linear-algebra

[cunning arch]

thank you so much

[acoustic path]

thx baklava. youre carrying the channel

[viscid kernel]

@cunning arch Im sorry to say, but I couldnt figure it out 😦

[cunning arch]

It's ok! Thanks so much for trying anyways

[wintry steppe]

ok so

i don't understand this

for instance, for the cokernel...

coker g = M/img(g) = {m + img(g)| m <- M}

how does this connect to idea of cokernel measuring how surjective something is bros

OOOOOOOh i get it nvm i am dumb

ignore me

[tacit girder]

Can someone help:
The pressure P is a linear function of temperature T. For T = 1, P = 3. For T = 2, P = 4 and for T = 3, P = 4.5. What is the best guess for P when T = 4?

[cloud thistle]

Are you familiar with linear least squares?

[tacit girder]

sort of yeah just started learning

[cloud thistle]

Are you familiar with the matrix representation of it?

[tacit girder]

yeah a bit

[cloud thistle]

You can find your coefficients of a line from calculating
$$\mu = (A^TX)^{-1}X^Ty$$
where $y$ are your pressure observations and $A$ contains your basis functions evaluated at each temperature $T$.

[stoic python]

Sebastian:

[slow scroll]

What is X?

[silver tree]

I have a question, that I can't seem to answer (edit - I have gotten, that there is no solution to the expression - is that true?)

You have triangle T1, that has the corners A = (1,1), B= (3,-1) and C = (4,3)
We also have another triangle T2 with the corners R= (2,-1), Q=(7,-2) and P = (5,-7)

Does a linear Map / transformation exist that satisfy the following expression:

f(T1) = T2?

A mapping of the situation is seen below

[wintry sphinx]

@silver tree no, but an affine transformation does exist

ehhh wait

maybe there is

in general, no

usually there has to be some translation

and translation is not a linear transformation

[silver tree]

I have showed a so-called scenario with only variables

Which gives no solution.

[wintry sphinx]

I don't know what you mean by that

[silver tree]

Uhm, Idk if I would violate any rules if I show it here.

[wintry sphinx]

but basically you need to show that the system of linear equations [A B C]^T [x, y] = [R Q P]^T is inconsistent

why would that be a violation of rules?

[silver tree]

I meant more like violation in discord.

[wintry sphinx]

strange

I can't imagine linear algebra being a violation of the rules of discord, so I really don't know what you're talking about

[silver tree]

I actually don't know myself - I'm very tired, I apologize.

Was thinking of something entirely different.

Anyway, with your linear equations [A B C]^T [x, y] = [R Q P]^T, I get this matrix (Just made augmented matrix with the linear equation) :

[wintry sphinx]

how'd you get a 3x4 matrix though

I expected a 3x2 matrix

oh wait nvm I see what you did

[silver tree]

Then I have to solve for this, right?

[wintry sphinx]

yeah you row reduce it

[silver tree]

Then I get this one:

[wintry sphinx]

so it is inconsistent

if you think about what the equations mean

the last one basically says that (0, 0) maps to some point that isn't the origin

[silver tree]

Which isn't possible, right?

And thus, it is inconsistent.

[rancid rain]

alright this problem is wacky, so I wouldn't be surprised if no one can help, but let's see

[spare crystal]

im kinda rusty on eigenstuffs but I think you try to find the eigenvectors with the given info

[rancid rain]

yeah the problem is i don't know how to find the vectors

[spare crystal]

well for the eigenvalue 2, there is one eigenvector

(hint its given in the problem)

[rancid rain]

sure but the 1 eigenspace is what's confusing me

[spare crystal]

you know any eigenvectors with eigenvalue of 1 are in V

so you have to find the vectors that span V

then you can say V = span{u, v}, which would make u and v eigenvectors

[rancid rain]

i guess you would have y and z as free variables so that x = y{-14/8,1,0} +z{-3,0,1} is that right?

[wintry sphinx]

find a basis for the eigenspaces

and then write out the diagonalized form of the matrix

multiply together

and presto

[viscid kernel]

Hmm so for example suppose det(A) = 0, you have the equation Ax = (4,7,1). If there exists a solution the solution would be the kernel + a vector right. Is that an affine subspace ?

[wintry sphinx]

well, if det(A) = 0, there's only one solution, but yes, it's an affine subspace

[cunning arch]

is it ok if i bump my question from this morning down one last time?

[wintry sphinx]

what question is it

[cunning arch]

I'm a bit lost on how to write the transformation matrix across a plane

Here's the full question. I know hwo to find the inverse matrix and whatnot, I'm just not sure how to actually get the transformation matrix

[soft burrow]

You can think of R^3 as generated by three vectors: the two vectors that generate the plane and the normal vector of the plane. A linear transformation is uniquely determinated by the images of the basis. So you want a transformation that leaves the two generators of the plane constant, and reflects the normal vector only. You can write this as something like

Av_1=v_1
Av_2=v_2
Av_3=-v_3 (we reflect this one)

but this is equivalent to writing the matrix equation A[v_1 v_2 v_3]=[v_1 v_2 -v_3], by definition of matrix multiplication. (edit: here [a b c] means the matrix with columns a, b and c.) You should be able to find A from here

[cunning arch]

Ah I'll sit on that and think about it for a bit, thank you so much! almost forgot normal vectors were a thing

[warped garden]

hey guys, i managed to do (a), found the basis, rank & nullity. how do I prove for (b) & (c)? am stuck

r(Au) + s(Av) + t(Aw) = 0, since the vectors are linearly independent, r=s=t=0.
A(ru + sv + tw) = 0
ru + sv + tw = 0
Since, r=s=t=0, u, v, w has the trivial soln for the homogenous system, hence they are linearly independent.

Anyone able to verify the above? ^

[native rampart]

Well, Don't conclude r=s=t=0 in the first step

If A(v)=0 v can be nonzero

So,You can't conclude that

[warped garden]

hmm, how does v being nonzero affect r=s=t=0?

[rose umbra]

how can i find vectored triangle's angles in 3d ?

[round coral]

@warped garden for understanding b) and c) remember the relationship between linear maps and matrices.

then you will clearly understand the question

b you can prove by this, for c look at the RRE form and find the null space of A. If the null space is 0 then the map is injective, and c holds but if the null space is not 0 then the statement of c is not valid.

[simple hornet]

hello why did they change y + y = z?

and why didn't they reassign the value of 2x, or the value y + x?

this is the definition we're using btw

[round coral]

F is just any field, maybe not R where y+y=2y, it can be any number in F so this is he best way of writing the sum of subspaces

If you knew F = R then things could have been like you said

[simple hornet]

ahh

Yes you're right

multiplication can be defined completely differently in any arbitrary field

[round coral]

Yep !! 1st chapter of Axler!

[simple hornet]

but why didn't they reassign the values of x?

yeap lol i've been trying to work my way through it

ohhh wait no i got it

[round coral]

doesn't matter actually, because x is just any element of F

[simple hornet]

x + x = x + x, since we have two of them we just assign them to be the same thing

it can be whatever, just that the elements are the same

y + x is different, so we give it the variable name y, and y + y is something else so we give it a different name

[round coral]

they took the sum of x+x as x not because the sum of x+x is x, but he just took it x , for simplicity or you would have too many x,y,z,w,..

[simple hornet]

yeahh

when they say (x,x,y,y) + (x,x,x,y) they just mean that you need to use the definition of addition on that field for any two elements that are the same

the x in the first list just has to be the same as the x in the second list right?

not just any arbitrary ones with no conditions whatsoever

sorry if im being a bit dense atm lol

[round coral]

maybe, maybe not. It doesn't matter actually, why are you thinking to that point, that is not what Axler is trying to explain here.

[simple hornet]

idk the notation is a little confusing ig

but i think i get the point, the sum of two subspaces is all the possible sums of all the elements in the subspaces

[round coral]

yes, that is what is given in the definition

you are right

[simple hornet]

im just not sure how to go about demonstrating

it

i guess

[round coral]

see ahead, do some examples, do the exercise

you will understand it clearly

[simple hornet]

yes ill try doing that

thank you

one other thing

if we decompose a vector space V into n different vector subspaces U, in that case the union of the vector subspaces is a vector space

right?

