linear-algebra

[verbal pivot]

hope im remembering it right lol

btw I don't know how to write that fancy L that is used to denote spaces of linear transformations

[wintry steppe]

probably \mathcal L

this is a nice fact

let me ponder its proof

[frosty vapor]

found it

yeah

so what i was worried about in the formulation was like

if U and W were just any old subspaces of V

as opposed to the ones where U oplus W = V

[wintry steppe]

hm i can prove that ker T^k is eventually constant, but i haven't got the specific statement you posted

feels like it'll use cayley hamilton

the sequence \dim \ker T^k is non-decreasing and bounded above by n so it must be eventually constant, and therefore ker T^{k} = ker T^{k+1} = ... for some large k

[verbal pivot]

Maybe there is a way to prove it using cayley-hamilton (although I can't really see it right now lol)

There's a proof in Axler's book (I learned it there), that I find pretty nice

[wintry steppe]

what page?

thank you so much, I wrote that when the eigenvalues are negative the origin is stable and when the eigenvalues are postive the origin is unstable. Then i provided some examples of slope fields to prove that claim

i might take a peek if i get lazy on the problem

i hate how my school went about post calculus topics in highschool

we did multivariable calculus first semester then did linear algebra second semester

so annoying

[verbal pivot]

The start of the chapter 8

[wintry steppe]

linear algebra after MVC

right?

idk it was so tough

im surprised i have a B in this class

[wintry steppe]

ugh of course it's a contradiction proof

makes sense

neat result and neat proof

thanks

[wintry steppe]

giving some examples \neq a proof

one way to see it is to try and put A into diagonal form or jordan normal form

[torpid rover]

is this where you ask linear algebra questions?

[wintry steppe]

yes

[dire thunder]

@wintry steppe привет

👋

[charred shell]

I have subspace for {W as x(3,2,1) where x is in R} C R3

could anyone please tell me how to find the orthogonal complement of W?

I tried and am getting

(-1/3,0,1) and (-2/3,1,0)
could anyone please confirm if it is correct

and the dimension of W is 1
dimension of Wt is 2

[dire thunder]

do we have usual inner product?

[stoic python]

Commander Vimes

[dire thunder]

so
3ax+2ay+az=0

[charred shell]

yeah that's what I did
put 3 2 1=0 and solved for it

[dire thunder]

a(3x+2y+z)=0

so yes

x y z is in orthogonal compelment if 3x+2y+z=0

this should be 2d space

find its basis

[charred shell]

I got basis for W as 1

and basis for Wt as 2

[dire thunder]

why you need basis for W

(-1/3,0,1) and (-2/3,1,0) this what you wound sounds ok

[charred shell]

sorry dimension*

[dire thunder]

check if any vector which is linear combination of those two is solution to 3x+2y+z=0

if it is then you are pretty much done

[charred shell]

as in if I plug back the stuff in W I'm getting 0 so it is true

right

okay, thank you!

[night bear]

$$ y=mx+b$$

[stoic python]

beautyiskeytosuccess

[night bear]

hopefully this helps you

[dire thunder]

well actually yes

you cound find plane normal to basis of W

but no real need

[charred shell]

Oh okay

[lapis fern]

This is how i had to do it

[wintry steppe]

why would i use gauss-jordan elimination over just gauss elimination?

$A =
\left[
\begin{matrix}
2 & 3 \
5 & 2 \
\end{matrix}
\left|
,
\begin{matrix}
7 \
1 \

\end{matrix}

\right.
\right]$

[stoic python]

zeffs

[wintry steppe]

for this matrix, why would would i use gauss elimination vs gauss jordan?

[native rampart]

You get a solution directly

Although Gaussian elimination is just one step longer

[wintry steppe]

my unis teaching is fuciing awful

[native rampart]

As far as I am aware,just Gaussian is for computing determinants and checking if a matrix is invertible

[robust tartan]

Explain with reasons whether the set S = {(x, y)|x = y} is a subspace of the vector
space or not?

[wintry steppe]

literally first lecture the instructor just starts doing row echelon operations without telling us wtf gaussian elimination is, it was first in second lecture that term was introduced

[native rampart]

Gauss Jordan is for solving equations

[robust tartan]

Please help with above question

[wintry steppe]

but i need to learn gauss elimination to also do gauss jordan?

[native rampart]

Gauss jordan is an extension of gaussian

So yes

https://youtu.be/2j5Ic2V7wq4

This might be helpful

[wintry steppe]

okay thx

when is back substition used? @native rampart

[native rampart]

After you get the upper triangular form

[wintry steppe]

is that only used when finding gauss jordan?

[native rampart]

Also this is the difference between gauss and gauss jordan

In gauss you do back substitution or you could stop with the upper triangular form if you want to compute determinant or smt

In gauss jordan,you do row eliminations from bottom to top which is kinda the same thing to get a diagonal matrix

[wintry steppe]

i though back substituon and back elimination was the same

[native rampart]

Yea,Back elimination is just a way of doing back substitution but with matrices

The part of the algorithm upto the triangular form is gaussian

If you do anything further,it is gauss jordan

[cobalt tartan]

Hello, any ideas on how to do this, given that definition 1.7 is the definition of the dot product ($cos\theta(u, v) = \frac{u \cdot v}{|u| |v|}$)? I've been stuck on this for several hours now

[stoic python]

Liria ^(;,;)^

[cobalt tartan]

This was my attempt at it

But at this point a horrific amount of algebra begins appearing so I would think that this is not the solution

Ah yea, I went and assumed that u, v, w are unit vectors in their respective directions, to make the math nicer

Any suggestions for a different method or any flaws in what I've gotten would be greatly appreciated

[cobalt tartan]

Like trying to expand that using the fact that w = su + tv (Assuming all unit vectors) just gives me this monstrosity

(where a = u dot v)

[wintry steppe]

divide both sides by -xy, then square, maybe?

[cobalt tartan]

Wait, why by -xy?

I could see trying to go and do this

but Id ont' think that helps very much

[dusky epoch]

what are you doing exactly?

oh

[cobalt tartan]

This is the actual question text, definition 1.7 is the dot product

[wintry steppe]

Oh, try to assume the equality is true, then derive another equality that is obviously true like 0=0

[dusky epoch]

no

[wintry steppe]

say $z=\frac{\vec{u}\cdot\vec{v}}{|u||v|}$
then $z = xy - (\sqrt{1-x^2})(\sqrt{1-y^2})$
maybe from there somehow

[stoic python]

Orangus

[dire thunder]

0(u,w)+0(w,v)=(0u, w)+(0w, u)=(0, w)+(0, u)=(0, w+u)=0

and 0(u,v)=(0u, v)=(0,v)=0

🤔

[cobalt tartan]

Pardon, how-to-get-help?

[dire thunder]

oh wait

it is theta

[cobalt tartan]

That's theta

Yea

[dire thunder]

i saw it as 0

[cobalt tartan]

understandable

[dire thunder]

ok so you have field as R

[wintry steppe]

then you have $(-\frac{z}{xy})^2 = (1-x^2)(1-y^2)$, try to cancel out everything

[stoic python]

Orangus

[dire thunder]

[stoic python]

Commander Vimes

[dusky epoch]

what

since when is theta bilinear

[cobalt tartan]

wait that does not look quite right? Also I have to use the dot product to do this

that too

I dont' thinkt hat we have that assumption

[dusky epoch]

θ(u,v) is the angle between u and v

@dire thunder you shit yourself there i think

[dire thunder]

since i thought theta is scalar and () is product

i am idiot ye

[wintry steppe]

you expand z, x, y to what they are in terms of u, v, w

[cobalt tartan]

that becomes a uh horrific amount of algebra

Like I showed here

[crude falcon]

In the scaling approximation part, is there any particular reason why you, in the first step, scale by the max absolute value of the vector, and in the next step you use the min value? Its because thats a convenient way to get the same vector?

