linear-algebra

[gritty sorrel]

there's one problem: how do we know that the determinant is the best way to do this?
@limber sierra does this question mean that the determinant as a tool is the only way to achieve this, or does it mean the determinant itself is unique from other determinants?

[limber sierra]

is the determinant

@gritty sorrel it's basically asking "is there another way to define a function with these properties?"

like ok let me ask another question

let's say i asked you for a function with the following properties:

i. f is continuous

ii. f(x) > 0 for all x

iii. f is not a constant function

there's a lot of functions that satisfy these properties

[gritty sorrel]

yeah

[limber sierra]

something like f(x) = |x| + 1, for example

or f(x) = x^2 + 1

or whatever

f(x) = 3x^2 + 7

etcetc

the theorem that "the determinant is unique" tells us that

if you define a function delta with these three properties

i.e. an alternating multilinear map with det(I) = 1

that map is the determinant

there is no other way to define the determinant

they'll all be the same

so, if you prove those properties about whatever definition you use

you're talking about the same function

[gritty sorrel]

and these rules give rise to the formula of the determinant

[limber sierra]

yep

[gritty sorrel]

wow

[limber sierra]

any formula that satisfies these 3 rules

is a formula of the determinant

it just so happens that there's 2 particularly convenient ones, leibniz and laplace

and you can prove that these both satisfy these 3 properties

[gritty sorrel]

i liked the philosophies you gave for (ii)

this is very precious information

is there a reference for this in case i forget?

[limber sierra]

this mathexchange topic https://math.stackexchange.com/questions/668/whats-an-intuitive-way-to-think-about-the-determinant gives a variety of intuitions, some of which resemble the one i gave

the fundamental "idea" here is that

we want a function that translates multiplicative information about matrices

into multiplicative information of real numbers

the way it does this feels kind of "magic", admittedly

it gets a bit abstract and is hard to fully grok at first

but some deep results tell us that this is the only way to do this

well, i should say

if you set up a formula that does this

it's equivalent to all the other formulas

for the determinant

[gritty sorrel]

yea its hard to grasp this all at once at first, but i can see where this is going

[limber sierra]

it's worth noting that the determinant historically arose as just a tool to talk about solutions to systems of equations

[gritty sorrel]

matrices were also historically the same right?

[limber sierra]

yep

but as mathematicians grew more interested in asking questions about, say, inverses of matrices

they realized that the determinant has a lot of useful properties

this idea of "translate multiplicative information about matrices into multiplicative information about real numbers" is particularly valuable when we want to justify characteristic polynomials

idk if you're familiar with those

but keeping that intuition in mind makes them feel a bit less "pulled-out-of-thin-air"

since the idea behind a characteristic polynomial is to find solutions to $A\mathbf{v} = \lambda\mathbf{v}$

[stoic python]

Namington:

[gritty sorrel]

i thought these were the characteristic equations of differential equations but apparently not

so im not familiar with them

[limber sierra]

and we can observe that the right hand side is a real number

while the left hand side is a matrix

and we're discussing multiplicative structure

so the idea to convert from the matrix understanding of multiplication to the real number understanding is immediately tempting

and the determinant (with an extra step in between) lets us do this

ah, no worries

just bringing this up to give further explanation of why thinking of the determinant as a "multiplicative correspondence" between matrices and real numbers is useful

[the "extra step in between" is rearranging the equation to (lambda I - A)v = 0 and then realizing that, by asking about when v is nonzero, we're asking about when (lambda I - A) has nonzero kernel, i.e. is not invertible. hence we're asking about when the determinant of (lambda I - A) is zero!]

anyway, if that's a bit too abstract/beyond what you're covering right now, no worries

i'm just hoping that it's clear that this definition does have deep reasons justifying its existence

[gritty sorrel]

if by kernel you mean null then yea thats right

[limber sierra]

yeah, "kernel" and "nullspace" are synonyms when talking about a linear operator

[gritty sorrel]

i self studied linear algebra in 'linear algebra done right'

[limber sierra]

haha, that text hates the determinant so

[gritty sorrel]

all i have to do in the future is to grab a pen and paper and write down these determinant properties and why do we want a function of these properties

and then convince myself

[limber sierra]

it's understandable to have undeveloped intuition there

[gritty sorrel]

yea, the author says they're unintuitive

[limber sierra]

they are

i don't disagree with that

i still think they're a valuable enough tool to introduce, but that's just my opinion

the formulas for det(A) still kinda feel like they "come out of thin air"

[gritty sorrel]

as an application, yeah, but what bugs me the most, if I wanted to be a mathematician for example and wanted to come up with theorems

i ask myself

[limber sierra]

even when you do all the math and justify why these properties are worth caring about

[gritty sorrel]

how do did these mathematicians come up with the definition of the determinant like what the hell

[limber sierra]

that's an interesting quirk of linear algebra in particular

linear algebra is so widely used across mathematics that the definitions are like

hyper-refined and focused

so it's sometimes hard to see where everything "comes from"

since it's been through the washing machine so many times

that it's almost unrecognizable from when it went in

on one hand, this makes it streamlined, which is pretty nice; but it can make it harder to find an intuition for some of the weirder stuff

and, as "weirder stuff" goes, the determinant is definitely pretty high on that list

[gritty sorrel]

yeah the cross product was pretty weird too until i understood some properties of the determinant

which is ( u x v) . u = 0

some texts dont bother to give motivation

i find that saddening

@A Waxwing Slain anyways, i really appreciate your time on this, this really helped, thank you

[wintry steppe]

Hi guys, I plan on starting a proofy linear algebra course in July. Aside from taking say Strangs OCW, how can I best prepare?

(What would you have focused on before taking linear algebra?)

[gritty sorrel]

@wintry steppe strangs OCW is more computational than proofy

[wintry steppe]

We take topics in LA with our calculus but it's computational

Hmm okay

[gritty sorrel]

i think linear algebra does not require prerequisites other than some mathematical maturity

if you're planning to start a proofy course then you definetly got that

[wintry steppe]

I'm bad at proofs but want to improve and haven't formally studied math in about 8 years. I'm working through Hammands Book of Proof

[gritty sorrel]

some basic set theory for 2-3 days will get you prepared

[wintry steppe]

If I'm not ready I can withdraw and take it next year, it's just part time.

[gritty sorrel]

then you can go through 'Linear algebra done right' it follows the style of axiom, definition, theorem, proof

and has good exercises

[wintry steppe]

Thanks, had my eye on that book.

