linear-algebra

[dusky epoch]

how can you tell whether or not the system Ax = 0 has nontrivial solutions based on the matrix A?

[young plinth]

I guess by setting it equal to zero that's what I tried and it said the answer was right. I had to get it out of matrix form to do that. That sounds counter intuitive though I don't know why you would set it equal to zero for non zero solutions.

[dusky epoch]

ok so the thing here is

doing it your way

you did algebra all the way up to z(k-4) = 0

and then you realize

if k is anything but 4, then you can divide both sides by (k-4) and end up having only z=0 as a solution to this equation

but if k is 4, then the equation becomes 0=0 and is satisfied by all real values of z

[young plinth]

Oh that make sense.

How would you solve this problem?

Is there a better way?

Is there a way by making it rref

[dusky epoch]

well you could rref the matrix and then require its last row to be zero

[young plinth]

I don't even know how to make that rref lol I find it easier to just convert it back into its system of equations

[wet stratus]

it's the same steps as for any matrix. just now with one variable

for example in the first step you would subtract k times the first row from the third row

[young plinth]

What do you mean for any matrix? Don't you take different routes for rref-ing matrices or is there a universal way for it?

I will have to check it out later got other classes to do

[wet stratus]

well the algorithm is always the same

the numbers may be different

[little vigil]

Could somebody elaborate a little bit on what this means? I've only done a first basic linear algebra course and none of the spaces i'm familiar with dealt with "matrices" as their spaces, rather all the spaces I'm used to were "built up" by vectors.

[wintry steppe]

a real vector space is a set of things that can be added and multiplied by real numbers

in "abstract" linear algebra, a vector is defined as an element of a real vector space

[little vigil]

So as long as we can multiply something, and add too it, with real numbers, it can build a space?

[wintry steppe]

read the definition of a vector space

[little vigil]

Okay

[gray dust]

the operations of vector addition/scaling must satisfy some rules, laid out in the definition

[little vigil]

As long as ,something a matrix in this case, satisfies the rules of a field it can be (or is?) a vector space

[little vigil]

Ok this is just the definition then, anything on which you can apply these fundamental rules, is a vector space?

[gray dust]

bad wording, a field contains what we call scalars, the set of reals R is an example of a field

M_nm, together with sensible vector addition/scaling rules (entrywise), is a vector space

[little vigil]

Is there some way to intuitively understand what it means for matrices to be a vector space? With three independent vectors I can easily "see" how that would construct some sort of space, but a matrix or multiple, don't feel so easy to visualize. Or is this rather something I'm just to accept as it obviously meets the definition of what a vector space is?

[wintry steppe]

list out all of the entries in one column instead of a grid

[little vigil]

That's true. Did not think about that.

That clears it up a bit. I did not consider that M_nm includes the "intuitive case", so it is not so different from what I'm used to after all. Thank you for the help

[oak stratus]

hey could someone help me with some linear algebra homework?

[wintry steppe]

if you have a question just ask

[tulip cedar]

"Non-zero" solution refers to the vector x, which has to be non-zero and yet be a solution to Ax=0. It's obvious that a system Ax=0 always has trivial solution A = 0 (A is the zero matrix) and x = 0 (x is the zero vector)
Here's criteria for when such non-zero x exist for matrix A:
If it's a square matrix, then it has to be non-singular for Ax=0 to have only trivial solution (i. e. if A is singular, then there exist non-trivial solutions)
If A has more rows, than columns, then having linearly independent columns (full column rank) ensures that Ax=0 has only trivial solution, and vice versa for when there's more columns than rows

Look up the definition of a kernel of linear operator, it's a set of all x for which Ax=0 is satisfied. Essentially in this problem you're asked to find such values of k, so that matrix turns singular and has non-trivial (aka non-zero x) solutions to Ax=0

[wintry glen]

Hey, can I get a confirmation for parts (i) and (j)? For (j), I believe that the first 3 entries determine the entire sequence, so I think {(1, 0, 0, 2, 0, 0, ...), (0, 1, 0, 0, 2, 0, ...), (0, 0, 1, 0, 0, 2, ...)} is a basis for W and so dim(W) = 3. For (i), I think every polynomial in W can be written as a_1(x - z) + a_2(x - z)^2 + ... + a_n(x - z)^n, so {(x - z), (x - z)^2, ..., (x - z)^n} is a basis for W and dim(W) = n. Are these right?

[dusky epoch]

what's the goal here?

to verify that these are subspaces, or to find their dimension?

[wintry glen]

Ah, sorry. The goal is to find a basis and the dimension for each of the subspaces W.

[dusky epoch]

aha. ok

in that case, yes, you are correct

[wintry glen]

Ah thanks.

[urban egret]

sorry maybe i should have sent my question here:

Hi everyone! I just need some help conceptually understanding something about vector matrix multiplication! In class, to figure out how much a matrix A acts in a certain vector direction n, we did nAn. Why is it like this and not just An? I get that one is a scalar resultant while the other a vector resultant, but I still cannot conceptually understand what we did here. I would really appreciate it if someone could explain! 🙂 --------For further context, we were trying to calculate strain in the direction n. So A was the strain tensor

[wet stratus]

When we are calculating nAn we are calculating the dot product of n and An. The dot product in a sense compares the direction of two vectors, if the dot product is 0 then the vectors are orthogonal and if the dot product is maximal (that is if you fix one of the vectors and consider the function x-> xu and check the maximum of that function for all x with the same length) then u and x go into the same direction. If it's minimal then x and u go into opposite directions. I assume this is why you use the dot product nAn and not just An. But I don't know the context of strain/strain tensor so maybe there is a different reason

[urban egret]

Ahhh righttt I didn't think of it that wayyy

That makes a lot of sense

I was a bit confused because a matrix doesn't really represent a direction

[serene solstice]

How do we solve this?

[uneven raptor]

I guess find the augmented matrix and row reduce

[tranquil steeple]

Set up the linear system (get expression for x1 and xn from first and last expression) then find an eigenvector x for which Ax=x that is eigenvalue 1 (just tested and it seems there are two eigenvalues that are 1) There is probably a nice analytical expression (probably some nice sine-expression since this is pretty much the Laplacian)

[latent hawk]

Matrix A satisfies the condition A^2 = A. What value should t be so that the matrix A - tE does not have an inverse matrix?

Can somebody give me a hint?

[whole peak]

t=0?

[latent hawk]

the answer key says that and also t = 1

But there is no explanation

[wet stratus]

can you tell me something about the minimal polynomial of A?

and the possible eigenvalues of A ?

[latent hawk]

No information about that whatsoever

Also, this is in the first chapter of the book

[wet stratus]

the condition is equivalent to A^2-A=0

[latent hawk]

Oh, is this a cayley hamilton problem?

hmm

[wet stratus]

I guess if you wanna call it that

[latent hawk]

So, a+d = 1 and ad -bc = 0

[wet stratus]

oh we also know that A is 2x2 ?

[latent hawk]

oh, yeah sorry

That was a given condition

forgot about that

[wet stratus]

then yeah you can solve it just by setting up equations for a,b,c,d

well ad-bc doesn't have to be 0

for example A=I the identity

[latent hawk]

oh yeah, true

Still stuck 😦

[tulip cedar]

you should apply the $A^{2} = A$ to $A-tI$

[stoic python]

Transparent_Elemental

[tulip cedar]

then you simply factor it and take an inverse explicitly

[latent hawk]

Factor $A^2 - tI$ ?

[stoic python]

Kinuly

[tulip cedar]

no, notice that if $A^{2} = A$, then if $A-tI = B$ you can left multiply the $A-tI = B$ by $A$

[stoic python]

Transparent_Elemental

[latent hawk]

So, A^2 - At = AB
meaning, A - At = AB
meaning At = tI?

But what does B have to do with anything about the inverse?

[tulip cedar]

$A-At = AB$ is correct, but factor it like this $A(1-t) = AB$

[stoic python]

Transparent_Elemental

[tulip cedar]

B is just notation trick to have an equation, we don't really care about what B is

now take explicitly inverse of both sides and show that the inverse on the left doesn't exist for some t (notice that on the left it's a matrix multiplied by a number)

[latent hawk]

Sorry, still confused.

