linear-algebra

[silent sandal]

If I have a real invertible nxn matrix A such that there is a non0 vector x having: <Ax, x> = 0. Can I conclude A has a complex non-real eigenvalue?

[wintry steppe]

all real square matrices have complex eigenvalues

this is from the fundamental theorem of algebra applied the characteristic polynomial

you don't need any extra conditions

[silent sandal]

Sorry

I meant, complex non-real.

Let me edit the question

Just found a counter-example

[wintry steppe]

yeah i was about to type one

1 2
0 1
and the vector (1, -1), if i hadn't made any mistakes computing

[silent sandal]

Ok; thank you.

[dreamy iron]

I have a QQ about projections….
are projections just linear functionals?

I have a basis of a dual space define as $$\beta^\star \equiv {e^1, e^2}; \
e^1 (x,y) \equiv x ;\
e^2(x,y) \equiv y$$

[stoic python]

ninnymonger

[dreamy iron]

Are all projection-functions elements of the dual space??

[dusky epoch]

what is a projection in your eyes

cause to me it's an operator from a space to itself

[turbid trellis]

Hey, I find linear dependence a bit confusing, how can it be that if {v_1, v_2, v_3} are dependent that {v_1, v_2, v_3, v_4} are also dependent ?

[sonic osprey]

How have you defined linear dependence?

[turbid trellis]

If the vectors can be summed to zero if not all coefficients are zero

Oh wait nvm, v_4 can be given the 0 coefficient right ?

[sonic osprey]

Yeah exactly

[turbid trellis]

Hmm, it still feels a bit odd

Is this something normal that becomes less odd over time ?\

Whole linear algebra doesn't feel that natural to me

[sonic osprey]

why does it feel odd?

[turbid trellis]

It's not intuitive to me, I don't understand the proofs in the book

They're just bloated fancy notation

They show the proof but don't answer the 'why' question

[sonic osprey]

I mean, to mathematicians, proofs are the 'why' this is true

Maybe watching 3blue1brown's linear algebra videos on youtube might give you some more geometric intuition for these concepts and why we care about them

These proofs usually come from geometric intuition, but sometimes this isn't explicitly stated

[turbid trellis]

I am watching his videos, they are very good

Okay well then the issue is that I don't understand the proofs haha

I'm kinda new to math, this is my first 100+ days that I'm working on it

[sonic osprey]

Yeah, then maybe your issue is on the rigor side and not the intuition side

[turbid trellis]

This is one of the proofs which confuse me

[sonic osprey]

What's confusing about this proof to you?

[turbid trellis]

I don't understand what he does in the last step* when he says j> 1, and

[sonic osprey]

They're just splitting it into two cases

j is the largest index for when c_j isn't 0 so they split into two cases

when j = 1 and when j > 1

[turbid trellis]

Yes but I don't see what he does in the equations below

j + 1 etc

Where does that come from if it's the last entry

[sonic osprey]

What's the definition of j?

[turbid trellis]

The largest index for which c_j != zero, however where do we decide zero

[sonic osprey]

What do you mean by that

[turbid trellis]

The coeffiecients aren

aren't zero themselves right, they can be anything

However here he assumes that one of them will become zero right ?

[sonic osprey]

The author defines j to be the largest non-zero index

its possible that none of the c_j are zero and you have that j = p

[turbid trellis]

Sorry I don't understand

[sonic osprey]

What exactly don't you understand?

[turbid trellis]

The pieces of the proof, they're not connected in my head, it just doesn't make sense

The last part mostly

[sonic osprey]

You need to find a non-zero index c_j so that you can divide by c_j basically in the last line

[turbid trellis]

Okay that I do understand

[sonic osprey]

In the first case, when j = 1, there's only one non-zero vector so you can't move the vectors to the other side and write v_1 as a linear combination of the others

So that's why you need to split it off into a different case

[turbid trellis]

Okay I understand that

But where do the zero coefficients come from ?

[sonic osprey]

Again, look at the definition of j

j is the largest index for when c_j isn't 0

[turbid trellis]

Ah like that

I think I get it thanks

[turbid trellis]

When I have a R2 vector, and want to rotate it by 90 degrees with a 2x2 vector [[0, 1], [-1, 0]] how can it be that the matrix transforms the vector ? It kind of confuses me how it rotates the two unit vectors, by adding two numbers ?

[dusky epoch]

2x2 vector

you meant 2x2 matrix

[turbid trellis]

Whoops mistake yes

[dusky epoch]

the mapping is $v \mapsto Av$

[stoic python]

Ann

[dusky epoch]

i.e. you multiply the vector by the matrix from the left

[turbid trellis]

Yes

I understand that, however It confuses me a little bit when I think about the transformation as moving the unit vectors i, j

[dusky epoch]

'moving' from $\mathbf{i}$ to $A\mathbf{i}$

[stoic python]

Ann

[turbid trellis]

Since [[x(x), y(x)], [x(y), y(y)]],

[dusky epoch]

(ditto for j)

[turbid trellis]

Yes

[dusky epoch]

i think you're overthinking it somewhat but can't pin down exactly how

[turbid trellis]

Since these are the meaning of the matrix values

I was wondering how a x of x, and y of x transforms the x in the vector

Do you understand what I mean ?

Like is my question not unclear

[lavish jewel]

that's kinda bad notation

i would suggest you write out the matrix in terms of sines and cosines of the rotation angle

then take some vector x, and write y = Ax

and see how the components of y depend on the components of x as well as sines and cosines

[lyric dawn]

hello, im wondering if x_k is a finite sequence what are the condition of x_k such that its discrete fourier transform is real and non-negative. i heard about Bochner’s theorem but as far as i understand, it is for function defined of the real of complexe space so feel like it's a little overkill. do you know ressources detailling this ?

[edgy tree]

What is the difference between Rank(A), dim(A) & Basis of A ? (Where A is a nxn Matrix)

[quartz compass]

well, what are the definitions?

[edgy tree]

From what i understand is that rank of A is the minimum set of linearly independent vectors

like if we had Matrix A nxm

Rank(A) would be equal min(n,m)

what am confused about here

is that, does Rank(A) = the basis of A ?

and as for dim(A) i don't know a certain definition

[hard drum]

ranks relate to matrices whereas when we talk about a basis, we mean a basis of a vector space, so they are quite different terms

[edgy tree]

Ahaa

[hard drum]

There is a link between all three but yeah

Do you know what the definition of a basis of a vector space is?

[edgy tree]

the set of linearly independent vectors in a vector space ? 😅

Doesn't a matrix form a vector space ?

or is a subspace of a vector space ?

[hard drum]

What do you mean by "form a vector space" here?

[edgy tree]

From what i understand

is that if we have a 3x3 matrix A for example

And we found that we have 3 linear independent vectors in that matrix

[hard drum]

By that do you mean e.g. the columns/rows

[edgy tree]

either rows or columns

[hard drum]

sure

[edgy tree]

like if we had 3x4 matrix

there would be 3 linear independent vectors

is that correct ?