[lavish jewel]

aren't they just making the smallest abs element into 1 at every iter?

[dusky epoch]

there is no real reason to scale one way or another

[lavish jewel]

yeah, at any rate, it's all made up. just try and keep the result from exploding after many iters

[crude falcon]

so its like a conscious decision? nothing to do with the algorithm?

[lavish jewel]

indeed

for a matrix with a dominant eigenvalue of multiplicity 1, repeated multiplication of a vector will yield vectors that are closer and closer (in their direction) to the dominant eigenvector corresponding to the dominant eigenvalue

[dusky epoch]

be consistent about how you scale basically

[lavish jewel]

i think as long as you keep track of what the vector was at the previous iteration, not even the scaling matters (i.e. you don't actually need to rescale at all if this could be done on paper)

but it isn't uncommon for the vector to blow up quickly, making your computer unable to handle it. if you have to rescale at every iter, you need to at least know the rescaled vector from the previous iter

at a glance, i'm pretty sure they choose the min abs element (or max abs or w/e) because this is less expensive than computing the actual norm. those square roots will pile up

[charred shell]

Hi, if I have a 3 by 2 matrix how do I find whether it is linearly independent or dependent?

Because I cant find it's determinant

[dusky epoch]

you don't because those concepts make no sense for matrices

did you mean finding whether or not its columns are linearly dependent?

[charred shell]

No I want to do QR factorization

And wanted to know if the matrix is linearly independent or not

By checking its determinant

Oh okay
So for it to be linearly independent or dependent number of col should be greater than or equal to number of rows

Right?

[native rampart]

The row vectors are linearly dependent

[dusky epoch]

QR factorization only makes sense for square matrices, does it not?

[native rampart]

You have 3 elements of R^2

[charred shell]

Oh yeah right

[charred shell]

Ah okay

Thank you both!

[wintry steppe]

Why are orthogonal transformations defined as <u, v> = <Tu, Tv> and not as <u, v> = <y,w> for every u in T^-1 y and v in T^-1 w ?

the latter seems make more sense to me in analogy to how continuous and measurable functions are defined for example

[dusky epoch]

you mean (∀y,w)(∀u,v)(Tu = y, Tv = w => <u,v> = <y,w>)?

i believe this reduces to <u,v> = <Tu,Tv> anyway...

[wintry steppe]

hopefully it does, but would be nice to have a consistent way to define structure maps...

[dusky epoch]

?

i don't see your point sorry

[wintry steppe]

structure-preserving*

[nocturne jewel]

what you said is just a longer way of saying <u,v> = <T[u],T[v]>

[wintry steppe]

yea its kinda confusing imo to teach it like that cuz then when you learn about continuous and measurable maps you dont understand why use preimage

when if im not mistaken what always matters is the preimage for structure preserving maps

[quartz compass]

I appreciate what you're trying to figure out, but I think there's no analogous property for orthogonal transformations that we have for continuous and measurable functions

when I think of continuous or measurable functions there's some sense of thinking about refining it to smaller sets within, but I don't see any analogous thing for an orthogonal transformation going on

[strange delta]

how comes this isn't linear?

is it typo or what

[dusky epoch]

no, this map really is not linear.

if it was, then f([0,0,0]) would have to be [0,0], which it is not.

[strange delta]

ohhh ok

thanks

[wintry steppe]

well orthogonal maps are injective so in a sense could maybe think of it as the trivial refinement.

[wintry steppe]

in row echelon form there is no need for every leading entry to be 1 right?

[dusky epoch]

indeed

only RREF requires that

[wintry steppe]

so this video is wrong ? https://www.youtube.com/watch?v=2GKESu5atVQ&t=109s

[lavish jewel]

already the thumbnail shows gauss jordan RREF

[wintry steppe]

yeah thats what i was thinking,

[strange delta]

is it possbile for the rank nulllity theorem to not hold?

if so does that mean there is no isomorphism between two maps?

[lavish jewel]

why are you talking of two maps

[strange delta]

i mean

a map

[lavish jewel]

the rank nullity theorem talks about the properties of one map

[strange delta]

i meant 1 map

[lavish jewel]

still, no

[gray dust]

we speak of morphisms between vector spaces, not between maps

[pliant helm]

for a dependent system of three linear equations of three variables , identify infinite number of solutions

Anyone suggest please

Project work on it

[wintry steppe]

@lavish jewel hey. do you remember this question ?

[lavish jewel]

mhm

wdym?

i can't help you, then

[wintry steppe]

Okay, here it is:

The zero vector maps every element to zero.

So this zero vector must be in the set of T.

But we're just saying that z(pi) = 0.

then yes it's in T.

but how do we know for z(2) , z(4), ....

[lavish jewel]

wtf

that doesn't matter?

read the problem again

if z(pi) = 0, it's in the vector space

nothing else is needed

[wintry steppe]

okay, that's the condition.

[lavish jewel]

yes

and the zero function satisfies it

so it's in the vector space

[wintry steppe]

okay. let me rephrase that.

I already know what you mean, but I think I misunderstood the concept.

[lavish jewel]

you're confusing yourself

sure

i have no idea what that image is

idk what you mean by above

[wintry steppe]

I know. but Vielen Dank. :)

[lavish jewel]

you're gonna have to learn how to 😛 or you'll have a hard time with linalg

[wintry steppe]

what did you mean?

[lavish jewel]

defining subspaces is kinda the whole point of linalg

you'll keep seeing this all course long

[wintry steppe]

No, I got all the point, don't worry.

It's only 2 days.

:)

[strange delta]

would this be correct?

[wintry steppe]

this makes no sense

why are you assuming {v_1, ..., v_n} is linearly dependent? you need to prove {f(v_1), ..., f(v_n)} is linearly independent

it's not clear what your steps after that are

you need to redo this

start by carefully writing out in full what it would mean for {f(v_1), ..., f(v_n)} to be linearly independent. then, prove it.

[strange delta]

how about now

[wintry steppe]

if you want to do a proof by contradiction you should add at the start that at least one of the scalars c_i is nonzero

but this is good

in fact if you get rid of the very first line it becomes a solid direct proof

very good

[vale plover]

I agree with you. The proof is ok but is not by contradiction

[hollow finch]

so my intro to linear algebra class didnt talk a whole lot about skew symmetric matrices other than their definition, but theyre pretty cool. what course(s) would go into more detail about them?

[wintry steppe]

unironically a lie theory course, they are the lie algebra of the matrix lie group O(n)

a pretty important example in that area

[hollow finch]

quite interesting although slightly disappointing that it won't likely be for a while. i dont think my college offers that. might study it on my own though; it sounds cool.

[wintry steppe]

How do i solve a gaussian elimination when the systems of equations have an unknown coefficient a ?

Like this

[native rampart]

Write that as a matrix and do
R_2->R_2+aR_1 and then R_3->R_3+2aR_1

So,You do it the same way as you usually do it

[wintry steppe]

is this correctly set up?

[native rampart]

Yes

Now row reduce as usual

[wintry steppe]

but in the assignment they want me to do the following operations $ar_1+r_2 \rightarrow r_2$ and $2ar_1+r_3 \rightarrow r_3$

[stoic python]

zeffs

[native rampart]

Yea, That's the usual one

[wintry steppe]

not sure if i should completely reduce then

[native rampart]

You can reduce up to that step,I think

[short coral]

Help me understand this solution https://math.stackexchange.com/a/438762

Please

What elementary operations are applied?