I've solved everything else in the problem set, this one has been making me stumble close to an hour 😓

[tulip cedar]

so we have that $A(1-t) = AB$, explicitly taking the inverse we have $[A(1-t)]^{-1} = (AB)^{-1} \rightarrow \frac{A^{-1}}{1-t} = (AB)^{-1}$, the inverse on the right can't be equal to the inverse on the left if t=1 because we divide by 0

[stoic python]

Transparent_Elemental

[latent hawk]

Oh ok.

I get the logic behind it thanks 🙂

There is something about that way that is solved which leaves me with a wierd feeling in my stomach lol

[slender yarrow]

Also what's E in your thing ?

Is it the identity or the matrix full of one's?

[latent hawk]

E is the japanese way of writing a symbol for the identity matrix

[slender yarrow]

Ah k

[wet stratus]

this proof is definitely weird and incomplete. what happens if A doesn't have an inverse. and so far we have only shown that AB has or has not an inverse. that doesn't necessarily tell us something about B having an inverse

assume that lambda is an eigenvalue of A. then you can check that lambda^2-lambda is an eigenvalue of A^2-A (in general p(lambda) is an eigenvalue of p(A) for any polynomial p). but A^2-A=0 so lambda^2-lambda is an eigenvalue of the zero matrix, therefore lambda^2-lambda=0

so lambda=0 or lambda=1

those are the only two possible eigenvalues of A

which means they are the only possible values such that A-lambda I is not invertible

note that this doesn't tell us anything about whether A or A-I actually are not invertible

[latent hawk]

I have no idea what's going on, but from the conversation, what I can infer is that this problem is very hard, and I shouldn't beat myself up too much over it 🙂

Thank you for your help though

[wet stratus]

well I solved it in the general case

for 2x2 we could do it explicitly

again by doing some equations for a,b,c,d and checking some stuff

[latent hawk]

I see (?)

[wet stratus]

check out for example wikipedia https://en.wikipedia.org/wiki/Idempotent_matrix

[crisp minnow]

Feeling especially stupid today and my 5 month old notes were written by some stupid guy with my name. The $r$ in $z = re^{i \theta}$ is just a scalar for the length of the positional vector of the complex number right?

[stoic python]

Joachim

[wet stratus]

yes

[crisp minnow]

Okay thanks

[wet stratus]

r is the length, e^(i theta) is the direction

[latent hawk]

That helps so much! thank you!

Finally settled the butterflies in my stomach lol

[tulip cedar]

if A is invertible it's simple to show that B doesn't have an inverse for t = 1, though I agree that in order for it to be a complete proof you have to review all the cases

[tranquil steeple]

That matrix actually has an interesting eigenstructure. Thanks for posting 🙂

[normal pilot]

Fellas, what theorem of elementary algebra is this referring to?

[dusky epoch]

factor theorem

[leaden tide]

If P(a) = 0 then X-a divides P

[normal pilot]

Alright I see I see, but what does he mean by W is a polynomial in x_n?

[leaden tide]

Recall that a determinant is a (multivariable) polynomial in the entries of the matrix

[tulip cedar]

If x_n was variable and x_k were constant for k <= n-1 then determinant with said coefficients is a polynomial in x_n

[quartz compass]

expand it out in the 2x2 and 3x3 cases if you need to convince yourself

[leaden tide]

(it only consists of multiplications and additions of the entries between one another)

If you fix all x_i's except x_n, you get a single-variable polynomial

[tulip cedar]

e. g. (x-2)(x-3) is a polynomial in x

[leaden tide]

And this, try an example to convince yourself of it

[normal pilot]

I seee, so in this step he fixes all x_i's and treats them as constants? With only x_n being the variable?

[tulip cedar]

yeah, but it depends on what follows after, sometimes it's just easier to think about complicated things like this as regular polynomials you're used to by imagining that all the roots are fixed

or there could be some proposition that relies on these roots to be fixed

[normal pilot]

But if that's the case, shouldn't W(x1,..., xn) just be the product of its factors? Where's this leading coefficient part coming from?

[leaden tide]

shouldn't W(x1,..., xn) just be the product of its factors
If you know a polynomial of degree 2 has roots 1 and -1, it could be equal to X²-1, but also 1-X², 2X²-2, etc

Polynomials that have exactly the same factors are equal up to a scalar multiple

Studying the leading coefficient is just one way to figure out that scalar multiple

[normal pilot]

Ooooohhhh, I see, that's news to me

Thanks a bunch!

[normal pilot]

Okay, last question hopefully, this page has given me more trouble than it's probably worth... How does expanding the determinant with respect to the last column show that a(x1,...,x_n-1) = W(x1,..., x_n-1)?

[stoic python]

Syst3ms

Syst3ms

[leaden tide]

@normal pilot

[normal pilot]

Oh my God, I think I finally understand... Just to confirm I get it, it's always true that you can represent a polynomial of degree n-1 in x in the form a(x-x_0)...(x-x_n-1), where x_i are its n-1 roots, and a is its leading coefficient, is that right?

@leaden tide

[leaden tide]

Indeed (assuming you can fully factor the polynomial like in ℂ)

[normal pilot]

Gotcha, gotcha, gotcha

Thanks a ton Syst3ms, you're a life saver

[oak stratus]

can someone help me with this question

because to me this problem makes no sense

especially d) because I dont understand how cars are going from west to east along Rue de Maisonneuve

[humble token]

Is this correct?

(The formula for the reciprocal of A)

[wintry steppe]

,rotate

[wintry steppe]

the end result is correct. didn't read it all

[humble token]

Thanks but smt doesn’t make sense to me

It’s that should the determinant of A inverse be 1 over the determinant of A?

But the final result gives a determinant of 1

So I am kind of confused

[wintry steppe]

it's fine

you're computing the determinant of the inverse wrong

if c is a scalar and B is an n x n matrix, det(cB) = c^n det(B)

so when you're computing the determinant of A^{-1} and pull out the 1/(det A), you need to square it

[humble token]

Oh right I forgot about this property

Sorry that’s my bad

For some reason, I thought it was just gonna be det(kA) = kdet(A) bc i did something before this where I had det( of A but one of its column’s is multiplied by k) = kdet(A) so I confused the both

[wintry steppe]

in kA you're multiplying every column by k, so you pull out a new factor of k for each column

hence k^n det A

[humble token]

Yup, I just mixed up smt

Thanks

[eager kestrel]

What is a semi direct sum of vector spaces? Or rather, Lie algebras?

[zinc timber]

what's semi direct sum?

[eager kestrel]

Um?

[wintry glen]

Is there a systematic way of doing (b) or am I just supposed to check everything.

[dusky epoch]

the cosets are in bijection with elements of V/W

(or rather, are elements of V/W)

[wintry glen]

Right. I get that {W + (0, 0, 1, 0), W + (0, 0, 0, 1)} is a basis for V/W, so does that mean that there are 4 cosets, since I can only form W + (0, 0, 0, 0), (0, 0, 1, 0), (0, 0, 1, 1) and (0, 0, 0, 1)?

[wet stratus]

yes. it may be helpful to just think of this using the dimension. as V/W has dimension 2 over the field F_2, it contains 2^2=4 elements

[wintry glen]

Ah.

[proud gate]

vector space is closed under addition and multiplication. But isn't this trivial if we already "defined" as $+: V \times V \to V$ and $\cdot: \mathbf{F} \times V \to V$?

[gray dust]

theyre really just equivalent phrasings

[zinc timber]

already asked in #groups-rings-fields

[gray dust]

man

[gray dust]

actually u only need W cap U={0}

now the proof of 1.44 is short. the forward direction should be apparent. for the backward direction, take two representations of the same vector in sum(U_i), ie sum(x_i)=sum(y_i), then show x_i=y_i for all i

[gray dust]

being a sum is independent of adding up to V

[clever totem]

Im trying to understand normal matrices and normal functions in general
if f is normal then
$$f \circ \hat{f} = \hat{f} \circ f$$
f hat is the conjugate matrix of f
i read that matrices with the same system of eigenvectors commute
is there some kind of intuition i can get from this for normal matrices and normal functions?