[round coral]

Sorry, I was wrong, the union of subspace is not a subspace

I made a mistake

[wintry steppe]

Guys how do this using subs method. I'm not allowed to use other methids.

[round coral]

what do you mean by subs?

[wintry steppe]

Substitution

[round coral]

convert it into RRE form then solve it, substitution too messy. I won't do that way.

[wintry steppe]

I'm supposed to use substitution method only. I'm not allowed to use other methods

[simple hornet]

yeah substitution is torture for this

oh wait

notice that if you add one to both sides on the top one and then equate the top and bottom ones

you might be able to get some headway there

[round coral]

yeah just do the traditional way then as cutiepie said

[simple hornet]

yes im a cutiepie

:)

[marble lance]

Let x_4 and x_5 be free variables. The use equation 3 to write x_1 in terms of x_3, x_4 and x_5. Substitute that into equations 1 and 2. Then make x_2 the subject of Equation 2 and substitute it into Equation 1*, then you can solve for x_3 in terms of x_4 and x_5. Then work backwards to get x_2 and x_1 too.

[simple hornet]

oh wow

[marble lance]

That's a lot of work, so have fun. You can ask if you're stuck

[wintry steppe]

Thanks man

Btw what do you mean by letting x_4 and x_5 be a free variable.

[simple hornet]

x_4 and x_5 can take on any value if they are free

what they are are of no consequence to the validity of the solution

[wintry steppe]

I get that but how does that make sense?

I don't get how a variable can take any value

[simple hornet]

if they were 0 or if they were 100 it doesn't change the solution

[marble lance]

The equations don't have a single solution. So x_4 and x_5 can have any value, and that will determine the values of x1 x2 and x3. So there are infinitely many solutions.

[viscid kernel]

@wintry steppe when doing gaussian elimination with an augmented matrix its same as substracting or adding an “n times “equation to another one

[simple hornet]

i believe when you multiply the matrices the free variables are zeroed out

[wintry steppe]

I haven't learnt Gaussian elimination yet

[viscid kernel]

You can kind of cheat, so if you put it in a matrix its easier to see, which row to subtract from the others. And then do it the same with those equations.

Ahhh

U still in highschool ?

[wintry steppe]

Just started Uni

[viscid kernel]

Aight, Good luck

[wintry steppe]

Thanks man

So when I take X_1 in terms of X_3 x_4 and x_5 I can give x_4 and x_5 any values right?. So can I just give them 0 to remove them from the equation?

[marble lance]

That will give you one solution but not all solutions

[warped garden]

anyone could give some hints on how to start (i)?

[wintry steppe]

Btw how do you determine what is a free determine. If the no. of equations< no. of variables, the extra variables become free variables?

[viscid kernel]

The free values are the variables ( who correspond with its columns ) become all zero after reduced row echelon form

[warped garden]

anyone could give some hints on how to start (i)?
i figured this out haha

[simple hornet]

hello is this a reasonable proof that the sum of subspaces is the smallest subspace that contains the constituent subspaces

[slate fox]

hey guys, I've been stuck on this question for a couple of days now. Does there exist a field F with characteristic p>0 and an element a such that a^p does not equal a?

any help would be appreciated

[native rampart]

Isn't a^p always 1?

[slate fox]

I dont think so?

[native rampart]

Nvm,I thought you used * instead of + in the notation

[slate fox]

any thoughts?

[native rampart]

Try a matrix field?

[slate fox]

hmm nothing is really coming to mind

<@&286206848099549185> any ideas?

[swift plaza]

Try looking in the field of rational functions F_p(X)

[slate fox]

but it doesnt have characteristic >0 does it?

@swift plaza

[swift plaza]

F_p(X) has characteristic p

[slate fox]

oh you mean rationals with coefficients in Zp?

sorry I'm new to this not very familiar with all the notations

[swift plaza]

Oh yeah F_p is the field with p elements i.e. Z/pZ or Z_p however you wanna write it

[slate fox]

oh I see

so I can just take x+1 in F_2(X) right

characteristic 2

but (x+1)^2 doesnt equal x+1

does that work

[swift plaza]

or even just x, but yeah that works

[slate fox]

oh wow that was simple

I just didnt know about F_p

thanks alot

[marble lance]

@stray granite where are you stuck?

[stray granite]

@stray granite where are you stuck?
@marble lance c)

[cunning arch]

I'm a bit confused on how to write the matrix transformation for a projection onto a line, because I don't think what I did here is right oops sorry for interrupting

[marble lance]

It's okay, just ask in a help channel or wait until we are done

@stray granite seems similar to the others...

(1,3,-3) = a(1,0,0) + b(0,-1,1)

Then solve for a and b

[stray granite]

should there not be 3 vectors in which give a linear combination of v ?

[marble lance]

There are only two basis vectors

So why would there be 3?

[stray granite]

because v = to a vector of 3 coordinates ?

[marble lance]

No, v = a linear combination of however many basis vectors there are

When you work in R^3, you have 3 basis vectors

But now you are working in a subspace of R^3 that only has 2 basis vectors

[stray granite]

so no need to use w in any way ?

it is just describing the subspace ?

[marble lance]

Yep, w is nothing

It's just the symbol they use to describe the elements of V

[stray granite]

ok

what about e) then ?

[marble lance]

Typo there with the 1=x

Should probably be 1, 1+x, 1+x^2

[stray granite]

possibly

error by book

[marble lance]

Then you do v =a(1)+ b(1+x)+c(1+x^2) and solve for a b and c

[stray granite]

it could be x though

the 1=x

[marble lance]

Yes

[stray granite]

maybe a*x

ok

how come you wrote a(1)c

typo ?

[marble lance]

Yes

[stray granite]

ok

[cunning arch]

unless you're still working on the problem, would it be ok if i bump my question down?

[marble lance]

Go for it

[cunning arch]

I'm a bit confused on how to write the matrix transformation for a projection onto a line x_2=(tan(theta))x_1, because I don't think what I did here is right

After, it's just matrix multiplication and showing that it's not commutative, which I can do, I just can't get past this first part of actually building the matrix

[lone gull]

my drawing is terrible but i hope you can see what does it says. basically i need to add vectors a'+c if i know b=1,a'=2 .i got sides b and c. problem is i dont know how to get x and y coordinates of a' and c so i can add them.

[stray granite]

since dimNull(A) + rank(A) = n

does that mean dimNull(A) = n - rank(A) ?

where n is the number of columns of A

[gray dust]

@stray granite yes

[stray granite]

ok

[wintry sphinx]

@cunning arch you went from a vector to a scalar somewhere in your calculation; also, it's sufficient to compute the projections of <1, 0> and <0, 1>

and indeed, those are the columns of the matrix

[edgy kraken]

how would I go about solving this

[mild igloo]

Ianswered 5 to this

is it correct

[austere cedar]

To the best of my knowledge if A is invertible, det(A) is non-zero, there will be only one unique solution.

[mild igloo]

alright

I looked it up and it said its the number of b in Rn space

[austere cedar]

No idea what any of that language means yet sadly.

[mild igloo]

like Ax=b

number of b in Rn

Rn is the dimension

[austere cedar]

You could also think of it like this, if det(A) is nonzero, there exists an inverse to A, A^(-1), so pre-multiplying by A^(-1), you get x = A^(-1)b.

[mild igloo]

im really bad at thinking of math like that

like my algebra is very weak

[austere cedar]

This playlist has a lot of good stuff in which is very easy to understand: https://www.youtube.com/watch?v=0iSvYwdVVPI&list=PLg2tfDG3Ww4us-ACGUOMdCiYXyg4ymxrK&index=104&ab_channel=TLMaths.

[mild igloo]

@austere cedar thank you. Also if you have a matrix row with all zeros equal to zero its infinite solutions right

[austere cedar]

You mean, if you write a system of n equations with n unknowns using an augmented matrix, then using row operations to produce a reduced row-echelon form matrix(Gauss-Jordan Elimination) and if you get a row of zeroes at the bottom of the Augmented matrix?