[wintry steppe]

is $R_2=R_2+aR_1$ the same as $R_2 + aR_1 \rightarrow R_2$ the same?

[stoic python]

zeffs

[native rampart]

*R_2 becomes R_2+aR_1

[wintry steppe]

yea so they mean the same right?

[native rampart]

Ys

[wintry steppe]

this should be it

[native rampart]

Ye

[wintry steppe]

okay thank u,

The next question is that i have to do this

i guess i could switch out a with the fraction but im not sure

what is meant by there isnt any solutions for it

[native rampart]

What's the matrix you end up with if you only do R_2->R_2+aR_1 and R_3->R_3+2aR_1?

$\begin{bmatrix}
1& 2 & 1\
0&0&1+a\
0&-1+4a&4+2a\end{bmatrix}$

[stoic python]

Buncho Dragons

[native rampart]

So,you move R_3 and R_2

Now you when you are doing R_2->1/4a-1 R_2 you are assuming 4a-1 is nonzero

Because if it were zero,you can't divide by 4a-1

[wintry steppe]

what do u mean exactly? After doing only $R_2->R_2+aR_1 and R_3->R_3+2aR_1?$

[stoic python]

zeffs

[wintry steppe]

i swap row 2 and 3

[native rampart]

What is your next step after swapping?

[wintry steppe]

R_2->1/4a-1 R_2 doesnt make sense to me, since a is now 1/4 why is there still an a?

[native rampart]

a is some real number

It could be 1/4 or it could be smt elze

[wintry steppe]

i see, but why does that mean there isnt any solution if a = 1/4?

because nonzero?

[native rampart]

Suppose a were 1/4,this becomes
$\begin{bmatrix}
1&2&1&| & 1\
0&0&\frac{5}{4} & | & 0\
0&0&\frac{9}{2}& | & \frac{1}{4}\end{bmatrix}

[stoic python]

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

[native rampart]

Which clearly has no solution

Since you could need 5/4z=0 and 9/2z=1/4 simultaneously

[charred shell]

is this channel free?

I'm guessing yes
I have to find the coordinate vector without using linear combination method and I'm not sure what the other method is

If anyone is aware of the other methods name could they please share it?

<@&286206848099549185>

[lavish jewel]

give more info

do you have a matrix? a set of functions?

[wintry steppe]

This question

What is this guy doing? What is this method for solving (a) named?

[charred shell]

I have a basis set
and wanted to find coordinate of the vector that was given to me
somebody just told me about the other method: finding inverse of basis set and multiplying it with the vector given.
I got it, thank you!

[wintry steppe]

<@&286206848099549185>

[lavish jewel]

the circle-looking thing denotes composition

for matrices, this is the same as multiplying

[wintry steppe]

but idk if its a type from the guy who made the answer but why is there an u ?

[lavish jewel]

find a matrix A that has the same effect as applying T to x twice

[wintry steppe]

shouldnt it just be T◦T(x) = T(T(x))

[lavish jewel]

sure

[wintry steppe]

typo*

[sterile plank]

Hi, anyone knows what this representation for matrix M means? P and Q are matrices too

[dusky epoch]

this is a block matrix

@sterile plank

the zeros are zero matrices sized appropriately to fit with P, Q and their transposes

[sterile plank]

oh, thanks. So I can expand it and take M as a normal matrix?

[dusky epoch]

?

not sure what you mean

[sterile plank]

is it a matrix of matrix?

[dusky epoch]

???

no, it's a matrix of numbers, only its structure is described as being assembled out of nine smaller matrices

[sterile plank]

got it, thanks

[sharp idol]

So I am having a really, really hard time showing that the characteristic polynomial of this matrix is indeed p(t). I tried just evaluating the determinant of C(p)-I\lambda, but I was struggling to make any progress. Can anyone help shed some light on this problem? Thanks!

[strange delta]

how would i show this?

[wintry steppe]

feels susceptible to induction

on n

[sharp idol]

So by using leibniz' formula (with respect to the last column), I was able to deduce that the only possibilities for c_i that are not 0 is exactly c_i\lambda^i.
However, I am having a hard time figuring out the signs of each term. They should be positive, but I am struggling a little bit on that part

[weak needle]

You can just do it brute force by evaluating the determinant like you tried. Basically evaluating using the top row, if we start at the lambda term any non zero selection will have to choose one of the ci’s after choosing i lambdas

[sharp idol]

wait, I think I got it

[weak needle]

Oh nevermind, you already did it

[sharp idol]

Yeah, I took the transpose and it helped a lot

[weak needle]

Well you could do lambda - the matrix and then you don’t have to worry about the sign

[sharp idol]

I am referring to leibniz' formula here

[weak needle]

I see

[strange delta]

when they say u is an elemnt of ker(p) do they just manipulate it because u = u- p(u) is just p(u) = 0

and how comes u is suddenly an element of im(id_v-p)

[verbal pivot]

Yeah, since p(u) = 0 and u = u - 0

[verbal pivot]

You have this type of result, u = (Id_v - P)u

[strange delta]

hmm can you actually do that because P is a linear map which confuses me

[verbal pivot]

do what exactly?

[strange delta]

isn't p(u) the map itself?

like you can't move 'u' around

[verbal pivot]

if u is a fixed vector, then p(u) represents the image of u through the transformation p, this is the same as saying that p(u) is another vector

of course, you should be careful to not confuse when they refer to the image of a certain vector or the whole transformation

[strange delta]

this feels so new to me like i've never been taught this

my exam is in like 3 weeks and i had no idea i could do this

[strange delta]

@verbal pivot also when we get the result u = (Id_v - P)u, how comes we know it's an element of the image?

[verbal pivot]

Recall the definition of the image set, if $x \in Im(T)$, where $T \in \mathcal{L}(V,W)$, then there exists some vector $u \in V$ such that $Tu = x$

in short, if you arrive at a result of a form $Tu = v$, then $v \in Im(T)$

[stoic python]

b2unit

[verbal pivot]

If you look at the example above, you have $u = (Id_{V} - P)(u)$

[stoic python]

b2unit

b2unit

[verbal pivot]

fixed a typo*

[strange delta]

so do we actually end up with

u = range(Id_v - P)

[verbal pivot]

by range() you mean Im()?

[strange delta]

yeah

[verbal pivot]

I don't think that's right, keep in mind that u is a vector, while Im() is a set, maybe you meant to say $u \in range(Id_V - P)$

[stoic python]

b2unit

[strange delta]

ah ok i think i get it now

thanks for the help

[wintry steppe]

Can someone help me out? I'm asked to find the wronskian for f(x) = sin(ax) and g(x) = cos(bx) where a and b are reals. Now I've worked it out and it is -bsin(bx)sin(ax) - acos(ax)cos(bx). I'm stuck when asked if these func are lin independent when a =/= 0

In my head there are a lot of possibilities one can consider

[weak needle]

At x=0 sin(ax) is zero but cos(bx) is not, so they are linearly independant as if a/=0 then the dependance relation has to have non zero coefficients for both functions

[nocturne jewel]

how did you conclude on them being dependent?