[gusty axle]

Yeah you can have a direct sum of two 1D subspaces of a 100D vector space and it's still a direct sum of the two subspaces

[wintry steppe]

Can anyone help with this problem?

Suppose $b \in \mbb{R}$. Show that the set of continuous real-valued functions $f$ on the interval $[0, 1]$ such that $\int_{0}^{1} f = b$ is a subspace of $\mathbb{R}^{[0, 1]}$ if and only if $b = 0$.

bruh, what is happening w the bot

here, managed to get a picture

[ornate smelt]

I'm taking linear algebra for the first time in a week. I want to get ahead and start reading the textbook. Anyone got any good linear algebra textbooks?

[slender yarrow]

@wintry steppe still here ?

[wintry steppe]

yes sir

[wintry steppe]

depends on the syllabus, if you are taking it an undergrad level probably Axler or Friedberg

[wintry steppe]

aude's recommendations are good if you're taking a theoretical course (though be warned that axler treats some topics later in the book in a rather unorthodox manner). is it a theoretical course, or something simpler (will you be focusing on matrices, row reduction, etc.)?

[wintry steppe]

it's been down for everyone

happens from time to time

what have you tried and where are you having trouble?

well we know that to show U is a subset of V we have to show that U contains the additive identity, i.e. 0, that U is closed under addition and scalar multiplication

in this case, I tried letting U equal the set of continuous real-valued functions f on the interval [0,1] such that int_0^1 f = b where b = 0

I suppose I would have to show that F(x) = 0 is part of the set?

that is indeed the additive identity

right so

if we let the integral of f be F

and b = 0

can we not do F(1) = F(0) hence F(x) = 0 is a member of the set since F(a) = F(b) for any a, b in R?

you really don't need to use the fundamental theorem of calculus to do this

this is the only method I know how to use xd

and I'm not sure it's correct

haha

[ornate smelt]

Its the first class in linear algebra

[wintry steppe]

that doesn't really clear it up, do you have a syllabus?

"first class in linear algebra" could be anything

[ornate smelt]

Yes I do sorry

Im not sure how to describe it because i havent taken ityet

[wintry steppe]

it might be more towards matrix theory in which case it will be a lot more computational than proof oriented

[ornate smelt]

Let me show syllabus

That's the recommended book. I tried it but it wasn't very good

[wintry steppe]

okay, definitely axler or friedberg is overkill for this

I gather you'd probably be fine with youtube alone

and some past papers from your university

[wintry steppe]

i do not follow. you said previously F denotes the additive identity, but what is f?

[wintry steppe]

f is a function in U?

also, "integral" should not be used interchangeably with "antiderivative"

[ornate smelt]

Why is Axler so good

[wintry steppe]

axler is terrible

[ornate smelt]

Oh

[wintry steppe]

jk

[ornate smelt]

Is friedberg a good book

oh

[wintry steppe]

lmao

[ornate smelt]

lol

[wintry steppe]

u dont need axler or friedberg for this

axler is fine until you get to the part on characteristic polynomials

[ornate smelt]

Im a physics and math major so I really want to understand linear algebra

Is it advantageous to read friedberg after I take this class?

[wintry steppe]

most likely

if u have another lin alg class in the future yes

[ornate smelt]

I will be taking another linear algebra class yeah

[wintry steppe]

why not

[ornate smelt]

But it's gonna be in a year

[wintry steppe]

but you're considering a particular one here, which you said yourself: F, the zero function

bruh

yes

[wintry steppe]

"antiderivative" is a function which differentiates to your given one, "integral" is a real number quantity

oh

wait so you're saying I let b = 0 and f = 0

ok, so you want to use lowercase f to denote your additive identity

sure

so you need to show that: if $f$ is the zero function on $[0, 1]$, then $$\int_0^1 f(x),dx = 0$$

[stoic python]

TTerra

[wintry steppe]

oh the bot's back up

poggers

i'm just that good

well, if f(x) = 0

int f(x) is c and c bounded from anything to anything is c - c = 0

ez clap

you can be more direct

f is its own antiderivative, so the integral is f(1) - f(0) = 0 - 0 = 0

(by the fundamental theorem of calculus)

but you are right

so we have showed 1/3

so the additive identity is in U. what must you now check?

that U is closed for addition

i.e. if we have f in U and g in U then f + g is also in U

yes

try not to use the fundamental theorem of calculus for this one. look up "properties of integral" if you need to

$\int_0^1 f(x) \dd{x} = -\int_0^1 g(x) \dd{x}$

[stoic python]

Aude

[wintry steppe]

yes sir?

sure. both sides are zero

it feels a bit odd adding in g(x)

like I am not sure we are allowed to do that

are we

i do not know what you mean

well

you haven't done anything incorrect so far

wait f is any arbitrary function in U

[wintry steppe]

you said what f and g are yourself. no problems here

yes so

wait they dont have to be distinct

can I just say 0 + 0 = 0

and call it a day

[wintry steppe]

be more specific than this

well, we know LHS is 0 hence RHS is also 0

thus

$\int_0^1 f(x) \dd{x} + \int_0^1 g(x) \dd{x} = 0 \implies \int_0^1 g(x) = 0 \implies 0 + 0 = 0$

[stoic python]

Aude

[wintry steppe]

Since $f$ integrated over the interval [0,1] is = 0 and since g = 0 when integrated over [0,1], and 0 + 0 = 0 we conclude that f + g integrated from [0,1] is also in U

[stoic python]

Aude

[wintry steppe]

I might be stupid

this argument is really hard to follow

the sum of the integrals is zero, okay. this implies the integral of g is zero, but we already assumed this, so why state it? then you wrote 0 + 0 = 0

can you show me how to go from the integral of f + g to zero?

you mean

actually integrate it?

please go from $$\int_0^1 (f(x) + g(x)) , dx$$ to $0$ and state explicitly what properties of integrals you use

[stoic python]

TTerra

[wintry steppe]

separation of integrals

oh

I just realized

it

haha

write it down

smart

$\int_0^1 f(x) + g(x) \dd{X} = \int_0^1 f(x) \dd{x} + \int_0^1 g(x) \dd{x}$

[stoic python]

Aude

[wintry steppe]

but we proved previously that $\int_0^1 f(x) \dd{x} = 0$ and we have assumed that $\int_0^1 g(x) \dd{x} = 0$

[stoic python]

Aude

[wintry steppe]

you did not prove that, you assumed it

hence $\int_0^1 f(x) + g(x) \dd{x} = 0$

[stoic python]

Aude

[wintry steppe]

why is my differential a big X

you wrote \dd{X}

[wintry steppe]

It seems i am in a circular argument

[wintry steppe]

you assumed right here that f and g are in U

f being in U means, by your own words, the integral of f over [0, 1] is zero

yes

[wintry steppe]

so say here "but we assumed that \int_0^1 f(x) dx = 0 and \int_0^1 g(x) dx = 0"

not "proved that..."

once you have that, this step is good

but we assumed previously that $\int_0^1 f(x) \dd{x} = 0$ and we have assumed that $\int_0^1 g(x) \dd{x} = 0$

[stoic python]

Aude

[wintry steppe]

ah yes

and so the integral of f + g is zero, meaning f + g is in U

cool

yessir

okay

now

a remark on language

the property $$\int_0^1 (f(x) + g(x)) , dx = \int_0^1 f(x) , dx + \int_0^1 g(x) , dx$$ is called "additivity of the integral", not "separation of integrals". the latter phrase may be confused with "separation of variables", so avoid it

[stoic python]

TTerra

[wintry steppe]

U is closed under scalar multiplication, i.e. if $k \in \mbb{R}$ and $f \in U$ then $kf \in U$

[stoic python]

Aude

[wintry steppe]

HENCE

$\int_0^1 k(f(x)) \dd{x} = k\int_0^1 f(x) \dd{x}$

[stoic python]