I'll grab a copy.

[shut kite]

hey, should i ask my linear question from here?

[shut kite]

is there anyone help me to solve the questions 😦

[solar osprey]

Any help with this

My thoughts were to define scalars a., b, c.

Write W in terms of a sum involving a b c

Then apply the projection formula

But like, the final answer would be in terms of a b c

I am not sure if that is a correct approach

[sage mauve]

you can use gram-schimdt to find an orthoganl basis for W

then you use the projection formula

[dusky epoch]

the basis they give you is already orthogonal

[solar osprey]

Yea but @dusky epoch @sage mauve

why do I need Gram-Schmidt

in the first place

Lets say it is not orthogonal

why do I need to make it orthogonal

[dusky epoch]

gonna make shit a lot easier

[solar osprey]

@dusky epoch When computing projw (v), its simply <v, w>w / norm(w)^2

but here w is not a simple vector no?

its given as a span

[dusky epoch]

that's when you're projecting on a 1-dim space

[solar osprey]

so should I take any posssible vector

[dusky epoch]

but here dim W is 3

not 1

proj_W(v) is a vector in W, and v - proj_W(v) is orthogonal to W

[solar osprey]

so do i like take scalars a b c and write

w = av1 + bv2 + cv3

where v1 v2 v3 are the vectors in the span

[dusky epoch]

the vectors in the basis

[solar osprey]

yea

[dusky epoch]

yeah well

yeah

and then <w,v_i> = <v,v_i>

for i=1,2,3

[sage mauve]

:p computing dot products is computationally difficult for me

[solar osprey]

Im still confused about the thing lol

[shut kite]

Is there anyone who can help me, I've some answers for kinda basic questions of linear but don't have the answer key.

If there is , please contact me. I need a bit help.

[shut kite]

suppose that A is a (3x3) matrix and det(A) = - 3 . What is the det (4A) ?

help <@&286206848099549185> 😄

-12 ?

or -3/4

[brittle juniper]

what's the reasoning behind those attempts?

[shut kite]

dont have answer key

[brittle juniper]

did you just say numbers that seemed kind of legit?

[shut kite]

im trying to work my exam and could not solve one

which is this , suppose that A is a (3x3) matrix and det(A) = - 3 . What is the det (4A) ?

-12 or -3/4 i guess

[pallid rampart]

What is your reasoning

[shut kite]

as i know i should do -3x4 or -3x-1/4

its hard for me to explain myself in english sorry

[pallid rampart]

Just like, work out an example using 2x2 matrices

[sage mauve]

hell 1x1 matrix

[pallid rampart]

[shut kite]

ahh now i found it

i was completely wrong

[pallid rampart]

So what’s the answer

@shut kite

[shut kite]

-192

i guess

[pallid rampart]

Ah yes

Well

Did you look at the answer

Lol

[shut kite]

just got help of the web

[pallid rampart]

Well

It helps to do this

det(4A)=det((4I)A)=det(4I)det(A)

Where I is the identity matrix

So you have 4*4*4*(-3)

[shut kite]

i ve one more question but u scared me

[pallid rampart]

How

[shut kite]

@pallid rampart have an idea ?

x=(2s-3t+5, -3-2t, s, 7+4t, t) found this but not sure

[lean spoke]

i guess you should take the inverse of the big matrix and multiply both sides with it

[pallid rampart]

[lean spoke]

then you gain the matrix x

[quartz compass]

bold move

[pallid rampart]

Why didn’t I think of that

Big brain

Lol

[sage mauve]

just do gaussian elimination as usual

[pallid rampart]

But um, you can’t invert a 3x5 matrix

As much as I would like to

[lean spoke]

right

sorry

[shut kite]

😦

trying

thx

@sage mauve x=(2s-3t+5, -3-2t, s, 7+4t, t) is that true, i dont have answer key

[sage mauve]

you should know how to turn your equation into a system of linear equations

plug it in

see if the system is satisfied

[shut kite]

i couldnt dont know english well

is there a calculator

[pallid rampart]

Let $X=(x_1,x_2,x_3,x_4,x_5)^T$, then we have \begin{align*}x_1-2x_2+3x_3+2x_4+x_5&=10\x_3+2x_5&=-3\x_4-4x_5&=7\end{align*}

[stoic python]

Whoever:

[shut kite]

thanks

everyone

[pallid rampart]

You haven’t solved your problem yet

[steady fiber]

maybe he got it

[pallid rampart]

Try to eliminate the x_3 and x_4 in the first equation

[main drum]

hello

[pallid rampart]

But ok

[main drum]

i have a question

is this subdiscord busy

so I had this one on my hw yesterday

and I checked it with my textbook which was also saying true to my understanding

can anybody offer an explanation of why it would be false?

[quartz compass]

make A the identity matrix, what are its eigenvalues?

can you choose x so that it's smaller?

[main drum]

1

the smallest eigenvalue would be 1 right?

if identity

[quartz compass]

yeah all eigenvalues are 1

so you just need to pick x so that Q is less than 1

[main drum]

how can a matrix be less than 1

[quartz compass]

Q is effectively a scalar

it's a 1x1 matrix

[main drum]

yea but how can Xt be a scalar

[quartz compass]

it isn't, it's a vector

[main drum]

yea exactly

oh i see

so what type of X would make Q less than 1?

a zero vector?

[quartz compass]

yeah perfect

[main drum]

and therfore false

?

because Q can be less

[quartz compass]

yeah, clearly Q=0 which is less than the smallest eigenvalue of A

[main drum]

ok

[quartz compass]

you're welcome lol

[main drum]

thank

[spiral sonnet]

A and B are matrices. How can I get AB = 0 is B can't be the 0 matrix?

[quartz compass]

if the columns of A are linearly dependent, you can pick B to force them to add to 0

[spiral sonnet]

I'm confused!

[half ice]

"AB = 0 is B can't be the 0 matrix"
Huh?

Let B = 0 matrix
Then AB = 0

Unless I'm misunderstanding?

[quartz compass]

I think you are

I think A,B are nonzero matrices by assumption

but AB=0

like given some fixed nonzero A, when can you find a nonzero B so that AB=0, that's how I read the question

[limber sierra]

what's A

just an arbitrary matrix?