[mild igloo]

yea

sorry

[austere cedar]

If so, yes, infinite solutions.

There's a neat thing when you do this, the reduced row-echelon form matrix is either an identity matrix or has row(s) of zeroes, if you have the identity, you can read off the unique solutions directly (no back substitution required, that would be Gaussian elimination with the augmented matrix being in any of the possible row-echelon form matrices, in that case you would have a unique solution, if you have a row of zeroes, to the best of my knowledge, you will have infinite solutions.

[frigid otter]

when I do this, is it as simple as row reducing the matrix?

[mild igloo]

that actually answers the next question

if you have an augmented matrix and one line is a scalar of another you also have infinite right

cuz the overlap is the solution set

and the overlap is infinite

[frigid otter]

row reduction with variables makes me uneasy

like I end up with this

[mild igloo]

if the determinant of a matrix is 0

it has no solution?

[frigid otter]

in what context

[mild igloo]

Ax=0

sorry I was asking a question not answering urs

[frigid otter]

yeah I know! that's okay. if the matrix has a zero determinant, I think it has none or infinite solutions

[austere cedar]

They call these systems, homogenous systems, you are guaranteed to have at least one solution, x = 0, they call this the trivial solution.

In summary, you either have exactly one solution, the trivial solution, or infinitely many non-zero solutions as well as the trivial solution of course.

To go back to what I was talking about before, since the matrix is now in row-echelon form (ref), you will have to use back substitution (Gaussian elimination), you could have used further row operations to get the matrix into reduced-ref (rref), so you can read the solutions off directly (Gauss-Jordan elimination), so that's just an option to you.

[mild igloo]

Vectors in R4 space with the last dimension as 0

is that sentence describing a 4x4 matrix with the bottom row being zeros?

[obsidian wren]

how do i know that semi-orthogonal matrices preserve the norm?

i wanna prove it for myself but i can't figure this out

[edgy kraken]

would this be true or false

cant get my mind around it

[dark gale]

with x-Px you are removing the part of x that is in W

[frigid otter]

@austere cedar can it get to rref? The best I got to was:

[austere cedar]

It is in rref.

[frigid otter]

with two zeros on the right, since it's an adjugate

[frigid otter]

so is the basis {1,0} {0,1}?

[austere cedar]

No, x1 = a/b and x2 = -1.

I don't know anything about basis', tbh.

[frigid otter]

it's actually this, since it's an adjugate

[austere cedar]

What are you trying to do, I should ask?

[frigid otter]

determine a basis for the subspace

[gray dust]

then why are you row reducing an arbitrary matrix in W?

[frigid otter]

is that not what you're supposed to do?

I mean it's the definition of W

[steady cargo]

Can anyone confirm if the general solution of this augmented matrice is correct:

(1 0  0       | 1          )
(0 0 g/(2-g)  | 1/(2-g)    )
(0 0 0        |  0         )

General Solution: {(1, 1/(2-g) - g/(2-g), g) g E R}

I feel like I need to add another variable cause matrice itself has a variable

For general solution I mean

[gray dust]

what you SHOULD do is find vectors that span W, then from those spanning vectors pick out a basis of W

for context the vector space you consider is M_2,3 ie the vectors here are real 2x3 matrices

[frigid otter]

I thought that was found by reducing the matrix definition, then from the leading coefficients you can determine if those vectors form a basis

[gray dust]

the main thing you're not grasping is the vector space we're in

for context the vector space you consider is M_2,3 ie the vectors here are real 2x3 matrices

[frigid otter]

hm. I think I'm missing something for this.

[gray dust]

M_2,3 is the set of all 2x3 matrices and this is a vector space where entire 2x3 matrices are its vectors, with rules of vector addition/scaling defined entrywise

[frigid otter]

okay so it's not a matrix of vectors

it's a vector of matrices lol

[gray dust]

eg (1 1 1; 1 1 1) is a vector in M_2,3

[frigid otter]

I don't know how to find the span of that :/ crap

[gray dust]

rewrite the general form of vector in W as a linear combo av1+bv2 where v1,v2 are vectors

[steady cargo]

I'm assuming that conversation is over, so I'll repeat myself:
Can anyone confirm if the general solution of this augmented matrice is correct:

(1 0  0       | 1          )
(0 0 g/(2-g)  | 1/(2-g)    )
(0 0 0        |  0         )

General Solution: {(1, 1/(2-g) - g/(2-g), g) g E R}
I feel like I need to add another variable to general solution because matrice itself has a variable

[frigid otter]

@gray dust Sorry to bother you again, but I'm back around to this question. Is this the veritable first step in the problem?

So the span would be the matrices on the right?

[steady cargo]

Pretty sure your last matrice needs to have -1 in 2,3 position no?

[gray dust]

check the 2nd matrix

[frigid otter]

oh

-1

you right

I wrote that down, but didn't type it out right lol

but the span is those two matrices right

[gray dust]

rewrite the general form of vector in W as a linear combo av1+bv2 where v1,v2 are vectors
calling those vectors v1,v2 you showed any vector in W is a linear combo of v1,v2 so W=span{v1,v2}

[frigid otter]

you got a hint for the next step lol

or is it just if the span is linearly independent, it's a basis?

[gray dust]

we call set of vectors that spans W linearly independent, not the span itself

[mild igloo]

if ther determinant is zero the matrix is dependant?

or independant

[gray dust]

we say the columns/rows of the matrix linearly independent, not the matrix itself. the cols/rows are lin indep iff det!=0

[frigid otter]

so by showing linear independence through c_1v_1 + c_2v_2 = 0 matrix, shows that it's a basis?

[gray dust]

showing the only solution to c_1v_1 + c_2v_2 = 0 is c1=c2=0 shows {v1,v2} is lin indep

if {v1,v2} is lin indep and W=span{v1,v2} then {v1,v2} is a basis of W

[frigid otter]

and thus... a basis?

nice

thanks a lot

[gray dust]

you're welcome

[exotic tusk]

are these not the same? I was under the impression that matrices are associative

[steady cargo]

👁️ I'd have thought the same

[muted holly]

They are associative but not commutative

So u cannot cancel the BB-1 since they're not next to eachither

[steady cargo]

🧠

[austere cedar]

To see it:\
$(BC)B^{-1} = B(CB^{-1})$.

[steady cargo]

If you knew CB=BC then you'd be right cause you could do (BC)(B^-1)
(CB)(B^-1)
(C)(BB^-1)

[stoic python]

stabulo:

[mild igloo]

is there a fast way to find the determinant of the inverse of a matrix

[soft burrow]

it's the inverse of the determinant of the original

since $\det(A)\det(A^{-1})=\det(AA^{-1})=\det(I)=1$

[stoic python]

derivada.schwarziana:

[mild igloo]

so 1/det(a)?

[soft burrow]

yes

[mild igloo]

thank you

[cunning arch]

(bumping my question down from this morning). I figured out 1, I'm just really confused about building the matrices for reflections and rotations in 3D

Here's the work I've done so far: The projection of a vector v, <x1,x2,x3> onto a plane is equal to that v - projection formula, right? And that formula needs the normal vector, whcih I'm not sure how to get

And then for 3, I'm just really confused in general. Not sure how to start building the transformation matrix for that one

It anyone could help, it would be really really appreciated

[graceful seal]

find the coordinates of the 4 points which have integer coordinates and are a distance of √5 from the point (1,2) hint 5=1^2+2,2

since 1

someone help me?

[cunning arch]

Update: I've figured out projection, if someone could just help me with reflecting, it would help a lot :)

[slow scroll]

@cunning arch what about reflecting?

assuming im understanding what you're supposed to do, one way describe a reflection over a plane, is to first apply a transformation that rotates the coordinate axes such that one of the axes is the normal of the plane. Then you just have to apply a simple reflection (where you negate a coordinate), and then undo the original transformation.

[steady cargo]

Can anyone help me with general solution

[cunning arch]

@slow scroll i'm just a bit confused on how to build the actual transformation matrix, and what the unit vector has to do with any of it

[steady cargo]

What I think was {(1, 1/(1-2s), -s/(1-2s), s) s E R}

[slow scroll]

if you were asked to make a transformation that rotates about some vector v in R3, could u do that?