Aude

[wintry steppe]

so we know the LHS is equal to

$\int_0^1 k(f(x)) \dd{x} = k \cdot 0$

[stoic python]

Aude

[wintry steppe]

hence kf(x) in U

$\qed$

[stoic python]

Aude

[wintry steppe]

cool

is that it

looks good to me

poggers thanks sir

mucho apprecionado

all of the exercises in this chapter are showing that something is or isnt a subset of something else

i would write it as: "assume that $k \in \bR$ and $f \in U$. then $$\int_0^1 (kf)(x) , dx = k\int_0^1 f(x) , dx = k \cdot 0 = 0,$$ where the integral vanished because $f \in U$. so $kf \in U$"

[stoic python]

TTerra

[wintry steppe]

just to make the reasoning absolutely crystal clear

[wintry steppe]

subspace

subspace apologies

you still have the other direction of the problem to prove

wait wut

"is a subspace of \bR^{[0, 1]} if and only if b = 0"

bruh

now I need to prove

left to right

yes

tears

this direction is much easier to prove

shall I use contradiction

up to you

you don't need to, but you probably could

direct proof is easier and shorter

wait

didnt we just do left to right

no

no we assumed b = 0

you did right to left. you assumed b = 0 and showed the set is a subspace

so now we need to force it into a condition where b must be 0

now you must assume the set is a subspace and show that b = 0

yes

use one of the conditions of being a subspace to show that b = 0

dont i just

int f(x) dx from 0 to 1

= b

f(x) = 0

ez

b = 0

$\qed$?

[stoic python]

Aude

[wintry steppe]

so f is the zero function (additive identity)?

yes sir

that's exactly the proof i had in mind

very nice

thank you sir

very nice

can I +rep you like on steam profiles

nope

but you can pay it forward

what does that mean in this context

oh

help someone else

well sure if someone is stuck in the first three uni channels i will

[oak stratus]

hey

can someone please show me how to calculate

Calculate the max which is P,
P = 18x + 24y + 20z
8x + 12y + 12z ≤ 1240
6x + 8y + 4z ≤ 700
4x + 4y + 4z ≤ 480

[dusky epoch]

are x, y and z also required to be nonnegative

[oak stratus]

yes

sorry I forgot to mention that

x,y,z ≥ 0

[raven chasm]

Is it possible to find out what a transformation is if we have the initial and final states of the vector?

[limber sierra]

if you only have a single vector, no.

if you have that info for all vectors, then that is the linear transformation

well actually, let me step back a bit

if you know your transformation is linear, then we have the following fact:

• a linear transformation on a vector space is uniquely and completely determined by what it does to a basis of that space

(the codomain doesnt matter here (unless it's {0}), only the domain does)

so if your domain is a 1-dimensional vector space and the vector you know isn't the 0 vector, then yes, that information is enough to determine it

but linear transformations on 1-dimensional vector spaces arent very interesting

if your vector is, say, from ℝ³, then you would need three (linearly independent nonzero) vectors to determine the transformation

[raven chasm]

Ahhhh I got you

Wow that actually made sense thanks 👍

[limber sierra]

as an aside, this is why we care about bases of vector spaces

and the general concept - look at part of a space to determine the behaviour of "special functions" on the whole space - comes up a lot in advanced algebra as a whole

not just linear algebra

e.g. an abstract algebra student would learn that the same thing holds for groups, substituting "linear transformation" with "group homomorphism" and "basis" for "generating set"

(as well as a slightly-different-but-practically-similar phrasing using the quotient by the kernel)

[wintry steppe]

Hi, sorry for being a bother but I am going through Gilbert Lang's linear algebra playlist. Although I know what free variables, degrees of freedom, etc are by definition, I can't figure out how they affect the column space. Does this mean I would have to revisit basic algebra? or should I continue with the playlist and it will click one day?

[lucid shore]

please

S X S -> S

What does it mean

How do I verbally make sense of it

[teal grotto]

function which takes two inputs from S and gives you one output in S

[lucid shore]

oh

thank you very much

[tulip cedar]

https://en.wikipedia.org/wiki/Cartesian_product here's what the S x S means exactly

[glacial terrace]

For (a), the eigenvalues are (2,2,1). Then I solved the systems: (i) (A-2I)v = 0; (ii) (A-I)v = 0;
For (i) we get x = y and z = 2x as a solution, so (1,1,2) is an eigenvector for A.
For (ii) we get y = 0 and z = 2x so, (1,0,2) is another eigenvector for A.
However, since we only have two linearly independent eigenvectors for A it means that we can't build a basis for R^3, and so A is not diagonalizable.

Can anyone check if my reasoning is correct?

[tulip cedar]

since you have to take an inverse of a matrix with linearly dependent vectors, yes, such operation can't be performed

[glacial terrace]

indeed

thanks

[narrow steeple]

Is anyone here familiar with linear algebra that I can DM?

[glacial terrace]

[real bone]

can someone walk me through this problem

i really don't get this jcf shit

it seems that the only eigenvalue of T is 0

[wintry steppe]

to start, the dimension of the eigenspace is the number of jordan blocks, so up to permutation you have two possibilities: a 1 x 1 block and a 3 x 3 block, or two 2 x 2 blocks

(the eigenspace is two dimensional)

[glacial terrace]

$\det(xI - P^{-1}BP) = \det(x(P^{-1}P) - P^{-1}BP) = \det(P^{-1})\det(xI - B)\det(P) = \det(xI -B)$

[stoic python]

mns

[glacial terrace]

ok?

[real bone]

How do we know this

[wintry steppe]

one way to figure out what it must be is to use the minimal polynomial. the characteristic polynomial of your matrix is t^4, so the minimal polynomial must be of the form t^n for some small n. here, n will be the size of the largest jordan block corresponding to the eigenvalue, so either n = 2 or n = 3. for n = 2 this doesn't annihilate your operator, but for n = 3 it does, so your jordan form must be a 1 x 1 block and a 3 x 3 block

[wintry steppe]

just calculate it lol

write out what T does to an arbitrary matrix

[real bone]

But I can't find a matrix for T

[wintry steppe]

why? M_{2 x 2}(\bC) has a perfectly good standard basis you can use

[real bone]

so i did that and i still can't figure it out

[wintry steppe]

using the standard ordered basis $$\left{\begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix}\right}$$ you should find that $$\begin{pmatrix} 0 & 0 & -1 & 0 \ 1 & 0 & 0 & -1 \ 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 \end{pmatrix}$$

[stoic python]

TTerra

[real bone]

ohhh

i somehow forgot that the space is 4 dimensional and thought it was 2 dimensional

[wintry steppe]

for my explanation you don't need to compute this matrix, but it does help

[real bone]

ok

don't i need to find the eigenvalues though?

[wintry steppe]

you do, and you can do that directly from the image you just posted

[real bone]

oh hmm

you mean this? #linear-algebra message

[wintry steppe]

yes

the image you just posted

do the work and you should find that the only eigenvalue is zero, and the eigenspace is $$\left{ \begin{pmatrix} a & b \ 0 & a \end{pmatrix}, a, b \in \bC \right},$$ two dimensional

[stoic python]

TTerra

[real bone]

i'm still not sure how you would get that

i don't understand how the only eigenvalue is 0

[wintry steppe]

i'll write out the first step for you

[real bone]

ok

[wintry steppe]

"assume that $$\begin{pmatrix} a & b \ c & d \end{pmatrix}$$ is an eigenvector with eigenvalue $\lambda$. this means that $$\lambda\begin{pmatrix} a & b \ c & d \end{pmatrix} = T\begin{pmatrix} a & b \ c & d \end{pmatrix} = \begin{pmatrix} -c & a - d \ 0 & c \end{pmatrix}."$$

[stoic python]

TTerra

[wintry steppe]

you get four equations, plus the fact that not all of a, b, c, and d are zero simultaneously

play around with it

[real bone]

but this just means T(M)=0 always

[wintry steppe]

what?

no

[real bone]

i'm so confused

[wintry steppe]

"only eigenvalue is zero" does not mean "linear transformation is zero"

[real bone]

if lambda = 0 then T(M)=0

[wintry steppe]

give me a moment

what is "M" here?