[spiral sonnet]

Yeah

A 3x3 matrix

[limber sierra]

then this is, in general, not solvable

for example, if A is the identity matrix

oh, so not an arbitrary matrix, but a specific one

okay

[spiral sonnet]

oh yeah

[limber sierra]

i mean you can write this as a big system of equations to solve if you wish

we have $\begin{pmatrix}-4&-1&2\0&-4&-8\2&2&2\end{pmatrix}\begin{pmatrix}b_{11}&b_{12}&b_{13}\b_{21}&b_{22}&b_{23}\b_{31}&b_{32}&b_{33}\end{pmatrix} = \begin{pmatrix}-4b_{11}-b_{21}+2b_{31}&-4b_{12}-b_{22}+2b_{32}&-4b_{13}-b_{23}+2b_{33}\?&?&?\?&?&?\end{pmatrix} = \begin{pmatrix}0&0&0\0&0&0\0&0&0\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

fill in the ?s yourself if you want, im too lazy

this gives us the first few equations in our system

[spiral sonnet]

Oh shit. I forgot about the addition part of matrix multiplcation

That makes sense now

[limber sierra]

$-4b_{11} - b_{21} + 2b_{31} = 0, -4b_{12} - b_{22} + 2b_{32} = 0$

[stoic python]

Namington:

[limber sierra]

and $-4b_{13}-b_{23}+2b_{33} = 0$

[stoic python]

Namington:

[limber sierra]

this is admittedly a tedious super computational approach, but its the most direct path that comes to mind

[spiral sonnet]

If I multiply the 2nd row the matrix A with the 1st column of matrix B. Does that give me b21?

[limber sierra]

uh, it should give you 0

$0b_{11} - 4b_{21} - 8b_{31}$

[stoic python]

Namington:

[limber sierra]

which should = 0

[spiral sonnet]

well yeah in general would that be how I get the a value for b21

[limber sierra]

oh you mean like

the position in the 2nd row and 1st column

of the product AB

yes

[spiral sonnet]

yeah

okay cool

[short magnet]

erm.. probably a stupid question but I'll ask it anyways.. heh :
let U be a subspace of V such that U != V .
if u is in U and v is in V\U (in V but not in U) .. I don't suppose we can say anything about where their sum will end up right ?

We can't say if u + v will be in V\U or in U, right?

[sonic osprey]

yeah

[short magnet]

ah, okay ! thanks :"3

hmm

I checked an online solutions manual and they made use of that :

[vital swallow]

that seems right. the above commenter was probably thinking of the subspace complement of U in V instead of the set difference V\U

[flint flicker]

I’m confused by this proof problem. I’m not sure where to go from here. My goal is to transform the right side of the equation into Rng(T). And therefore prove T is onto

[sick dragon]

Does a linear system with more unknowns than equations always have an infinite number of solutions?

[short magnet]

Yes, I believe so.

[sonic osprey]

No, it can have no solutions

[sick dragon]

Can you explain it please?
The vocab of this class is tripping me up

[short magnet]

0.0

[sonic osprey]

0x + 0y = 1, has more unknowns than equations, but it has no solutions

[short magnet]

Example (?)

But the coefficients are 0 ..

[sonic osprey]

So?

[short magnet]

Nothing...

[sick dragon]

are 'terms' coefficients?

[sonic osprey]

You can come up with other examples, x + y + z = 1, x + y + z = 2, has more unknowns than equations

[short magnet]

Oh

It depends on if it's homogeneous or not...

If it is homogeneous then it will certainly have infinite solutions right?

[sonic osprey]

what does homogeneous mean

[short magnet]

(more variables than equations)

No constant term

ax + by = 0 is homogeneous

ax + by + c = 0 is inhomogeneous.

[sonic osprey]

oh sure, that's true

[sick dragon]

well i still dont know anything

ok so a are coeffecients for the variables

b is the solution

both are constants

[wintry steppe]

how do i do this

[wintry steppe]

nm i got it

[kind summit]

Hey, working on a personal project and ran into a scenario I think I need to do a change of basis on. I have a curve

$A(t)=ai+bj+ck$

[stoic python]

Gestalt Shift:

[kind summit]

I also have three unit vectors that are orthogonal, B(t), C(t), D(t)

I want to project the curve A(t) onto the plane containing B(t) and C(t)

The way I'm going about doing this is by doing a change of basis so that B(t), C(t), and D(t) are <1,0,0>, <0,1,0>, and <0,0,1>

Then applying that change of basis to A(t)

And projecting A(t) to the plane by "removing" D(t)

Could someone confirm if my matrices are set up right to do this if I send them?

I believe this falls under linear algebra, as I'm only taking the class next semester, sorry if it doesn't

[wintry steppe]

can someone help me understand this proof

i dont see how this proves that it cant be zero

maybe all the a constants are trivial and the b's aren't

?

[limber sierra]

the definition of linear independence is that, for ANY set of coefficients $a_1, a_2, \dots a_n$, if $a_1s_1 + \dots + a_ns_n = 0$, then $a_1 = \dots = a_n = 0$ necessarily

[stoic python]

Namington:

[wintry steppe]

can't that still be true

cause if they're all zero they're just gone

[wintry steppe]

yeah

[slender hull]

That was impressively fast circling

[limber sierra]

i'm not sure what you mean

the point is that the b_1, b_2, etc b_k aren't all 0

so we have a found a set of coefficients a_1, a_2, ... a_n

that are not all zero

yet a_1s_1 + ... + a_ns_n = 0

in particular, this "set of coefficients" is

b_1, b_2, ..., b_k

and then all the rest 0

[wintry steppe]

o

why does existence of non zero B's show that a's are not all zero

cant they cancel out some way

[limber sierra]

review the definition of linear independence

if something is linearly independent, that means that

if you have $a_1s_1 + \dots + a_ns_n = 0$

you MUST have $a_1 = a_2 = \dots = a_n = 0$

[stoic python]

Namington:

[limber sierra]

no negotiating

[stoic python]

Namington:

[limber sierra]

but we know that $b_1s_1 + \dots + b_ks_k = 0$

[stoic python]

Namington:

[limber sierra]

so we have a set of coefficients

such that it adds to 0

but those coefficients aren't all 0

[wintry steppe]

oh so b is one

[limber sierra]

contradiction

[wintry steppe]

er

B is a

[slender hull]

yes the b’s are just saying “this is a different possible choice we could use for some subset of the a’s”

And it’s a contradiction because that would imply you could find some choice of a’s which aren’t all 0

[wintry steppe]

ah ok that makes sense

thanks guys ☑️

[sick dragon]

isn't this just x_1? i wish i had a teacher for this shitshow of a class

[wintry steppe]

i think its 1 and 0

leftmost item in each row

[golden magnet]

isn't this just x_1? i wish i had a teacher for this shitshow of a class
@sick dragon

you're not alone :(

[sick dragon]

Yeah, it's x_1

[wintry steppe]

Anyone can help?

