linear-algebra

[limber kiln]

So what do I do with this? Like i was thinking quad formula, but the discriminant is a negative

[quartz compass]

you can find the sum of solutions by expanding (x-a)(x-b) and observing what the coefficients look like

[limber kiln]

I'm confused as to how to get this to (x-a)(x-b)

[quartz compass]

it's a quadratic that's factored, it has solutions a and b

don't worry about it for the moment if you're still confused by that, just expand out (x-a)(x-b) and we'll discuss it

[limber kiln]

so like (x-1)(x-(-2))?

[quartz compass]

no, just exactly what I have written with a and b

[limber kiln]

ohhh

so literally just (x-a)(x-b)

[limber kiln]

so like what coeffients do you mean by this?

[quartz compass]

sorry that was like 20 minutes ago I gotta go

[wintry steppe]

mero

[quasi vale]

@limber kiln for any quadratic ax^2 + bx + c = 0, the sum of the roots/solutions is c/a.

[limber kiln]

so in this case the sum is 2/1, which is just 2

correct?

[quasi vale]

Yeah

plus this isnt the correct channel

[limber kiln]

it was on my linear algebra practice exam

so thats why im asking it here @quasi vale

[quasi vale]

ok

[limber kiln]

just checking, should I reduce this again to 1/2(R2) so that i have a leading one?

[nocturne jewel]

should: Yeah, probably
Need to: no

[limber kiln]

How would I do this without reducing, im bad at doing things like basic solutions

[nocturne jewel]

you'd still just get y=0

you just dont need to formally do the ERO

2y=0 and y=0 are the same lines

[limber kiln]

so the solution itself is just y=0, i never clued into the fact that the right side was all zeroes

[nocturne jewel]

yeah, y's your pivot and x and z are free

[limber kiln]

so like (x, 2, z) = 0?

[nocturne jewel]

y=0

[limber kiln]

ah

[nocturne jewel]

so [x,0,z] is your general solution

for x,y in R

[limber kiln]

but y=0 is the basic solution

[nocturne jewel]

no

the general soln are vectors.

Ax=0, where A is that matrix, x is the general [x,y,z] vector, and 0 is the 0 vector

[limber kiln]

Ok, so let me get this straight, the answer to the question is [x,0,z]

[nocturne jewel]

yes, or written as a standard plane, x[1,0,0]+z[0,0,1]

aka the xz-plane

[limber kiln]

ok thank you, I am struggling through this course

Thankfully the only linear algebra course for my cs degree

[teal grotto]

it should be true that if V and W are normed vector spaces, V is finite dimensional, and T : V --> W is linear, then T is lipschitz, right?

[slow scroll]

yes. ||T|| < infty lipschitz

[teal grotto]

how though

[slow scroll]

Uh I forgot let me think

[teal grotto]

like, if v = a1v1 + a2v2 + ... + anvn in V, and ||v|| = 1, then i would need to put a uniform bound on |a1| + ... + |an|

oh

the set of all points v in V such that ||v|| = 1 should be compact

and i feel like the function which takes v to the sum of the absolute value of its coefficients should be continuous

[slow scroll]

can't you just say on the unit ball, if $v = \sum_i \alpha_i v_i$ for a basis $v_1, \dots, v_n,$
$$|Tv| = |\sum_i \alpha_i T(v_i)| \leq \sum_i |T(v_i)| $$
Take your lipschitz bound to be $L = \sum_i |T(v_i)|$

[stoic python]

kxrider

[teal grotto]

no because thats not true right? should be
$$\lVert T(v)\rVert \leq\sum_i|\alpha_i|\cdot \lVert T(v_i)\rVert\leq\left(\max_i\lVert T(v_i)\rVert\right)\cdot\sum_i|\alpha_i|$$

also, nice background

[stoic python]

c squared

[teal grotto]

wait. this is kinda it tho right? because the abs value coeff sum should be a norm on V, regardless of a choice of basis, and all norms on finite dimensional vector spaces are equivalent

[slow scroll]

yea, it definitely suffices to check on the l1 metric. And i think the unit ball thing i was trying to do is unnecessary anyway (even though this does give a weaker condition to check continuity).

[zinc timber]

@teal grotto $\norm{Tx-Ty}=\norm{T(x-y)}\leq \norm{T}\norm{x-y}$

[wintry steppe]

[nocturne jewel]

🤨

[wintry steppe]

im just thinking

[teal grotto]

that’s true

i was basically just having trouble arguing it

but i think i got

it

[zinc timber]

what did I say something wrong?

[zinc timber]

that's the definition of induced norm anyway

[wintry steppe]

i'd like to comment this only works when ||T|| is finite, but you get that from the domain being finite-dimensional

idk

i didnt read the convo

[zinc timber]

yeah domain finite

otherwise it's not Lipschitz

[wintry steppe]

right, you could get an unbounded linear transformation

those are funny

[zinc timber]

[wintry steppe]

[limber kiln]

What is this garbage

what does this mena

mean

[wintry steppe]

it means find a function T: R^4 -> R^2 with T(0, 0, 0, 0) = (0, 0)

but which isn't a linear transformation

[limber kiln]

imma be real with you chief, I have no clue how to do that, or like, even where to start

[zinc timber]

find a function

honestly any nonlinear will work

then subtract f(0,0,0,0)

or take the easy way

[wintry steppe]

you just have to come up with one, it's not really the kind of question susceptible to hints

i guess you could try to find a function T: R -> R with T(0) = 0 that isn't linear, and try to do something similar for R^4 -> R^2?

that's a possible starting point

[limber kiln]

ok

[zinc timber]

can I get an example where A neq 0, but exp(A) = I

[dusky epoch]

does that even exist

[blissful vault]

Just want to make sure. The definition of inner product here only allows for linearity (additivity and homogeneity) in the second slot when F = |R, right?

[zinc timber]

idk really, that's the whole question

[dusky epoch]

for F=C you still get additivity in the second argument but homogeneity is replaced by conjugate-homogeneity

so yes

[blissful vault]

Thanks. Was wondering why it was so specific.

[wintry steppe]

what is exp A

exp every component?

[zinc timber]

exponent of A

[zinc timber]

no

[wintry steppe]

yeah but if A is 2 x 2

what is e^A

[zinc timber]

,w matrix exponent

[wintry steppe]

yikes

[zinc timber]

sad

$e^A = I+A+A^2/2!+\cdots$

[wintry steppe]

o

does such an A exist is the question

no lol?

unless

all those other A cancel somehow

so if sum A^i/i! = 0?

[zinc timber]

[worldly bear]

how do you even raise an exponential to a matrix

i remember learning about it in differential equations but it was all online so i didn’t understand any of.

[tidal wharf]

if the matrix is diagonalizable then you diagonalize it since exponential of diagonal matrix is simple

otherwise you can try Jordan form or other decompositions that work even when the matrix is not diagonalizable

[frosty vapor]

[rotund jetty]

https://youtu.be/O85OWBJ2ayo

[pliant atlas]

Can any one solve it

[dusky epoch]

this is not linear algebra, and also we don't do your homework for you.

[lavish jay]

So what's the logic of this /?

[pliant atlas]

@dusky epoch xd

[crystal hound]

This is just a definition of the cross product

That gives you a vector that is orthogonal (perpendicular) to the vectors you’re cross producting

In LPs for geometric method Is your optimal soln point the first or last point your objective line touches when “sliding” the line up or down?