[cunning arch]

Av_1 = v_1
Av_2 = v_2
Av_3 = -v_3 (this is the one that we reflect)
like here's my understanding so far

uhhh

like i know the rotation matrix, but i'm not sure

[slow scroll]

hm i have an idea, one sec

[cunning arch]

alright

[slow scroll]

does this stuff make sense to u?

[cunning arch]

yes

[slow scroll]

ok, i assert that you can use what is there to get the reflection of v (the orange vector) across p

[cunning arch]

so just negative v-u

[slow scroll]

close

v-u is normal to the plane while v is not, so that would not make sense as a reflection, but you have the right idea

[cunning arch]

2(v-u)? lol sorry i'm not sure

because i thought everything just stays the same, and the normal vector just flips to be negative

[slow scroll]

ur so close lol. we want the red vector

[cunning arch]

lol i'm not sure what else i can try

[slow scroll]

its v - 2(v-u)

[cunning arch]

wait let me process that

wait can you explain why there's a v in front?

wait subtracting vectors

[slow scroll]

so, intuitively, you start with v
subtracting off one v-u gives you u.
subtracting off another v-u gives you the reflection of v across p

[cunning arch]

i thought subtracting vectors had to be tail-to-tail?

OH WAIT

sorry i'm trying to visualize this so if we subtract v-u from v, all i can see is yeah it's the length of u, but not on the pane

[cunning arch]

isn't that adding vectors then

OH NEVERMIND

LIGHTBULB MOMENT!!!

thank you so much, i think i can use that to build it

[slow scroll]

alright, nice, npnp!

[steady cargo]

Can anyone confirm if the general solution of this augmented matrice is correct:

(1 0  0       | 1          )
(0 1 g/(2-g)  | 1/(2-g)    )
(0 0 0        |  0         )

General Solution: {(1, 1/(2-g) - g/(2-g), g) g E R}
I feel like I need to add another variable to general solution because matrice itself has a variable
Can anyone help me I asked this question few times over past 3hours ago and no one helped

[slow scroll]

doesn't look right

[steady cargo]

Thank you for responding first off and in what sense?

oops

matrice missed a number

[slow scroll]

ah okay hm

what is the g supposed to be in the matrix? You treat it as a free variable in your solution, but it can't be both some fixed value, and a free variable in the solution

[steady cargo]

The question itself is bit weird basically I had a matrix by a vector matrix = a vector matrice

i replaced the vector i used for multiplication with the sum of two vectors one in its general solution

hence the g

That's what I was confused about yeah I'm not sure how to formulate a general solution

[slow scroll]

do u have a screenshot of the original problem?

[steady cargo]

For that matrice

Yes

[slow scroll]

it would help if i could see it

[steady cargo]

part d

[slow scroll]

alrighty, so this is the augmented matrix you would have wanted for part a: $$\left[\begin{array}{ccc|c} 1&1&2&0\0&1&2&1 \ 1&2&4&1 \end{array}\right] $$

[stoic python]

kxrider:

[steady cargo]

do you mean part b?

[slow scroll]

yea

[steady cargo]

perfect yes

[mild igloo]

if I have a homogeneous linear equation with a determanant of zero I have infinite

or exactly one or infinite

[steady cargo]

👁

[mild igloo]

sorry thats terrible wording

[slow scroll]

lol this channel is occupied rn, wait or ask in a #questions channel or something. your question is not worded well enough for me to answer concisely

[steady cargo]

general solution for part c:
{(-1, 1-2t, t) t E R}

I think?

That is what I went with anyways for part d

[mild igloo]

Ill reword and I dont mind waiting

I understand other people need help too

If det(A) = 0 how many solutions does Ax=0 have

[slow scroll]

okay harbh, i agree so far

[steady cargo]

Perfect

part d i said x = c then so
( -1 )
( 1-2t)
(t )

but subbed t =0

since that still satisifes Ax = b

Then added that vector to y which is
( 0 )
( -2g)
( g )

with a resultant vector of
(-1 )
(1-2g)
(g )

and multiplied that by A and put it equal to b

[slow scroll]

right, so you have for any t in R,
(-1, 1-2t, t) = (-1, 1, 0) + t(0, -2, 1)
by the definition of vector addition

[steady cargo]

Yes

[slow scroll]

(-1, 1, 0) is a solution to Ax = b,
(0, -2, 1) is a solution to Ax = 0

[steady cargo]

wait did i not need to solve matrice once i made t = 0

thought they are different variables?

t and g

[slow scroll]

im not sure why you introduced g lol

[steady cargo]

t came from previous matrice solution to part b

and g came from part d when it equals to vector with all zeroes

[slow scroll]

so you did rref and stuff to solve Ax = 0?

[steady cargo]

Yes for a vector of
( 0 )
( -2g)
( g )

well general solution
{(0, -2g, g) g E R}

that vector plus vector i made t=0 in gave me vector with g in it

[slow scroll]

ok, so when you did part c, you got this solution:
{(-1, 1-2t, t) t E R}

the algorithm for solving systems Ax = b, gives you a solution of the form,
(-1, 1-2t, t) = (-1, 1, 0) + t(0, -2, 1)

(-1, 1, 0) is sometimes called the "particular" solution.
t(0, -2, 1) for any t in R, are just all of the solutions to Ax = 0. You don't have to do any additional work to get this.

"t" and "g" here represent any real number. For every real number t, there is a solution t(0,-2,1) = (0,-2t, t) to the equation Ax = 0.

[steady cargo]

I feel like I may have messed up with assuming that is the vector..

[slow scroll]

so what is your "c" and "y" here?

[steady cargo]

Hmm what confused me was the fact it said general solution of y and what do u mean?

[slow scroll]

so, (0, -2, 1) is one solution to Ay = 0. To get all of the solutions, we note that A(ty) = t(Ay) = t0 = 0, so any scalar multiple of (0,-2,1) is also a solution to Ay = 0. So the general solution to Ay = 0 is
"t(0,-2,1) for any t in R"

[steady cargo]

I get what gou mean by any scalar multiple is a solution but not A(ty) = t(Ay) do u mean if we treat t as a scalar?

[slow scroll]

yea, t is a scalar. its just a fact of matrices (more generally, linear transformations), that scalars can be "factored" out in this way.

[steady cargo]

I haven't heard of linear transformations but I think I get you

[slow scroll]

u probably will at some point haha.

[steady cargo]

So I don't need to give matrice where x is equal to y + c and true lol

the question is so weird

[slow scroll]

so we have a solution to Ax = b
I just explained how the general solution (everything that x is allowed to be), starting with what you got: {(-1, 1-2t, t) t E R} can be expressed as
x = (-1, 1, 0) + t(0, -2, 1) for any t in R.

[steady cargo]

Yes

I'm following that

[slow scroll]

moreover, u know that A(-1,1,0) = b and A(t(0,-2,1)) = 0 for any t in R.
So what is "y" and "c" here?

[steady cargo]

y = t(0, -2, 1)
c = (-1, 1, 0)

how come u subbed t in for g?

this just makes it clearer what i did

g = s

[slow scroll]

{(0,-2s, s) : s in R} = {(0,-2t, t) : t in R}
the letters you pick make no difference here t, s, g, or whatever else are just representing completely arbitrary real numbers.

I just decided to stick with the same letter you used in {(-1, 1-2t, t) : t E R}

[steady cargo]

Ah okay

and I forgot : thankyou

[slow scroll]

y = t(0, -2, 1)
c = (-1, 1, 0)
and this is pretty much correct. When you write out the solution, its important to specify "for all t in R"
Then ur golden
npnp

@mild igloo the answer to ur question is infinitely many

[steady cargo]

oh damn

so this? 100% unnecessary

[slow scroll]

yea

[mild igloo]

not exactly 1 or infinitely many

just infinitely many

[steady cargo]

Lol good to know, thankyou v much!