I tried a proof by contradiction

but couldn't do it in the end

This is hint:

This is problem

I supposed it is finite dimensional, and so I let {a1, ..., an} be a basis

but I couldn't use the hint to get to a contradiction

I tried using the span definition of a basis but maybe I missed something there

assume it's N dimensional

yep

then any set of N+1 vectors is linearly dependent

let a be a real number

ah

{1, a, a^2, ...., a^n} is a set of N+1 vectors

right

do you see what to do now?

Let me think

{1, a, a^2, ...., a^n} is a linearly dependent set of vectors

what does that mean

what does that tell you about a

I see

so a^n can be written as linear combo of the others

which contradicts the fact that some real number are not algebraic

so a^n can be written as linear combo of the others

how do you know that?

it's true, but it's not immediate

let me look at dependence definitiona gain

again*

I'm not sure, because it says {a1, ..., an} are dependent then not all ci's (fields) are 0

that's the definition, yes

but then what if a^n has a coefficient of 0, like how to get around that

there exists a linear combination sum_i c_i a^i where not all c_i are zero

it doesn't need to be a^n

it can be a lower term

you'll still get an equation of the form c_0 + ....+ c_k a^k = 0

for some c_k != 0

oh thats really smart

thanks

[obsidian jackal]

Given this is line l, I am to find w1 and w2 where V=w1+w2 and w1 is parallel to l and w2 is perpendicular to l

So w1 is (1,2,5) and w2 is the dot product of w1 and w2 equals zero

Is this right?

[broken hawk]

what is V?

anyway (1,2,5) is parallel to l and a vector w₂ where w₂·(1,2,5)=0 would indeed be perpendicular to l, yes

[obsidian jackal]

@broken hawk okay thank you very much

i forgot to mention but V = (2,1,3). It seems like it is a vector @broken hawk

[lean spoke]

hello i can't find the values of x y or z and i was thinking if i could give one of them a value myself and write the other 2 variables with respect to it(I saw something like that when i google about this question) if you explain me how to find eigen vectors in this question or tell me what i am doing wrong i will appreciated

i will be*

[quasi vale]

It would be better if you turned your last matrix which is in echolon form to reduced echolon form

But anyways, as you can see, we have x and y as our pivot variables and z is our free variable.

You can let z = t, where t is any real number. From the second equation, you get 8y = 6z or y = (3/4)t, since z = t

Plug this in equation 1, we have -12x + 2(3/4)t + 3t = 0

Which gives us x = 0.375 or 3/8t

You can now write your set of vectors X = <x,y,z> as X = <(3/8t), (3/4)t, t> or just X = t<(3/8), (3/4), 1>. All the eigenvectors of eigenvalue 13 will be formed by the vector <3/8, 3/4, 1>. @lean spoke

For example, you can let t=0.5 and you'll get <3/16, 3/8, 1/2> which is an eigenvector of eigenvalue 13.

[smoky lagoon]

this is sort of a linear alg question and not something I actually need to know for class

but we learned about how recursive sequences can be expressed with diagonal matrices and stuff

for example the fibonacci sequence

something I'm wondering is what the square root of the fiboacci recurrence looks like in this matrix form, and if it would even work at all.

as shown in this paper

never mind

square roots arent linear

oops

[dry spear]

in an ordered basis how do you know what order the vectors should be?

like if im finding an ordered onb for a matrix

[subtle walrus]

what do you mean "should be"? i don't think it really matters in most cases

[dry spear]

ill show you an example

for C_s<-b i had the first 2 columns switched

does that matter?

[broken hawk]

it doesn’t matter at all but things wrt to the basis (e.g. the way a matrix looks) will be different

[pale shell]

Is column space the same as rank?

[gray dust]

no

[pale shell]

Difference?

[gray dust]

we call the dimension of the col space the rank

[pale shell]

Okay I get that so if we have a matrix

And it does a transformation into say two dimensions

So a plane

Then it is rank two

?

[gray dust]

sure

[pale shell]

Ohhh

So if col space is dim

I MEAN RANK

oof

IF Rank is dim(colspace)

How to define colspace

[gray dust]

shouldn't you have a definition of col space before asking how col space relates to rank?

[pale shell]

I am ask it one definition because I thought they were

SAME THING.

[gray dust]

col space=span of col vectors

[pale shell]

So you know how a matrix is basically a transformation of space right?

Then the COLUMNS.

Of that matrix

If you take all liner combinations

Then you get col space.

Or what?

[gray dust]

you're parroting what i said

[pale shell]

Okay sorry im trying to learn ok

So wait

That means

That lets say I HAVE. Some matrix. And I find what col space is of it. Then the answer I get is somefing likeR^2 or R^3 or subspace or line or zero vecror right?

[gray dust]

if the matrix has all real entries then its col space will be a subspace of some R^n space

[pale shell]

OKAY.

So that can also be

THAT R^n

I think I am making progress

So the col space can just be that same R^n as it is transformatin?

[gray dust]

your wording is hard to follow

[pale shell]

Im sorry what I mean to say is

If the col space can be a subspace of the R^n that the matrix is transforming

Then can the col space of that matrix be R^n as well?

Or can it only be a subspace?

[gray dust]

the R^n that the matrix is transforming
for example idk what you're saying here

[slow scroll]

if you have a linear transformation T: Rn to Rm, then the column space of T is always a subspace of Rm. Whenever the column space IS Rm, then T is surjective

[pale shell]

I mean the dimension that the matrix lies in

So R^n which the matrix exists in

[slow scroll]

T would be a mxn matrix...

[pale shell]

So can the col space be that R^n

Not you I am talking to robotu

[gray dust]

kx try not to answer earl's questions if they don't make sense

[pale shell]

Im sorry ok

Im trying

[gray dust]

ok remember from high school algebra when you learned about functions and their domains/codomains/images

or at least i'm hoping you remembered that

[pale shell]

Mhmmhmh

Yes

But listen I get what you mean

CAN THE COL SPACE SPAN ALL OF R^N

that is my only concern

or can it only span a smallar subspace

[gray dust]

i'm trying to fix your vocab because it's all meaningless word salad now. matrices represent linear maps between vector spaces

[pale shell]

Okay

Sorry

I know

Matrices represent tramsofrtmations

[gray dust]

linear maps

[pale shell]

Linear transformations.