[hard drum]

Also if you just define the cross product for i,j and k (with i x j = k, k x i = j, i x i =0 etc etc) then you get this by linearity

[halcyon spindle]

I started by arriving that since AB is mxm matrix I was able to figure out that trace(AB) = $\sum_{k=1}^{m}\sum_{j=1}^{n}a_{k,j}b_{j,k}$. Also we have BA is nxn matrix and I found out that trace(BA) = $\sum_{j=1}^{n}\sum_{k=1}^{m}b_{j,k}a_{k,j}$. With that I just needed to show trace(AB) - trace(BA) = ($\sum_{k=1}^{m}\sum_{j=1}^{n}a_{k,j}b_{j,k}$) - ($\sum_{j=1}^{n}\sum_{k=1}^{m}b_{j,k}a_{k,j}$) = 0.

[stoic python]

Plegasus

[halcyon spindle]

Showing it equal to 0 is where I am stuck at, basically a manipulation of the sum. I been trying to figure out how to do this.

[hard drum]

the only difference is order of summation which can be swapped here (they are finite sums)

[halcyon spindle]

Ok I see, I look more into swapping for finite sum to understand why it works.

[hard drum]

One way is just to note that the sum is over all a_k,j b_j,k such that 1 <= j <= n and 1 <= k <= m

so they really are the same thing

[lavish jay]

Why he multiplies to subtract ?

What does it bring in terms of ease?

[lavish jay]

why not jus subtract

[nocturne jewel]

Cause it gives a way of finding inverses.

A is invertible iff it's the product of elementary matrices.

and elementary matrices are invertible, and their inverses are elementary matrices themselves

[lavish jay]

and if A is a matrix A^-1 . A = I ?

[nocturne jewel]

if A is invertible, yes.

[lavish jay]

ok

thank you

[warm zealot]

I see. I can see fortran being popular. I mean the syntax isnt awful. but yeah

[spring fulcrum]

hey, i'm trying to find the rotation matrix of angle -45° around the axis (0, 1, 1)
i found this formula on wikipedia, but i was wondering if there's maybe an easier way to do this?

i thought maybe we could use the fact that the axis doesn't get rotated so R(0, 1, 1) = (0, 1, 1), therefore (0, 1, 1) is an eigenvector of R ? but idk how to continue from there

or should i just learn the formula

[fringe fjord]

If you want to make your own derivation, one strategy could be to use Gram-Schmidt (or cross products or whatever) to extend (1/sqrt2)(0,1,1) to an orthonormal basis, giving a basis change to a system where your rotation goes around the x-axis.

[ionic belfry]

hello

can anyone help me with a linear algebra question

[spring fulcrum]

thank you! it's exactly what i needed

[oblique prairie]

@ionic belfry you’re wasting everyone’s time by not saying your question in the first place

[ionic belfry]

oh my bad

well

the question is

Solve the following matrix equation

i dont know how to write the numbers on here @oblique prairie

its a matrix equation where

X[2x3] = [2x3] matrix

how can you get X from that?

[nocturne jewel]

What's the actual question.....?

[ionic belfry]

This

@nocturne jewel

how do you get this?

[nocturne jewel]

I mean a priori you know X is 2x2

[ionic belfry]

i know

[nocturne jewel]

so just write X as a general 2x2 matrix then set up a system of eqn

[quartz compass]

the middle column tells you good information

it means -1 of the first column + 0 of the second column

[nocturne jewel]

oh true

[quartz compass]

so [1;3] is the first column of X

[ionic belfry]

wait how

[quartz compass]

makes determining the second column easier, cause now you just have 1 variable to determine

in each entry

it might help to write out X=[a,b;c,d] then multiply it by the column vector [-1;0]

I use , to mean next entry in row and ; to mean new row to be clear

[nocturne jewel]

$X:=\begin{bmatrix}a&b\c&d\end{bmatrix}$

[ionic belfry]

hey can we go on a call

[stoic python]

Mosh

[ionic belfry]

??

ohhh

oh okay

thank you

[tranquil steeple]

it can be other matrices too

[dusty pond]

Can someone explain what this would be? I'm not exactly sure here

[gray dust]

compare SoT vs ToS

[dusty pond]

yeah?

[gray dust]

one of em is 0, which one (prove it)? do u think the other's also 0? if so prove it, if not give a counterexample

[finite fiber]

HI 🙂 I'm seeing definitions of the tensor product of vector spaces that start with the free product of vector spaces, and quotient it out by some linear relations, namely by the ideal generated by b_{a * v, w} - b_{v, a * w} = 0, b_{a * v, w} = a * b_{v, w}, and b_{v, w} + b_{v', w} = b_{v + v', w}, where I use b_{x, y} to mean the basis vector in the free product of vector spaces V and W, where x and y are any vectors in V and W, respectively.

My question is: Instead of starting from the free product of vector spaces, could one start from the product? That is, can I just take the usual product of vector spaces (the categorical product, if one wishes to use that terminology), and quotient it by some relation, to get again the tensor product?

[wise oriole]

Is this correct

[dusky epoch]

"B is forthcoming in V"?

[wise oriole]

Idk the word in English

[dusky epoch]

what language did you translate from?

[wise oriole]

Dutch

[dusky epoch]

right

in english we say B spans V, or B generates V

[wise oriole]

Oh generates ok ty and is it correct ?

[dusky epoch]

yes

[wise oriole]

Oké thanks (:

[halcyon spindle]

My problem in the book says construct a nonzero matrix A such that A^2 = 0. I was able to do it with [(a, 1)^T, (-a^2, -a)^T]. Now is there something more going on with this matrix in terms of linear transformation?

[dusky epoch]

$\bmqty{a & -a^2 \ 1 & -a}$

[stoic python]

Kanga Gang Annihilator Ann

[dusky epoch]

this?

[halcyon spindle]

yeah

[dusky epoch]

hm

i can't think of a non-contrived geometric interpretation rn

[gentle marten]

Hi, can anybody help explain to me why is this valid

so that's the question

so the way I would do this

is by doing this

But then i stuck and i go to a Professor to ask for the next step

He gave me this

but I don't understand why ||y||^2 can become 2, isn't it suppose to be -2

IIyII^2*

[lavish jewel]

is what supposed to be -2?

[dusky epoch]

in $\bC^n$ the squared norm of a vector $(z_1, z_2, \dots, z_n)$ is in fact $\sum_{k=1}^n |z_k|^2$ and NOT $\sum_{k=1}^n z_k^2$.

so the squared norm of (i, i) is not -2, but +2.

[stoic python]

Kanga Gang Annihilator Ann

[gentle marten]

oh I see

ok Thank you for clearing that

[honest escarp]

can anyone help with this, i cant seem to be getting anywhere

[gray dust]

apply T to the usual basis of M_n

[shut orbit]

can someone explain me why in the matrix that have i variable different from S= and getting '-', for example 1 + i become 1 - i.

thankyou!

[lavish jewel]

without further context, my impression is that they are projecting onto the set

one takes projections using the inner product, which in C^n can be written as y^* x, where ^* denotes complex conjugate transpose

[slim geyser]

Guys how do i calculate 25ˆ23 mod 55

[dusky epoch]

do you know in general how to calculate exponents modulo something?

also this is not linear algebra

[slim geyser]

My mistake, it was given as hw in my lin.alg class. I guess i know in simple cases, but i cant manage to do this one

[dusky epoch]

here is a hint: calculate 25^1, 25^2, 25^4, 25^8 and 25^16 mod 55 in that order

[tranquil steeple]

I would just do mod(big(25)^23,55) in julia :--)

[slim geyser]

okay i guess theres probably a fast way to calculate 25ˆ4 mod 55 that i dont know of

[dusky epoch]

each number in this sequence is the square of the last.

what i am suggesting is for you to square, reduce mod 55, square, reduce mod 55, etc.