[slow scroll]

npnp

[robust pond]

could i ask for help interpreting something

[slow scroll]

not exactly 1 or infinitely many
infinitely many. to be precise, when det(A) = 0, the null space is a nontrivial subspace. And assuming ur matrix has real entries, this subspace contains infinitely many vectors

[mild igloo]

thank you

[slow scroll]

np

[grizzled folio]

Quick question:

Let's say I have the equation $V=ABCW$ where all the values are matrices. If I want to isolate $W$, would I rewrite the equation as $C^{-1}B^{-1}A^{-1}V=W$. Is this the correct way to rearrange and isolate?

[stoic python]

Joaquin-Revello:

[grizzled folio]

Or would it be $A^{-1}B^{-1}C^{-1}V=W$

[stoic python]

Joaquin-Revello:

[native rampart]

First is correct

Let's say I have the equation $V=ABCW$ where all the values are matrices. If I want to isolate $W$ rewrite the equation as $C^{-1}B^{-1}A^{-1}V=W$.
This

[stoic python]

DrunkenDrake:

[grizzled folio]

thank you

[native rampart]

Just write down the rotation matrix?

And note R(x)R(-x)=I, where x is the angle

[grizzled folio]

ok, got it - thanks!

[robust pond]

I have a dumb question I think

I understand that in this equation, the right hand side is scalar multiplication

and in order to get to the form we want we need a matrix

I'm curious why the identity matrix is what you multiply lambda by

man im on a roll tonight, nevermind

i figure out everything i ask the second i ask it

[tacit marsh]

How do you describe a non-linear tranformation of points?
IE if a transformation of (1,1) to (1,1) (2,1) to (2,1) (3,3) to (3,20) and (5,5) to (30,20)

[wintry steppe]

The identity matrix is used so that the equation makes sense when factoring out x. (like in algebra, when factoring out x from x, you replace it with a 1. Think of the identity matrix as a "1" for matrices (thats why its called the identity)).

[unique jasper]

Can anyone give some examples where axioms 1, 4 and 6 work, but one of the others does not because this video https://youtu.be/QMwwplztQfY
Shows 1,4, and 6 but not an example of the rest.

Also does anyone know of good videos to study chapter 3 (vector spaces) of linear algebra with aplications by Steve leon. My professor didn't upload half of the lectures so I am really confused.

[slow scroll]

Can anyone give some examples where axioms 1, 4 and 6 work, but one of the others does not because this video
define addition on R2 the normal way and scalar multiplication on R2 to be something like c(a,b) = (a^c, b^c). Then axioms 1,4, and 6 hold but 7 doesn't

@unique jasper

[unique jasper]

Thanks @slow scroll , do you know good videos or ways to study this unit. My professor neglected to upload the lectures for the first half of this unit. I get the material for the half he uploaded, but I am still really confused on the half he did not upload. Mostly theory and proofs.

[slow scroll]

hmm, maybe khan academy or gilbert strang videos on mit ocw

[unique jasper]

Ok thanks ill get on that.

[slow scroll]

huh, im not even sure if Gilbert Strang talks about vector spaces in his video lectures. I can't seem to find one about it. I know khan academy definitely has videos tho

[austere cedar]

Imo, the MIT videos are horrible for linear algebra.

I'd honestly recommend going through Paul's online notes on Linear Algebra. I can give you the download if want it.

[unique jasper]

@austere cedar ill take what I can get

[austere cedar]

They used to be on his site but no longer someone sent them to me, I've been going through them speed run and it's the best stuff I've seen so far, in conjunction with 3Brown1Blue and TLMaths matrices stuff. The rest is utter garbage I've found.

[unique jasper]

3brown1blue is good but their videos don't go to in depth and there are only 15 for the whole course.

[austere cedar]

I agree.

[slow scroll]

random, but I want to point out that it seems like axiom 1, 4 and 6 are "good" enough, because they are good enough whenever you are testing whether a subset of a vector space is a vector space. But that takes for granted that you already have addition and scalar multiplication defined.

Hence why I had to define a different kind of scalar multiplication to give an example of ur question

[unique jasper]

@slow scroll ill keep that in mind when I'm taking the test. Thanks

[slow scroll]

np

[warped garden]

anyone able to give me some hints to start with (i)? been on it for an hour....

[dusky epoch]

take w in V^perp

and v in V

show A^Tw is orthogonal to v

[warped garden]

am I able to assume here that V is orthogonal?

[dusky epoch]

what?

what does that even mnea

mean*

[warped garden]

can I assume V is an orthogonal set?

[dusky epoch]

V is a subspace

so no there is absolutely no way in hell it is an orthogonal set

this all is literally just definition-pushing for the most part

[warped garden]

welp was confused previously

am i right to say dot(v, w)=0?

[dusky epoch]

that follows from the definition of V^perp of which w is a member

[stoic python]

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

[thorny hemlock]

<@&286206848099549185>

$\frac{s + \frac{7}{5}}{s^2+2s +3}$

[stoic python]

Yes:

[thorny hemlock]

is that correct ?

[summer wagon]

Could someone help me figure out this question? @everyone

[fading plaza]

Hello @summer wagon 🍞

[summer wagon]

Hi

[fading plaza]

This looks like a drawing exercise

[summer wagon]

Yea

Is it possible if you could draw on a piece of paper?

[fading plaza]

I might wanna type it out instead.

[summer wagon]

Oh okay 🙂

[fading plaza]

So, T(e1) is the blue one

[summer wagon]

T(e2) is the black one

[fading plaza]

For the 0.5 in the T(0.5,-1) it will be half of the T(e1)

And for negative one, it would be the opposite direction of the T(e2)

And using tip to tail method, you can find the final vector :D

[summer wagon]

Oh ok, could you gimme 1 min? I will find the final vector and let u know

So it will be (0.5, 0) and (0. -1) right?

[fading plaza]

Yea

[summer wagon]

Okay, could you tell me why would t(e1) be 0.5 and t(e2) be -1?

[fading plaza]

Hmmm....

Standard coordinate vectors

https://textbooks.math.gatech.edu/ila/linear-transformations.html

I think it's defined this way

[summer wagon]

Thanks

I was practicing for my midterm exam next week, I got 2 for questions. Could you help me figure them out?

Let T(x, y) = (x - y, 2x + y, y, y) and S(x, y, z, w) = (x + y + z + w, y + z + w). Find the standard matrix of S . T and determine if S . T is one-to-one. For practice clearly show your work.

[unique jasper]

Can anyone explain to me why this makes sense. I just don't get why you wouldn't distribute to u1 as well but only to u2

[austere cedar]

Oh, I remember this page, I think it's just to how the scalar multiplication is defined for that particular question.

They proposed that, scalar multiplication is defined so only u2 is multiplied by c.

[unique jasper]

Oh

[austere cedar]

Which meant an axiom of vector space (which one is it needs to have distributive properties) is not true for how it's defined.

[unique jasper]

It was part of the same example

[austere cedar]

Do you understand it now?

[unique jasper]

So let me run this back. What this page is saying is that the scalar distributive properties is defined by (u1, cu2) so therefore if we were to distribute it would not satisfy axiom h

Is that what its getting at or am I just extra slow today

[austere cedar]

Yes, if you follow the proposed rule, then if you try to verify distributive properly, which needs to be true for it to be a vector space, you see it's not true, therefore since it's failed at least one of the axioms, it's not a vector space.

No, it's not you, from what I've seen this stuff is horribly described everywhere, Paul's is the best I've come across.

People will tend to just throw definitions at you, with no intuition etc, so makes it very hard to learn.

[unique jasper]

It doesn't help that my professor didn't upload a lecture on vectors spaces and subspace only linear indépendance

[austere cedar]

I'd imagine these notes will be far superior that what you'd be given based on what I've seen anyway.

[unique jasper]

Yeah its better feels better to me. Again thanks for these notes.

[austere cedar]

No problem, just passing them on as someone passed them onto me.

[unique jasper]

These are the notes I got for vector spaces

[steady cargo]

I lost 10% of my grade for not showing it and not sure if its worth the appeal

Not graded towards anything

but just annoying seeing it on a homework

:(

I just did what question told me

Gotta be more careful it seems

Very fair

Thank you

[hollow finch]

Is there a name for the invertible matrix P such that B=PAP^-1 or B=P^-1AP (where A and B are similar matrices of course)?