In which the origin remains fixed and grid lines are parallel.

Same thing.

[gray dust]

sigh ok

[pale shell]

Look I am just looking for help and I am not trying to frustrate you

I just want to be exactly clear on the notions of col space and rank

[gray dust]

here's where i steal what kx said

[pale shell]

Okay

[gray dust]

say you have a linear map T whose domain is R^n and codomain is R^m, then the image of T, which is equivalent to the col space of the matrix that represents T, is a subspace of R^m

[pale shell]

Okay. I get all of that

But

Let me clarified two things

One: so the col space is the span of the column vectors which make up matrix
Two: can a subspace of R^m also be R^m

[gray dust]

a vector space is a subspace of itself

[pale shell]

So one is correct thne?

[gray dust]

1 is taken directly from what i just said

2 is correct

[pale shell]

Thank you

[gray dust]

also if you're talking about subspaces i hope you've got the defn of subspace down pat too

[pale shell]

I will eat dinner

And then

I will come back for null spaces and kernels

[pale shell]

Can someone please explain null space and what a kernal is

[half ice]

Take a linear transformation T.
T takes vectors from a vector space A, and puts them into a vector space B

Now, some of these vectors in A will be mapped to B's 0 vector

These vectors make a subset of A that we call ker(T)

Understand up to there? @pale shell

[pale shell]

Uhh

What do you mean by T tzkes vectots from a vector space

And puts them into vector space b

Do you mean like a matrix

[limber sierra]

T's domain is a vector space A, and its codomain is a vector space B

or, T is a map A -> B

[half ice]

Oh no okay.
So a linear transformation is a function. It takes vectors from one vector space and puts them into another

[pale shell]

Which just transforms every vector

[half ice]

It ALSO is linear, but that's beside the point atm

[pale shell]

Yes and thats a matrix a linear transformation right?

[limber sierra]

linear transformations correspond to matrices when the correspondence makes sense, yes

[pale shell]

Yes

[half ice]

Forget matrices at this very second. We want to think of them as functions right now

[pale shell]

Ok same thing

So what would the null space be

And also the kernels

[limber sierra]

the null space and kernel are the same thing

[pale shell]

o

[half ice]

T will map some of A's vectors to B's 0 vector

Whatever vectors get mapped to 0, are the null space of T

[pale shell]

Omg so the null space is just a set of vectors, in which when they are get transformed, become the zero vector?????

[limber sierra]

yes

[pale shell]

OMG THANKS

[limber sierra]

these vectors happen to form a subspace of A

[pale shell]

I LOVE YOU

[limber sierra]

hence the term "null space"

[half ice]

Yeah don't miss that important part. This set of vectors is also a vector space itself

[pale shell]

Liner algebra is so interested

[limber sierra]

if you prefer the matrix representation, suppose $M$ is the matrix representing the linear transformation

[stoic python]

Namington:

[pale shell]

Mmhmhm

[limber sierra]

then, the null space of $M$ is the set of solutions to the equation $M\mathbf{v} = \mathbf{0}$

[stoic python]

Namington:

[pale shell]

0 vector.

Not zreos

Just zeo vector rigght

[limber sierra]

yes

that's why the 0 is bolded

[pale shell]

So interesting things

But

[limber sierra]

0 indicates the zero vector here

[pale shell]

How to calculate

And is there one solution

[limber sierra]

well, if you have a matrix representation, then the naive thing to do (which works fairly well) is to multiply Mv

[pale shell]

OR MANY

[limber sierra]

the null space is... well, a vector space, so it'll "generally" have many many vectors

usually one asks for a basis of the null space, or something similar

[pale shell]

Uhhh wdym by vectis space

Also do you mean INFINITE VECTORS

[limber sierra]

if A and B are infinite-dimensional, yes (unless there are no solutions besides the 0 vector)

[half ice]

It's probably not fully covered in your course. Basically, a vector space is a place where you can add vectors together, and scalar multiply them

Without breaking anything

[pale shell]

Waxis i am confused on what is that oean

Ok wait I am now confused again

[limber sierra]

as mentioned, your course probably isnt covering this in full detail; the key thing here is that the null space forms a subspace of A

so it'll have a bunch of vectors in it

usually

[pale shell]

Soooo what not breaking anything

What is BROKEN.

Is the zeros vector ALWAYS A SOLUTION

[limber sierra]

to Mv = 0, yes

[pale shell]

also infinite dimensions isn’t real.

[limber sierra]

what

[half ice]

The zero vector is always in the nullspace, yes

[pale shell]

.......

How can you have infinite dimensiosn that makes no cense

[limber sierra]

well, R is infinite dimensional over Q

[half ice]

Infinite dimensional vector spaces are plentiful. All functions on the real numbers is an easy example

[limber sierra]

but this is beyond the scope of your course it sounds like

[pale shell]

What.

What is q akd r

[limber sierra]

we're not using "dimension" in the context of spatial dimensions here

Q and R here denote the vector spaces consisting of the rational numbers and the real numbers, respecctively

again, don't worry about it, beyond the scope of the course

for all the vector spaces you're likely to see, yeah

there will either be exactyl 1 or infintiely many elements of the null space

[pale shell]

Is my course dumbed down or is that most courses or what.

[limber sierra]

(and if it's "exactly 1", then that element is 0 the zero vector)

proof-based linear algebra courses explore the concept of vector spaces in full generality

most computational ones only look at "common" vector spaces

R^n, C^n

with scalars from the field R or C or w/e

[half ice]

95% of people usually only need the applied course

[limber sierra]

and subspaces thereof

[pale shell]

Woah that is a lot of uhhh

Confusions but

Let me just try to unpack this

[limber sierra]

anyway you might've heard the result before, "homogeneous systems have either 1 solution or infinitely many solutions"

and that's what's going on here

either your null space is just 0 the zero vector

or it includes 0 and infinitely many other vectors

in either case, it forms a subspace of the vector space A

[pale shell]

yes but before he said all these things vector spaces NO BREAKING THINGS.

What is it breaking.

[limber sierra]

well, here's an example of a structure that isn't a vector space

[pale shell]

Yes

Time to see it

[limber sierra]

consider the set of vectors $\begin{pmatrix}a\1\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

where a is a real number

[pale shell]

Whats bad about.