[hard drum]

Not necessarily, I mean even just take the matrix with all zeros

[edgy tree]

Okay

[hard drum]

I think I see where the confusion is coming from and it's probably from having had a lot of ideas introduced at once

[edgy tree]

yeaahh

[hard drum]

So let's forget matrices for a second and just think about what a basis of a vector space is

[edgy tree]

Okay

[hard drum]

Do you know what the definition is? It's pretty fundamental to this

Well, before that, do you know what the definition of linear independence is?

[edgy tree]

the set of vectors that can amount to zero through linear combination, iff all the constants are equal to zero

[hard drum]

I'd make sure to say a set of vectors but yes

that's what it means for a set of vectors to be linearly independent

[edgy tree]

because there can be multiple bases for the vector space right ?

[hard drum]

Exactly

[edgy tree]

Okay

[hard drum]

So now the definition of a basis of a vector space V?

[edgy tree]

is all the linear independent sets that are in vector space V

[hard drum]

It's not, no

a basis of a vector space V is a linearly independent set B of vectors which also spans V

So every element of V can be expressed as a linear combination of elements of B

(and uniquely, as is a good exercise to prove)

[edgy tree]

but can't multiple bases span V ?

[hard drum]

do you have any questions about that? I can give some easy examples if you'd like

Yup, there can be infinitely many bases of a vector space

[edgy tree]

Okay so basically that's the formal definition.

[hard drum]

Ye, so could you give an example of a basis of R^3?

[edgy tree]

{(1,0,0)(0,1,0)(0,0,1)}

[hard drum]

Exactly

the three vectors are linearly independent and span R^3 so if you get that fine

[edgy tree]

got it

[hard drum]

But {(0,1,0),(0, 0,1)} is linearly independent for example

But not a basis

[edgy tree]

yeah okay

[hard drum]

And similarly {(1,0,0),(1,0,2),(0,1,0),(0,0,1)} isn't

Do you get why those two aren't bases?

[edgy tree]

because they don't span V

Vector space V

[hard drum]

The first doesn't span R^3

[edgy tree]

this one ?

[hard drum]

Nah

[hard drum]

This one

[edgy tree]

okay

[hard drum]

This doesn't span R^3

[edgy tree]

exactly

but why ?

[hard drum]

Well think of a vector in R^3 that can't be written as a linear combination of (0, 1,0) and (0,0,1)

[edgy tree]

i don't think i know

[hard drum]

Well let's consider a general linear combination of those two

[edgy tree]

(0,1,1)

[hard drum]

a(0,1,0)+b(0,0,1) = (0,a,b)

So every linear combination of those two is in that form

[edgy tree]

okay

[hard drum]

hence (1,0,0) would be an example

[edgy tree]

ahaaa

[hard drum]

or (3,0,1) etc

[edgy tree]

so the x dimension we cannot have in a vector

[hard drum]

yeah

And so to make it a basis we need to add another vector etx

[edgy tree]

through the linear combination of only two linear independent vectors

out of the 3

[hard drum]

Now why isn't this a basis of R^3?

[edgy tree]

or n in general

[hard drum]

Well not the x dimension necessarily but ye we can't get them all

[edgy tree]

yeah i get what you mean

[edgy tree]

Because one of them is linearly dependent ?

[hard drum]

Well the set of vectors is linearly dependent

Can you think of a way to show that?

[edgy tree]

(1,0,2) = (1,0,1) + (0,0,1)

(0,0,1) is redundant ?

[hard drum]

You can remove (0,0,1) and still have a basis yeah

We have a linear combination that add to zero with non zero constants

[edgy tree]

yeah

[hard drum]

Anyway I need to go, but this leads to the idea of dimension

[edgy tree]

Okay

[hard drum]

As it turns out every basis of a finite dimensional space is the same size

which we call the dimension

So yeah, maybe look at that and review what I just said, hope that helps!

[edgy tree]

Thank you for your time

[hard drum]

Np

[wintry steppe]

i got a question: im asked to determine the general form of a matrix that represents linear applications from R4 to R3 such that Kerf=Vect{v1,v2} where v1={1,1,0,-1}, v2={0,1,-1,0} and Imf is represented by the equation x+y-z=0

how do i go about this

i created a matrix A, then i imposed that f(ei) is in Imf so i got A= [ a1, ... , a4; b1, ... , b4; a1+b1, ... , a4+b4]

but i dont know how to use the information given on the kernel of f

[umbral birch]

im currently trying to solve this problem but im having some trouble, can someone help?

[hard drum]

What have you tried so far?

@umbral birch

[umbral birch]

@hard drum row reducing the matrix and i found the eigenvectors

[hard drum]

Oh so you found two eigenvectors for lambda = -2 but want to find an orthonormal basis of that space? Which two eigenvectors did you get?

[winter summit]

wat networking problems can u solve with matrices like there is the number of connections

[umbral birch]

i got (0 1 0) and (2 0 1)

[hard drum]

Oh, so those are already orthogonal! All you need to do now is normalise them to make an orthonormal basis

(and I mean the first already is normalised)

[teal grotto]

the matrix representation will depend on the bases chosen for R^4 and R^3

[wintry steppe]

it definitely means in the standard bases

[umbral birch]

@hard drum how would u normalize a vector again

doesnt that mean the magnitude is equivalent to 1?

[hard drum]

It means the magnitude is one yes

So simply find the magnitude of the vector and divide the vector by that

[umbral birch]

ohh right

[hard drum]

Although it is probably also pretty important to know what to do in the event that your original vectors were not orthogonal too ig

[wintry steppe]

Canonical bases (i think thats how you say it in english) but its the normal (1,0,0,0)...(0,0,0,1)

[teal grotto]

the second and third columns are the same.

the last column is the sum of the first and second columns

(1,0,0,0) and (0,0,0,1) and (0,0,1,0) are not in the span(v1,v2) so the first, third, and last columns are in the image of f.

this should be enough i think

[wintry steppe]

yes it is, thanks!

[wintry steppe]

I'm very confused on this question

how would I find u_1 or u_2 i dont understand

orthogonality means their dot product is 0 and parallel means its just multiple of that vector

[nocturne jewel]

project u onto v and you get a vector parallel to v

[wintry steppe]

wait really?

[nocturne jewel]

yes...

[wintry steppe]

how do you know that

[nocturne jewel]

cause that's what vector projections do

$u_1=(\hat{v}\cdot u)\hat{v}$

[stoic python]

Mosh

[wintry steppe]

if you subtract that from 'u' that would be orthogonal no?

[nocturne jewel]

yes

since the projection vector will have it's tip right below the tip of u, since it's a projection

$u=u_1+u_2\implies u_2=u-u_1$

[stoic python]

Mosh

[fallen scaffold]

can someone please check if these true or false look good for hw?

[last holly]

which one do you have doubts about?

[wintry steppe]

oh shit

i get it

[fallen scaffold]

a, b, c, e, f

[dire thunder]

@fallen scaffold ok so why you think a is true

[fallen scaffold]

rank is the number of nonpivot columns and since its 4x6 then it has 4 columns

[teal grotto]

rank is the number of pivot columns and it’s a 4 row by 6 column matrix.