[lavish jewel]

following what ann says, the usual approach is to make a table of powers of 2 of the desired number using the given modulo and compose the larger exponent from this

[slim geyser]

could i play on the fact 55=5x11 which are both prime?

[dusky epoch]

no need

,w 625 mod 55

[stoic python]

Results provided by WolframAlpha

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

Requested by Ann#0413

[dusky epoch]

,calc 20^2 mod 55

[stoic python]

Result:

15

[dusky epoch]

etc.

[lavish jewel]

the name is "fast exponentiation algorithm" or "binary exponentiation" depending on where you look it up, btw

i think crypto uses the former

[slim geyser]

aha okay. But this seems to only work if i have some easily squarable numbers such as 25, what happens if i have something not so pretty

[slim geyser]

thanks! i was searching for it couldnt find it

[dusky epoch]

squaring is not very hard

if you know how to do long multiplication then squaring should not pose any issue to you

[slim geyser]

okay thanks

[wise oriole]

A element R {0} is this a correct notation. I wanna say that A can't be a zeromatrix.

[gray dust]

A element R {0}
???
A can't be a zeromatrix.
just say this

[dusky epoch]

$A \in \bR^{m \times n}, A \neq 0$

[stoic python]

Kanga Gang Annihilator Ann

[dusky epoch]

if you really want symbols

[wise oriole]

okay thank you very much

but was my notation wrong?

[dusky epoch]

yes it was

[sage ibex]

I think they were trying to say ℝ^(m × n) ∖ {0}

[sleek sundial]

Is there any class that's directly after linear algebra that isn't abstract algebra

[nocturne jewel]

Linear Algebra II

[sleek sundial]

._.

[nocturne jewel]

depends on the university.

[sleek sundial]

Is that like graduate level or something? We used axler and treil if that gives any idea of where we are

don't have a 2nd linear algebra class :/ only next one is graduate but i doubt i can take it or am qualifided

[nocturne jewel]

Nope, my university has the option of R^n LinAl and vector space LinAl as seperate courses.

[sleek sundial]

i guess im never seeing it again lol unless for some reason i decide to be a math major XD

[wise oriole]

is this a good proof

[restive hinge]

maybe not a problem worthy of the channel but am i messing up? i assume i have not found a counterexample to the cayley-hamilton theorem lol

i just can't find what the mistake is

ok i think WA is just buggy

[zinc timber]

counter example of the cayley-hamilton theorem?

[restive hinge]

according to WA maybe 🤪

[zinc timber]

it's a theorem for a reason

[restive hinge]

well WA clearly found a counterexample so 🤪

ferdinand frobenius seething in his grave

[zinc timber]

lol what

need to look deeper but I'm asleep

[restive hinge]

i think its just W|A bugging

i don't think anything to worry about

[zinc timber]

maybe mis interpreting the input

idk

💤

[nocturne jewel]

Nope, just did the calculation

still works perfectly fine.

You likely just messed up putting it into wolfram.

[zinc timber]

oh ye

[nocturne jewel]

Yeah you did

[zinc timber]

missing bracket

[nocturne jewel]

missing brackets for the (A-I) factor

[zinc timber]

wasn't me

[nocturne jewel]

You referred to the person that thought they debunked cayley-hamilton

[restive hinge]

ah yea

there we go

i knew it was something stupid

i guess dumb of me to assume W|A made a mistake instead of me

but i looked at it like 8 times

[restive hinge]

i promise i didn't actually think that lol

[storm sandal]

Hey everyone, I'm a bit confused on linear programming. I was able to understand the basics, but I was wondering if someone could help me. Yes, this for homework. I don't expect you to do it for me, I simply want to understand it. Every resource I've used just doesn't help with my understanding...

Prove that any LP optimization problem can be transformed into the following form:$\\minimize 0 \cdot x\\subject to Ax = b, x \geq 0\\$If the LP is feasible, then it has an optimum value of 0. If the LP is not feasible, then it has an optimal value of infinity

Explain what is the dual of this LP. (I only sort of understand duals, but not really... the book I'm using is far too verbose and densely packed for me to gleam any information from for this)

Prove that if the primal is feasible, then the dual has an obvious optimum solution. (again, my book is incredibly verbose on this, and it uses several backwards-referencing proofs in order to prove this)

Compare the time bounds of the given algorithm with that of one you could build, which could solve any LP without knowing a dual solution (The algorithm provided is $O((m + n)^k)$ time. I think for this, simplex would be used?)

[stoic python]

Foxify

[pliant kayak]

Can someone give me an example to this definition?

this should be the definition of the span of S, which also is the set of all possible linear combination of S. Does someone have an analogy or a connection btw the two definitions? Or just an example of the definition I sent?

I don't have the right image in my mind

[dusky epoch]

whoever wrote that ^ is misreading the set builder notation

fancy V is the set of all W such that S is contained in W and W is a subspace

[pliant kayak]

yes exactly

[dusky epoch]

anyway to give a somewhat simple example

consider the space R^3, and S = {(1,0,0)}

ie S consists of just one vector, the unit vector pointing along the x-axis

its span will be the x-axis itself

however, many subspaces of R^3 contain the x-axis

you can imagine them geometrically as planes (if they're dimension 2) or the entire space (dimension 3)

[fickle citrus]

Hello, I think you should try to look into which parts you don't get (as that is not clear from what you are asking)

What book are you using, by the way? I would believe the standard is Bertsimas & Tsitsiklis

Anyway LP is somewhat not-quite LinAlg, but there is no place for it other than here perhaps

Explain what is the dual of this LP.
The dual should be gotten programmatically. I personally wouldn't say I understand duals, but you should be able to follow instructions in getting the dual of a standard LP.

(one way is by Lagrangians, so there's that if you'd like to derive it)

No one in #discrete-math will answer LPs

#discrete-math's topics does not include OR either

[pliant kayak]

yes this makes sense, I just don't get the fact, that if {(1,0,0)} is in W, why is the span of S (the x-axis) the intersection of every W

like how does that implication work

(if it's one)

[dusky epoch]

span(S) is what all these subspaces have in common, to say things very very informally

[pliant kayak]

... but why? (this is exac what I don't get) Is it just per definition? (sorry for the spam)

[dusky epoch]

yes

yes, it's just by definition

[pliant kayak]

okay then, thanks a lot and sorry for bothering! ✨

[lime zinc]

from given condition it is clear that A^7-I is its characteristic equation. given that
A^7-I =(A-I)(A^6+A^5+A^4+A^3+A^2+A^1+I)

so minimal polynomial will be A^7-I , and all the eigenvalues are distinct, so option D should be true but given that option A is true, (likes like they want to say that A is not diagonizble over R ) any hint ?

[quasi vale]

@lime zinc Every operator on a real dimensional vector space has atleast one eigenvalue. Let a be the eigenvalue, so we know there exists a non-zero vector v such that Av = av. We know A^7 v = I v = a^7 v. From this, we have a^7 = 1. Since a is real, we only have one real solution and that is a=1. Now use the fact that A^6 + .... + I = 0 to come up with a contradiction.

[rich badger]

need help here

[dusky epoch]

"a linear operator of R -> 4"?

did you mean "a linear operator on R^4"?