[limber sierra]

this is certainly a subset of R^2

but the problem is

if we add two vectors from this set

say $\begin{pmatrix}3\1\end{pmatrix} + \begin{pmatrix}2\1\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

it takes us outside of the set

we get $\begin{pmatrix}5\2\end{pmatrix}$

[stoic python]

Namington:

[pale shell]

Ok

[limber sierra]

so vector addition "doesn't make sense" when we only look at this subset

[pale shell]

Then one xzmple of bector space please

[limber sierra]

hence it's not a vector space

a vector space would be any structure you're probably "used to" working with

$\bR^2$ for example

[stoic python]

Namington:

[pale shell]

Ohhh

Ok so like i am getting tbis

[limber sierra]

but also certain "special" subsets there

[pale shell]

Wow so this is so interesting to me

[limber sierra]

anyway, the key result is that null spaces always form one such "special" subset

[pale shell]

Like to kernels and subspace and nullspace

What is the special subset of null space

[limber sierra]

the null space itself

i.e. null spaces are always vector spaces

[pale shell]

..

[limber sierra]

and specifically subspaces of their domain

so if you have a real 3 by 3 matrix, then the null space of this matrix will be a subspace of R^3

what this means is, we can ask questions like "find a basis of the null space of this matrix/transformation"

[pale shell]

So tbat is where it is making contain of ONE ZERO VECTR, ONE CLOSED UNDER VECTOR ADDITION, CLOSED UNDER SCALAR MULTIPLY.

right.?

[limber sierra]

that's what you need to check to verify that something is a subspace, yes

[pale shell]

Mmm

Final question.

Oh btw

It is closed under both those ops right?

[limber sierra]

thats... what we just said

[pale shell]

Final question

[half ice]

Vector spaces have like 8 rules. You won't learn them fully in an applied course. But there's a short list of 3 rules for proving something is a subspace

[pale shell]

How to calculating one kernel/null space please.

(vector space dont have 8 rules also)

Their just must be a notion of adding vectors and multiplying by scalars for a vector spacis,

[limber sierra]

that's if you already know the operations behave nicely

ie you're taking a subset of an already-known vector space

anyway that's a side tangent

[pale shell]

ye sorry bout tht

my real question is calculate one null space

[limber sierra]

there are a few different methods

some more convenient in certain situations

to find a null space of a matrix, it's generally most convenient to just multiply it out

let me give an example

[pale shell]

I like eoled

• exampis

• examples

[limber sierra]

let's say we have a matrix $\begin{pmatrix}3&-1\-2&1\end{pmatrix}$ and want to find the null space of this matrix

[stoic python]

Namington:

[pale shell]

yes time to find one null space of that

[limber sierra]

this will be the set of solutions to the equation $\begin{pmatrix}3&-1\-2&1\end{pmatrix}\mathbf{v} = \mathbf{0}$

[stoic python]

Namington:

[limber sierra]

that is

[pale shell]

Mhmhm

Making sences

[limber sierra]

the set of solutions to the equation $\begin{pmatrix}3&-1\-2&1\end{pmatrix}\begin{pmatrix}v_1\v_2\end{pmatrix} = \begin{pmatrix}0\0\end{pmatrix}$

bleh

[stoic python]

Namington:

[pale shell]

oh noe

Yes still making so much sence

[limber sierra]

so let's just multiply out the left hand side

using matrix multiplication rules

[pale shell]

Mhmhm which are easy if you lnow just ONE TRANSFORM.

Because a matrix is actuzlly the new BASIS VECTORS

[limber sierra]

$\begin{pmatrix}3&-1\-2&1\end{pmatrix}\begin{pmatrix}v_1\v_2\end{pmatrix} = \begin{pmatrix}3v_1-v_2\-2v_1+v_2\end{pmatrix} = \begin{pmatrix}0\0\end{pmatrix}$

[stoic python]

Namington:

[pale shell]

left side is î and right side is j hat

[limber sierra]

so now we just solve this system

this gets us $3v_1 = v_2$ and $2v_1 = v_2$

[stoic python]

Namington:

[pale shell]

Time for elimation

[limber sierra]

so the only solution will be $v_1 = v_2 = 0$

[stoic python]

Namington:

[limber sierra]

so here, the null space is just the zero vector

[pale shell]

ONLY THE SERO VECTOR.

but that only give one sol’n

[limber sierra]

right

so here the null space is {0} and therefore a basis for it is the empty set

[pale shell]

Whzt aabout he other ones if they are just hiding

[limber sierra]

[pale shell]

Like this one time I was solving this sytem

And I got this one solution

But their wzs also other solutions too

And they were just hiding

[limber sierra]

anyway, let me give another example

[pale shell]

Verify difficult

Yes other example please

[limber sierra]

lets say we had the matrix $\begin{pmatrix}3&-1\-2&0\end{pmatrix}$ instead

[stoic python]

Namington:

[pale shell]

This one was just zroes

[limber sierra]

so very similar to the first example, but i changed one of the entries

now we need to solve $\begin{pmatrix}3&-1\-2&0\end{pmatrix}\begin{pmatrix}v_1\v_2\end{pmatrix} = \begin{pmatrix}0\0\end{pmatrix}$

[stoic python]

Namington:

[pale shell]

Thats why i like liner zlgebrz because it is just zeroes everywhere and that is my favorite number

[limber sierra]

again, the most direct method is to multiply out the left hand side

$\begin{pmatrix}3&-1\-2&0\end{pmatrix}\begin{pmatrix}v_1\v_2\end{pmatrix} = \begin{pmatrix}3v_1 - v_2\-2v_1\end{pmatrix} = \begin{pmatrix}0\0\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

so this gets us the system $3v_1 - v_2 = 0, -2v_1 = 0$

[stoic python]

Namington:

[pale shell]

Noooo

Its all zrroes again

[limber sierra]

yep, same story

let's do one more example

i promis ethis one won't be all zeroes

[pale shell]

Ok lets do one without so many zeris

[limber sierra]

let's find solutions to the matrix equation $\begin{pmatrix}3&-1\-3&1\end{pmatrix}\mathbf{v} = \mathbf{0}$

[stoic python]

Namington:

[limber sierra]

$\begin{pmatrix}3&-1\-3&1\end{pmatrix}\begin{pmatrix}v_1\v_2\end{pmatrix} = \begin{pmatrix}3v_1 - v_2\-3v_1+v_2\end{pmatrix} = \begin{pmatrix}0\0\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

solving these equations gets us $3v_1 = v_2$ and... $3v_1 = v_2$

[stoic python]

Namington:

[pale shell]

Zero vector agizn

[limber sierra]

not quite

[pale shell]

