told us to try

[broken hawk]

uh, first of all, what is your definition of the determinant

cause there’s a few

[rocky hill]

the area of a region after a transformation 😄

[dusky epoch]

idk why having the 0 vector is listed as part of the definition of a subspace tbh it follows from the other two properties

[broken hawk]

okay, that makes it a lot easier than any of the formal definitions then

[rocky hill]

I don't know any formal definitions

I just know the mechanics of taking determinants

[broken hawk]

they’re a lot nastier

I mean you can define the determinant via one of the algorithms for computing it

[rocky hill]

can detAB = detAdetB be done directly? or does it need induction or something?

[broken hawk]

or, what we did, as the unique function $det: M_{n \times n}(\mathbb{F}) \to \mathbb{F}$ which is multilinear, alternating and fulfills $det(I_n) = 1$

[stoic python]

Sascha Baer:

[broken hawk]

this is extremely technical, of course

uh, so, I think you can probably just make a geometric argument based on your definition

but I can’t think of a good one off the top of my head

basically you wanna argue that:
if you first transform the unit cube with B, and then the resulting parallelopiped with A, then the area change is the same as teh product of the area changes of A on the unit cube times B on the unit cube

[rocky hill]

ye

[broken hawk]

for that I would probably employ the fact that shearing leaves area invariant, so you can move the parallelopiped after A back to some cuboid first

I don’t remember at all how we proved this ourselves tbh

I think we showed two things

1. degenerate matrices have determinant 0

2. the statement holds for elementary matrices

if one of the two matrices is degenerate, then their product is too, so the 0 case follows from 1

if neither are, then they’re a product of elementary matrices

for elementary matrices our formal definition worked particularly well, of course

[rocky hill]

gotcha. But yeah the geometric argument doesn't feel too rigorous lol

[broken hawk]

because it isn’t, really

I mean you can make it rigorous, I‘m sure of it

but I don’t think I could

not without a bunch of thinking at least

[rocky hill]

oh no, thinking!

lol

[broken hawk]

it’s the worst, innit

[rocky hill]

mmk thanks for the tips

cheers

[broken hawk]

I think it makes intuitive sense at least

which is always nice

[wanton spoke]

Best way to show it is probably to write both A =(A1,A2,...,An) and B similarly with Ai, Bi column vector then use the fact det is multilinear and alternate

Or wait, this doesnkt seem a good option my bad

[winter reef]

So let phi be an endomorphism of V and let W be a subspace of V Such that for any alpha in W phi(alpha) is in W. How do I show, that if psi is transformation from W to W, then charachteristic polynomial of psi divides char. polynomial phi?

I tried to think that if it didnt then there would be an eigenvalue that is in W but not in V and somehow Get to cobtradicition but not sure if thats the way

[brittle juniper]

Are you working in finite dimension?

@winter reef

[winter reef]

Yes

[brittle juniper]

yay

[winter reef]

So like it looks pretty obvious but not sure how formally do I justify it

Probly has to do with Michael Jordan matrix

[brittle juniper]

Are you sure you've written all the info?
if V=R³, W=span((1,0,0),(0,1,0)), phi=id and psi=some rotation of W with no real eigenvalue, it doesn't work though all the requirements are filled

[winter reef]

Wait, psi=phi|W (cut to W? Not sure how to call it) = W- >W

[brittle juniper]

ah, the restriction of phi on W

okay then you can choose a good basis such that the matrix of phi will have a big block for what happens in W, and you can show that big block is the matrix of psi

[winter reef]

ok, thanks.

[brittle juniper]

something like that

this took way too long to make

$\text{Mat}(\varphi)=\left[\begin{array}{ccc|ccc}
&&&&\
&\text{Mat}(\psi)&&&A\
&&&&&\
\hline
&&&&&
\
&0&&&B&\
&&&&&
\end{array}\right] $

[stoic python]

Tuong:

[sand hedge]

wat dat

[slow scroll]

Ok this makes sense to me but I have no idea how to actually write the proof. Because V is a subspace of X, dimV <= dimX. And if we have some transformation B: V -> Y, "there is no way it can map to a space of dimension higher than dimV." The part in quotations is the part I don't really know how to prove. Any ideas?

[half ice]

Let Vsp be a spanning set for V. Then AVsp is a spanning set for V after transformation.

|Vsp| ≥ dim(V) ≥ dim(AVsp)

As a spanning set always transforms to a spanning set, but a basis may not be a basis

Which makes algebraic what you were trying to say, you can't transform into a higher dimension

@slow scroll

[winter reef]

Wtf tuong thats some PhD Latex level

[slow scroll]

you said "...then AVsp is a spanning set for V after transformation." AVsp doesn't necessarily span V tho, right? It could span some space of lower dimension?