[rich badger]

yes

[dusky epoch]

okay, and what is giving you trouble here?

[rich badger]

yeah, I'm lost on the first step

I can't find the basis for the vector

[dusky epoch]

the what?

[jagged solstice]

I don't get what is the difference between a row vector and a column vector. The book i m studying from says that "The row vector is the 'transpose' of the column v".
What does the author mean by "transpose" here?

[dusky epoch]

do you know what a matrix transpose is?

it's that but for n × 1 and 1 × n matrices

[jagged solstice]

chaning the rows with the columns?

Like this?

[jagged solstice]

How does this change the vector? If it doesn't change the vector, I suspect that there won't be needed a differentiation written like this

[dusky epoch]

whether a vector is a row or column matters when it comes time to multiply it by a matrix

if your vectors are columns they go on the right of matrices, if they're rows they go on the left

[jagged solstice]

Ohhh, thank you

And the representation on the x, y, z .. n axis is not changed for the vector, right?

[nocturne jewel]

nope, the ith coordinate is still the one in the ith position

[strange lion]

How do you find a transformation matrix of this?

a=1,b=2,e=5,j=10,i=9 suppose. but theres still 3 variables: x,y,z

[lavish jewel]

the easiest way is usually to study what happens to the canonical basis

you transform each of the e_i vectors (columns of the identity matrix), see what you get out

[strange lion]

Shall I put 3 columns for the 3 variables? Then how do I multiply with the column vector of (x,y) ?

[quartz compass]

my mind kind of immediately factors out the column vector [x;y;z] from a matrix

as it is, doesn't look like R is a linear transformation unless that's a typo with z left out, cause it doesn't send 0 to 0

[nocturne jewel]

should be R[x,y,z] I think.

[strange lion]

That's what I was thinking. Otherwise it's not possible right?

That does make sense. It must have been R(x,y,z)

[lavish jewel]

oh i didn't even notice that until you pointed it out haha

[quartz compass]

I'm not really sure what the question is, the way it's written in the screenshot I kind of assumed the question is asking "write if these statements are true or false"

seems peculiar to assert R(x,y) = ... is a LT. otherwise

[lavish jewel]

indeed. where did you get those values for a,b,e,j,i, anyway?

can you show the full thing?

[halcyon spindle]

For the first part the sentence what does it even mean "can be considered as a different representation of the same space"? The author tell me what it means in the second part but I am still trying to understand what is meant in the first part.

[strange lion]

There's actually another matrix below that. The question asks to find some composite linear transformation stuff

Here it is

[gray dust]

the actual underlying sets of isomorphic spaces may be different but as far as vector space structure goes theyre essentially the same space

[halcyon spindle]

I see, so that would mean I can use all the stuff I have about V and the isomorphism between V and W to do the same thing with W. Likes bases, linear independence etc? if that make sense.

[gray dust]

yes

[halcyon spindle]

Thanks.

[gray dust]

ex: C & R^2, if considered as vector spaces over R, are isomorphic (map x+iy to (x,y))

so we can see C as a 2d space

[storm sandal]

Hello, all of it. I understand hardly any of it despite hours of studying. For some reason nothing in here clicks with my brain. I'm using CLRS (Introduction to Algorithms)

Last math course I took was several years ago, and we never talked about anything closed to linear algebra/linear programming

[storm sandal]

Lagrangians is at least something specific I can look up... ;-;

Man, masters are stressful

I've been studying for days, assignment is due today, yet I don't understand a single thing on it

Ugh, I wish I had taken a higher-level math other than just algebra

[golden reef]

How do i turn the equation of the line into parametric equations?

the line 3x+2y+3z = 1, 7x +5y + 9z = 4

would it just be the intersection of the two planes is the line?

[quasi vale]

i think they mean a line parallel to the line which is formed by intersection of the planes given

yeah

[golden reef]

oh i see

[sudden narwhal]

Someone knows where I went wrong ? Wolfram alpha finds something else

[quasi vale]

second line, in the determinant, last entry should be x+1

[dawn ocean]

you dropped a negative at -(3+x-1)

i can't tell if the last entry of your matrix A in the diagonal is positive or negative

[fickle citrus]

CLRS is an Algorithm book, not a linear programming book.

I realise I don't really have an undergraduate textbook for LP, hmm

I have springer access and so after looking through
https://math.stackexchange.com/questions/20643/linear-programming-books
I looked at Vanderbei's book and it seems like a good introduction
https://link.springer.com/book/10.1007/978-3-030-39415-8
(sadly it uses MATLAB of all languages)

Bertsimas and Tsitsiklis works as well, so you could go for that.

Since you don't get everything I actually think starting from basics will be good. If not, Wikipedia which does have the same basic material as the books but certain seems like, extremely fast.

[storm sandal]

Indeed, it's an algorithm book. I'm taking Computer Science, and it's the only resource we have for the class.

Unfortunately, this is going to be a one-and-done thing that's due tonight. Despite spending the week trying to learn this and using as many resources as I could find, it's just not clicking. I'll probably lose 10% of my grade from this one assignment 🤣 I'm just not willing to pay a huge amount of money for a single assignment on something that may or may not help me (and especially since it's due tonight), and the "basics" (most basic things I could find on these topics) of this don't really click for me either

So, I think I'm a bit screwed for this assignment

Thanks for the help though, unfortunately I'm just too dumb

[fickle citrus]

Right, I'll at least ask you the simplest thing

[storm sandal]

I appreciate you looking through and trying to help me find resources

[fickle citrus]

Which is if you can take the dual

of that 0 * x LP

[storm sandal]

(Which I've looked at dozens of resources and I still don't know how to take the dual... I'm just kinda dumb on this particular subject I guess. Didn't have nearly this much issue with any of the other topics :/)

[fickle citrus]

Right, basically the dual is like a transposing of the entire LP, much like how a matrix is transposed

[storm sandal]

Whenever I try to find resources everything they say goes in one ear out the other, and nothing clicks for me

Ok, so turning a maximum into a minimum? Is there any other form of transposition?

[fickle citrus]

In typical not-pathological cases, dual of a dual is the same LP

Yes but simply max/min change is nothing exciting

For example a simple case is that $\min c^Tx$ = -\max -c^Tx$

[stoic python]

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

[storm sandal]

I assume that the c and the x are the variables here? What is the T?

[fickle citrus]

Transpose

Inner product

c is inner producted with x

The typical LP is
$$\begin{aligned}\min &&\langle c, x \rangle \
\text{subject to} && Ax\leq b\end{aligned}$$

[storm sandal]

What is the A?

[stoic python]

ShatteredSunlight

[fickle citrus]

So c is just a linear coefficient (so linear algebra comes in

A is a matrix, it is typical matmul-ed onto x (which is why linear algebra comes in again)

and b is a vector

The $\leq$ symbol is done element-wise. All rows of $Ax$ must each satisfy a corresponding element of $b$

[stoic python]

ShatteredSunlight

[fickle citrus]

Anyway all LPs are written this way or can be converted to this way

[storm sandal]

Trying to remember the matrix math from years ago... And we never learned any linear algebra so I'm being thrown into the deep end on that

[fickle citrus]

According to Wikipedia, this is the Dual

[storm sandal]

I think I vaguely remember some of the matrix stuff we learned

[fickle citrus]

You should brush up on LA if you want to do CS

[storm sandal]

I see 🤔

[storm sandal]

I had planned on learning it on my own after my masters since I plan to get into that sort of more theoretical territory, but this course just kinda threw it at me with no preparation beforehand

[fickle citrus]

I mean, now that I think about it, it's not immediately obvious why LA is useful in CS

[storm sandal]

I've heard it useful in a lot of algorithmic applications

Is that correct?