[lavish jewel]

that whole sentence was riddled with issues

that it's 4x6 has nothing to do with it having 4 pivot columns

[dire thunder]

rank is minimum of dimension spanned by row/columns tho

@fallen scaffold so yes essentially you are right but not pivots

[fallen scaffold]

right so rank is number if pivot columns

[dire thunder]

@fallen scaffold why b) is false?

[dire thunder]

i mean there is connection

but like

,w rank of matrix

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Commander Vimes#4407

[dire thunder]

💢

[lavish jewel]

lol

[teal grotto]

lol

gave you the rank of a random matrix

[lavish jewel]

rank is related to the number of linearly independent columns in the matrix

[dire thunder]

In linear algebra, the rank of a matrix A is the dimension of the vector space generated (or spanned) by its columns.[1][2][3] This corresponds to the maximal number of linearly independent columns of A. This, in turn, is identical to the dimension of the vector space spanned by its rows.[4] Rank is thus a measure of the "nondegenerateness" of the system of linear equations and linear transformation encoded by A. There are multiple equivalent definitions of rank. A matrix's rank is one of its most fundamental characteristics.

[lavish jewel]

then you recall the definition of a spanning set

and that a subspace with the same dim as the ambient space is the same ambient space

[teal grotto]

matrix multiplication is conventionally done on the left (of vectors). here X is a vector, which is conventionally confusing

[lavish jewel]

the rule of thumb is that "all vectors are column vectors"

and then the product you are asked for is only defined if you put the vector on the right of the matrix (same as coycoy said)

[fallen scaffold]

but we just assume there are 4 columns right since it is 4x6

[lavish jewel]

there is nothing to assume

they gave you all the information

[dire thunder]

i mean like ok

[lavish jewel]

4 rows, 6 columns, and since the rank is 4, there are 4 linearly independent columns

[dire thunder]

i mean

if there are 4 independent columns

it then turns out that there are indeed 4 pivot columns

[fallen scaffold]

so (a) is true

[dire thunder]

ye, a) is true

we where just checking whether you understand why

[lavish jewel]

the columns are the spanning set of the image of the matrix. you can make a basis out of 4 of them. this means the matrix image is dim 4, meaning it is the same subspace as R^4

[dire thunder]

@fallen scaffold why you think b) is false

[fallen scaffold]

wait im looking at my notes

its because if 0 is an eigenvalue the nullspace is nontrivial and the matrix would be non-invertible

[dire thunder]

yes

in other words there is nonzero vector x s.t Tx=0x

so you are correct

@fallen scaffold how did you find vector for c)

[fallen scaffold]

orthogonal projection

[dire thunder]

yep then if there are no miscalculations should be fine

ok for e) it is just computations check them

for f) it is correct

but why

[gloomy nebula]

Would saying T(x)=AxB be correct then? The ordering of multiplying is important

[dire thunder]

@fallen scaffold like ok why you think f) is correfct

[fallen scaffold]

i was thinking because they were in the same subspace

[teal grotto]

can you just try and make sense of how that would even work first? not meant to be sarcastic, just asking how you expect your definition of T(x) to behave

[dire thunder]

actuall reason is that

[stoic python]

Commander Vimes

[dire thunder]

where brackets denote dot prduct

that means that coordinates of vector wrt this basis are actually just dot products @fallen scaffold

[fallen scaffold]

ok so (f) is true

[gloomy nebula]

wait you're right. It would have to either be xAB or ABx. In this case they are asking for ABx but they are telling me to do xAB

[teal grotto]

thought this was only for orthonormal bases

[fallen scaffold]

i know (d) is definitely true

[teal grotto]

xAB doesn’t make sense with the common definitions of matrix/vector multiplication.

and they are telling you, given a vector x, first multiply it by A, then multiply that result by B. so you first act on x by left (standard) multiplication by A to get Ax, and then multiply Ax by B, so BAx

[dire thunder]

h ye

for othonormal

[gloomy nebula]

If x is 1 by 3 instead of 3 by 1 then it's valid

[dire thunder]

anyway @fallen scaffold seems fine but recheck everything

i gtg

[teal grotto]

even still, that wouldn’t be the right product

[fallen scaffold]

thanks

[wintry steppe]

Hi, how do you put the linear application D in the basis 1, x, x^2?

[dire thunder]

@wintry steppe find images of basis vectors

[lavish jewel]

that's pretty much always the trick, you can express any poly of order <= 2 as a linear combination of the basis vectors 1, x, and x^2

so check how each of those are transformed by D

[wintry steppe]

Thanks! @dire thunder @lavish jewel

[wintry steppe]

I fail to see why this matrix isnt a vectorial space?

or more precisely a subspace

[lavish jewel]

do you mean exactly that matrix or matrices of that form?

[dusky epoch]

a matrix is not a vector space

[lavish jewel]

what if it was the 0 matrix

[gray dust]

seems ann wants cera to distinguish between such matrices & the set of such matrices

[delicate sphinx]

what do double straight lines around a vector mean?

[dire thunder]

@delicate sphinx presumably its norm

but like more context

[delicate sphinx]

uh whats a norm

[dire thunder]

length of vector

[turbid trellis]

Hey, there are questions that say 'Show that this is that, or show that this maps to that'

I suck at these, is there some training which should be done before ? They feel like proofs but even more confusing

And when I look at the answers they're just so unclear

[last holly]

try breaking it up into subproblems, if you need to show smaller things first

and build it up

[turbid trellis]

The thing is that the questions aren't problems at first, they ask me to proof the most basic properties, however proving them is just so confusing since they're intuitive

Compare it to, 3 + 3 = 6 or something like that, only in linear algebra

[last holly]

yea so actually that's a great example, you can look at tao's talk of the peano axioms to see how addition and multiplication are built up from 0

it's not as easy as it seems (if you haven't done it before), but at the same time it's not bad

[turbid trellis]

Is it useful to do these things ? I'd rather just skip the proofs and learn linear algebra as quick as possible, I'm limited in time

[last holly]

what are you learning lin alg for?

[turbid trellis]

Machine Learning on the long term, and things like Inverse Kinematics in the short term

Also image processing and 3D engines

[last holly]

ok, so you do want to know it well then

yes it is super useful then

it takes more than a semester to even get the basics down

unless you're coming in with a boatload of technique

[turbid trellis]

I look at the proofs and they don't even make it past my eyes

They trigger nothing and I don't even understand what they're doing

The only thing they do is make me sleepy

[last holly]

so maybe more proof technique first? or find something that gets you to get inspired about the motivation for proofs in the first place

a lot of the best insight is gained from getting comfortable with trying to write proofs and being able to dissect them to an extent

[turbid trellis]

I'll watch some videos on proving

However my brain indeed doesn't see why it should spend energy on learning them, causing me to get sleepy

[last holly]

like just take statistics as an example. it's got a set of completely unintuitive results which if one doesn't have technique it's easy to get lost in

[turbid trellis]

I'll try, thanks

[last holly]

@turbid trellis one last idea

like if i have a project i want to sink my time into, say half-year or 2 year project

if i have no way of proving a minimum end-result

it means i have a higher percentage of risk in investing that time

[turbid trellis]

Proving something reduces the risk of wasting time ?