[half ice]

@slow scroll
Sorry I mixed up what I was trying to say, AVsp is a spanning set for AV

Note that this can still span a space of lower dimension

A set of 4 vectors can span a 2D space

[slow scroll]

hmm i need to think for a bit

[broken girder]

help

[dusky epoch]

@broken girder what have you tried so far and where are you stuck?

[broken girder]

Haven’t really tried anything but I think it’s about quadratic forms right?

[dusky epoch]

do you know what it means for a quadratic form to be positive-definite?

do you know the definition?

what's giving you trouble here?

[little cairn]

Can anyone here think of interesting projects or potential research papers involving adjacency matrices? Perhaps their usefulness in computing?

[broken girder]

@dusky epoch positive-definite quadratic form means it always outputs a number ≥ 0

[dusky epoch]

no.

that's positive-semidefinite.

[broken girder]

oh my bad. its > 0 for any non-zero vector

[dusky epoch]

yes... so what's the problem exactly

you've now correctly stated the definition

and surely you've proved "if and only if" statements before?

[broken girder]

yup

so in this question, we assume A to be a complex matrix right?

is <x,x> = x^H.B.x a quadratic form?

my textbook only talks about quadratic forms in the form of x^T.A.x, with real matrices A only

[dusky epoch]

<x, y> = y^H B x is a sesquilinear form on C^n

and you can prove that even though A is a complex matrix that may very well not be real, x^H B x is a real number for all x

[broken girder]

i see

so when we say a complex matrix is non-singular, it means that the determinant is any complex number but 0 right?

[dusky epoch]

the definition of "non-singular" is field-independent

a matrix over a field is non-singular iff its determinant does not equal the field's zero

[broken hawk]

however, that doesn’t make layovah’s statement false

[dusky epoch]

true, it is not false. just... poorly worded

[broken hawk]

why? the determinant of a complex matrix is a complex number

[broken girder]

B = A^H A .. the matrix B is hermitian correct?

[broken hawk]

yes, prove it

(easy proof)

[dusky epoch]

@broken hawk the wording makes it sound like the statement is true only in the case when the base field is C

maybe i'm reading too much into it

[broken hawk]

that’s also true though :P
if the matrix is over 𝔽₂, then the determinant of a nonsingular matrix cannot be a complex number :P

[dusky epoch]

"that" = what?

[broken girder]

how do i do matrix (or matrix-vector) multiplication if the matrix is complex, and the vector is complex? same strategy for the real case right?

[dusky epoch]

matrix multiplication is field-independent!

[broken hawk]

it’s all defined exactly the same way

the only thing that changes is the inner product

[broken girder]

ok

fuck

[broken hawk]

which is now sesquilinear instead of bilinear

[broken girder]

no wait so in real case, the 'standard inner product' we use is the dot product. so for complex case, do we use the standard inner product for C^n ?

[broken hawk]

but any matrix operations work the same way no matter what field (real numbers, complex numbers, finite fields…) the entries come from

[dusky epoch]

that's what i said

[broken hawk]

the standard inner product for ℂⁿ is the same as that for ℝⁿ, except that one of the two vectors (conventions differ) has to be conjugated

[broken girder]

ok yeah so it is pretty different

[broken hawk]

not really

[dusky epoch]

why do you even need the standard inner product tho

[broken girder]

im trying to expand out x^H B x

[dusky epoch]

lmao what

[broken girder]

component-wise

[broken hawk]

don’t do that

[dusky epoch]

like explicitly entrywise?

yeah... waste of time lol

[broken girder]

ok what do i do

[broken hawk]

write B as AᴴA

[dusky epoch]

use the definition of B

[broken hawk]

and group things in a clever way

[dusky epoch]

B = A^H A

[broken hawk]

then use some fundamental properties of ᴴ

and of the inner product

[broken girder]

guys my teacher literally didnt teach us any of this

[dusky epoch]

$x^H A^H A x$

[stoic python]

Ann:

[broken girder]

so i have to self-learn

[broken hawk]

(you’re trying to show positive definiteness, right?)

[dusky epoch]

they are yes

[broken girder]

i dunno any properties of A^H

[broken hawk]

yea, so here’s a hint:

not of Aᴴ

of the operation ᴴ

[dusky epoch]

if you need to, unfold ^H into its definition

[broken hawk]

here’s two hints (together they essentailly solve the exercise):

[dusky epoch]

just a side note how the fuck is there a Unicode version of ^H?