[fickle citrus]

More like applications

no clue about algorithmic

As a very 'raw' subpart, I'd say LA is useful for understanding R^n

And R^n is useful for any positioning system

For example, Ray tracing

The hot new part of machine learning uses linear algebra of course

If you'd like to do ML, then LA is not just 'useful,' it becomes a critical component

Because right now I see no pathway for non-linear machine learning without any linear component, at least right now

[storm sandal]

I've heard of this, but from the very basics of ML math I've seen, I hadn't seen anything like this, so that's good to know

https://tenor.com/view/noted-meme-but-looney-tunes-gif-22310685

[fickle citrus]

Anyway back to this, notice the change in position of b, c. A just becomes A^T. Note y non-negativity that gets added in.

As I said, just a transpose.

[storm sandal]

So if I understand correctly

A transpose is swapping the columns and rows with one another, correct?

[fickle citrus]

Yes

For a matrix

[storm sandal]

Ok I see...

I don't recall anything about linear coefficients though

[fickle citrus]

linear coefficients means no funny coefficients

it's just linearly multiplied

It's just a number

[storm sandal]

Ahhh ok

[fickle citrus]

So $c^Tx$ is $\sum_{i=1}^{n}c_{i}x_{i}$

[stoic python]

ShatteredSunlight

[storm sandal]

I see. That's a bit of a weird notation, but it makes a lot more sense now

[fickle citrus]

By the way LP means the program is linear in x, nothing to do with c

c are just linear constants

But they could be quadratic part (independently)

For example if you want $a^2x_1 + ax_2 + bx_3$, the (a,b) is constant

[stoic python]

ShatteredSunlight

[fickle citrus]

This is an LP since the decisions $x$ are linear

[stoic python]

ShatteredSunlight

[storm sandal]

I see!

[fickle citrus]

Nonlinear programming isn't very exciting if it is non-convex, unless you like guessing or numerics maybe

[storm sandal]

But the linear coefficient of that restricts a to be a real number, correct? (I hope I'm not misremembering that term)

[fickle citrus]

Well I guess nonlinear optimisation happens a lot but the issue is theoretically there is no guarantee of optimality unless you find bounds/envelopes (again, another pain)

[fickle citrus]

Yes LPs are real.

So the coefficients cannot have imaginary components

By the way I would wonder how a complex generalisation looks like, but either way you need total ordering (to do 'min' meaningfully) so ... hmm

[storm sandal]

So for the minimize $0 \cdot x$

Subject to $Ax = b, x \geq 0$

What transformations would need to be made to any LP optimization problem to get to that point? I can't see a way to get from point a to point b here. What transformations are available to me in order to prove this?

[stoic python]

Foxify

[storm sandal]

Also, the feasibility statement of

If the LP is feasible, then it has an optimum value of 0

If the LP is not feasible, then it has an optimal value of infinity

doesn't make any sense to me for this optimization problem

[storm sandal]

I still don’t see how infinity is not feasible…

Tried doing some research but couldn’t find anything

[storm sandal]

After a nap I'm feeling a lot more present. So would this mean that the dual of this is $0^Tx$ subject to $A^Ty = 0, y \geq 0$ as $c$ is $0$?

Going off the Wikipedia article you mentioned

If so, what is y then? 🤔

Wait...

Is y the minimization (in this case)?

[stoic python]

Foxify

[storm sandal]

Wait I think it would be $b^Ty$ subject to $A^Ty \geq 0$ as that's what it looks like in the primal column of that table you were mentioning

[stoic python]

Foxify

[storm sandal]

Since the 0 doesn't have that $T$ exponent, would it make it $b \cdot y$ subject to $A \cdot y \geq c$? Or is this not something you can do?

[stoic python]

Foxify

[storm sandal]

Am I just being stupid? 🤣

[storm sandal]

I was wondering if anyone could help me understand Problem 4a

[viscid lagoon]

Let $V$ and $W$ be vector spaces over $\mathbb{K}$, and let $e_1, \ldots, e_s \in V$ for $s \in \mathbb{N}$. Define the linear mapping $f : V \to W$. Consider $f$ injective $(1)$ and $f$ surjective $(2)$. Does the following implication hold? $$\operatorname{span}\left(f(e_1), \ldots, f(e_s)\right) = W \implies \operatorname{span}\left(e_{1}, \ldots, e_{s}\right)=V$$ My guess is that it is true for $f$ injective, but incorrect for $f$ surjective. I might have found a counter-example for surjectivity, but haven't proven it for $f$ injective. I just wanted to somehow make sure that I'm not trying to prove something incorrect.

[stoic python]

lewis

[torn stag]

@viscid lagoon I agree.

[viscid lagoon]

Actually, does it really hold for injectivity.

oh wait, nvm, I think it does

Will I need to use the fact that $f$ injective if and only if $\operatorname{ker} f = {\vec{0}}$

[stoic python]

lewis

[fickle citrus]

y is the decision variable of the dual.

[fickle citrus]

"INF" is just a special value for non-feasibility.

For non-feasibility, it makes sense to assign infinite cost. For min cost, max cost is infinite, which is a special value and the most important thing for that is that it is defined by the relations INFTY > x for any x in the real numbers. (There is just about no other use/point to infty showing up afaik)

For max profit, the infeasible special symbol should then be -INFT

[gray dust]

nah

[viscid lagoon]

I just realized, yeah

Is there an elegant way?

I have a solution, but I'm not sure whether it's the best

[gray dust]

can u show it

[viscid lagoon]

I haven't written it down

wait

@gray dust

but yeah, im not quite sure whether it's correct, or not

wait, sorry, there a major issues with this, ill correct it

[gray dust]

howd u get spanV=W

[viscid lagoon]

that should be it

So what I have is:\
$\operatorname{span} \left(f(e_1), \ldots, f(e_s)\right) = f(\operatorname{span}(e_1, \ldots, e_s)) = W \implies f(V) = W$ since $ \operatorname{span}(e_1, \ldots, e_s) \subset V$, so $f$ is even surjective. Due to injectivity, we receive that $V = \operatorname{span}(e_1, \ldots, e_s)$

[stoic python]

lewis

[viscid lagoon]

now, this should be it

sorry

@gray dust

[gray dust]

didnt expect u to apply f to sets but it works

[viscid lagoon]

is there another way?

[gray dust]

element chasing

[viscid lagoon]

so show that $V \subset \operatorname{span}(e_1, \ldots, e_s)$

[stoic python]

lewis

[viscid lagoon]

cuz the other way is trivial

[gray dust]

yes

[viscid lagoon]

So let $v \in V$

[stoic python]

lewis

[viscid lagoon]

how do i go on to show that it is contained in the span

[gray dust]

use assumption

[viscid lagoon]

so $f(v) \in f(\operatorname{span}(e_1, \ldots, e_s))$

[stoic python]

lewis

[viscid lagoon]

?

[gray dust]

id ditch the set computations from above

use the assumption directly

[viscid lagoon]

$f(v) \in \operatorname{span} \left(f(e_1), \ldots, f(e_s)\right) = W$

[stoic python]

lewis

[viscid lagoon]

?

[gray dust]

what does this mean?