In math, it makes you remember better ?

[last holly]

well i'm just using general non math analogy, if you want to make it concrete it could be proving an algorithm

then you take the months to program it knowing it will work

but if you didn't check and you take the months, it's wasted time

[turbid trellis]

I understand

[last holly]

so yea a lot of the most powerful tools are not sexy at first

but once you see their power, it's easier to relate imo

[turbid trellis]

I will give proving a try and restart reading the book

[whole zodiac]

axler >>

[faint dune]

I cant intuitivly explain why lik = - aik/akk works.

This is an algorithm to solve gauß-elimination but with developing the lu decomposition.

I dont understand why the divison works in general, is there any page I can read the idea? If I do matrix multiplication of course it works..

[gentle igloo]

I find that knowing proofs of the basic concepts is more rewarding later in the process of learning a math topic. In the beginning the proofs might seem unmotivated and useless, however later you will likely see how useful and essential the basic proofs were to proving the topic at hand. Knowing the basic proofs also allows you to think and reason about linear algebra by yourself. For example, if someone asks you a linear algebra question which you have never thought of, but you have a great understanding of the basic proofs, then chances are you will at least be able to reason your way to an intuitive answer. Hope this helps motivate your linear algebra adventures!

Numerical Linear Algebra by Trefethen and Bau has a good gaussian elimination section. I feel that it gives some intuition with the explanation.

[faint dune]

The book is very expensive and not online in my uni library ;_;

[gentle igloo]

if you just look up numerical linear algebra trefethen and bau pdf there should be one

but i’ll send a screenshot of the pages

[faint dune]

Ok nvm I got it!

@gentle igloo .

[gentle igloo]

ok sick

[hollow mesa]

I want to prove that, given a 7x4 matrix A and a 4x7 matrix B, where B's rank is 6, that AB is not invertible. Any help or guidance is greatly appreciated

[weak needle]

A 4x7 matrix of rank 6 can’t exist

[hollow mesa]

ugh sorry I am very new

to linear algebra

a 4x7 matrix of whatever rank actually

but I think I figured the problem out though, thanks though

[weak needle]

Np

[fallen scaffold]

can someone help me for (b)?

[wintry steppe]

Is there any way to deal with determinants like this?

I think there must be a formula that can be deducted from there

[teal grotto]

you have four linearly independent evals from part a. find the kernel of (xI - A) for each eval x

[fallen scaffold]

these are my e-vectors from part a

i found the eigenvalues and eigenvectors in part a

[wintry steppe]

So we can say that in general the determinant of that type of symmetric matrices is:

$\Delta = [a + (n-1)b](a - b)^{n-1}$, right?

[stoic python]

Elfire

[wintry steppe]

Where n is the order of the matrix

a is the element in the diagonal

And b is the othef element the matrix has

[teal grotto]

these just so happen to be orthonormal. the matrix Q will be the matrix with each column as one of these vectors

[fallen scaffold]

so thats it for part b?

[fallen scaffold]

do you mean orthogonal?

[teal grotto]

yes

[teal grotto]

well D is the diagonal matrix with the first diagonal entry as the eval corresponding to the first column in Q, the second diagonal entry is the eval corresponding to the second column in Q,… and so on

[teal grotto]

let J be the n by n matrix of all ones. look at the general form of (x-a)I + aJ where I is the identity matrix and a is some scalar

[fallen scaffold]

these were the eigenvalues i found

i put it as D

but what would be Q?

[teal grotto]

this

[fallen scaffold]

so Q would be the eigenvectors right?

[teal grotto]

yes. but each column has to correspond with the eigenvalue in D, like i described here

[fallen scaffold]

ok

so to do this problem, i just need to prove Qt * D * Q = A

[teal grotto]

yes

[forest quiver]

Yo I have a quick question

When I am explaining something in linear alg

[stoic python]

s[]gh[/dfsh';.dsf'jdf'hlk;'erhl

[forest quiver]

Or is [x_1 x_2 x_3] always a vector through origin

like the actual arrow pointing

I am pretty sure I can interpret this as a point in R^3 as well right?

I am asking because when describing the intersection between two planes in R^3

We can represent this with the form x= p+tv

where x, p, and v are vectors and t is a scalar

and I wrote down "The intersection of the planes can be described as a straight line, namely a line that passes through POINT p and is parallel to VECTOR v"

I feel like this is all-over-the-place

ugh

[fallen scaffold]

@teal grotto do you mind helping with this?

how do i write the matrix equation?

[fallen scaffold]

thanks

[fallen scaffold]

<@&286206848099549185>

[dusky epoch]

@fallen scaffold do you still need help with this?

[fallen scaffold]

nah im good now

[wintry steppe]

Hi

I already solved tis

this*

But I wonder if there's a faster way than applying properties

Because the computations are a bit boring lol

$\begin{vmatrix}
1 & 12 & 123 & 1234\
2 & 23 & 234 & 2341 \
3 & 34 & 341 & 3412\
4 & 41 & 412 & 4123
\end{vmatrix}$

[stoic python]

Elfire

[wintry steppe]

The result is 160

But I want to know if there's a faster way to calculate determinants like these

[nocturne jewel]

Laplace expand? But I don't see a quicker way other than a CAS

[dusky epoch]

you can do row ops to reduce the det of this matrix to that of
[1, 2, 3, 4;
2, 3, 4, 1;
3, 4, 1, 2;
4, 1, 2, 3]

well not row ops but col ops

subtract 10 times col 1 from col 2

100 times col 1 from col 3

etc

@wintry steppe @nocturne jewel

[lavish jewel]

you could then swap rows 2 and 4 to get a circulant matrix and replace your boring calculations with exciting ones using vandermonde vectors of complex exponentials

[wintry steppe]

I already tried that, thanks

Hi

I have another question, well, it's to see if what I did was correct

$\begin{vmatrix}
x & a & b & c & d\
a & x & b & c & d\
a & b & x & c & d\
a & b & c & x & d\
a & b & c & d & x
\end{vmatrix} = (x-a)^2(x-b)(x-c)(x-d)$

Is this correct?

[stoic python]

Elfire

[wintry steppe]

After doing some row transformations, I obtained that formual

formula*

I don't know if that is correct

[wintry steppe]

I know it is incorrect

But I don't know how to solve it then

[manic wedge]

So I put some values in x for which the Determinant is 0 and thereby I found the decomposition

but I didn't explicitly calculate the Determinant, I made the rows/columns linear dependant. You know that then, Det must be 0

[thorn robin]

@wintry steppe Did you figure it out yet?