[wintry steppe]

change of basis matrix

[robust pond]

why dimension 2

i thought free variables in the rref form determined the dimension of the eigenspace

oh, maybe i forgot to subtract the value dangit

oh, but that still doesnt make sense

shouldnt it be 1?

[robust pond]

no i was confused because theres one free variable in A-4I

but thats not what determines the dimension

its the rank of the REF

of A-lI

okay, cool

thanks

🙇‍♂️

no, 2 is the correct answer

thats why im confused

me too

theres only one dimension of freedom

is it because theyre repeated?

that doesn't gain you an additional dimension does it

it just adds generalized vectors

do you want i can post the problem again

[edgy kraken]

how would i do this

[stoic python]

slimvesus:

[robust pond]

should be -4I?

oh im sorry

its -4

oh

you lose a row

gotcha

sorry my brain is fried from doing this for 2 hours

thanks again

[edgy kraken]

anyone have ideas about mine?

[robust pond]

you may wanna repost @edgy kraken

[rustic panther]

Hi all, need help with this

[knotty blaze]

x*=[b1+b2+b3...+bn]

@edgy kraken

[ionic nimbus]

can someone explain me how is this possible??

[ionic nimbus]

just this sentence

is here linear algebra master?

[limber sierra]

what's $\mathcal{R}$ denote here? range?

[stoic python]

Namington:

[limber sierra]

row space?

[ionic nimbus]

yes

[limber sierra]

yes to which one

[ionic nimbus]

row space

but I think it doesn't matter here

I have a problem to understand this

they are independent => they are all 0 => the are dependent

why it isn't false

[limber sierra]

yeah it seems like theyre abusing terminology here

unless youre taking it as convention that the 0 vector is linearly independent with everything, which would be... bizarre

[ionic nimbus]

so what now?

I think i don't take it as convention

I learned that set of vectors is dependent when there is 0 vector

but on the other hand, if all those vectors become 0 then we have 1 vector

and independence is not internal feature

maybe it is not false because of clavius rule

or I make vicious circle mistake

I'm trying to understand it for 8 hours

[limber sierra]

my guess would be that that should be rephrased as "linearly independent or zero"

but that argument doesnt really make sense either

since as far as I can tell, we have no reason to assume that

or its at least unjustified

[ionic nimbus]

those vectors are from subspaces

and there is direct sum

so linearly independent or zero make sens

i think

but I'm not conviced that there is problem

they are independent or zero => they are all 0 => the are dependent

but if they are dependent how can we know that they are all 0 ?

and I think this is where vicious circle comes

but I have not encountered such a situation before

I mean

they are independent or zero, using independence we have that they are all zero so they are both indepedent and zero, it is false, if we don't use independence we won't figure out that they are all zero

do you understand my thought?

[limber sierra]

the reasoning is

"either they are linearly independent are 0"

``but we have nonzero scalars $\alpha_0, \beta_0, \alpha_0 + \beta_0$ such that $(\alpha_0 + \beta_0)v_{\mathcal{R}} + \alpha_0 v_B + \beta_0 v_A = 0$"

[stoic python]

Namington:

[limber sierra]

"this means they cant be linearly independent (that's literally the definition of linear dependence), so they must be all 0"

this is just $((A \lor B) \land (\neg A)) \implies B$

[stoic python]

Namington:

[limber sierra]

no fancy logic going on

"it's either raining or snowing. but it's not raining. so it's snowing"

[ionic nimbus]

yes

you'r right

so the key was " linear independent or zero " instead of linear independent

so thank you very much

[wintry steppe]

So I just started learning Gaussian elimination. I took down the note of an example my teacher did. Can someone please explain why x_3 and x_4 became free variables

[limber sierra]

you mean here?

no matter how we change x_3, x_4, we can "adapt" x_1, x_2 to make the equations still true

so the values of x_1, x_2 are determined ("bound") by what values the "free" variables take

as a general rule, free variables are any variables that are not pivots (After doing gaussian elimination)

if youre familiar with that term

[wintry steppe]

I am not familiar with pivots. Sorry

[limber sierra]

okay well

once you've "finished" gaussian elimination

i.e. ended up in a row-echelon form system

the leftmost variable in each row is a "pivot"

if a variable is not the leftmost in any row, it's a free variable

[wintry steppe]

Oh I see

But I don't get why x_3 becomes free variable

In a system of 3 equations

[round coral]

If we have three distinct primitive pythagorean triples, (x_1,y_1,z_1), (x_2,y_2,z_2), (x_3,y_3,z_3) , how to show that these 3 are linearly independent in R^3?

<@&286206848099549185>

[slow scroll]

I think it suffices to show that a linear combination of two primitive pythagorean triples is not a pythagorean triple.

well, of course it would suffice to show that, but i think that is the best approach to this problem

[wintry steppe]

Reduce to the case when $z_1 = z_2 = z_3 = 1$ by multiplying the vectors by $1/z_1, 1/z_2, 1/z_3$ respectively. So you have three points $(x_1, y_1, 1), (x_2, y_2,1), (x_3, y_3,1)$ where $(x_i,y_i)$ is each a solution to $x^2 + y^2 = 1$, so on a circle. They're 3 distinct points on the circle because you started with distinct primitive pythagorean triples

and last hint: showing that these points are linearly independent is equivalent to the statement that a line intersects a circle on a plane at most 2 distinct points.

[stoic python]

ℓ-adic:

[round coral]

Yeah! That is really good way of doing it.

Thank you.

[wintry steppe]

You're welcooome.

[round coral]

But I had a doubt when you multiply each (x_i,y_i,z_i) by 1/z_i then shouldn't it divide x_i and y_i as well and if you first take z_1=z_2=z_3=1 , how is that a primitive pythagorean triple

[wintry steppe]

they're no longer pythagorean triples after you divide by z_i. instead they become rational points on the circle (x_i,y_i), x_i^2 + y_i^2 = 1

(with that extra 1 stuck in in the last coordinate)

[slow scroll]

unless u are allowed to assume this, you would still need to show that after you fix one entry of each vector in a collection of 3-vectors, then as long as the other two entries are distinct, then the collection is linearly independent.

[wintry steppe]

x^2 + y^2 = z^2 iff (x/z)^2 + (y/z)^2 = 1.

[round coral]

I understand it now. Thank you for explaining me

[wintry steppe]

you have (x_1, y_1, 1), (x_2, y_2, 1), (x_3, y_3,1) with (x_i)^2 + (y_i)^2 = 1 for all i. Showing that they're linearly independent is equivalent to showing that (x_1 - x_3, y_1 - y_3, 0), (x_2 - x_3, y_2 - y_3,0), and (x_3, y_3, 1) are linearly independent. This is equivalent to showing only that the first two are linearly independent, and this in turn is equivalent to showing that a line intersects a circle at at most 2 points.

[slow scroll]

nice.

[round coral]

@wintry steppe Thank you very much.

[wintry steppe]

You're welcome. One last thing: Do you see why the points (x_i/z_i, y_i/z_i) for i = 1, 2, 3 are distinct? This depends on the original points (x_i,y_i,z_i) consisting of distinct primitive pythagorean triples. One should provide an argument for this, albeit easy.

@round coral

One can prove the following more general lemma to show this: If (x_1, ..., x_n) and (y_1, ..., y_n) are points in Z^n with gcd(x_1, ..., x_n) = gcd(y_1, ..., y_n) = 1 then either they're linearly independent over R or if they're dependent, then (x_1, ..., x_n) = +/- (y_1, ..., y_n).

There are many easy ways to prove this, so I'll leave it here. DM if you need more help.

[open pivot]

I have a practice question I don't understand. I have the answers and the first observation in the answer is that 1 is an eigenvalue of A with eigenvector (1,0,0).

I'm not sure why we know 1 is an eigenvalue with that eigenvector

[dusky epoch]

A * [1;0;0] = [1; 0; 0]

[open pivot]

Was that supposed to be obvious to me? aha yikes... I didn't notice that

[dusky epoch]

A * [1;0;0] is the first col of A

[viscid kernel]

@open pivot I found it

[wintry steppe]

If there's no requirement for X to be finite, shouldn't this say "finite or countably infinite"?