[weak needle]

They are independent because if you look at csin(ax)+dcos(bx)=0 where c and d are non zero, and substitute x=0 you get a contradiction

[nocturne jewel]

you said they're dependent and independent at the same time

[weak needle]

No they are independant

[nocturne jewel]

"has to have non zero coefficients"
"they are independent"

are contradicting statements

[weak needle]

It is a proof by contradiction

[nocturne jewel]

well you didnt say that

[weak needle]

Assuming they are dependent we get a contradiction

[nocturne jewel]

$\alpha\sin(ax)+\beta\cos(bx)=0 \ x=0\implies\beta =0 \ x=\frac{\pi}{2b}\implies\alpha = 0$

[stoic python]

Mosh

[weak needle]

The second line is not necessarily true, if a=2b for example

[wintry steppe]

Yeah I've figured it you take x = pi then they are linearly independent

Using the wronskian test

Also for those who are around, how can I show that this is not equal to the eigenspace

I get that it's not, in a practical setting

But here I'm only given a 3x3 matrix A with a thrice-repeated eigenvalue having only 2 linearly independent eigenvectors

[weak needle]

I think this will work

[sharp merlin]

how do i simplify (-2 x 10^7 i) X ( - 0.01i + 0.01j )

[wintry steppe]

Wow thanks nice proof

[weak needle]

Welcome

[nocturne oracle]

thank

[lapis river]

help on these two would be much appreicated

for #3 also please let me know how you got to your answer

[native rampart]

map 1 to (1,0,0), x to (0,1,0) and x^2 to (0,0,1)

[lapis river]

can you show that in terms of T(p) = {...}?

[native rampart]

T(a+bx+cx^2)=(a,b,c)

[lapis river]

wouldn't you need to define a,b,c as a function in terms of the Original Polynomial?

I don't think I'm able to just show it as (a,b,c)

[lapis river]

for #4 would [1 2 3 4, 5 6 7 8, 9 10 11 12, 1 2 3 4] be a good answer?

[lavish jewel]

sure

well, it has to be 4 x 5, yeah?

that's 4x4

but you got the idea

[lapis river]

oh word

still need help getting a proper transformation for #3 if anyone is around

something like

t(p(t)) = [ p(something1) , p(something2), p(something3)], as an equivalent of the identity matrix probably

[lavish jewel]

the matrix would have to be something like the operations that yield a taylor approx around 0

like the nth partial derivative of p(t) divided by n! and evaluated at x=0

(this should also be the basis of the dual space)

[lapis river]

are there any transformations that I can show in the matrix that can be simpler when onetoone and onto?

[lavish jewel]

purely injective would require the image to have a dimension >= than the preimage

could be any full column rank matrix

what i would do i just say what the basis is and just do it as matrices

unless they explicitly require to do it otherwise

integration should also work for the one to one

for onto, you can do differentiation without the 0th order derivative

[short coral]

Please explain this. I am not able to understand

[native rampart]

Do you know generalised elementary transformations?

[short coral]

Yes. But I didn't get which transformations are being applied

[native rampart]

Can you list down the ones you know? I am not familiar with what counts as a generalised elementary transform

First step is R_2->R_2+AR_1, which can be broken down into a bunch of elementary transforms

[short coral]

Swapping 2 rows or columns and putting a negative sign at front

Adding two rows or columns

Multiplying a row or column with a non zero number

[native rampart]

So,Like regular elementary transforms?

[short coral]

Yes

[native rampart]

I think if you show R_2->R_2+AR_1 is a generalised transform for all matrices A,this becomes easy

Prove same for columns too

[short coral]

I got how the third and fourth part. They swapped first and second column and put a negative sign

I didn't get how the second part changes to third part

[native rampart]

You do C_2->C_2-C_1B

First to second is R_2->R_2+AR_1

[short coral]

Rank of the matrix at the extreme right will be n + rank(A), isn't it?

[native rampart]

Yes

n>=rank(B)

[strange delta]

when finding the image of this map i did row reduction and found out the basis for the image is the first row, so would the image be x(1,2,3), where the 1st , 2nd and 3rd column respond to x,y,z respectively

[lavish jewel]

what do you mean the image would be x(1,2,3)?

[strange delta]

like the column 123

[lavish jewel]

yep, but what's x there?

[strange delta]

idk wouldn't x be there for the image

[lavish jewel]

i mean

[strange delta]

because the first column is basically all the x terms

[lavish jewel]

the image is rather (x-y+2z)*[1,2,3]

[strange delta]

where did you get x-y+2z from?

o

oh

but wouldn't the basis of the image correspond to the image itself

bit confused by (x-y+2z) is getting multipled by 1,2,3

[lavish jewel]

the image is what the input gets mapped to

the basis of the image is a set of vectors that spans every possible image

[strange delta]

yeah

the origonal question was "Determine the image ofgand find a basis for the image ofg."

[lavish jewel]

well, the image is (x-z+2z)[1,2,3]

[strange delta]

so im just a bit confused by it's [x,2x,3x]

[lavish jewel]

a basis is [1,2,3]

tbh the second line there looks wrong to me

you're saying the image of the rref matrix is the same as the image of the original matrix

which is not true

you can test this yourself

,w {{1,-1,2},{2,-2,4},{3,-3,6}}*{{1},{1},{1}}

[stoic python]

Results provided by WolframAlpha

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

Requested by Edd#5182

[lavish jewel]

,w {{1,-1,2},{0,0,0},{0,0,0}}*{{1},{1},{1}}

[stoic python]

Results provided by WolframAlpha

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

Requested by Edd#5182

[lavish jewel]

see

,w {{1,0,0},{2,0,0},{3,0,0}}*{{1},{1},{1}}

[stoic python]

Results provided by WolframAlpha

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

Requested by Edd#5182

[lavish jewel]

all those things are different, don't mix them up

[strange delta]

hmm

ok

how did you determine the image was multiplied by (x-z+2z)

[lavish jewel]

you look at the matrix and notice that all the columns are scaled versions of the first one

and recall that matrix multiplication of the form A[x,y,z] is a linear combination of the columns A_i of A, of the form A_1 * x + A_2 * y + A_3 * z

then notice that A_2 = -1 * A_1, and that A_3 = 2 * A_1

substitute back into the above linear combination

you get A_1 * x + A_1 * -y + A_1 * 2z

factor out the column A_1

[icy harness]

I am learning game development and I have a vector used to describe camera direction. Why is the the x direction in the vector the cos(yaw) * cos(pitch), and not just cos(yaw)? If it helps I am using this tutorial https://learnopengl.com/Getting-started/Camera.

[wintry steppe]

Hello guys!!

Can anyone please find part e for me?

Range L

Thankss

[nocturne jewel]

you can split the general image vector into av1+bv2+cv3, then just define the basis for Im(L) to be {v1,v2,v3}

[wintry steppe]

@nocturne jewel This is how I did it, is it readable for u? But I still cant seem to determine how I must write Range L. Thanks so much for ur time

[nocturne jewel]

it's [1,-3,0] for the 2nd basis vector

[wintry steppe]

You’re right

So is my range correct? Should I rewrite it?

[nocturne jewel]

if you fix the basis vector then yes, that's a basis for Im(L)

[wintry steppe]

Thanks so much! So Image L = (a+b-c ...) part

And i ignore the xyz?

[nocturne jewel]

Im(L)=span{[1,0,1],[1,-3,0],[-1,6,1]}

[wintry steppe]

❤️❤️❤️❤️

Although on chegg some expert said dimension of range L is 2

Hmmm, I need a bit of help with linear algebra.

This is cheggs answer but they didnt get the image

[nocturne jewel]

that's a different thing.. and also that'c chegg

[wintry steppe]

Oh, is this still taken?

[wintry steppe]

Okayy thxx

[wintry steppe]

I just meant that for me the dimension of range L is 3

So thats why I mentioned chegg

[nocturne jewel]

yeah the dimension of Im(L) is 2, since the span I wrote is a dependent set

[1,0,1]=2[1,-3,0]+[-1,6,1]

so you do have to remove a vector from the spanning set I gave to make it a basis

[wintry steppe]

My god, our teacher never mentioned this

This is a previous of his test

Thanks so much but my head is gonna explode now, I am going to reread everything u said like 5x so I can get the idea in my head. Thx ❤️

Hey dude.

any set that includes zero vector is linearly dependent.

we could get rid of that.