Huh

[limber sierra]

well, some of our equations here are redundant

we basically end up with just the equation

$3v_1 - v_2 = 0$

[stoic python]

Namington:

[pale shell]

Wait so v1=(1 1) and v2 = (3 3)

BUT WHAT

SEPRATE VECTORS

[limber sierra]

no, remember that $v_1, v_2$ are individual entries in a vector $\mathbf{v}$

[stoic python]

Namington:

[limber sierra]

they're not vectors

but this DOES tell us that

v_2 is 3 times the size of v_1

[pale shell]

Right forgot it

Got to be it hmmmm

[limber sierra]

in other words, since $3v_1 = v_2$

[stoic python]

Namington:

[limber sierra]

we can write

[pale shell]

V1 e is one third

V2 equal one

Finish

[limber sierra]

$\begin{pmatrix}v_1\v_2\end{pmatrix} = \begin{pmatrix}v_1\3v_1\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

that is one vector in the null space, yes

but there are more

[pale shell]

Uh oh

Ho many

[limber sierra]

infinitely many

[pale shell]

WHAT

[limber sierra]

for any choice of v_1

[pale shell]

HOW DO YOU KKWOW THAT INFINITE

[limber sierra]

you can construct a vector $\begin{pmatrix}v_1\3v_1\end{pmatrix}

[stoic python]

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

[limber sierra]

and this vector will be a solution to the matrix equation

[pale shell]

so what this tell it is that

[limber sierra]

in this case, though

finding a basis for this space

isnt very hard

probably the most "obvious" is setting v_1 = 1

[pale shell]

How do you know

The other ones

Aren’t inifnite

[limber sierra]

then we get the basis $\left{\begin{pmatrix}1\3\end{pmatrix}\right}$

[pale shell]

And tjis one is

[stoic python]

Namington:

[limber sierra]

well, the other ones, we only got 0 as a solution

[pale shell]

Oh

[limber sierra]

that's only one vector

but in this case

[pale shell]

Therefore

[limber sierra]

we got a solution for any choice of v_1

[pale shell]

It must only be

[limber sierra]

so the other ones are finite, since there's only one solution, 0

but in this case, there's a solution for any choice of v_1

[pale shell]

One solution or infinite many

[limber sierra]

so there's infinitely many solutions

right

[assuming your vector space has infinite cardinality]

[pale shell]

what.

[limber sierra]

ignore that if you havent seen finite vector spaces

[pale shell]

waif so how do we know zero is only a solution to the other kernels we did

Whzt if there are other ones and we just didn’t spot it

[limber sierra]

we proved that the system's only solution is the 0 vector

for example, when we had this system

-2v_1 = 0 implies v_1 = 0

so we have -v_2 = 0

[pale shell]

How to know prove it thzt only zero

[limber sierra]

which of course means v_2 = 0

[pale shell]

OH I SEE

OK OK THANK

I get now

Wow so thznks so much

I now understand column space, rank, null space, and kernels

Such interedting things

[thorn hornet]

I don't know if it belongs here, but my problem is based on graphs. How do I calculate the number of different ways a graph can have from one point to another whereby the graph contains of branches that are connected to other branches etc.. The Problem is the amount of following brachnes can vary, s not every branch has the same amout of following branches.

Example: Green point is start point, Red point is end point
blue lines are branches

There are many different possibilities to get to red. How do I calculate them?

[vital swallow]

that could be a combinatorics question, but is likely more of a slick programming type of question. my intuition is that this would generally be very difficult

[thorn hornet]

So there is no ways to calculate it by hand?

[vital swallow]

possible, sure. but time consuming. What I meant was: I don't think there's any useful mathematical insight

[thorn hornet]

I only need to calculate one such map

one that has 22 possible ways. But I don't know how I da it by hand

[waxen spear]

Hello everyone, I got stuck at this question. Can someone help me? I tried to use Cayley-Hamilton theorem but I couldn't figure out how to find adj(A). Thanks ☺️

[vital swallow]

you know that p(A) = 0, so A^3-7A^2+5A-9=0, or A^3-7A^2+5A=9

[thorn hornet]

I don't know if it belongs here, but my problem is based on graphs. How do I calculate the number of different ways a graph can have from one point to another whereby the graph contains of branches that are connected to other branches etc.. The Problem is the amount of following brachnes can vary, s not every branch has the same amout of following branches.
Let's say I have a tree with branches that each have 2 further branches. How do I caculate that for a layer l

[waxen spear]

@vital swallow yes, I figured that out

but I don't know how it is related to adj(A)

I know that adj(A) = A^-1 * det(A)

[vital swallow]

you can use the above relation to find a formula for A^{-1}

[pale shell]

A subspace is closed under vector addition and scalar mutliplication

Does that mean that

If you add two vectors you will just get another vector ORRRRR another vector IN THE SUBSPACE

[vital swallow]

in the subspace

[waxen spear]

@vital swallow Could you give a little hint about it?

[vital swallow]

how to find A^{-1} from the relation A^3-7A^2+5A=9 ?

[waxen spear]

A(A^2 - 7A + 5) = 9

9*A^-1 = A^2 - 7A + 5I

[vital swallow]

yep

given the characteristic polynomial, you also know what det(A) is

[waxen spear]

is it 9?

-9

[vital swallow]

hm... if it +9 or -9 depends on which definition of characteristic polynomial you are using

[waxen spear]

Dimension of matrix is 3 so -1*det(A) = -9

Then det(A) = 9

Is it correct?

[vital swallow]

It depends on the definition of "characteristic polynomial" being used

[waxen spear]

Do you mean this?

[vital swallow]

yeah. some people define it as det(A-lambda I) and that changes the sign of the determinant

[waxen spear]

Oh, got it

[vital swallow]

-9 = p(0) = det(-A) = (-1)^3 det(A) ==> det(A) = 9, like you said

so now you have the determinant and the inverse, so you can form the adjoint

[waxen spear]

Thank you :D Got it

[somber wigeon]

If V is an infinite-dimensional vector space with a positive definite inner product, can I have U a subspace of V that isn't V such that its orthogonal complement is {0} ? I know it is not possible when V is finite

I think I have such an example but it seems counter intuitive

so I thought maybe I was wrong

(the example is the subspace of polynomial with real coefficients, its inner product being the integral from 0 to 1 of the product of the 2 polynomial, I think I have evidence the subspace of polynomial with root 0 has {0} as its orthogonal complement)

[vital swallow]

that doesn't fit here @wintry steppe

[wintry steppe]

Ok

Where then?