[real bone]

an arbitrary matrix same as (a b c d)

or is (a b c d) an eigenvector

[wintry steppe]

then refer to this

[wintry steppe]

read what i wrote

"assume that ... is an eigenvector..."

[real bone]

oh i get it

wait no i don't get it

why does c have to be 0

in the eigenspace

[wintry steppe]

have you done this?

i'm not going to write out the casework here for you

[real bone]

ok

sry

1 sec

[wintry steppe]

i will say though, that it might be helpful to look at the bottom left entry

with the solution i wrote in mind, you could assume that c is non-zero and try to get a contradiction

[real bone]

ok so we have lambda c = 0 so either lambda = 0 and/or c = 0. if c = 0 (and lambda != 0) then note that lambda a = -c = 0 so a = 0, but this means d=-a =0. then the whole matrix is 0. thus, both lambda and c must be 0

well if c is nonzero and lambda c = 0, then lambda = 0 and lambda a = -c = 0 so c=0

which is a contradiction

either way we get both lambda and c are 0

then also a-d =0 so a=d and b can be arbitrary

i see

[wintry steppe]

you forgot about b here, b could still be anything

[wintry steppe]

this here is correct

[wintry steppe]

but it is not clear how you've gotten \lambda = 0

(once you've shown \lambda = 0, though, the rest of the explanation for a = d works)

you just need to say a little bit about b in the first line and you're all set

[real bone]

if lambda != 0 then lambda a = -c = 0, so a would be 0, lambda d = c = 0 and lambda b = 0 - 0=0, so b=0

[wintry steppe]

sounds good to me

[wintry steppe]

casework done, we've proven this

[real bone]

yeah

[wintry steppe]

now

the goal is to figure out which of these two possibilities it must be

for this, i used the minimal polynomial in my message following that one

[real bone]

yeah the minimal polynomial is x^2

[wintry steppe]

x^3.

[real bone]

wait wa

ok lemme do it again

[wintry steppe]

T^2 is not zero, but T^3 is

[real bone]

how do you find this directly without the matrix form

is that even possible

[wintry steppe]

by the definition of T

you are literally given T

you can compute T^2, T^3

(T \circ T and T \circ T \circ T)

[real bone]

yeah T^2 = XE^2-2EXE + E^2X

[wintry steppe]

i think that it would be easier for you if you used this

[real bone]

ohh

[wintry steppe]

don't do the hard work twice

[real bone]

ok

thats why i thought that would be really bad

[wintry steppe]

it's not terrible since E^2 = 0, leaving T^2(X) = -2EXE

[real bone]

ok yeah now i see x^3

ah

[wintry steppe]

but you'd probably like to avoid multiplying matrices if possible

[real bone]

ok but T^2 has -2c in top right

[wintry steppe]

yes

[real bone]

so T^3=0

[wintry steppe]

yes

[real bone]

ok

[wintry steppe]

the problem is done

[real bone]

its 3 and 1?

[wintry steppe]

yes

[real bone]

i actually don't know how to determine that

[wintry steppe]

the multiplicity of an eigenvalue as a root of the minimal polynomial is the size of the largest jordan block corresponding to it

here, it's 3, so you have to have a 3 x 3 jordan block

[real bone]

our instructor gave us this

which i guess i can use

[wintry steppe]

you can use that too

[real bone]

but it doesn't seem very intuitive

[wintry steppe]

it's not

the minimal polynomial solution i gave is a lot more intuitive (imo)

at least it's straightforward to use this

\lambda is zero and you already know what the powers of T are, so the answer i wrote pops out from this pretty quickly

[real bone]

oh its basically the same thing except your method is finding which power of the operator is 0 which is the same as finding when the dimension of the nullspace stabilizes

[wintry steppe]

something like that

the minimal polynomial is the only way i can remember this stuff

[wintry steppe]

the point i stopped paying attention in LA is probably the point where we got formulas like this

[real bone]

yeah our instructor gave us this example for similar type of problem

i didn't understand what he meant by nilpotent parts of both jordan blocks for 1 square to 0

[real bone]

i think i'm going to use this method lol

i can't understand the head or tail of this stabilizing of dimension of kernel stuff

[wintry steppe]

you are not alone in that

this is the ugly part of JCF

[real bone]

ok thank you very much

[wintry steppe]

almost every single JCF problem i've ever seen posted here, when it's not explicitly "find a basis which puts operator into JCF", can be solved using the minimal polynomial

[real bone]

i want to try to do it the other way after being informed by the minimal polynomial way since i actually want to understand what is happening lol

[wintry steppe]

and if you take an abstract algebra class and get to see the module/PID decomposition theorem, you'll be thinking of things using minimal/characteristic polynomials anyways

[real bone]

i've taken an abstract algebra but they didn't cover modules or PID lol we did have minimal polynomials in field theory tho

not in ring theory

idk why

[wintry steppe]

usually abstract algebra does groups/rings/fields with modules optional

my class did modules at the expense of doing almost no field theory

weird class

thankfully the module theory has been more useful to me

[real bone]

yeah my university has the undergraduate version of abstract algebra then lets us take the graduate version of abstract algebra after which covers the stuff in detail i think

idk tho haven't taken the 2nd one

[wintry steppe]

if you're interested in learning the technical bits of JCF stuff, friedberg's book has your back

i remember the exposition there being pretty clear

[real bone]

oh

we are using LADR but the instructor doesn't follow it very well

e.g. he skipped from eigenvalues (chapter 5) to generalized eigenspaces and JCF (chapter 8)

[wintry steppe]

when i took LA the prof put some pretty detailed summaries on JCF, finding JCF, etc. on the course website: http://www.math.toronto.edu/payman/mat247/main.html

i personally found it pretty useful so i'll share

[wintry steppe]

wtf lol

[real bone]

ok tyvm oh u went to utoronto!

i was considering that but then chose not to because international fees are way too high 😔

[wintry steppe]

it's way too expensive for international students lmao

i'm from toronto so it's not that bad, but the international fees are truly ridiculous

[real bone]

yeah its ridiculous but its not like i'm paying half that amount for an in state public university here in the states anyway 😢

[glacial terrace]

$\det(xI - P^{-1}BP) = \det(x(P^{-1}P) - P^{-1}BP) = \det(P^{-1})\det(xI - B)\det(P) = \det(xI -B)$

this ok?

[stoic python]

mns

[wintry steppe]

yes it is fine

[glacial terrace]

More

p(T) is a map from V to V? I.e. p(T) : V -> V ? Because I am not understanding its action

[wintry steppe]

it is indeed a map from V to V

T^k means T composed with itself k times

[glacial terrace]

and its action on the vectors of V? Is it p(T)v = (a_oI)v + (a_1T)v + ... + (a_nT^n)v ?

[wintry steppe]

yes

[glacial terrace]

One more thing

F[x] is the set of all polynomials with entries in F. But do we restrict the order of the polynomials or it can be any n as in a0 + a1x + a_nx^n ?

[wintry steppe]

it's just the set of polynomials with entries in F. there is no restriction on the degree.

[glacial terrace]

so the basis for F[x] is indeed a set {1,x,...,x^n} with n + 1 elements

[wintry steppe]

no

that would be for the subspace of F[x] consisting of polynomials of degree n (or less)

[glacial terrace]

so this space has infinite dimension?

[wintry steppe]

yes

a basis is {1, x, x^2, x^3, ...}

{x^n : n \geq 0}

[glacial terrace]

I see

thanks once again!

[median forum]

does anybody have any idea how to do a)?
|.| btw represents elementwise absolute value and inequalities betweem those mean elementwise inequalities

[clear iron]

Question, given a unitary transformation T, it follows that <Tu,Tv>=<u,v>
however in the same light, <T*u,T*v>=<T(u),T(v)> right?
since <Tu,Tv> = <u , T*(T(v)) = <v,u>=<u,v>=<Iu,v> = <TT*u , v>=<T*u , T*v>

basically both T and T* preserve dot product , and not only T

[zinc timber]

Unitary matrix implies TT* = T*T = id

[leaden tide]

@clear iron What you're saying amounts to the fact that T* is unitary when T is

Which is kinda like saying T⁻¹ is invertible when T is

[clear iron]

yea exactly
that's what my little sketch was based on

[clear iron]

well.. yea
but it is correct tho.. no ?