In this case lambda sub 4 is 0.

[nocturne jewel]

The image vector given a quadratic ax^2+bx+c is given. You can write that vector as a linear combination of 3 vectors [1,0,1];[1,-3,0];[-1,6,1]. For that to be a basis, it needs to span, which it does by definition, and be independent. It isnt independent since [1,0,1]=2[1,-3,0]+[-1,6,1], so I remove one vector and get a basis for Im(L)

[wintry steppe]

Then we have the equation -20 lambda sub one + lambda 3 = 0.

According to that we can get rid of one of these matrices, right

?

[nocturne jewel]

what are you trying to do?

[wintry steppe]

I'll just reduce this form to its basis.

[nocturne jewel]

ok so yes, remove the 0 vector

[wintry steppe]

because it already makes the set linearly dependent.

[nocturne jewel]

and clearly [-20,-20]=-20[1,1], so remove one of those vectors too

[wintry steppe]

yeah, yeah.

but wanna get the information from the coefficients.

that's what I asked.

[nocturne jewel]

you dont need to though

[wintry steppe]

I don't need to exactly.

Just because I'm wacko.

that's why.

[nocturne jewel]

$\begin{bmatrix} -20&1&0&0\-20&1&1&0\end{bmatrix}$

[stoic python]

Mosh

[misty sparrow]

Hey, I've got a question, is it possible to present this subspace in the form of a linear span, if it's presented in a form of a linear equation system? And if possible, how do I do it?

like I've got 2 linear subspaces, one of them is presented in the form of a linear span of this set of vectors, and I'd like for the first one to be in the same form if possible

[stoic python]

Mo1729

[misty sparrow]

oh okay, thank you very much! turns out I actually only needed to find the basis of the subspace, and I thought I needed to find a linear span first 😅
But thank you, good to keep it in mind

[vale plover]

In the case of this one, is the intersection of 2 planes. You just need the director of that line

[inner hull]

Corrected version: So the above is the kernel of the linear map: $R^3 \to R^2 : (x_1,x_2,x_3) \to (9x_1-6x_2+3x_3, 6x_1-4x_2+2x_3)$. Now comes the question of finding a basis for the kernel. We know the matrix representing the linear map has rows: $r_1,r_2,$. So any vector $v$ in the kernel must have $v . r_1=v . r_2=0$. So the kernel is orthogonal to the space spanned by the rows. We then find the basis of the rows $r_1,r_2$ and then orthogonally extend using gram schmitt.

[inner hull]

This is wrong. See above for the correct version

[stoic python]

Mo1729

[inner hull]

Hi, sorry the answer i gave was wrong. But i have corrected it now. see the correct answer

[misty sparrow]

oh I see

thanks again!

[inner hull]

This is wrong.

[brittle roost]

how to show that |a| x |b| x cosX = a1b1 + a2b2 + a3b3 where a and b are two vectors and X is angle between them

[spiral star]

that follows from geometry you have probably last seen in middle school :)

pythagorean theorem and the law of cosines

[brittle roost]

can you give a reference or show it to me pls

[spiral star]

you just have to draw a triangle

and apply the law of cosines

you have to start out with defining the norm of course

the formula for euclidean distance follows from pythagoras

but once you have that, its just plugging into the law of cosines

|b-a|² = |a|² + |b|² - 2|a| |b| cos(alpha)

now use the bilinearity of the inner product

|b-a|² = <b-a, b-a> = |b²| - 2<a,b> + |a|²

cancel |a|² + |b|² on both sides

and divide by 2

[spiral star]

<a,b> = |a| |b| cos(alpha)

[brittle roost]

i want to show this

[spiral star]

i just gave you the proof no?

[brittle roost]

that is 2d so remove a3b3

[spiral star]

the dimension is irrelevant, you are always in the plane with two vectors

and you already know the formula for <a,b>

this was derived from pythagoras

it applies in all dimensions

[brittle roost]

i havent learned this...

[spiral star]

thats the law of cosines

its elementary geometry

https://en.wikipedia.org/wiki/Law_of_cosines

[brittle roost]

what country are you from, i didnt learn this stuff in elementary school

[spiral star]

not in elementary school but definitely before college

its a standard trigonometric theorem

like 7th grade math or something idk

[brittle roost]

i learned trigonometry in 11th grade

[spiral star]

fine then 11th grade

you should have seen this before

and if not, then read up on it, you cant prove it without

[brittle roost]

okay

i will try

[spiral star]

if you want a geometric proof then this is the way

[brittle roost]

yes i want geometric proof

[spiral star]

then read up on the law of cosines :)

[ebon river]

Does someone know an elementary but rigorous proof of "Row-Reduced Echelon Form of any matrix is unique"? We were given the following proof in my math methods class but it makes no sense to me

[inner gyro]

@ebon river if you're comfortable with induction, this article's proof is not too hard to follow:
https://www.maa.org/sites/default/files/Yuster19807.pdf

[ebon river]

Thanks!

[lapis fern]

Quick QQ: what’s the canonical set builder notation for a line in R^3, and also a plane in R^3?

I’m making it up on the spot and Im thinking:

$$\text{line: } \left {\sigma \mathbf{v} \mid \sigma \in \mathbb{F} ,, \mathbf{v}\in \mathbb{R}^3, , \mathbf{v}\neq \mathbf{0} \right }$$

[stoic python]

Videlicet

[lapis fern]

$$\text{plane: } \left {\sigma \mathbf{v} + \alpha\mathbf{u} \mid \sigma,\alpha \in \mathbb{F} ,, \mathbf{v}, , \mathbf{u} \in \mathbb{R}^3, , \mathbf{v},, \mathbf{u}\neq \mathbf{0} \right }$$

[stoic python]

Videlicet

[lapis fern]

It seems long winded, but i think that’s right

[lavish jewel]

these are ok, but you need to specify that v is lin indep from u

furthermore, these planes and lines you describe pass through the origin

for something more general, you need to consider an offset as well

[crude falcon]

can an eigenvalue have multiple eigenspaces with differents eigenvectors? or it always has an unique eigenspace with its unique associated eigenvectors?

[limber sierra]

an eigenvalue only has a single eigenspace, but that eigenspace may not be one-dimensional (i.e. may necessarily be the span of multiple vectors)

the eigenspace will, by definition, contain all eigenvectors associated with that eigenvalue

[lapis fern]

Oh! I need them to be subspaces off R^3. So Im okay with that.

How do throw in linearly independent in the set builder notation?

[lavish jewel]

since you want a plane in R^3, it suffices to say u =/= beta v

[dusky epoch]

i mean... do you really want to overload your notation this much

[lavish jewel]

this also excludes either of them being the 0 vector

[dusky epoch]

i think itd be better to just say in words that you want {u,v} to be LI...

[lavish jewel]

that's probably more sensible indeed

[lapis fern]

Oh! Nice. Tyvm

[olive halo]

does Av = T(v) under this scenario

trying to check my understanding

[lavish jewel]

looks like $A \phi_\beta(v) = T(v)$ to me

[stoic python]

Eρρa

[olive halo]

i thought its phi_betaT(v) not just T(v)

[lavish jewel]

yeah mb

[olive halo]

so would A(v) = T(v)

[lavish jewel]

no

[olive halo]

still kinda confused

oh

[loud halo]

Aphi_beta(v) is a column vector

[lavish jewel]

A can only be applied to vectors in the basis beta

[loud halo]

T(v) is an element of V

T(v) = phi_beta^(-1)(A(phi_beta(v))

$Tv = \phi_\beta^{-1}(A\phi_\beta(v))$

[stoic python]

ryc

[lavish jewel]

yep, there'S the phi beta i missed

the two are related via phi beta. they're not the same thing tho

[olive halo]

so how would i prov that if phi_beta(v) is an eigenvector of A with eigenvalue lambda then v is an eigenvector of T corresponding to lambda?