[vital swallow]

I think number theory

@somber wigeon infinite dimensional vector spaces might fit better under analysis-and-pde

but that sounds like a reasonable example

[eager burrow]

@somber wigeon your example could work, yeah! in infinite-dimensional vector spaces, these unintuitive things can happen when you're working with spaces which are not complete; for example, the space of polynomials with respect to your inner product is not a complete one

that's why in infinite dimensions, you'll have to pay close attention to topologies and stuff like that; if you don't know much about that, you probably shouldn't care too much right now, but if you ever start learning about this, then be prepared for a whole lot of norms, inner products and topologies that pop up in infinite dimensions

[fathom flax]

How can I proove spectral theorem?

[pastel saddle]

Can anyone help me out with this one?

[vital swallow]

what have you tried, Jcrdan?

[slender hull]

For Jcddan’s problem does T*=-T imply that each element of T is purely imaginary since only the conjugate purely imaginary numbers is -1 times the original?

[vital swallow]

$\begin{bmatrix}0&1\-1&0\end{bmatrix}$ would be such a map

[stoic python]

Alek:

[vital swallow]

wait..

yeah

[dusky epoch]

if T is antisymmetric then iT is symmetric

[pale shell]

How is the null space of some matrix in R^n always a subspace of R^n?

[dusky epoch]

what do you mean by "matrix in R^n"?

[pale shell]

Linear map of that R^n

[dusky epoch]

do you mean an n by n matrix

[pale shell]

Sure

[dusky epoch]

ok, do you know the definition of a subspace, and do you know the definition of Nul(A)?

[pale shell]

Yup

[dusky epoch]

Nul(A) = {x in R^n | Ax = 0}

[pale shell]

Yeah so the set of vectors which are transformed to zero vector

[dusky epoch]

can you show that the following is true?

• the zero vector is in Nul(A)
• for any two vectors x,y in Nul(A), their sum x+y is again in Nul(A)
• for any vector x in Nul(A) and any scalar c, the vector cx is again in Nul(A)

[pale shell]

For two and three I don’t get why

One I do

Like what if you have say a null space of the zero vector and [3 -2] (vertical)

[dusky epoch]

are you trying to refer to the set that contains the vectors [0 0] and [3 -2] and nothing else?

[pale shell]

Yeah for the kernel

[dusky epoch]

that is not the kernel of any matrix

[pale shell]

......

Null space=kernel

[dusky epoch]

yes

the set {[0 0], [3 -2]} is not the nullspace of any matrix

[pale shell]

No look I’m referring to a specific random case where I don’t get why that is true

If you add that second vector to itself you won’t get another vector inside the set

[dusky epoch]

the set you presented is not the nullspace of any matrix and hence cannot be required to be a subspace of, in this case, R^2

any matrix's nullspace is closed under addition and i was about to explain exactly why

[pale shell]

For it to be closed under scalar multiplication and vector additions means it has to have an infinite number of vectors if it has anything more than the zero vector

[dusky epoch]

yes, that is exactly the case

any vector space with dimension 1 or greater contains an infinite amount of vectors

[pale shell]

Subspace*

[dusky epoch]

a subspace of a vector space is a vector space in its own right

[pale shell]

Okay but I don’t want to abstract it too much

I just want to focus on the kernel

[dusky epoch]

a little bit of abstraction is necessary

[pale shell]

So then it would have to contain an infinite number of vectors if there is anything more than the zero vector?

Sure

[dusky epoch]

you're overthinking it

take x, y in Nul(A). then Ax = 0 and Ay = 0. therefore A(x+y) = Ax + Ay = 0 + 0 = 0. therefore x+y in Nul(A).

[pale shell]

Yeah sorry about that

Okay makes sense

But

That means there must be either zero or infinite null space

• zero vector

[dusky epoch]

yes, that is exactly the case. the nullspace of a matrix either contains only the zero vector (in which case we call it trivial) or it has dimension >=1 and hence contains infinitely many vectors.

for closure under scaling:

take x in Nul(A) and c in R. then Ax = 0. Then A(cx) = c(Ax) = c*0 = 0. therefore cx in Nul(A).

[pastel saddle]

@vital swallow I think I figured it out, is this right?

[dusky epoch]

this argument looks correct

[pastel saddle]

Sweet, ty

[vital swallow]

yep

[dusky epoch]

@pale shell does what i said up there make sense to you?

[vital swallow]

depending on audience, you may want to clarify slightly more why lambda is imaginary

[pale shell]

Yes

Thank you for explaining that

[dusky epoch]

okay, any other questions before i finally go to sleep?

[pale shell]

I was just confused if there was a finite amount of vectors but that was cleared up

No, thanks for your help.

[pastel saddle]

Can someone help me with the first line where I have the "?"

[limber sierra]

what's your confusion

$S((X+Y)S^{-1}) = S(XS^{-1} + YS^{-1}) = SXS^{-1} + SYS^{-1}$ by distributivity

[stoic python]

Namington:

[pastel saddle]

Oh yeah lol. I knew it was something simple I was missing ahah. Thank you

[spiral sonnet]

If the determinant of this matrix is 5

how can I use that to find the determinant of this matrix?

[steady fiber]

$\det(A) = \det(2M) = \det(2IM) = \det(2I)\det(M) = 2^4 \det(M)$

[stoic python]

PorosInMyAshe:

[spiral sonnet]

how'd you go from det(2M) to det(2IM)?

oh wait nevermind, but where did 2^4 come from

[steady fiber]

multiplying by the identity matrix does nothing

think about what the determinant of a 4x4 identity times 2 would be

it's very easy to calculate by hand if you want

[spiral sonnet]

oh lol bad math

I see

[steady fiber]

it's just 4 2s along the diagonal

[spiral sonnet]

multiply diagonal

[fervent moon]

hello guys, anyone wo can maybe help me understand a problem better?

[stuck tide]

sure

hmu

[left hill]

Um can I make a vector matrix that's composed of [e^(-2t) -e^(-2t)] into [ 1 -1] by dividing through e^(-2t)
Or can I not do that since it's a function of t

[vital swallow]

I think we need more context

[sonic osprey]

You can't do it

Your scalars have to be real numbers

[left hill]

Darn
Ty

[gloomy arrow]

There are distinct eigenvalues so they have to be linearly independent correct?

[pallid rampart]

Yes

[left hill]

if the determinant of the wronskian equals 0 does that mean that the set is linearly independent?