[viscid lagoon]

uh

it means that f(v) can be written as a linear combination

of elements in the span

[gray dust]

explicitly write that combo

[viscid lagoon]

$f(v) = \sum_{j=1}^{s} \lambda_j f(e_j)$

[stoic python]

lewis

[gray dust]

can u see the end from here

[viscid lagoon]

oh wait, of course

we can apply linearity

and due to injectivity

we get that v = the sum

so it is particularly a linear combination of the span

of e_j

[gray dust]

yes

[viscid lagoon]

yeah, thank you

[gray dust]

np!

[hollow void]

any property? or i should find det of this. it would take alot time

[dusky epoch]

well the determinant is clearly 0 so that will be of no help to you

you might however want to do some tricks with the charpoly

[sick sandal]

maybe start by finding the characteristics polynomial

[jade fractal]

Hello everyone, can you help me with this problems. Thank you in advance.

[tranquil steeple]

You can find permutation such that you know you have the same eigenvalues as the block diagonal matrix ```
1 1 0 0 0
1 1 0 0 0
0 0 1 1 1
0 0 1 1 1
0 0 1 1 1

then just compute the eigenvalues of the blocks

[dusky epoch]

oh that is pretty clever

[hollow void]

how to kn ow that quickly. that it is zero

[dusky epoch]

you literally have three identical columns right there lol

[hollow void]

shitt

[hollow void]

clever bro.

i cant comprehend this que. please explain. please

[nocturne jewel]

do you know fundamental theorem of invertible matrices?

[hollow void]

det non zero

[nocturne jewel]

that's one of the TFAE statements

[hollow void]

[nocturne jewel]

there's like 20.

[hollow void]

real?

det

[nocturne jewel]

det is still only 1 of the statements

det(A) non-zero iff A is invertible.

[hollow void]

wth im learning this first time

what is approach?

[strange iron]

do you know any good sites with exersise of integrals with solusion

[nocturne jewel]

#calculus

integrals aren't LinAl

[hollow void]

@nocturne jewel bro what's is product of 2 eigen vectors

[nocturne jewel]

what?

you mean inner product...?

[glossy swan]

Anybody know the best linear algebra playlist on YouTube to learn it? Prof Leonard doesn’t have one

Wanted to study it before I take it winter semester

[dawn ocean]

@hollow void he was talking to the person below your pic not you

[hollow void]

sorry

ill not interfere

interfear

[dawn ocean]

i think i can help you with the question if you still need it tho

[hollow void]

yes

please

please

[hollow void]

yes

[dawn ocean]

all that is left is to figure out if it's a circle or ellipse. i susspect an ellipse you just need to rearrange some stuff to get it into a recognizable form

i hope you can see what i did @hollow void

[prime snow]

Hiya I have a quick question about this definition/example, are the elements x_n sequences in this case?

i would post this in the advanced section but I don't have access to it

[nocturne jewel]

yes

[dawn ocean]

you just react to the message with a green checkmark in #get-advanced-access

[nocturne jewel]

space of sequences forms a vector space w/ those operations

[hollow void]

wow absolute genius

[prime snow]

so is it a vector space of sequences?

[nocturne jewel]

yes

[prime snow]

brilliant that clarifies loads thanks!

[nocturne jewel]

If you need justification for it being a vector space just run through the axioms

commutativity and associativity hold from comm and asso of F, 0 vector exists, all sequences have an inverse, etc

[hollow void]

what indicates. if det is less than zero

[dawn ocean]

i completely misread your message

[hollow void]

order is 100, so i =10 , j=10,

answer should be d

[wintry steppe]

what are i and j

[fringe fjord]

Is d your answer or someone else's claim about which answer is correct?

[zinc timber]

exactly 15 means what? 15 distinct or counting multiplicities as well ?

[dusky epoch]

we're talking about degree here @zinc timber

[zinc timber]

I feel like the answer to this will matter

[nocturne jewel]

exactly 15 means... not 14 and not 16, exactly 15

[zinc timber]

why didn't i think of that, very enlightening

[nocturne jewel]

Well that's what you asked.

[fringe fjord]

The question asks for the degree of a polynomial. There's no usual concept of "multiplicities" there.

[nocturne jewel]

(x-r)^2 and (x-r)^3 both only have 1 distinct root.

[torn stag]

@viscid lagoon Your solution is what I had in mind. The fact that $f$ is injective and surjective means it is an isomorphism, so it is obvious then that $e_1, \dots, e_s$ span $V$.

[stoic python]

IlIIllIIIlllIIIIllll

[zinc timber]

ya ig I was tripping,

but like $\mqty[\imat{5}]$ has one\textbf{distinct} eigen value, min poly degree is 1, not 5

[zinc timber]

that's why I don't see how the answer is 100

[fringe fjord]

It isn't.

[zinc timber]

ya that's what I thought

[halcyon spindle]

I need to show if AB are invertible then A is right invertible and B is left invertible. Where B : V -> U, A : U -> W are linear.

I been trying to figure how how to show this, it would be a lost easier if I can assume either A or B is invertible to show the other other factor is left/right invertible. Though I am not sure if I can assume that since the invertibility of AB does not imply either A or B is invertible.

any take on this?

[jade fractal]

Can anyone tell me how to covert vector equation into general equation (R3)?

Or from canonical equation to general equation

[fringe fjord]

No you can't assume that. Try writing down what it means for AB to be invertible, though.

Then use the fact that composition is associative.

[halcyon spindle]

Thanks, I must be tired or something I don't know why I did not see it. I even wrote down what AB being invertible means.

[fringe fjord]

Can you show that here?

[halcyon spindle]

The last sentence I just wrote down. I literally had the answer in my face.

[fringe fjord]

Exactly.

[spring fulcrum]

$Ker(A) \oplus Im(A) = E \Leftrightarrow Ker(A^2) = Ker(A)$ \
is it sufficient to prove that $Ker(A) \cap Im(A) = {0} \Leftrightarrow Ker(A^2) = Ker(A)$ ?

[stoic python]

Sacha

[spring fulcrum]

this is the definition i have for the oplus symbol (direct sum?)

what confuses me is why did they put it equal to E

nvm i think i get it

the symbol was used to mean "V is in direct sum with W", it didn't mean the actual direct sum

[zinc timber]

[hollow void]

Someone else (me)
Is claiming

[hollow void]

Well now I'm more confused 😳

[hollow void]

Row column

[wintry steppe]

in the problem below ive solved part (a) by doing the following { i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) } <--- unit vectors for R^3 then made { d = i + j + k, u = i + j } took the dot product of { dv = | | d | | | | v | | cos(theta) } rearranged it { arccos( dv / (| | d | | | | v | |) ) = theta } then just solved for theta in part (b) its a bit tricky the vector v looks like {(0, 1, 1)} or just {j + k} however its coming off of vector u so would it be { v = (j + k) - u } or just { v = (j + k) } in the case {v = (j + k) - u}, the dot product is zero which means the vectors are orthogonal in the other case they are not they look pretty orthogonal to me the solution in the book is trying to assert otherwise NOTE: ive put { } around the mathy parts to try to make this question a bit more readable any help would be appreciated thanks

[lavish jewel]

as you say, d = [1,1,1] and v = [-1,0,1]

and their dot product is -1 + 0 + 1 = 0, so they're indeed orthogonal

[wintry steppe]

Alright, so the solution is wrong then. 'not mine'

thanks @lavish jewel

[jade fractal]

Can anyone help me with this problem?

[drifting scarab]

@jade fractal What's you definition of a scalar product?

for the functional space

also did you already define the hermite polynomials in class

[jade fractal]

I think we learned

[sharp shell]

Can someone explain why the amount of free variables is equal to the dim of the Null space of a matrix?