[loud halo]

Use this identity to help you show that Tv = lambda v.

[olive halo]

I had Aphi_beta(v) = lambdaphi_beta(v) from the given

[loud halo]

We have that $A\phi_\beta(v) = \lambda \phi_\beta(v)$

[stoic python]

ryc

[olive halo]

ye

then i multiplied both sides by phi_beta^-1

[loud halo]

We also have that $Tv = \phi_\beta^{-1}(A\phi_\beta(v))$, and using the eigenvector identity, this is $\phi_\beta^{-1}(\lambda \phi_\beta(v))$

Yes

[stoic python]

ryc

[lavish jewel]

phi_beta is an isomorphism, not something you multiply btw

[loud halo]

Well, you apply it on both sides I guess.

[olive halo]

oh

[loud halo]

That's not really the same as multiplying

It's not a matrix

it's a linear transformation

[olive halo]

for this one can we move the inverse inside the parentheses since lambda is only a constant

[loud halo]

Yes

The lambda pulls out

[olive halo]

but otherwise we can't shift the inverse phi_beta around right?

[loud halo]

Because $\phi_\beta^{-1}$ is linear

[stoic python]

ryc

[loud halo]

well

sure

but like

then you have $\lambda \phi_\beta^{-1}(\phi_\beta(v))$

[stoic python]

ryc

[loud halo]

this is just lambda v

f^(-1)(f(x)) = x

for any invertible function

ok, so we started with Tv and simplified it to lambda v

[olive halo]

does phi_beta*T(phi_beta^-1(v)) = A then

[lavish jewel]

that looks wrong

[olive halo]

say im trying to prove the forward direction here but what does it really mean to be an eigenvector in this case

[lavish jewel]

that's exactly what it is

they're telling you Ay = lambda y iff T( phi_beta^-1(y)) = lambda phi_beta^-1(y)

[olive halo]

so to prove the forward direction i would have T( phi_beta^-1(y)) = lambda phi_beta^-1(y)

then multiply both sides by phi_beta

[lavish jewel]

"multiply"

[olive halo]

so it becomes phi_beta*T( phi_beta^-1(y)) = lambda(y)

sorry not multiply

apply it to both sides i guess

[lavish jewel]

mhm

[olive halo]

how would i simplify the lhs?

[lavish jewel]

hmm i think this is the same direction as before tho

since we're starting with y being an eigvec of A

[olive halo]

thats what im trying to prove

[lavish jewel]

you should start with phi_beta^-1(v) being an eig vec of T

[olive halo]

i started here

[lavish jewel]

aight

[olive halo]

then applied phi_beta to both sides

[olive halo]

then got to this point

[lavish jewel]

all right, yeah

in the previous one, we used this

[lavish jewel]

this here

so let v = phi_beta^-1(y)

then we get T (phi_beta^-1(y)) = phi_beta^-1 (A phi_beta( phi_beta^-1(y) ) )

apply phi_beta to both sides, and we get that phi_beta( T (phi_beta^-1(y)) ) = A y

[olive halo]

ah ok

[lavish jewel]

which you already showed is also = lambda y, as we wanted

[severe heron]

https://cdn.discordapp.com/attachments/826040884574683187/845429036921782312/Screenshot_20210522-035922.png

can someone help me with this? I can't figure out what happens when 1 is an eigenvector of A

[lavish jewel]

the condition that adding up all the elements of A yields 0 is the same as 1 A 1 = 0, where 1 are vectors of 1s

you can use the symmetry of A and associativity of the products to show the last part... presumably

[severe heron]

yes but if 1 is an eigenvector of A then after diagonalising, we can't be sure that all the eigenvalues are 0

if you decompose A into $UDU^T$ and let $w = U^T\mathbf{1}$, you are left with $w^TDw = 0$. Some entries of w maybe 0

[stoic python]

Schrödinger's cat

[lavish jewel]

if 1 is an eigvec of A, it would have to have eigenvalue 0. that's about all it says, as far as i can tell

[severe heron]

okay

[wintry steppe]

Guys.

I've found the general solution.

But the particular solution is different.

the answer is:

the thing is she picked other values that are fit for the constraints for the vector b.

I picked another b.

Therefore my particular solution is different.

Is it true what I said?

she picked b = (1,-1,3)

I picked b = (1,1,3)

still satisfies. would that cause problems @loud halo

[loud halo]

this is how A is defined, no?

look back at what the matrix for a linear transformation is

[lavish jewel]

the general solution to what?

[zinc wasp]

I need serious help I don't get this and it is due sunday

Please help me understand it

Or at least how to make the equation

Wellultiple equations

[wintry steppe]

@lavish jewel

To this.

[lavish jewel]

the matrix is rank defficient, so the general solution would be the projection of x onto the row space plus any scaled version of a vector in the null space

[wintry steppe]

I don't know what projection means yet.

but:

[wintry steppe]

so according to b, our particular solution also will change?

[lavish jewel]

yeah

[wintry steppe]

so there's no exact solution to an arbitrary b.

right?

[lavish jewel]

each b has a different particular solution

or maybe none

[wintry steppe]

guys, i have a diagonal infinite matrix. How can i prove that if the diagonal is bounded, then the linear application is continous?

[queen fox]

Hi guys, can anyone point me in the right direction here? I want to get the two vectors in red and blue in world space, how do I get those if I know v1 (in the pic)

The green raster thing is all vectors perpendicular to v1

[wintry steppe]

oh thanks! so my solutions are also not wrong.

[karmic obsidian]

hello, can someone tell me if my proof is correct ? thank you !

[wintry steppe]

yes

[karmic obsidian]

thank you very much 😋

[acoustic zodiac]

How do I find the orthogonal projection of a subspace?

[lavish jewel]

find an orthonormal basis for the subspace and then do a vector projection onto those basis vectors

[acoustic zodiac]

I did that, but I'm not getting the correct answer. To be more specific, I want to find the matrix representation of the orthogonal projection of the subspace of R2 that has components that sum to 0 (so its basis is (x,-x).

The answer has a 1/2 factor, but I don't know where it's coming from

Ah shit I got it, I just wasn't using normalized vectors

Woops lol

[wintry steppe]

[lavish jewel]

to be fair, the vectors in the basis need only be orthogonal. when you compute the vector projection, one needs to normalize the vectors anyway. the result is the same if this is done correctly, since the vector projection is "in the direction of" the vectors you project onto.

[acoustic zodiac]

Yeah but I missed the last part of the question, that it has to be represented in an orthonormal basis

[split condor]

is there some kind of result for that ABA^T is full rank

if B is square and invertible

and A is nonsquare but full row rank

[lavish jewel]

you can use the last two results and substitutions to show it

on second thought those are not enough

[lavish jewel]

yeah this is false in general

if B has no property other than being invertible, it's not true

you can build trivial examples using B = identity with permuted rows

then just using a few vectors from the canonical basis as A^T can give rows with only 0s as a result

trivially in R^2, let a = [1,0] and B = [0,1 ; 1,0]. the result is [0], since B was constructed to take A^T and map it into its own orthogonal complement

you can do something similar in higher dims following the same procedure

[split condor]

aw darn i see that

thanks

[cerulean mantle]

Anyone want to help me translate the Jacobi elliptic functions into a markup language. 😅

[sage ibex]

How did you get rank(A) = rank(AIB)?