[broken hawk]

1. ||you want to try and bring this into a form that’s the inner product of a vector with itself||
2. ||xᴴAᴴ = (Ax)ᴴ||

dunno but it’s on my keyboard so I’ll use it :P

[dusky epoch]

what is your kb layout

[broken hawk]

Neo2

[wintry steppe]

lol

[broken hawk]

in QWERTZ mode on my laptop

but in normal mode on my desktop

this somehow is not confusing

[dusky epoch]

oh.

got a friend who uses it as well hm

as it happens to be, he is German

[broken hawk]

I have one issue with it

[dusky epoch]

and that is?

[broken hawk]

which is that the combination compose+d+l should be retroflex d, but they accidentally mapped it to not only that but also to ㎗, and the latter takes precedence

[broken girder]

the question is to show that it satisfies the positive-definite postulate iff A is non-singular

so far i've got: x^H.B.x = x^H.A^H.A.x = (Ax)^H.Ax

[dusky epoch]

yes

so now

do you know the general strategy for proving an IFF statement?

[broken girder]

yup. show it in both ways

[dusky epoch]

okay so well

can you show that if A is non-singular and x is not zero then (Ax)^H (Ax) is positive?

once you're done with that, can you prove that if (Ax)^H (Ax) is positive for all x nonzero, then A is non-singular?

[broken hawk]

is that . you have there intended to be matrix multiplication?

[broken girder]

yup

[broken hawk]

okay, then it makes more sense

I thought it would be dot product, in which case the notation wouldn’t have made sense

so, (Ax)ᴴ.(Ax) is the inner product of Ax with itself, right?

[broken girder]

oh yeah

[broken hawk]

and why is A required to be nonsingular?

[broken girder]

so if A is non-singular, then Ax would be non-zero for non-zero vectors x right?

[broken hawk]

yes!

[broken girder]

how do u know that this is true in the complex case?

[broken hawk]

the sensible definition of non-singular is “has trivial kernel”

that is, only 0 is in the kernel

[broken girder]

i guess the invertible matrix theorem is field-independent?

[dusky epoch]

yes, it is

[broken girder]

eh i got an appt. for smthn else, ill be back! thx for the help guys

[dusky epoch]

if the proof of a theorem in linear algebra does not use any properties specific to a particular field, then the theorem is field-independent

[broken hawk]

most of linear algebra is, tbh

I remember there were a few things that required the field to have characteristic other than 2

and the whole topic of inner products is a bit special

similarly, diagonalization stuff is different in algebraically closed fields (e.g. ℂ) than ones which aren’t (e.g. ℝ)

like, in ℂ every matrix has a jordan normal form, but this requires the fact that every matrix in ℂ has eigenvalues, which requires that ℂ is algebraically closed

[sharp lodge]

Where did this proof go wrong?
rankAB
(By the rank nullity theorem)
=n-nul(AB)
(If q_C is an orthogonal projection onto kerC, then nulC=rank(q_C))
=n-rank(q_(AB))
(Projection onto V and then U is just projection onto the intersection)
=n-rank(q_A q_B)
( rankAB≤rankA )
≥n-rank(q_A)
Which is false.

q_(AB) isn't q_A q_B

Hey there

[jagged pendant]

hey

[sharp lodge]

Why is my brain telling me they're equal. Basically the issue is that kerAB≠)kerA intersection kerB)

i think right?

[jagged pendant]

well

Ker A and Ker B may not even intersect

so that's one indication

like

A could be W→U
B could be V→W
then AB: V→U

ker(AB) is a subset of V

but ker(A) is a subset of W

so it doesn't make sense to take their intersection

(assuming no relationship between V,W,U)

(other than being over the same field)

[sharp lodge]

I'm considering they're all V-> V

Basically proj_KerAB is in KerB.
But they're not equal. Some of left are not in right. And the exceptions are not necessarily in kerA. So if B sends some v that is in cokerB and kerA, to cokerA then: The composition of the right projections won't be equal to the left projection. Which is enough to mess with dimensions and mess up the proof.
By cokerC i just mean some complement of kerC. Cant remember how is actually defined.

Ok, thank you!

[sharp lodge]

Go ahead

[random hollow]

https://i.mlgimg.xyz/9272537c.png That's probably very basic for you guys, I'm trying to understand the situation: We do have a ship that has a velocity of (1,2) and we do have a current flowing, I guess it's water, who add a velocity of (1,1). So the final velocity for me would be the result of the addition of those two vector ? Where I am wrong, I can't understand.

the possible answer are https://i.mlgimg.xyz/1e3b7e30.png

they don't make any sense for me

[random wasp]

@random hollow

[random hollow]

ahhhhhh vector projection

[random wasp]

jup

[random hollow]

but why, jesus maybe my english is just bad

[random wasp]

you can sorta check your answer via the diagram 😛

[random hollow]

In fact I can't understand the meaning of this value in this case, what does that represent ? I was imagining a ship moving, so it's would be the addition of the vector. Now that I understood it was vector projection that I had to use, I know how to do it, but I don't understand how you knew the question was looking for that, what does that vector represent ?