[pallid rampart]

Why should X be countably infinite?

The theorem requires X to be separable

[wintry steppe]

I meant that if X is finite, say for example R3, it can't have countably infinite subspaces with increasing dimension

[pallid rampart]

I don't think the inclusion is strict there

also please say X is finite dimensional

[wintry steppe]

also please say X is finite dimensional
Wym? There is no requirement for X to be finite elsewhere in the book

[pallid rampart]

I meant X is finite is not the same thing as X is finite dimensional

[wintry steppe]

Ah, sorry

[pallid rampart]

So the inclusion is not strict

[marble lance]

Does that not contradict the dimension requirement?

[pallid rampart]

true

oh i wasn't looking clearly

hmm yeah i see the problem now

[wintry steppe]

I was thinking this could this work with something like X = R as a vector field over Q, with the basis vectors being some irrationals

Constructed as described in the theorem... but idk if that would produce a countably infinite set of basis

[pallid rampart]

Yeah I think it's probably wrong

Maybe they meant dim X_n≤n?

Or maybe they meant infinite dimensional X

[marble lance]

Where is this from?

[wintry steppe]

Ciarlet - Linear and Nonlinear Functional Analysis with Applications

🤔 it doesn't seem to be a typo from later mentions of the theorem

Constructed as described in the theorem... but idk if that would produce a countably infinite set of basis
Actually nvm - since we're taking the closure, the set of basis in the theorem doesn't have to be the set of all (uncountably infinite) basis 😅

Makes sense now

[nocturne jewel]

Anyone good with permutation matrices? been trying to find extra information to solve a question but to no avail

I think I understand this part of the question: It's the identity matrix with the k and k+1 rows swapped right?

[native rampart]

What is P_1,P_2...?

[nocturne jewel]

the ones from a im assuming

so P_1 swaps 1st and 2nd row, P_2 swaps 2nd and 3rd, etc

[native rampart]

Are you familiar with bubble sort?

[nocturne jewel]

Yeah i know it's basically do bubble sort on the identity matrix

but i cant put pencil to paper on how to prove it

[native rampart]

Do it the same way you prove bubble sort

[nocturne jewel]

the problem is idk how to prove it lol

Like i understand the logic fine, i just dont know how to express it yknow?

[native rampart]

Use Induction

[nocturne jewel]

on k or n?

[native rampart]

Let's say all permutations of {1,2,3....(n-1)} can be written in this form

Now take {1,2,3,4...n} and move n to the point the permutation is supposed to take it to

[nocturne jewel]

Oh wait

If I start with the Identity matrix, I can apply P1,P2,P3...,P(whatever) until the 1st entry is in the right place

[native rampart]

Yes

[nocturne jewel]

and then i do that but start at P2

then P3

etc

[native rampart]

But after getting the thing to the first row,you have n-1 points left

So, It's true by induction hypothesis(After relabelling)

[nocturne jewel]

no i want to get the nth row entry in place 1st right?

cause if i did 1st column entry right, a subsequent entry could be lower which affects the position of the 1st entry

[native rampart]

Just never use P1 again

[nocturne jewel]

Right

so apply all the P's in order, then all the P's (bar P1)

then P's except P1,P2

[native rampart]

Yes

[nocturne jewel]

kk ty

@native rampart sorry for the ping but just one logic thing. The (n,n) entry of I will be in (j,n) after 1 "pass" of P matrices right?

[native rampart]

Yes

[nocturne jewel]

kk ty again

[gritty kelp]

can anyone explain this question to me? I dont really understand eigenvalues, or anything about them.

[quartz compass]

imagine the simplest possible matrix with these as eigenvalues

what would that look like for starters?

[gritty kelp]

it would be a matrix with these values on the main diagonal, right?

[quartz compass]

yeah good exactly

so now what do powers of this matrix look like?

[gritty kelp]

the powers of this matrix would look like the matrix with each of its elements raised to the same power, yes?

[quartz compass]

yeah exactly

so when you take the limit you can imagine taking the limit of each element raised to the nth power

in order to be the 0 matrix, all of these must go to 0

[gritty kelp]

ok, so the limit would be false because the 5/4 isnt going to go to 0

yes?

[quartz compass]

exactly

[gritty kelp]

thank you so much

[quartz compass]

yeah you're welcome, and similarly if it wasn't this simple

if it wasn't diagonal, you could diagonalize it, then the the power would factor through and give the same thing too

[gritty kelp]

I have one more question that I am really struggling with:

for part a, I know that the eigenvalues of P are going to be either 1 or 0 but I am not sure how to determine how many would be 1 or 0.

as I was writing that last message I realized what might be the answer to part b, but I am not sure if it is correct so if someone could just confirm my answer then that would be splendid.

I think the two vectors that define the space is spanned by would by eigenvectors that correspond to eigen values of 1, and then there exists vectors orthogonal to the given vectors with eigenvalues of 0. Is this correct?

[quartz compass]

yeah that's right

[gritty kelp]

Awesome, and so that means that half of my eigen values are 1 and half are 0?

is this true for all projection matrices? What if it was 5-dimensional?

[quartz compass]

yeah

think of it as all the orthogonal parts getting mapped to 0, and the parts that are projected on just stay where they're at

so to them it looks like an identity matrix with eigenvalue 1

[humble elbow]

Is it true that $(AB)^\tau=B^\tau A^\tau$ if yes how does one prove it?

[stoic python]

27182818284tropy:

[humble elbow]

Here τ is the transpose over the anti diagonal

I mean it‘s straight forward given the dimensions of the matrixes, but I couldn’t work out how to abstract the transpose over the anti diagonal to any dimensions and couldn’t find anything online.

[quartz compass]

you can rewrite it in terms of something called the exchange matrix

which is just a permutation matrix

an identity matrix with the 1s on the antidiagonal

[dusky epoch]

$A^{\tau} = QAQ$ where $Q$ is the antidiagonal identity matrix

[stoic python]

Ann:

[dusky epoch]

$Q^2 = I$

[stoic python]

Ann:

[humble elbow]

Huh alright thanks

[unique jasper]

Could anyone explain to be how I would prove this?

No.10

[dusky epoch]

go through all the axioms checking each one carefully

note down which ones fail

[unique jasper]

Ok so wouldn't 5 to 8 work because of the first term. We would just have to disprove axioms form 1-4 right?

[dusky epoch]

i have no idea how your book numbers the axioms

you're gonna have to give me your list

[gritty kelp]

I have this question here, and I thought I knew how to answer it, but I am pretty sure my answer is wrong. Mainly I am wondering, is there a specific A that I should be finding, or is there more than one right answer?
I know that P is computed through A, and projects onto the column space of A. So I thought that if I just make a matrix with two of the columns equal to those already given, and then two more columns that are linearly dependent on the first two columns, then I would get a matrix A that has the column space spanned by the given vectors.
However, when I tried to compute $(A^TA)^-1$ I found that an inverse didnt exist.

[stoic python]

Blu3_bear:

[rustic panther]

Need help with this xD

if you can expand f(au+bv) = af(u) + bf(v)

uhm

🤔

any tips? @warm briar

x = au, y = bv?

oh we divide both sides by u

x/u (rational number) = a?

nope

x/k = au/k

x/k = (a/k) *u

wait how so

is the idea to express it in a/b (the form of rationals)?

hmm

how do I get from au to x, and bv to y

[wintry steppe]

Hello everyone, I'm trying to understand this question. The first picture is the question and the second is the offical answer. I'm not sure where the 2^3 come from in the answer.

[dusky epoch]

$\det(cA) = c^n \det(A)$ where $n$ is the size of $A$

[stoic python]

Ann:

[dusky epoch]

you can make sense of it by considering det(cI)

[wintry steppe]

ah thankyou so much

[queen gate]

Hi, (I don't know if it's the good channel for my question, so, sry if I made a mistake)...
My problem is very simple to understand :
If I have an integer i, such 1 <_ i < k.
How can I compute a prime factorization for each i between 1 and k, is there any simplification to have a form of logic ?