Assuming I is the identity matrix

[blissful orchid]

yes

[nova yoke]

A|B is just the augmented matrix , | is not Identity matrix but a vertical line

[blissful orchid]

i think i saw it on stack overflow where rank A is = rank( A|B)

[nova yoke]

the name for it is Rouche-Capelli theorem

[blissful orchid]

i think i can discard that and just put Rank A < n

[lavish jewel]

for b in the span of the columns, the rank thing holds

[sage ibex]

Ah I see

[torpid otter]

May I ask, what is the difference between a norm, a magnitude, and an absolute value?

[sage ibex]

And how did you get that rank(A | B) < n?

[blissful orchid]

so i believe that rank(A|B) is not relevant now that I think about it

[sage ibex]

I would have tried to prove the contrapositive here btw, that is pretty easy

[lavish jewel]

they all refer to some sort of length

[blissful orchid]

so my answer is correct then? or my logic to the question

[lavish jewel]

and for the 2-norm case (euclidean distance), they're all the same

[sage ibex]

Not sure what the proof is

[torpid otter]

Well, I already know that. But, isn't there any difference between them, if at all?

[blissful orchid]

just to prove that a nxn matrix of ax=b that has more than one solution is not invertible

[sage ibex]

Absolute value usually is the absolute value function on the reals or complex numbers, magnitude usually is the standard norm, and norm can mean any norm

[sage ibex]

Yeah but which proof are you referring to when you ask "so my answer is correct then?"

[blissful orchid]

not really a proof but just my explanation towards why the matrix is not invertible

[nova yoke]

if you conclude rank A < n from more than 1 solutions and infer det(A)=0 from this , it is correct

[sage ibex]

Yeah but I don't get how you inferred those 2 things

[torpid otter]

Well, given something like a metric, on $\bR$, the length output by the distance function would be of an absolute value, but it still represents length (or distance, as the name of the function suggests). If we put a norm/magnitude, they would be the same thing, so why invent all these terms? Unless, there is a difference, is there? If so, what is it?

[stoic python]

tryme

[native rampart]

Here's a much simpler approach,left multiply both sides with the inverse

Assuming A is invertible

[blissful orchid]

rank A < n in mxn means the det(A)=0. thats what I found on stack overflow that helped me with the asnwer

[sage ibex]

Norm is on a vector space, it is not just a metric distance

And usually the metric distance is just called distance I think, not abs value or magnitude

[torpid otter]

Like, isn't $|y - x| = \sqrt{(y-x)^{2}}$?

[stoic python]

tryme

[nova yoke]

if you're referring to rank A < n the proof for it involves linear combination but it is an established theorem , if you're talking about det(a)=0 you can prove this since 1 row will be of the form [0 0 0] in the matrix so determinant has to be 0

[sage ibex]

The other thing is what I'm not sure about. How did you get rank A < n

[torpid otter]

Sure, so you're saying it depends on the object?

[torpid otter]

A metric on R is an abs value, on R^n it would not be, for obvious reasons.

[sage ibex]

No, just that norm is a specific kind of metric that is compatible with vector space structure

But magnitude means the standard norm on R^n

[sage ibex]

This one

But there are many norms on R^n

[torpid otter]

Sorry, maybe I'm stupid. Can you rephrase?

[sage ibex]

One of the rows won't be 0, you'll have to work with the reduced row echelon for that I think. Also the established theorem part is what I'm asking about, if he's appealing to some theorem it's fine but I just saw a direct inference in the proof

[sage ibex]

Do you know the definition of a normed vector space?

[nova yoke]

reduced row echelon and row echelon has the same rank so how is that relevant here?

[torpid otter]

Yes.

[sage ibex]

A may not be in row echelon right? Like A could be
1 1
1 1

[blissful orchid]

i think this is what i used

[sage ibex]

And then Ax = b won't have a unique solution even tho A has no 0 row

[sage ibex]

Right so there are multiple norms on R^n

One is the Euclidean norm that you mentioned

Another is "add the absolute value of each coordinate"

[nova yoke]

I think you are misunderstanding what I am trying to say , A does not necassarily have to have [0 0 0] but if rank(A) < n then it can be transformed such that one row is of the form [0 0 0] at least

[sage ibex]

I mean you didn't say transformed lol

[sage ibex]

Only one of these 2 norms is called the magnitude

The Euclidean one

[nova yoke]

my bad

[sage ibex]

Any other norm is just called a norm

[radiant yarrow]

How do I prove this?

[dusky epoch]

take a parallelogram with vertices 0, a, a+b and b

[blissful orchid]

so then should i just use the example and leave out the rank A< n stuff. It seems that the example is sufficient enough

[dusky epoch]

its center, and also the intersection point of its diagonals, will be (a+b)/2

[sage ibex]

Example only checks one case

You need to prove that whenever there are multiple solutions, A is not invertible (or at least that's the statement you seem to have written?)

[nova yoke]

you're good with the rank A < n stuff as long you give the appropriate statements

[blissful orchid]

it shows that whenever there is a row with all 0's, it has infinite solutions with a free variable, then the determinant will be 0 , showing that it is invertible

[torpid otter]

As in: [d(x,y) = \Bigg(\sum^n_{i=1} (x_i - y_i)^2 \Bigg)^{\frac{1}{2}}] ?

[nova yoke]

as he said this is just 1 case

[sage ibex]

Yes

[stoic python]

tryme

[blissful orchid]

but wouldn't it work for all cases since whenever there is a free variable and a row with all 0's it will always be infinite amount of solutions plus a determinant thats 0?

[torpid otter]

What about the absolute value then? (Sorry, last question.)

[nova yoke]

you have to show that there is a free variable , and you literally get det(A)=0 while showing that which you did

[sage ibex]

I think that just means absolute value in ℝ and ℂ

[torpid otter]

I assume, it is equal to the metrised norm/induced metric?

[nova yoke]

like adding an example to that is just more work

[torpid otter]

Yes, so for those fields in R^1 and C^1 respectively?

[sage ibex]

Yeah

But I haven't seen anyone use absolute value for higher dimensional vectors

[torpid otter]

Oh well, I saw it used in my Topology book (Conway's book), so I thought I would ask, he used an absolute value and the one I gave you above interchangeably.

I was not sure about the definitions that I know in my head, nor the categorizations of them, so I thought I would ask.

[sage ibex]

oh I see, I guess it's also used as another word for standard norm then

[torpid otter]

Yeah, probably.

Thanks btw!

[sage ibex]

[torpid otter]

I don't have access to professors nor people who study the same things I do, because my grade in school is not expected to study such things, and so I can't ask anyone about these things, and my only place to refer to such questions is the internet.

But yeah, thanks again. 😄

[radiant yarrow]

Isn't that what they're asking to prove? How can I just claim that?

Wait

[dusky epoch]

well if you wish

one of the diagonals is {t(a+b) | 0 ≤ t ≤ 1}, the other is {sa + (1-s)b | 0 ≤ s ≤ 1}

you can prove that they intersect at (a+b)/2, with t = s = 1/2

[radiant yarrow]

Okay thanks

[twin cipher]

https://math.stackexchange.com/questions/3526712/algebraic-multiplicty-of-an-eigenvalue-greater-than-its-depth

im not understanding this statement in the answer, how do we show that this annihilates A? (i.e. substituting A into the polynomial, we get the zero matrix)

[native rampart]

Do you know Primary decomposition theorem?