[wintry steppe]

Yeah

I finally found that the result is

$(x+a+b+c+d)(x-a)(x-b)(x-c)(x-d)$

[stoic python]

Elfire

[thorn robin]

Yup cool

[manic wedge]

very cool

the a,b,c,d were easy to see but the -(a+b+c+d) make a problem when you add the columns together, the sum vector is 0 and therefore they are linearly dependant

[neon fable]

Hey everyone, so as I was attempting to solve this, I get the part of how to show what a basis or not for R^4 . I just need additional support in finding the coordinate of the vector relative to this basis. I look at the answers sheet, and I just got confuse since they use the inverse when putting these four vector in matrix. I know the coordinates are scaler use in linear combination of the particular vector, but since there basis, should their be no combination of each other?

[zealous junco]

All they ask is to represent (1,0,0,0) , etc using the alphas you got there

And write the lin combo compactly as a coordinate

Though of course you can form the change of basis matrix from standard basis to alpha basis once you got the 4 coordinates by taking the coordinates as columns

And the inverse of that would be the change of basis from alpha to the standard basis representation

[wintry steppe]

of that form, and the answer was that it doesnt because it lacks the null matrix, but i dont see that...

[lavish jewel]

are you told anything about these a_i and b_i?

[wintry steppe]

no

i did manage to prove that a1b2-a2b1 cant be 0

and i also know that kerf=Vect{v1,v2} where v1=1,1,0,-1 and v2=0,1,-1,0

[wintry steppe]

and throught the 3rd line i can see that the equation of Imf is x+y-z=0

[lavish jewel]

i'd have to see the original problem to see what's missing

[wintry steppe]

its in french

the 2. asks of us to determinate the general form of matrixes with the next criteria, which is this matrix:

and the 3. asks if theses matrixes represent a vectorial space with the general rules of composition

which i cant prove, but the answer is no since the null matrix is missing, but i dont see how i can prove that

[lavish jewel]

aha

the thing is if you let A be a 3x4 matrix of zeros, it doesn't satisfy the conditions you were given

the null space of the transformation is no longer spanned by v1 and v2, since it should rather be rank 4 (you'd need 2 more vectors). related to this, the image of f is no longer plane x+y-z=0

it's just the 0 vector in R^3

this means the set of matrices with these properties does not include the 3x4 matrix of all 0s

[wintry steppe]

why? couldnt you get the zero vector with the ocmbination of v1 and v2?

[lavish jewel]

yes, but not ONLY those 2

the null space of the 3x4 matrix of all 0s is all vectors, i.e. all of R^4

[wintry steppe]

oh i see, it has to be exclusively with those 2?

[lavish jewel]

so saying the kernel of f is spanned by v1 and v2 is wrong

yeah

[wintry steppe]

okay i understand

[lavish jewel]

v1 and v2 would only span a subspace of the nullspace

not all of it

[dusky epoch]

wait dont matrices of this form form a vector spcae tho

it looks like {that matrix | a1, a2, b1, b2 in R} is a vector space

[wintry steppe]

was there any other way to see this? with other info we got?

[lavish jewel]

can't it not be 0? it asks for the image to be of dim 2

[dusky epoch]

spanned by [1 0 0 1; 0 0 0 0; 1 0 0 1], [0 1 1 1; 0 0 0 0; 0 1 1 1], [0 0 0 0; 1 0 0 1; 1 0 0 1] and [0 0 0 0; 0 1 1 1; 0 1 1 1]

i think

this may even be a basis

yeah it is a basis

dim(that space) = 4

[wintry steppe]

but dim kerf=2

so rkf=2 also since its in R4

[dusky epoch]

nyeh

hghhrkgh

fuckdsg

what are the conditions on the a's and b's for the matrix to have the required kernel and rank properties

[wintry steppe]

:my brain 24/7

[wintry steppe]

i found that a1b2-a2b1 cant be 0

checked rn, yea thats the condition

[lavish jewel]

can't you just check that it doesn't have a 0 element?

[dusky epoch]

well like

that condition makes it not a vector space anymore

idk like this seems straightforward but overcomplicated by yall

hhrgrghr

maybe im too spacey to see it rn

who fucking KNOWS

(╯°□°）╯︵ ┻━┻

[lavish jewel]

it seemed pretty simple to me. can't be rank 2 and include the 0 mat

[wintry steppe]

why?

[dusky epoch]

idk

this feels like it hsould not warrant a week of discussion

what even like,

jeez

fuck

a

[lavish jewel]

it didn't, we just started rn

[wintry steppe]

yea thats the proof but why is that

guys im not even 1st year im struggling with this shit

[lavish jewel]

because the 0 mat doesn't satisfy the definition

the image of a transformation defined with a matrix of all 0s is a point, the 0 vector

this is of dim 0

not 2

[wintry steppe]

oh, okay

yea i see it now

[lavish jewel]

with that alone, it's not in the set

[wintry steppe]

of course

didnt click until now but yea its obvious

[lavish jewel]

i might've misread the french, but that would be what i would say if i understood it right

[wintry steppe]

its good, thanks both of you

[lavish jewel]

it says the image is the plane x + y -z = 0 yeah? the point 0 is not that plane (although it is ON it)

[wintry steppe]

but then in what situation would you have a 0 matrix that would fit any dimension bigger than 0?

would you have to have some scalars in the matrix?

besides the variables

[lavish jewel]

never, the catch here was that they specified a dimension

[wintry steppe]

oh i see

so dim kerf would always have to be 0? for it to be a vectorial space?

[lavish jewel]

that's the dim of the vector space of the 0 element, yeah

dim, not dim ker f

dim ker f would be the size of whatever the domain is

but you don't normally see definitions based on a specific rank

[wintry steppe]

i dont see how a matrix could be a vectorial space then since dim would always be bigger than 0 for the null matrix? or am i missing something

[lavish jewel]

the dim of what

[wintry steppe]

of that null matrix

for all the conditions to be met for it to be a vectorial space, we need the null matrix, and woudlnt that matrix be of dim bigger than 0 in all cases?

[lavish jewel]

what do you mean by dim here

dim is used to denote the number elements in a basis for a subspace

[wintry steppe]

well if i understood correctly the proof here, the problem was that dim ker f was 2 and that the null matrix was supposed to be 0?

i mean for it to be valid

[lavish jewel]

you'Re mixing up two different things there

if you have a 3x4 matrix of all zeros, then dim ker f = 4

separately, if you treat the 0 matrix of size 3x4 as a vector space, this vector space is of dim 0

it depends on whether you are looking at the matrix as a transformation or as an element in a subspace

dim im f = 0 as well

[wintry steppe]

so the problem was that dim ker f was 2 and not 4? so since 3x4 amtrix of 0s is dim 4, it couldnt be in ker f?

[lavish jewel]

whenever you say "the matrix is of dim #", this doesn't make sense

[wintry steppe]

the transformation illustrated by that matrix

would that make sense

[lavish jewel]

still no

you have to specify if you're talking about the image or the null space

or kernel, if you prefer that over null space

[wintry steppe]

dim ker f, as defined by them, is 2, but the null matrix would have to be in dim ker f=4, so that doesnt add up. Is this correct?

[lavish jewel]

yeah

well

not "in dim ker f"

[wintry steppe]

"of dim ker f"?