[round coral]

what is k? an integer?

Seems like you are writing a program, and want an algorithm, am I right?

This is a simple algorithm.

1. While i is divisible by 2, print 2 and divide i by 2.
2. After step 1, i must be odd. Now start a loop from j= 3 to square root of i. While j divides i, print j and divide i by j. After j fails to divide i, increment j by 2 and continue.
3. If i is a prime number and is greater than 2, then i will not become 1 by above two steps. So print i if it is greater than 2.

one thing I forgot to put here is that there is a loop of i from 1 to k if k is an integer and if not then from 1 to the greatest integer of k

[queen gate]

@round coral k is an integer, yep, and I my objective is to create an algorithm, but not a simple algorithm, I want to find a "fast" formula to create the algorithm

[round coral]

Check my out, this is the standard one.

Actually there are different algorithms we use depending upon how large your k is.

Check this out https://cp-algorithms.com/algebra/factorization.html

You will see lots of algorithms and the times they take

[lavish bolt]

When dividing a polynomial with a degree of n by (x-1)^2, should I expand the expression and then do the division?

[dusky epoch]

wrong channel, but there is a shortcut.

[lavish bolt]

Which channel should I post in?

[dusky epoch]

#precalculus would be most fitting for polynomial division specifically
otherwise there are 10 question channels that are for literally whatever

[lavish bolt]

Thanks!

[lapis ginkgo]

does anyone know how to solve this?

[fresh vector]

you know the algorithm?

[lapis ginkgo]

its fine I figured it out

thank you though

[rustic panther]

Hi there can someone kindly help me with question 1?

[marble lance]

@rustic panther What are the conidtions f has to meet to be a linear map?

[round coral]

Yes Lunasong is right

[rustic panther]

adding or scalar multiplying it before and after the function is applied should yield the same answer

[round coral]

Use these conditions

[rustic panther]

not sure how to manipulate what's given though 🤔

[marble lance]

So they already give you the first condition

Now you need to show f(ax) = af(x) where a is a rational number

[round coral]

the second condition can be proved from first quite easily

you need a little manipulation.

[rustic panther]

@marble lance i.e. express a as $\frac{\mathbb{Z}_1}{\mathbb{Z}_2}$?

@round coral what's the second condition sorry?

[round coral]

f( kx) = k f(x) where k belongs to field

here field is Q

[rustic panther]

can you tell me what the first line of working is?

kinda get what you're saying but I have no idea where to begin

[round coral]

first prove f(nx) = nf(x) where n is a natural number

[marble lance]

The way I am thinking of involves showing it for positive integers first

[round coral]

By the way, it is famous Cauchy functional equation

you can even find out its solution for fun

[rustic panther]

uhm

f(0) = f(0+0) = f(0) + f(0)?

grasping at straws here I'm afraid (

[marble lance]

What does that give you?

[rustic panther]

f(0) = 0?

[marble lance]

You need to be working with f(nx)

That's true

But you don't need that rn

[rustic panther]

f(xv + yv) = f(xv)+ f(yv)?

[round coral]

Ah! You are going the wrong way, you don't need to do that. Just do as we said, read the messages before.

But if you want to play with the equation it is good exercise and a lot of fun.

[marble lance]

Now you need to show f(ax) = af(x) where a is a rational number
@marble lance

first prove f(nx) = nf(x) where n is a natural number
@round coral

[rustic panther]

so I replace a with n/m for example?

[marble lance]

I'm going to cry

[rustic panther]

sorry 😦

[marble lance]

First prove f(nx) = n f(x)

For natural numbers n

So stop working with a

Until you proved that

Why is x in R?

[rustic panther]

in V?

[marble lance]

And why is f(x) = 1x?

Uh, yeah

You need to show it equals n f(x)

So just f(x) = 1*f(x)

f(x) and x are not even in the same space

[rustic panther]

assume k*f(x) = f(kx)? for some k in N where k>=1?

can I do this?@marble lance

[marble lance]

@rustic panther do you mean prove it with induction? Sure

[rustic panther]

is there a quicker way?

[marble lance]

f(nx) = f(x+x+... +x (n times)) = f(x) + f(x) +... + f(x) (n times) = n f(x)

[rustic panther]

ahhh

so now I go
n*f(x+y) = n[(f(x) + f(y)]
= n * f(x) + n * f(y)

[marble lance]

I don't understand what you are doing

Now you need to generalize it to all rational numbers

[rustic panther]

do I replace the n with a/b?

[native rampart]

Hint:q((1/q)x)=x

[marble lance]

Let p(x) = ax^2 + bx + c, and see what F(p(x)) looks like

That will tell you Im(F)

How can you write it as the span of a set?

[round coral]

@rustic panther Bro , it seems like we are solving your exercise

Sorry, I am a bit blunt, my weakness

[rustic panther]

Uhm do you have examples of similar questions with model answers by ant chance?

Nah it’s good

[round coral]

I personally study from Axler. It is a good book.

Friedberg Insel and Spence is also good, heard a lot about it.

Axler has very good questions for practice, and he gives a lot of examples also to hone understanding

[fickle shale]

Could someone help with this

[marble lance]

@fickle shale which question? And a part of Q3 is covered up

[fickle shale]

Q4

Only

Yeah my bad @marble lance

[marble lance]

Okay, what do you think?

[fickle shale]

Im on my way home rn do you mind assisting in like 30

Sorry

[marble lance]

Idk if I'll be around, but someone else might be. So just ask again when you're ready

[fickle shale]

Yeah no problem thanks anyway

[brittle orchid]

Hi guys, I've attempted this a few times but can't seem to find the correct "combination" if you like to send the matrix to 0

all the elements are going to zero except for 1 of them, is there a quicker way to do this other than to compute each and every possible combination of the minimal polynomial?

6

no?

[native rampart]

Well,in this case you can restrict T into 2 T invariant subspaces(which are linearly independent and direct sum is V) and and find minimal polynomial in each and multiply them back together to get the minimal polynomial

So,it's (t-11)^2(t-8)^2

[brittle orchid]

I'm a little lost on the explanation 😅

[native rampart]

Nvm, That's wrong

[brittle orchid]

your answer was surprisingly right

[fickle shale]

sorry @marble lance for the ping

but you here?

[marble lance]

Yes, but I'm about to sleep, sorry

[fickle shale]

no problem

thanks

can anyone help with q4

[wintry steppe]

@fickle shale What do you think? Go through it one by one. a)?

[fickle shale]

well what i was thinking is that V is Infinite so U can either be Infinte or Finite

If U if Finite

[wintry steppe]

What do you know about the relationships between dimensions of V, U, and V/U?

[fickle shale]

honestly

nothing comes to mind

i was looking thorugh bmy notes

couldnt find it

[wintry steppe]

Ok you can actually lift a basis of V/U to a linearly independent set in V.

Does this make sense to you?

[fickle shale]

Yes

[wintry steppe]

Ok so dim U + dim (V/U) = dim V should be a reasonable thing to expect, yes? Regardless of dim U, dim (V/U) being infinite or not (if one of them is infinite declare that the sum is infinite)

[fickle shale]

Yep

[wintry steppe]

Ok so look at part a), if you have U and V/U finite dimensional then dim U + dim V/U is finite so you'd expect that dim V to be finite also, no?

[fickle shale]

ye

[wintry steppe]

So a) cannot be correct. Agree?

[fickle shale]

wow

thanks

[wintry steppe]

It's similar for the rest of the parts.

[fickle shale]

makes so much sense

a) False, dim U + dim V/U is finite but dim V is Infinite
b) True, dim U + dim V/U is infinte so dim V holds

c) true
d) False

@wintry steppe sorry for ping

[wintry steppe]

Maybe a better choice of the words would be "possible" vs "impossible" in the context of this problem

[fickle shale]

thanks wording was never my strong point

[wintry steppe]

np English isn't my first language either. At any rate I understood what you meant by true/false

Ok so you wrote b) is possible, so now you need to provide an example showing that this is possible (it's in the direction, so)