[twin cipher]

no D:

oh oh, yes

yes i do

lol

[native rampart]

Every vector v can be written as v_1+v_2+...v_k where each v_k is in generalized eigenspace with eigen value \lambda_k

[twin cipher]

uhh what

i was thinking of this primary decomposition

[native rampart]

H

Was referring to this

This is probably a special case of that

[twin cipher]

the post i linked to is trying to prove this statement, so we cant use that lol

[native rampart]

Ok,That answer doesn't work then

[twin cipher]

what book is this from?

[native rampart]

Hoffman Kunze

[twin cipher]

hmm okay, ill have a look at that sometime, but im trying to understand this as it is

[native rampart]

Do you know Cayley Hamilton?

[twin cipher]

yup

[native rampart]

Maybe show the char polynomial of the restricted operator is (x-\lambda_i)^m_\lambda

And the minimal polynomial is (x-\lambda)^d_\lambda

[twin cipher]

hmm ok

okay, its obvious that the minimal polynomial is what you stated

but theres no reason why we should raise it to the m\lambda-th power for the characteristic polynomial (since we've already restricted the operator to the generalized eigenspace)

wait, are you taking m_lambda to be multiplicity of lambda in the characteristic polynomial, or the minimal polynomial?

[native rampart]

char polynomial

I think matrix shenanigans will work

[twin cipher]

im getting that $\lambda$ appears $d_{\lambda}$ times along the diagonal of $A|{E\lambda}$

[stoic python]

xy

[native rampart]

https://math.stackexchange.com/questions/1249707/connection-between-algebraic-multiplicity-and-dimension-of-generalized-eigenspac

[twin cipher]

okay, so $dim E_\lambda = d_\lambda$

so $\lambda$ appears $d_\lambda$ times on the diagonal of $A|{E\lambda}$

[stoic python]

xy

[twin cipher]

by this paragraph:

[native rampart]

Consider Identity,(x-1)=0 is the minimal polynomial so d_\lambda=1

1 appears n times along diagonal

[twin cipher]

ahhhhh

okay, so $\lambda$ appears $m_\lambda$ times along the diagonal

[stoic python]

xy

[twin cipher]

i was misreading $d$ in your link to be $d_\lambda$ and stuff

[stoic python]

xy

[twin cipher]

so combining these two facts we immediately get $d_\lambda \leq m_\lambda$

[stoic python]

xy

[twin cipher]

wait, so its also not true in general that $dim E_\lambda = d_\lambda$

[stoic python]

xy

[native rampart]

Yes

[twin cipher]

i see, thanks man

[high rover]

If the linear transformation T:R3→R2is defined as
T(x1,x2,x3)=(x1−x2,2x3), then the nullity of T will be what?
??

[lavish jewel]

at a glance, any multiple of (x1, x1, 0)

[maiden rock]

heyo

is there a nonlinear RREF method?

not sure how to approach, my current idea is just to take ln of everything but then that messes up the a & c terms

[lavish jewel]

you can pairwise subtract one equation from another

then do some substitutions for the exponentials, and take ln

[maiden rock]

is there any matrix method to do this?

[lavish jewel]

not off the top of my head

the idea is to do these operations to first turn it into a linear problem, so i wouldn't think so

[maiden rock]

I don't really see how subtracting would help

I would be left with ln(a) + ln (e^something - e^something)

[lavish jewel]

so you substitute

it gets rid of the c

[maiden rock]

oh

full on manual

fair enough

I was just curious if there was a more elegant/matrix method to do this

but makes sense, thank you!

[zinc wasp]

Still trying to finish my project anyone available?

[wintry steppe]

[zinc wasp]

I could use help on figuring out this.

I want to know if it is right but mostly need help on question 6

I believ the others are correct

[red prawn]

@zinc wasp To find x1 using Cramer's rule, you would form a new matrix from U, let's call it U_1, by replacing the 1st column of U with the given y. Then compute det U_1 / det U = x1

Here's a bunch of examples:
https://youtu.be/qmjapjGxf2s

[zinc wasp]

TY I finally figured it out

[frosty vapor]

coomer's rule

cramer's rule was the first thing i learned about matrices when i first saw them on the bus ride to a competition in like the 10th grade and man was i not impressed by it

so disgusting to do by hand

no way was i gonna be able to do that fast enough

its p aesthetic though when u write it out

[zinc wasp]

yeah it sucks by hand

[frosty vapor]

wish someone told me what matrices were back then i might have gotten interested sooner

but noooo they had to motivate nothing

pane

[radiant yarrow]

What does Linear space V over a field F mean?

Like, I recently learnt about fields

What's a space?

And what's "over a field" mean here?

[verbal pivot]

Are you using linear space as a synonim for vector space?

[radiant yarrow]

Yes

Sorry

What does a space mean here?

[verbal pivot]

I'm not sure if I get your question about space (or maybe I'm not able to answer), but for the other question, if you have a vector space V over a field F, it means that the elements that you use for scalar multiplications of the vectors are in the field F, this means that if you have a vector $v \in V$, then by definition of vector space $\lambda v \in V$ where $\lambda \in F$

[stoic python]

b2unit

[verbal pivot]

Examples of F can be the reals or the complex numbers, I think that you can also use finite fields (?) but I don't know about that

[radiant yarrow]

I understood what a field meant. But what's a space here?

[verbal pivot]

I don't think I can answer that :(, If you were to ask what's a vector space then I could answer you, but if you ask what an space is then I'm not really sure, but I don't have that much math knowledge, maybe someone else knows?

[radiant yarrow]

Like field has a very different meaning in maths than physics and english. I'm assuming space might as well be having a different meaning. So I can't just intuit something whose meaning that I don't know of

[verbal pivot]

mmm, I think that if you want to build up intuition about vector spaces, then the best thing you can do is to start working with them, like seeing examples, doing exercises, etc

[limber sierra]

"space" has no meaning by itself

linear spaces (also known as vector spaces) do, however

a linear/vector space is a set of vectors paired with a field

your scalars come from the underlying field

for example, R^2 as a linear space over Q would be like standard R^2, but you only have rational scalar multiples

so you cant multiply a vector by sqrt(2) for example

note that R^2 over Q is infinite-dimensional

even though standard R^2 (which is a vector space over R) is 2-dimensional

anyway, "space" as a word means nothing a priori, but if working in a linear algebra context, 99% of the time itll mean "vector space" (AKA linear space)

its used to name a bunch of things in math though

like topological spaces, which generally arent vector spaces

[radiant yarrow]

Ohh

[radiant yarrow]

What does infinite dimensional mean?

[limber sierra]

there does not exist a finite basis.

[radiant yarrow]

Basis? like no minimum value or something?

[limber sierra]

uh

ignore that bit then

[radiant yarrow]

Okay

[limber sierra]

the point is that changing your underlying field changes many properties of the structure of the vector space

[radiant yarrow]

Ohh

So a vector space is defined as set of vectors which are scalar multiples of a given field, changing the field changes the properities of the vector space?

[limber sierra]

er, not quite

[radiant yarrow]

[limber sierra]

a vector space consists of the following 4 pieces of data:

• A field (the elements of this field are called "scalars")
• A set (the elements of this set are called "vectors")
• A way to add 2 vectors together
• A way to multiply a scalar by a vector

the addition and multiplication need to follow various axioms

but thats the idea

[radiant yarrow]

Got it

[limber sierra]

you can change the scalar field at will as long as scalar multiplication still makes sense

[radiant yarrow]

Ohh

[limber sierra]

for example, you can regard ℝ[x] as a vector space over the field ℝ

[radiant yarrow]

where x is an arbitrary vector multiplied by scalar belonging to R?

[limber sierra]

er

[radiant yarrow]

What is it that I'm missing?

[limber sierra]

ℝ[x] means the set of polynomials in x with real coefficients