[lavish jewel]

i would write it something like

matrices in the set are required to have dim ker${f}=2$, but dim ker${0_{3x4}}=4$

[stoic python]

Edd ✓

[wintry steppe]

okaay

[lavish jewel]

the dimension of the kernel of the transformation defined via the 3x4 matrix of all zeros is 4, but the matrices in the set have a kernel with dimension 2

[wintry steppe]

okay, tysm !!

[woven haven]

Hey, all! Slight question about a proof I've drafted for my linalg course. Here's the question:

I've drafted the following proof:

About the second part of the proof: how can I guarantee that the number of solutions has not changed?

I know that both lambdas are nonzero, but how do I know that w or v was not mapped to a zero vector?

Oh wait, lambdas are nonzero

I might have answered my own question.

I'm a bit confused.

[nocturne jewel]

You've shown that a_1l_1 and a_2l_2 are 0, but even an independent set can have the trivial scalars give the 0 vector

you want to show that only the trivial will give 0

[woven haven]

Does that not follow from the fact that both lambdas are nonzero? I suppose I'm confused on how I'd demonstrate the uniqueness.

My intuition says that, if {u,w} is linearly independent, then a_1 = a_2 = 0 is the unique solution, so once I make the transformation, T(u) and T(w) are still nonzero, but they have the same coefficients attached, so a_1 = a_2 = 0 is still the only solution.

[teal grotto]

you can’t use the same scalars from au + bv = 0 in aT(u) + bT(v), as you have implicitly done here.

[woven haven]

Forgive me if I'm just repeating the argument.

Ah.

Rats.

[teal grotto]

just use the fact that $aT(u)+bT(v)=a\lambda_1 u + b\lambda_2 v$ for all scalars $a,b$

[stoic python]

coycoy

[teal grotto]

if u and v are lin. independent, then you’re done

sort of. you assumed that a_1 and a_2 were already 0 in part two of you proof, which is wrong.

[woven haven]

I think I'm understanding more

So in the first part, I should go one step farther and expand T(u) and T(w) as $\lambda_1 u$ and $\lambda_2 w$, respectively, and since {u,v} are linearly independent, the coefficients must be zero--but neither lambda is 0, so the a's are 0.

[stoic python]

Chris24

[woven haven]

Ugh, if only I had gone a single step farther: that seems much more. . . complete.

[teal grotto]

try doing a similar thing assuming that T(u) and T(v) are linearly independent.

[frosty vapor]

and then finally see if you can get a cute geometric interpretation

[woven haven]

It makes sense to me intuitively, in terms of stretching and compressing the basis vectors; Im just falling into some faulty logic.

[teal grotto]

you can do this in a really clever way using the fact that the eigen values have inverses

[woven haven]

We have not covered the invertibility of eigenvectors in class, yet. We took a class period to review for the third exam

Which I did quite well on, thank goodness.

[frosty vapor]

you are given that lambda 1 and lambda 2 are nonzero which is nice (i think we also need them to be distinct to be guaranteed invertibility of T tho)

maybe i forgor lmao

[teal grotto]

it does not make sense to invert vectors. all i’m saying is that lambda1 and lambda2 are non-zero, so they have inverses

[woven haven]

Oh, eigenvalues

I misread. Apologies.

[teal grotto]

no worries

[woven haven]

I'm confused on where the insufficiency is with the second part (abundantly clear in the first).

If {T(u), T(w)} is linearly independent, then $a_1T(u) + a_2T(w) = 0$ has only the trivial solution.

[stoic python]

Chris24

[woven haven]

So if we substitute that u and w are nonzero eigenvectors, then *their coefficients are still 0.

grammar mistake

[teal grotto]

you have not really shown that if
au + bv = 0, then a = b = 0, but it’s pretty close. you just kind of substitute the values of T(u) and T(v), which doesn’t give you any new information, since we already know those are lin. ind. by assumption

[woven haven]

I definitely agree that it doesn't give you new info, but the given info seem sufficient to me.

I will think about this. Just getting trapped in my way of thinking.

[stoic python]

coycoy

[wintry steppe]

$\left{\begin{array}
x + 3y - az = 4\
-ax +y + az = 0\
-x +2ay = a+2\
2x -y -2z = 0
\end{array}\right.$

[stoic python]

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

[wintry steppe]

Well, imagine there's an x there

That's the last exercise in my LA book before starting with vector spaces

The problem here is that I started discussing it with the Gauss methd-od

And after finishing and getting some equations that rn are hard as it's late

[woven haven]

@teal grotto You're a goddamn genius.

[wintry steppe]

I noticed I could've applied Rouché-Frobenius theorem and basically calculate both determinants

So basically

I don't have any question

I just wanted to express how I hate this type of exercises

[green vine]

hum, if I didn't get help could I mention the helper role?

[teal grotto]

lol no where near it. thanks tho, glad i could help

[woven haven]

Believe I've solved my problems! Thank you very much @teal grotto @nocturne jewel @wintry steppe @frosty vapor !!!!

[wintry steppe]

🤨

[frosty vapor]

ty tterra

yw

[teal grotto]

ty TTerra

[wintry steppe]

[torn hornet]

Ty tterra

[sleek sundial]

is the product of 2 rank 1 matrices also rank 1?

[weak needle]

No

Eg: diag(1,0) times diag(0,1) is the zero matrix. (Diag(x,y) is the 2x2 matrix with x and y on the diagonal)

[forest quiver]

but there is a free variable here

OH wait

x_3 = 0

but theoretically

I could make a free variable

by multiplying 3rd row by 0

0x_3 = 0 has infinite possibilities

[teal grotto]

you never multiply rows by 0

that ruins the whole point of gaussian elimination

[forest quiver]

but what is stopping me?

For my solutions hre

here

if I multiplied row 3 by zero

it would just be a line right?

but the actual solution would be at x_3 = 0

idk this is confusing

[teal grotto]

it breaks the rules/principles of row reduction when you do that

[forest quiver]

I mean it doesn't break the principles

I am just removing an equation

that I have

for no good reason

[teal grotto]

it do tho

[forest quiver]

I am solving for another system when I do that

here it would be a system of Row 1 and row 2

so yeah it's wrong

I get it

[brisk python]

31 ka answer kya hoga? Skew symmetric?

[stoic python]

s[]gh[/dfsh';.dsf'jdf'hlk;'erhl

[forest quiver]

Is this correct notation?

I wanted to say that the solution can be any single point on the line that goes through [5, -2, 0] and is parallel to the vector [4, -7, 1]

[lavish jewel]

i would rather write something like w = [5,-2,0] + v, v \in span{[4,-7,1]}

or simply w = [5,-2,0] + t [4,-7,1], t \in F

idk if it's standard to add a set with a vector as you did

[forest quiver]

oh where t is a scalar?

Ok

But is that how I use this symbol?