OR maybe

this value is the actual velocity of the ship before the current velocity was applied ?

so like if the current was (0, 0)

[random wasp]

could totally be wrong

from what I've read, it's asking you how much of the ships velocity is going in the direction of the water current

nothing about the effect of the water current

you only need to care about the direction

[random hollow]

how much of the ships velocity is going in the direction of the water current ok that actually make a lot of sense

big thx @random wasp

[random wasp]

np

[sharp lodge]

Another question
If v is an eigen vector of e^A is it an eigen vector of A?

Quasi proof:
If v is an eigen vector of e^A then by (magic) v must be an eigen vector of e^(tA). In that case by (magic2) the eigen value of v for e^(tA) must be e^(ct)v, where e^c is the eigen value of e^(A).
Now if you take the derivative of e^(tA)v=e^(ct), and evaluate at t=0. You obtain the desired result.

But what are the two magic steps? I guess JNF or even Cauchy's functional equation help but this proof feels so complicated.

[sharp lodge]

):

[wintry steppe]

@random hollow hey that’s the mathematics for ml coursera thing 😄

[timid ivy]

@sharp lodge This problem reduces to showing that e^A and A have at least one equal Eigenspace, doesn't it?

[sharp lodge]

@timid ivy Yes, then induction all the way. If I could write A=ln(e^A) as a power series i guess that could work too.

[random hollow]

@wintry steppe yes you're right, the course from london imperial college

[timid ivy]

Hm, I know that if $T$ is an invertible linear transform then $T$ and $T^{-1}$ have the same eigenvalues and the same eigenspaces

[stoic python]

markus:

[timid ivy]

I just mean e^x and ln(x) are inverses of each other but problem is none of them are linear

So that wouldn’t help

Nice question tho @sharp lodge

[dusky epoch]

T and T^-1 do not have the same eigenvalues lmao

[timid ivy]

No, if T has $\lambda$ as an eigenvalue then $T^{-1}$ has $\frac1\lambda$

[stoic python]

markus:

[timid ivy]

If I remember correctly. Thank you for correcting that anyway

[sharp lodge]

Oh, I don't mind if they are linear or not. Say p(C) is polynomial (series) of the matrix C. Then if (c,v) is an eigen pair of C then p(C)v=p(c)v.
So if (c,v) is eig pair of e^A, then: ln(e^A)v=ln(c)v. Now if god exists we'd be able to say that ln(e^A)=A, but he doesn't so it gets a bit dicey I gues

@timid ivy

[ebon cloak]

v and u are vectors and A is an nxn matrix

but what happened here

[timid ivy]

$v^TA = v^T(A^T)^T = (A^Tv)^T$

[stoic python]

markus:

[timid ivy]

I see @sharp lodge

I dont know but maybe it has somethig to do with Cayley-Hamilton?

[viscid apex]

let $v_{1},...,v_{n-1},u,w$ be vectors on $\mathbb{R}^{n}$.
assume $x_{1}v_{1}+ ...+x_{n-1}v_{n-1} = u$ has one solution, and
$x_{1}v_{1}+...+x_{n-1}v_{n-1} = w$ has none.
prove that ${v_{1},...,v_{n-1},w}$ is a base on $\mathbb{R}^{n}$

i thought i solved it right and now i think im walking in circles and not sure

i can show all vectors are different thru the first tidbit of info and then i tried to go through showing canonical representation of both equations but it all bungled up and im not sure anymore

[dusky epoch]

was the second equation meant to have w and not u on the RHS

[stoic python]

kakamaika:

[viscid apex]

yes sorry

[dusky epoch]

btw minor language note: vectors in R^n, and {...} is a basis for R^n

[viscid apex]

alright thanks not my first language

[dusky epoch]

what is your first language actually

if you don't mind sharing

[viscid apex]

hebrew

[dusky epoch]

...ok nevermind i don't speak hebrew 😛

[viscid apex]

we say "on" and not in

[dusky epoch]

alright so

[viscid apex]

actually we dont i mistranslated nvm

[dusky epoch]

consider the span of ${v_1, ..., v_{n-1}}$

[stoic python]

Ann:

[dusky epoch]

and in particular consider what its dimension is

[viscid apex]

dimension as in in what field does it live?

or its matrix