I just cant see the connection

[nocturne jewel]

suppose you have Ax=0 w/ A being nxn

Further suppose there are n-m pivots in the RREF (m free variables)
You can write each of the pivot variables as a function of the free variables
Subbing these isolated pivots into a general vector then expanding gives a subspace, the nullspace of A, as a span of vectors.

The dimension of this span will be m.

[jade fractal]

Can someone tell me how to find a plane that passes through one line and is parallel to another

[lavish jewel]

take two points on the line that must be contained in the plane. then pick one of the two and displace it by the vector that the plane must be parallel to

this gives you 3 points that you can use to find the plane's normal

[jade fractal]

yes, I already found plane’s normal

and the equation like ax+by+cz+d=0

how to find d

[lavish jewel]

an alternative formulation of the plane equation is n . (p - p0) = 0

where the . is dot product, n is the normal, p is a generic point x,y,z, and p0 is a fixed point on the plane

so in your formulation above, d = -n . p0

[jade fractal]

n is the normal of the plane yes?

[lavish jewel]

yes

[jade fractal]

and p and p0 are the point of the line passing through this plane?

like if we have L= <x,y,z> + t<a,b,c>

is p0 <a,b,c>?

[lavish jewel]

p0 is any poit in the plane. in your example, <a,b,c> is not necessarily in the plane

<x,y,z> is though

[jade fractal]

okay, therefore we can take any point like (1,0,0)?

[lavish jewel]

that depends on whether you want the plane to contain the line or only intersect it once

[jade fractal]

can you tell me is it the correct way to write like that?

[jade fractal]

I guess it intersects it once

[finite fiber]

Given two linear maps f, g: V -> V, is there a nontrivial condition for when ker(f+g) is in ker(f) \cap ker(g)? The other inclusion is obviously true, of course. I have so far that if (f+g)(v) = 0, then either v is in ker(f) (and thus in ker(g)), or v is not in ker(f), and thus f(v) = -g(v), so -g(v) (and obviously g(v) as well) is in Im(f), so v is in the intersection of the images. but this last part is a pretty weak result.

[wintry steppe]

I dont quite see why v has to be in in Im(f). But your second case implies that the intersection of images is non-trivial , so a sufficient condition would be that the intersection of images is trivial. However its definitely not a necessary condition, since if f=g then it's also true

[finite fiber]

Sorry, not v, but g(v).

[dapper jolt]

hey, can i get some help on #13 please?

[analog nacelle]

what ideas do you have so far?

@dapper jolt

[dapper jolt]

ok so like

i got 12

so i'm thinking go by contradiction?

if we have U1, U2, U3

if U2 contains U3 or something it just reduces to 12

but i'm not really sure how to proceed afterward

[analog nacelle]

starting with the only if direction is probably a good place to start
starting with V_1,V_2,V_3 as subspaces with V_1 \subset V_3 and V_2 \subset V_3

[dapper jolt]

oh yeah i got that

[analog nacelle]

okay cool

[dapper jolt]

like V1 u V2 u V3 = V3

and V3 was a subspace by assumption

[analog nacelle]

mhm

[dapper jolt]

but i'm stuck on the other direction

which seems to be the crux of the proof

[analog nacelle]

Now, assuming the union is a subspace.
Then contradiction is a good way to go about solving it

one way to start is assuming neither V_2 or V_3 contain the other and take x \in V_2\setminus V_3 and y \in V_3\setminus V_2

[dapper jolt]

we can make this assumption because of the previous problem right

[analog nacelle]

yes

then bc you aren’t working in F_2 (where this fails), you can choose a,b \in F such that a-b=1
now it’s just showing ax+y and bx+y are in V_1

[dapper jolt]

wait what's the a-b thing?

and why do we need to show that ax+y and bx+y are in V_1?

[analog nacelle]

so you can show V_2 \ V_3, V_3 \ V_2 \subset V_1

the a-b thing is a formality (related to the parenthetical in the question), you can just choose like a-b for a,b\in R or something

[dapper jolt]

hm okay so

if x is in V2 \ V3

and y is in V3 \ V2

x + y is in V2 \ V3 union V3 \ V2 = V2 union V3 ?

or is that incorrect

[analog nacelle]

it doesn’t help us show that they’re both contained in V_1, what I’d do is assume that ax+y and bx+y are not in V_1
note that x \notin V_3 and y \notin V_2

[dapper jolt]

right x is not in V3 and y is not in V2

do we have to do something like (ax+y) - (bx+y)?

that gives us x

[analog nacelle]

yes, that’s a good observation

now you just need to show that both components are in V_1 so then you can show x\in V_1

[the same argument works for x +ay and x+by to show that y \in V_1]

[dapper jolt]

okay hm

[analog nacelle]

yeah. so what does it mean if ax+y and bx+y are not in V_1?

[dapper jolt]

they're in V2 union V3?

[analog nacelle]

yes

or, ax+y \in V_2 or ax+y \in V_3

[dapper jolt]

right

oh

[analog nacelle]

so how can you pull a contradiction out of that

[dapper jolt]

okay so

let's assume it's in V3

let's also assume bx + y is in V3

then by closure under addition

(ax + y) - (bx + y) = x should be in V3

which is a contradiction

[analog nacelle]

yes

[dapper jolt]

so ax + y is in V1

and we can do a similar thing for y

wait actually question; is it okay that we didn't consider if ax+y is in V2?

[analog nacelle]

no, but you can just do ax+y-ax for example

[dapper jolt]

oh right

so it's simpler that way

[analog nacelle]

bc closure by scalar product

[dapper jolt]

and then we can do the same thing for y

and then we're done?

showing that x and y are each in V1

[analog nacelle]

you just need to check that it’s still okay when V_2 \cap V_3 are nonempty

[dapper jolt]

wait what does that mean?

[analog nacelle]

so then you can show that V_2 \cap V_3 \subset V_1

[dapper jolt]

(what does \cap mean :p)

[analog nacelle]

intersection

[dapper jolt]

oh okay

[analog nacelle]

bc if the intersection is empty, you’re done, but it may not be empty

[dapper jolt]

is that because we assumed x was in V2 \ V3

and y was in V3 \ V2

[analog nacelle]

yes

[dapper jolt]

do we assume there's an element z in V2 union V3?

[analog nacelle]

yes, as if there are no elements in their union, they’re both empty

so trivially, they are both contained in V_1

[dapper jolt]

oops i meant intersect haha

z in V2 intersect V3

[analog nacelle]

oh yes

then z+y is not in their intersection, as z+y-z would otherwise be in V_2, which is a contradiction

[dapper jolt]

okay so

z+y is not in their intersection because y is not in V2

[analog nacelle]

yes

[dapper jolt]

i'm not quite following the second part

[analog nacelle]

yeah sorry, I messed up the phrasing, I just said what you said about why z+y is not in the intersection

[dapper jolt]

ah yes i see

[analog nacelle]

but now you just need to conclude that z\in V_1

by contradiction

above

[dapper jolt]

okay so assume z not in V1

[analog nacelle]

but then z+y-y\in V_1 bc z+y is

so

[dapper jolt]

wait okay so

z+y is not in V2 intersect V3

does that imply z+y is in V1?

hm how is that true?

if it weren't in V2 union V3 i'd understand why it'd have to be in V1

does it hold true for intersections though?