[stoic python]

s[]gh[/dfsh';.dsf'jdf'hlk;'erhl

[forest quiver]

I am new to it

I like using it tbh

instead of =

[lavish jewel]

they don't mean the same

= is "this is exactly what this thing is"

$\in$ means "this thing is in this set"

[stoic python]

Edd ✓

[lavish jewel]

or in what you wrote, did you mean m is in the span of that vector, and then afterwards you add the other vector?

at any rate, what you had written is either wrong or ambiguous

[forest quiver]

Right exactly

and I wanted to show that

m could be any point

on that line

[lavish jewel]

then you wrote it wrong

[forest quiver]

It's not like m is ONE SINGLE point

on the line

m CAN be anything as long as it's on that straight line

hmm trying to figure out how I can write this better

I'll take what you said and try to work with it

[lavish jewel]

i gave you 2 fully written out ways of doing it

[forest quiver]

Right but I wanto to learn how to use that E thing

[lavish jewel]

i used it up there too

[forest quiver]

\in ?

oh yeah

by F did you mean R ?

sorry to drag this on

[lavish jewel]

yeah, whatever your field of scalars is

[forest quiver]

So I don't say that my variable solution is \in a line?

Yeah that doesn't make sense does it

[lavish jewel]

you can do it, just not like that

[forest quiver]

Yeah you are right

I just feel like = insinuates that there is only one solution

and it's on that line

or wait

no

it means that the line is legit the solution

which it is

I undersatnd

[lavish jewel]

it doesn't, if you let there be a parameter

[forest quiver]

oH RIGHTTT

im so dumb

[lavish jewel]

just like when you use = in a function and it describes a curve of some sort

[forest quiver]

I see what you mean

[lavish jewel]

= is binary relation

[forest quiver]

right

[lavish jewel]

e.g. between the x's and the f(x)'s

[forest quiver]

binary meaning either it's equal or not?

[lavish jewel]

\in tells you one element is inside a set

no

binary meaning there are 2 elements

[forest quiver]

right

[lavish jewel]

when you use \in, there is only 1 element

not 2

[forest quiver]

interestiung

[lavish jewel]

\in relates an element to a set

= relates 2 elements

[forest quiver]

Span{vector} is a set right?

[lavish jewel]

yes

[forest quiver]

because of the { }

[lavish jewel]

you could write it with () if you wanted

[forest quiver]

ok

[lavish jewel]

it's rather due to the definition of span

[forest quiver]

Im looking back at my textbook

and they wrote it like you

with the t

as parameter

[lavish jewel]

that's a common one

the parametric form of a line

[forest quiver]

yeah

parametric vector form

represents just a line

if im not mistaken

because there is always a parameter involved

but yeah thank you

my textbook doesn't use /in, but I want to learn how to start writing fancy math

I guess I shoul djust stick to the stuff I know/in this book right now

[lavish jewel]

you should first learn what the symbols mean 😛

https://en.wikipedia.org/wiki/Set-builder_notation

[forest quiver]

Thank you

These lessons come too late in life

I wish they introduced functions as sets and not f(x)

[stoic python]

s[]gh[/dfsh';.dsf'jdf'hlk;'erhl

[forest quiver]

I think it gives much better intuition

for what a function actually is

just relating numbers from inputs, to numbers in output

[dusky epoch]

it doesnt have to be numbers

[forest quiver]

I was thinking about that

[dusky epoch]

numbers are just a convenient thing to define functions on

[forest quiver]

Hmmm

So variables too?

I guess that works yeah

[dusky epoch]

i mean

a function has a domain and a codomain

both of these are sets

these sets can contain anything you want

[forest quiver]

anything?

[dusky epoch]

yes anything

[forest quiver]

wow

[dusky epoch]

you could have a function like age: {all humans} -> N which assigns to each human their age in years

[forest quiver]

Along the way I guess I lost track of what a function actually is

it doesn't have to be just numbers

just a defined relation between two sets

[lavish jewel]

it's a binary relation, at that

[forest quiver]

right

[lavish jewel]

which is why you see stuff like y = f(x)

like i had said above

[dusky epoch]

it's a right-unique left-total binary relation

[forest quiver]

one last question before I get to bed

can there be 2 inputs?

[dusky epoch]

wym

[forest quiver]

like not just relating 1 set of inputs to 1 set of ouputs

[dusky epoch]

functions of two or more arguments which are written as f(x,y) are usually taken as having a product set as their domain

[forest quiver]

Oh so there can be more ways

[dusky epoch]

like if you have the first input from A and the second input from B and the codomain is C then your function has signature A×B → C

[forest quiver]

Oh I see

That is so much better than f(x)

[dusky epoch]

f: A×B → C

and then FORMALLY you could write f( (a,b) ) for the value of f at (a,b) but we just write f(a,b) for brevity

[forest quiver]

alright thank you both

[empty hemlock]

Function application does not need parentheses.

[icy pecan]

I’m reviewing my Linear Algebra and had trouble proving these:
1.) Given A, B are square matrices and AB = I, prove BA = I
2.) Given A, B are square matrices and AB = I, prove B=A^-1

For #1, I tried B=BI=B(AB)=(BA)B but I’m stuck because here if I want to conclude that BA=I then I have to assume B has an inverse

For #2, I’m thinking that I can use the result of #1 to say AB=I and BA=I and so by definition of inverse matrix, B=A^-1

oh “I” is the n x n identity matrix, and A, B are n x n

[half ice]

Indeed, I don't think 1 is true without inverses

Will try to get counter example

For 2, just multiply both sides by A^(-1) on the left

[icy pecan]

Doesnt that assume A has an inverse?

[sonic osprey]

1 is true, one way is to use determinants and the fact that det(A)det(B) = det(B)det(A)

[half ice]

The question has an A^(-1) in it, so A has an inverse

[icy pecan]

Oh I wrote the question, what I meant to say is prove that A and B are inverses of each other

[half ice]

Oh rofl yeah they are, aren't they

[icy pecan]

Well im trying to prove it

[half ice]

Sorry, I had a braindead moment.

[lavish jewel]

for the first one, can't you multiply by A from the right and then use associativity of matrix products?

[half ice]

If AB = I
Then by definition B is an inverse of A

[lavish jewel]

i'm pretty sure this alone is not true

since some matrices are invertible only from one side

[weak needle]

For 1) you can’t expect a solution that skips past stuff very specific to matrices, it is false in a general noncommutative ring after all

[weak needle]

Not true, but you are right that that is a possibility up until you prove 1)

[lavish jewel]

wdym not true?

[weak needle]

If a matrix is right invertible it will be left invertible

[lavish jewel]

for square matrices yes, but in general no

[weak needle]

But that is what the question is about

[half ice]

The question assumes square

[lavish jewel]

aight

then yeah, mb

[teal grotto]

only if its square tho, right?

[half ice]

But yeah weird shaped matrices are weird

[weak needle]

Yeah

[icy pecan]

Can you show me what you mean?

[lavish jewel]

you have square matrices and that AB = I. let's multiply both sides from the right by A, giving ABA = A. this means (AB)A = A, and A(BA) = A

[weak needle]

Again, this is true in a general noncommutative ring, it doesn’t say anything about the invertibility of A

[teal grotto]

hint:||BA = BIA = B(AB)A = (BA)(BA) = (BA)^2||

[lavish jewel]

you can add in stuff like the rank of a matrix product being <= min rank(A), rank(B), making both of them full rank

then AB = BA = I

[icy pecan]

does this mean that BA must = I?