[leaden tide]

It is

It's just not very deep

[tropic bridge]

bruh

im having trouble researching

info

about "lineal combination"

anyone knows where i can find this for dummies D:?

[wet stratus]

you probably mean linear combination

and there should be lots of stuff on the internet about that

are you looking for something specific?

[clear iron]

I know
I just tackled a question that ruled that property out when asked about properties of unitary Transformations

[wet stratus]

a linear combination of elements $v_1, \ldots, v_n$ in a vectorspace $V$ is a sum $a_1v_1+\ldots +a_nv_n$ with scalars $a_1, \ldots, a_n$ of your field (for example real numbers)

[stoic python]

Denascite

[tropic bridge]

yeah that linear, maybe that was the word that i was missing

thanks bro !

[glacial terrace]

What would be distinct linear factors?

like (x-k1)(x-k2)...(x-kn) with ki different from kj for all i,j in {1,...,n} ?

I can imagine this means that a linear operator T is diagonalizable if its minimum monic polynomial consists of (x-a1)(x-a2)...(x-an) for ai the eigenvalues of T

[dim epoch]

yeah this is true

since the eigenvalues are the roots of m_T

[glacial terrace]

Indeed

So linear factors are the (x-a) whereas a non linear factor is (x-a)^b with b != 1

[dim epoch]

i think you usually call the second one linear factor with multiplicity b

multiplicities and all that become important for JNF and that stuff

[hard drum]

Yes, another way to see this is just to consider what the min poly of a diagonal matrix is

[glacial terrace]

So I got a bit confused with this now

the theorem is refering to linear factors which are of the form (x-k) , i.e. , with b = 1 in (x-k)^b

right?

Hum ok, I see it. I needed to check something for the case of a diagonal matrix with two eigenvalues equal

[clever totem]

$$for every n \in \mathbb{N}_{\geq 2}$$
give a Matrix A such that
$$A \in \mathbb{C}^{n \times n}$$
$$A is diagonalizable, A is nilpotent$$
meaning that
$$there is a basis of eigenvectors, A^t=A$$

[stoic python]

~Martin

[clever totem]

i am not good at thinking in C, which is why i have problems with this task

[wet stratus]

what's the easiest matrix you know

[clever totem]

the easiest is the identity

oh wait

sry i wrote it wrong

[wet stratus]

I would say the easiest is the 0 matrix but yeah identity works too

[clever totem]

nilpotent not idempotent

so there exists k such that A^k=0

$$for every n \in \mathbb{N}_{\geq 2}$$
give a Matrix A such that
$$A \in \mathbb{C}^{n \times n}$$
$$A is diagonalizable, A is nilpotent$$
meaning that
$$there is a basis of eigenvectors, \exists k: A^k=0$$

[stoic python]

~Martin

[wet stratus]

if you want to use text inside $$this,$$at least use $$\text{\text}$$

[stoic python]

Denascite

[clever totem]

$$\text{for every } n \in \mathbb{N}_{\geq 2}$$
give a Matrix A such that
$$A \in \mathbb{C}^{n \times n}$$
A is diagonalizable, A is nilpotent
meaning that
there is a basis of eigenvectors, $$\exists k: A^k=0$$

[stoic python]

~Martin

[clever totem]

not pretty but works i guess

cant i just say the null matrix is diagonalizable and nilpotent?

[wet stratus]

if you want inline math you can use $n\in\bN_{\geq 2}$ for example

[stoic python]

Denascite

[wet stratus]

yes

[clever totem]

ok it never said that i can not use the null matrix so lets go with that
but if i wouldnt want to do that, i think in R there would be no other matrix right?

[wet stratus]

why do you think that

[clever totem]

i read that somewhere tbh

[wet stratus]

if you want, try proving it. what is the minimal polynomial of a nilpotent matrix. and what do you know about the minimal polynomial of a diagonalizable matrix?

[clever totem]

the min poly of a diagnalizable matrix falls into linear factors of power one

for nilpotent matrices the min poly is x^k where A^k=0

if a matrix is nilpotent and diagonalizable then the min polys are the same

[wet stratus]

so which min poly does the matrix have?

[clever totem]

x^k=(x-0)^k
so k=1
so m(x)=x

[dusky epoch]

what is the original question again

[clever totem]

$$n \in \mathbb{N}_{\geq 2}$$
give a matrix for every n such that A is nilpotent and diagonalizable

[stoic python]

~Martin

[clever totem]

given that $A\in\mathbb{C}^{n\times n}$

[stoic python]

~Martin

[dusky epoch]

nilpotent and diagonalizable implies zero, no?

[wet stratus]

and if m(x)=x is the minpoly then what is the matrix?

that's what we are currently proving, yes

[dusky epoch]

aint this like, obvious
the only eigenvalues that a nilpotent matrix ever has are zero

[wet stratus]

I guess yeah that's easier lmao

[clever totem]

yeah but the task looks like they want a non zero matrix

so you are sure that there is no such non zero Matrix in C^nxn?

[dusky epoch]

there sure is not

does the task SAY it wants a nonzero matrix tho

[clever totem]

no it does not say the matrix must be nonzero

[dim epoch]

That task seems very familiar lol

you said you were german right Martin

[clever totem]

yeah

Freiburg ehem

[dim epoch]

[real bone]

how do i prove this 💀

wait can i just multiply them and show that its the identity

that would be mega troll and almost certainly not the way i'm supposed to do it

[wet stratus]

why would that be mega troll

[real bone]

well its like

i feel like i'm supposed to show it without multiplying them

because this is the 2nd part of a question that asked

to find the eigenvalues and minimal/characteristic polynomial of inverse

[wet stratus]

I guess yeah you probably have to use that. given that J_m(lambda)*J_m(lambda^-1) is not the identity anyway

[real bone]

wait what

[wet stratus]

it only says that the inverse has that jordan normal form

[real bone]

oh

good point

[wet stratus]

it doesn't say that the inverse is equal to it

[real bone]

wait do you have any ideas on how to do it

[wet stratus]

well use the min/char polynomial. it tells you quite a lot about the jordan normal form

[real bone]

once i show that the inverses thing is true then its a simply matter of taking the direct sum of the jordan blocks right

[wet stratus]

something like that probably, yeah

[real bone]

interesting

[real bone]

are each of the jordan blocks linear transformations

if so, then what is the domain and codomain and stuff

i'm still slightly confused

[real bone]

nvm ignore my musings

[raven chasm]

In 3D space since each direction adds a new span, all vectors can be linearly independent?

Or are the vectors for the vectors for the x and y direction always linearly dependent?

[median forum]

what do you mean by that?

[raven chasm]

So like

3D space is just the 2D span moved throughout the Z direction. If vectors I hat and j hat both make up the span that’s being moved through the Z direction, are I hat and I hat always linearly dependent?

[median forum]

what do you mean by always what choice is being made here?

are they the \hat i and \hat j from the canonical basis?

[raven chasm]

Yes, assuming that both of those vectors are basis vectors. From what I think I understand, if two vectors can be represented as a resultant vector, they’re linearly dependent. So if 3 vectors are describing space, are two of them guaranteed to be linearly dependent? My original question was if x and y were always linearly dependent in 3 dimensional space since they can be represented as a resultant vector

But thinking it out, I’m suspecting that’s wrong and now I’m wondering if at least 2 vectors in 3D space are linearly dependent

[median forum]

yes
if they are "basis" vectors, then by definition they have to be independent

a basis is like a "minimal" set of vectors that "describes" space
here minimal means, it needs to be the minimum number and describe means to span

n

since in 3D space "all three" vectors in any basis are LI
any two of them are LI

[raven chasm]

And combining the span of both basis vectors would describe the 2-D plane making them linearly dependent to one another to describe said plane?