[teal grotto]

yea. BA cant be the zero matrix.

so it has to be I

[icy pecan]

Ok the property of the square of a matrix being equal to itself is unique to the zero matrix and identity matrices?

[teal grotto]

no.

[weak needle]

Matrices are not integral domains or anything A^2=A does not imply A is 0 or 1

[icy pecan]

Ah darn I forgot what Rank is lol, been a year since i took LA

[teal grotto]

? why not. BA = (BA)^2 means that BA(BA - I) = 0

[weak needle]

And then what

[teal grotto]

BA - I has to be identically 0

since BA cannot be zero, otherwise you get a contradiction, assuming that AB = I

[weak needle]

If XY=0 is either X or Y zero?

[lavish jewel]

the matrices are full rank, so you can't really get a 0 matrix out of that

[icy pecan]

I understood each of these words individually

LOL

[teal grotto]

oh shish. i realized. for some reason i was like, clearly BA(BA - I)(x) = (BA)x (BA - I)x.
but this is gibberish

[weak needle]

True but that is a different argument (and one that finishes off the proof at the very start too)

[lavish jewel]

😛

[weak needle]

Basically a I’m saying that there are examples of non commutative rings that have elements a and b at ab=1 but ba/=1. So that means that if we want to prove this property for matrices we have to use something specific to matrices like determinant or rank or something

[icy pecan]

OHHH thank you

Understood that perfectly now lol

[weak needle]

Nice

[earnest ferry]

When a scalar ,k is multiplied with a function f(x), we represent the resultant function as (kf)(x). But when me multiply f(x) with (k+m), where m is also a scalar, which is a better representation of the resultant function : ((k+m)f)(x) or (k+m)f ?

I stumbled upon this doubt while proving the set of real valued functions is a vector space.

[native rampart]

I guess 2nd

It's mostly personally preference

[earnest ferry]

Thanks for clearing up my confusion

[quaint sage]

I don't know if this is going to make any sense, but here goes:

So if I have an equation of a plane and a point on that plane, do I have a new coordinate system in that plane as well?

If I say that N is a vector defining my new coordinate system, and as such as it is in the Cartesian system, I want all the basis vectors to be orthogonal one another, how do I find the other two vectors? I realize there will probably be a "orientation factor", by which the other two vectors can rotate.

Maybe I am missing something obvious here

[half ice]

Gram Schmidt process takes any vector space and gives an orthonormal basis of it

[quaint sage]

I'll look into that, thanks!

[leaden tide]

*any basis

[half ice]

*an orthonormal basis

[leaden tide]

it starts with any basis

and turns it into an orthonormal one

[half ice]

Ah fair point yeah you need a basis already

I mean, for a plane that's really not a big deal. Just make sure the two vectors you start with aren't colinear

[forest quiver]

Are y'all still discussing ?

There's a problem that has been killing me since this morning

[half ice]

Jesus call the police

[forest quiver]

[half ice]

We can't deal with murder around here

I'm not funny. What's the problem?

[forest quiver]

There's a bit of reading

Is that cool?

[half ice]

Ye

[forest quiver]

And the solution is

I don't know much about economy so that might be part of the problem,

The one thing I don't understand is

[half ice]

Tf is this jibberish kek

[forest quiver]

XD ikr

But my question is

Why do we equate the sector's overall output to what it receives from other sectors

Like what????

I am never going to study economics

[half ice]

I have no clue at all. This is nonsense

[quaint sage]

sounds a lot like communism to me

[forest quiver]

You think I shouldn't spend time on this?

I kind of want to move on

[half ice]

They're just trying to force a system of equations where the lines must sum to a whole

[forest quiver]

But it pains me to leave some stuff out

It also pains me to think about economics (somethign I don't understand in the slightest)

like a sector can sell itself a percentage of its output???

[half ice]

I think they're trying to say "The coal company uses 40% of the electric company's output, and 60% of the steel company's output"

But they're being really vague

[forest quiver]

Yeah I understand that

[quaint sage]

Which makes no sense because steel takes coal as an input in the real world

[forest quiver]

That's not the problem

[quaint sage]

just being pedantic

[half ice]

Actually no, that doesn't even make sense. Then what are pc, pe, ps?

[forest quiver]

no you have it right

My problem is why do we equate THE TOTAL OUTPUT OF A SECTOR to the OUTPUT % IT RECEIVES FROM THE OTHER SECTORS

Does it have something to do with the sector PURCHASING the material from the other sectors?

because it says SELL

[half ice]

That's exactly it, I see no reason why we'd do this. It implies that everything these companies make stays entirely in this system

[forest quiver]

Right

that's very weird

You know what

[half ice]

Which, is a dumb assumption kek

[quaint sage]

no value added in this scenario

[forest quiver]

im skipping this

[quaint sage]

maybe it's a fair assumption for macro economics?

[forest quiver]

just fucking giving me the biggest headache all mornign long

how can a sector buy a share of its own output?

like that makes no sense

There are a ton of these types of problem in the exercises too

[half ice]

I also have taken zero economics so I am not certain if this is based off a real model, but you aren't learning economics atm

[forest quiver]

yeah this is from linear alegbra applications chapter

like I understand nodes / chemistry stuff

here

but not economics

its not a math problem

its an economics problem

'so thats why i dont give a damn

thank you for the chat

[half ice]

Cool cool. Make sure you can do sensible word problems, leave this one in the dust

[forest quiver]

[half ice]

Feel free to ask if you have any others

[forest quiver]

nah that's about it

but god damn

linear algebra has some insane real world applications

like I can balance chemistry equations

with vectors

my chemistry teacher is going to shit herself next year when I show her what I can do

Also, network flow stuff

like nodes and inputs/outputs

to model trafic IRL or electric currents

is pretty damn cool as well

[half ice]

Lin alg is bae

[forest quiver]

just before I go

what do you think is the most useful math class

to take

[half ice]

Actually lin alg

[forest quiver]

calculus?

I took it and itw as pretty good as well

[half ice]

Calculus is fun too

[forest quiver]

[forest quiver]

This is linear dependence???

But they aren't in the same line

I might be thinking of R^2 though

Oh I get it

in R^2, two linearly dependent vectors are aligned

In R^3, three linearly dependent vectors can form either a line or a plane

It's still linear

[quaint sage]

I wish I understood differential equations 😦

Some really interesting problems can be modeled as DEs

[wraith patio]

how would i start part d https://puu.sh/HZn7J/283cb20eff.png

[half ice]

Start with seperable DEs. They're very easy and handle a lot of cases

[wraith patio]

<@&286206848099549185>

ik it says part c twice. i mean the fourth one down that should be part d

[rugged arrow]

<@&286206848099549185> guys can you answer this question

@pine dragon can you please help

My exam starts soon

[pine dragon]

bruh why pping me

i dont even know you

[low halo]

Please help

[hollow finch]

Find what f(0) is in terms of the determinant definition of the characteristic polynomial

[wintry steppe]

following up on nix, for the second part, do you know the cayley hamilton theorem?

it comes up

[low halo]

No