[median forum]

you mean LI?

[raven chasm]

Wouldn’t i hat be relying on j hats position to describe a vector in 2-D space?

[median forum]

what do you mean by relying?

the vector in 2d space is uniquely described by a linear combination of $\hat i$ and $\hat j$

[stoic python]

Fractal

[raven chasm]

You can’t have a vector with position (3,2) without both i and j right?

Right!

So

[median forum]

that is already implied by the notation (3,2),right?
its 3 units of i and 2 of j

coordinate notation is generally made to be compatible with the basis you are choosing

[raven chasm]

What would make a vector linearly dependent

[median forum]

a vector is always LD

linear dependece and independence is always relative to a set of vectors

if your set of vectors has a single vector, then it is LD

and that set is LI whenever you cannot write any of the vectors by a linear combination of the others (in a vector space to be clear)

[median forum]

sure, you could abuse terminology a bit and say the vector v is LI to the vector w (meaning the set {v,w} is LI)

[restive raft]

A set containing a single nonzero vector is linearly independent

[median forum]

oh ok
true
a set with a single non-zero vector should be independent
my mistake

[wintry steppe]

when i looked at fractal's message, i just KNEW someone was going to come in and correct them

there's always someone who has to come in and say non-zero

[real bone]

am i getting trolled hard and missing something or is this just an ass problem? (i know i haven't done the jcf yet)

oh i'm getting trolled hard

i missed a condition

me being bad at math again

wait nvm its fine

the condition that i missed doesn't matter

wait can someone confirm that i'm on the right track here

[median forum]

actually, they missed the non-zero part
I said LD when should have said LI

[winter pond]

Hey can someone help me with this?

I’m thinking it’s B am I correct?

[wintry steppe]

Why do you think it's B

[limber sierra]

what's the zero vector of (i)?

[winter pond]

Because it is equipped with vector operations over its R

[limber sierra]

that means very little

[winter pond]

Oh wait

[limber sierra]

what if we multiply a vector from (ii) by an arbitrary real number?

say by pi

[winter pond]

Then the vector is irrational

[limber sierra]

so scalar multiplication doesnt work

since it's not closed under it

[winter pond]

So it doesn’t work for part ii

Ah I see

[limber sierra]

and as for (iii), look through the axioms of addition in a vector space, and check which ones subtraction doesnt satisfy

if any

||a-b is not equal to b-a||

[winter pond]

Oh I get it now.

[limber sierra]

sooo none of them are vector spaces

[winter pond]

Option is F. Finally.

[limber sierra]

(i) lacks a zero vector, (ii)'s vectors dont work with its scalars, and (iii) does not have commutative addition

[winter pond]

Got it

So let’s say if a 2x2 matrix has eigenval 3 and 5 should the matrix be $\begin{pmatrix}3 & 0 \ 0 & 5\end{pmatrix}$

[stoic python]

Tony Chiba

[wintry steppe]

not necessarily

what about $$\begin{pmatrix} 3 & 1 \ 0 & 5 \end{pmatrix}?$$

[stoic python]

TTerra

[wintry steppe]

it must be similar to the matrix you wrote, though.

two distinct eigenvalues, two by two matrix, so you've got a diagonalizable matrix on your hands

[winter pond]

Oh got it as long as either b or c is 0

[limber sierra]

that isn't the case either.

[winter pond]

Oh

I thought as long as b or c is 0 the eigenval is always the same

And the matrix is diagonalizable

Anyways

I’m not sure how to show T is a linear transformation

[dusky epoch]

verify that it satisfies the definition of a linear transformation

[winter pond]

So X is just a random 2x2 matrix?

[dusky epoch]

careful throwing around the word "random" like that

but yes, X stands for a 2×2 matrix, being the input of T

[winter pond]

Okay so X is a transformation matrix that I have to find

[dusky epoch]

no

you are not "finding" anything

you are verifying that T is a linear transformation

to which end you should be applying the definition of linearity

which you should know, or it should be in your notes

[winter pond]

Hmm lemme see

Is it smth like this?

[dusky epoch]

your verification of additivity is ok

but for homogeneity you made an even number of mistakes that cancelled each other out

also try not to make your opening parentheses look the same as c's

[wintry steppe]

an even number of mistakes that cancelled each other out
lmao

i love it when this happens

although their proof of homogeneity looks fine to me

i don't see where the mistakes lie

[dusky epoch]

(cAX - XA) is how i read the thing right before cT(X)

[wintry steppe]

ah

yeah it is a little confusing

[dusky epoch]

it looks like i fell prey to their conflation of c with (

[wintry steppe]

another reason never to use c for scalar

[glacial terrace]

How so?

like the "intersection" between all those polynomials is empty?

for k = 3 we would have t1 = (x-a2)(x-a3), t2 = (x-a1)(x-a3) and t3 = (x-a1)(x-a2) then t1 "cap" t2 "cap" t3 = empty as there will be a factor missing the (sigma_i)

[wet stratus]

in a sense, yeah

[balmy spire]

Yeah pretty much. They'll have n-2 common factors pairwise, but altogether they won't have any.

[wet stratus]

the (normed/monic) factors of t1 are {1, (x-a2), (x-a3), (x-a2)(x-a3)}, similarly for the other ones and then the intersection is just 1

[glacial terrace]

[glacial terrace]

Consider T : R^2 -> R^2 defined by T(x,y) = (y,0)

Then the matrix representation of T on the basis {(1,0),(0,1)} is (0 & 0\1 & 0)

Notice that T^2 = I. The characteristic polynomial is x^2 which coincides with minimal polynomial. Hence T is not diagonalizable as the minimal polynomial is not a product of distinct linear factors

What's wrong in here?

Notice that I didn't used the fact that the scalars are over C as it will for sure influence the result. Therefore, my question is where it should've been used?

[pallid rampart]

The characteristic polynomial is not x^2

[glacial terrace]

my bad, I wrote the matrix representation wrong. Edited

[pallid rampart]

Well no

Characteristic polynomial is the smallest polynomial p such that p(A)=0

Notice it's p(A)=0 not p(A)=I

T^2 = I means that T^2 - I = 0 so the minimum polynomial is x^2 - 1

[glacial terrace]

that's the minimal polynomial

[pallid rampart]

Oh oops sorry

But in any case the minimum polynomial divides the characteristic polynomial

So x^2 can't be the characteristic polynomial either

[glacial terrace]

The characteristic poylnomial is det(xI - T)

and indeed the minimum polynomial divides it

so it could be x or x^2

and only x^2 yields the zero matrix

[pallid rampart]

Oh ok I see what's the issue

For T(x,y)=(y,0), T^2 is not the identity

It's 0

[glacial terrace]

+1

indeed, thanks

[pallid rampart]

So sorry just woke up and mixed up a lot of things

[pallid rampart]

But also yeah I just remembered this theorem says the minimum polynomial not characteristic polynomial

[glacial terrace]

Is (x^n-1) a linear factor?

[winter harbor]

x^n - 1 is the product of distinct linear factors over C.

remember roots of unity stuff

[glacial terrace]

ohh

[wintry steppe]

degree n, so it couldn't possibly (outside of the trivial n = 1 case, in which yes it is linear)

[umbral elk]

So I was reading about how to find eigen values. How come this equation solves the determinant? Like, what's the basis for this method?

Ping me if anyone knows. TIA

[leaden tide]

How come this equation solves the determinant?
As in, why are eigenvalues the roots of dot(A-X*I) ?

@umbral elk

[umbral elk]

Not this equation but the ones that simplifies the det.
V= Lambda
V³ - [sum of diagonal elements]V² + [sum of diagonal minors] - |A| = 0

[leaden tide]

Ohh right

This is more general than this actually, for an $n\times n$ matrix $A$, the characteristic polynomial is $$X^n - \text{tr}(A)X^{n-1}+\ldots + (-1)^{n-1}\text{tr}(\text{Com}(A))X + (-1)^n\det(A)$$, up to a factor $(-1)^n$ depending on the definition

[stoic python]

Syst3ms

[leaden tide]

$\text{Com}(A)$ is the matrix of minors of $A$

[stoic python]

Syst3ms

[leaden tide]

Let me briefly explain the reason for this

We use det(XI-A) to define the characteristic polynomial, just multiply by (-1)^n if needed

The constant term is (-1)^n*det(A) because it's what you get when you evaluate det(XI-A) at 0

The dominant term is X^n because it's only reached for the product of diagonal elements

As for X^(n-1)

I'll finish the explanation

no worries

you didn't need to delete, now i feel bad

[umbral elk]

Lol, I want to take a screenshot of this so I deleted

[leaden tide]

ah ok

Do you happen to know the definition of the determinant as a sum involving the symmetric/permutation group?

[umbral elk]

I don't think I do. No I don't. Sorry. I'm just skimming over some important topics. 🤐

[leaden tide]

Don't worry, it's not very digestible anyway

$\det(A) = \sum_{\sigma\in S_n} \epsilon(\sigma) \prod_{i=1}^n a_{i,\sigma(i)}$

[stoic python]

Syst3ms

[leaden tide]

The sum ranges over all possible permutations of n elements

What epsilon is doesn't matter too much here

All that matters is that it's a sum of various products of entries, indexed by permutations

In XI-A, the terms that contain X are precisely the diagonal terms

To get a term in X^(n-1) by this formula, we'd need a permutation with exactly n-1 fixed points

But if it sends n-1 numbers to themselves, it has to do the same for the last one, otherwise it wouldn't be a permutation

All of that to say that the only contribution to the X^(n-1) term is when you expand the product (X-A_11)(X-A_22)...(X-A_nn), and nothing else

And the X^(n-1) coefficient of such a polynomial is famous, and it's -(A_11+...+A_nn) = -Tr(A)

[umbral elk]

wow

[leaden tide]

And let me recall how we prove the coefficient of X is Tr(Com(A)), i.e the sum of the diagonal minors

[umbral elk]

Ohh

[leaden tide]

Okay, so the proof I remember uses something that sounds really cursed : the derivative of a determinant

[umbral elk]

Oh god

[leaden tide]

The reasoning is that the X-term of a polynomial is the constant term of its derivative

So if we get a nice-enough formula for the derivative, we can just evaluate it at 0 to get the answer

[umbral elk]

Oh damn, that's neat tho

I've one more question to ask

[leaden tide]

The determinant derivative formula is not that bad

[umbral elk]

What level is this

[umbral elk]

Like, it's not complicated but the calculation is too long.

[leaden tide]

If you study determinants in the general case you'll probably come across this

Not even

Let me do it

(might take a while to write it all though)

[umbral elk]

Yes, i have encountered derivatives of det when I was reading about them

[leaden tide]

If you have a function of the form $t \mapsto \det(A_1(t),A_2(t),\ldots,A_n(t))$, where $A_i$ are column-valued, the derivative is $\det(A'_1(t),A_2(t),\ldots,A_n(t))+\det(A_1(t),A'_2(t),\ldots,A_n(t))+\ldots \+\det(A_1(t),A_2(t),\ldots,A'_n(t))$

[stoic python]

Syst3ms

[leaden tide]

There we go

It's like a product rule of sorts

If you apply this to the function that is the characteristic polynomial, each of the terms will look something like $\begin{vmatrix}1 & -A_{12} & \cdots & -A_{1n} \ 0 & X-A_{22} & \cdots & -A_{2n} \ \vdots & \vdots & \ddots & \vdots \ 0 & -A_{n2} & \cdots & X-A_{nn}\end{vmatrix}$

[stoic python]

Syst3ms

[umbral elk]

Ohh yes

[leaden tide]

However, that first column makes it so you just get a diagonal minor when evaluated at 0 (up to a (-1)^(n-1) factor)

[umbral elk]

I remember now

[leaden tide]

Each of the terms gives you a different diagonal minor, so all in all you get Tr(Com(A))

[umbral elk]

The derivative of det part.

[leaden tide]

In the 3x3 case, you fortunately can use these special cases only

Likewise in the 2x2 case

[umbral elk]

That's nice

[umbral elk]

Is this university level?

Like, bachelor's or what?

[uneven blade]

I was watching the essence of linear algebra by 3b1b, and I have a question. Grant mentioned that whenever a transformation takes place, and the transformed grid, as long as it remains parallel, evenly spaced, and has a fixed origin, the transformed vector will still maintain the same properties. I just wanted to know why this is the case.

Like in this image why does a transformation that remains parallel, evenly spaced, and has a fixed origin not affect v = -1(i-hat) + 2(j-hat). why does it remain the same?

[dusky epoch]

as you've written it right now (though it appears kind of word-salady), it need not be the case that -1i + 2j remain as -1i + 2j even after being transformed. it may so happen by coincidence, but it need not.

what it will be no matter what, however, is -1T(i)+2T(j).

(as long as T is linear of course)

[leaden tide]

Depends on the country, and at what level you study dets

[uneven blade]

Yes, grant said the transformation had to follow rules for -1i+2j to remain the same. But why do those rules make -1i + 2j stay the same?

[dusky epoch]

can you share the video and the timestamp

[wet stratus]

can you link to the specific timestamp? I think you might have misunderstood something. for some linear transformations there will be vectors that remain the same. not for all transformations and those vectors will in general be different

[dusky epoch]

bc i really feel as if you are misunderstanding and/or saying this incorrectly

[uneven blade]

https://youtu.be/kYB8IZa5AuE?t=244

[rough pilot]

Hey guys, any ideas for the proof of the second statement? (P1 and P2 are polynomials and T is a linear transformation V->V)

[uneven blade]

like in the 2-3 minutes following this

[dusky epoch]

hold on

let me transcribe exactly what is said

If we play some transformation and follow where all three of these vectors go, the property that gridlines remain parallel and equally spaced has a really important consequence: the place where v lands will be -1 times the vector where i-hat landed plus 2 times the vector where j-hat landed. In other words, it started off as a certain linear combination of i-hat and j-hat, and it ends up as that same linear combination of where those two vectors landed.

a far cry from requiring -i+2j to up and stay the same! as you can see the 'physical' vector undergoes quite a change.

this is what grant is saying

[uneven blade]

yes the physical vector undergoes the change, but the formula still remains the "same"

[umbral elk]

sophomore hs. That's where they teach the basics. Like, really basic stuff.

[dusky epoch]

in truth, it is this preservation of linear combinations that is actually how we define what it means for a transformation to be linear. and the grid line thing is secondary to it.

[uneven raptor]

unpack the notation. write P1 and P2 explicitly

[rough pilot]

I mean, I did, but I can't solve it

It's a double summation

[toxic apex]

How to check for injectivity and surjectivity? check if dim(ker)=0 and dim(img)=dim(co-domain)?

[winter pond]

dim(ker)=0 injective

[winter pond]

And dim(img) = dim of transformed matrix

[slender yarrow]

@winter pond did you manage the 2 last questions in your pic ?

[winter pond]

Yeah I was able to but I’m not exactly sure

[slender yarrow]

if you want to post your workings we can check

[toxic apex]

@slender yarrow I've tried it. Can you please check if my working is fine

[slender yarrow]

sure

[toxic apex]

@slender yarrow

[slender yarrow]

It's alright yeah

There's simpler ways for the "is T injective/surjective ?" parts though

For injectivity, you just showed ker(T) is not 0

That's pretty telling in itself right @toxic apex

[toxic apex]

Yup. Realised after solving it😂 . Thank you @slender yarrow

[clever totem]

does the modul Z/6Z have a basis?

my idea is {1}

a friend said 1*6=0

but 6 is already 0

[hard verge]

can someone explain the part that i underline pls

how does 1l = cos(alpha)1x + sin(alpha)1y

[winter pond]

Now I’m looking at it. I did it terribly wrong. 😭

It was an assignment and I submitted 😭

I got a restriction on b. What I did was when I equate it to 0 I moved XA to the other side and equate them

So AX=XA

[slender yarrow]

That could work also if you get the same results in the end tho