[analog nacelle]

it implies because V_3 \ V_2 \in V_1 and V_2 \ V_3 \in V_1

[dapper jolt]

does that mean we don't even have to introduce z?

oh wait sorry

[analog nacelle]

no, but for completeness it’s good to show that the intersection is in V_1

[dapper jolt]

so V3\V2 in V1 and V2\V3 in V1 implies that V2 intersect V3 in V1

i'm not too experienced with proofs, but is this like a relatively common deduction to make?

[analog nacelle]

I’m bad at math give me one sec, it’s because z+y-z = y which is in V_1

[dapper jolt]

noooo you're amazing c: tyvm for your help :D

[analog nacelle]

yeah, I just got mixed up and realized the union of setminus is not the intersection, but the reason I mentioned most recently is why it’s in V_1

bc V_1 is closed under addition

[dapper jolt]

okay so

[dapper jolt]

can you like, work in the "reverse" (given an element, find two elements that add to get that element, and conclude that those two elements are a part of the subspace?)

actually i'm a bit confused now

we just added and subtracted z

i have to go now but thank you so much for your help :)

[left nebula]

im not seeing an axiom that says vector-scalar multiplication has to be commutative

does it not need to be?

[analog nacelle]

wdym

[left nebula]

theres no axiom in my textbook that says av = va for all a in F and v in V

but i believe vector spaces do need to satisfy that, dont they?

[analog nacelle]

yes, I think this is just a consequence of scalar multiplication rather than vector space structure

well

if the field you’re working in isn’t weird

so no, it’s not a requirement for a vector space

but also va isn’t really a common notion

and most of the time, it will be the case that av=va

[left nebula]

so not in all vector spaces?

[quartz compass]

I don't think there's a definition of 'right scalar multiplication' in the vector space axioms in the first place

[analog nacelle]

yeah

[left nebula]

i see

so my question doesnt really make sense then lol\

[analog nacelle]

I think if you have V=\C (the complex numbers), right multiplication is v \dot conjugate(a) or something like that

[quartz compass]

but if you put a norm on your vector space you have $|a v| = |a| |v|$ which are scalars which commute so $|av|=|a||v|=|v||a|=|va|$ if you were to define it

[stoic python]

Merosity

[quartz compass]

personally I have never thought about it, I would just say av=va and not sweat it haha, maybe other people have some different idea, I'm definitely not the end-all-be-all

[gray dust]

dw i trust u

[quartz compass]

I worry that people trust me 😮

[gray dust]

too late if u refuse my trust ull break my heart

[quartz compass]

ok, then in that case I take this as permission to lie without repercussion 😎

[gray dust]

sparing feelings, thats the holiday spirit

[zinc copper]

isnt the second part only true when $\operatorname{car}K\neq 2$

[stoic python]

𝓛ittle ℕarwhal ✓

[zinc copper]

cause otherwise symmetric and alternating forms coincide?

but not every multilinear form is alternating or symmetric in carK=2 i dont think

actually they just might be

[dapper gorge]

Why is he talking about sizes and allows for negative values?

How do you then make the proper choice for the order of the vectors in the determinant (sign is changed)

So confusing...

ok I think I get it, but still confusing

[wintry steppe]

Can anyone explain it i didn't get it after 2 line ?

[dusky epoch]

which second line?

[wintry steppe]

Of Proof

[dusky epoch]

there are vectors u_1, ..., u_n in S \cup {v} such that sum[i=1,n] a_i u_i = 0 for some nonzero scalars a_1, ..., a_n.

is this the first part you don't get?

[wintry steppe]

Yeah i got it

[dusky epoch]

...

i don't understand.

i'm trying to locate where your doubt is

and your reply is not helping with that at all

[wintry steppe]

Actually my que is how can they multiply by a_1 inverse we dont know whether it is zero or non zero

[dusky epoch]

then why did you not say this is your doubt in the first place

why make me hunt it down

anyway, they say explicitly that they assume all the a's are nonzero

[wintry steppe]

Where ??

[dusky epoch]

for some nonzero scalars a_1, a_2, ..., a_n

[wintry steppe]

Then it should be for all a_i !=0 i€[1,n] right ??

[dusky epoch]

yes and that is what is said...

[wintry steppe]

Okay thank you !!

[wintry steppe]

Sorry it wont happen again

[haughty berry]

Out of curiosity though, this isn’t a good proof, no? Because the scalars can be 0, the definition of linear dependence is that at least one of the scalars must be nonzero, not all of them, right?
So you have to prove that a_1 is nonzero (otherwise if a_1=0, since S is linearly independent, all of the other scalars must be zero), and then you can multiply by the multiplicative inverse of a_1.
But the proof didn’t have that little part, and I don’t understand why…

[dusky epoch]

you can throw out all the terms with a_i=0

linear dependence guarantees at least one term will remain

[haughty berry]

Right, but that doesn’t mean you’re not throwing out v, right?

[dusky epoch]

yeah it does. if you had thrown out v you would have ended up with a linear combination of vectors from S only

thereby contradicting its LIness

[haughty berry]

Yeah, but I don’t understand why the proof didn’t add that part

The wording of the proof and the proof itself seem kind of confusing

[lavish jewel]

it should be clear enough from the notation though, since the u_i are taken from S union {v}

if you prefer, you could've written the linear combination as a weighted sum of vectors s_i in S plus a scaled version of v and gotten the same result, it's just a substitution. the way it is worded, the order of the u_i is not important either

[haughty berry]

I don’t really understand your explanation?
I understand the proof, I just don’t think that it’s reasoned well because it’s missing a large chunk of the proof. Sure you can say that it’s clear, but the whole point of a proof is to prove something mathematically, not intuitively

[lavish jewel]

it's proven well there

you can do it directly by taking the definition of lin (in)dep. let $s_i \in S$, where $S$ is a lin indep subset of v.s. $V$. then $$\sum_i a_i s_i \neq 0$$ unless all off the $a_i$ are 0. if $S \cup {v}$ is lin dep, then $$\sum_i a_i s_i + w v = 0$$ for some some nonzero scalars $a_i$ and $w$. we rewrite this as $$\sum_i a_i s_i = -wv$$ so that $v$ is a linear combination of the $s_i$.

[stoic python]

19eddy4

[lavish jewel]

it's the same thing except what was shown in the image makes a substitution u_i \in S \cup {v} and exploits commutativity of addition (fair enough for a vector space over a field)

idk if that makes it clearer for you

[haughty berry]

Again, I understand the statement and its proof.
But you’re adding explanations as to why the scalar of v isn’t 0, which isn’t present in the proof

[lavish jewel]

it's part of the definition of S being linearly independent

there is no need to state that

if the scalar for v is 0, so are all the others

[nocturne jewel]

Self-teaching quadratic forms:

Suppose I have 2 quadratic forms $f(x,y)=x^TAx$ and $g(x,y)=x^TBx$. Is there a way to find the solution set w/ just the matrices?

[stoic python]

Mosh

[nocturne jewel]

(w/ x being the vector [x,y] in the form)

[zinc timber]

by solution do you mean x'Ax=c?

[nocturne jewel]

all vectors x st f=g

[zinc timber]

x'(A-B)x=0

[nocturne jewel]

oh yeah

ty

[zinc timber]

now depending on the rank you get yr sol

[nocturne jewel]

Ye but I can solve x^T(A-B)x=0 w/ multivar at that point

since it'll be the level curve of that quad form

[wintry steppe]

If I have some plane = ax0 + by0 + cz0 + d = 0 and a point parallel to this plane, call it P(x1, y1, z1). How do I find an orthogonal vector from the plane to point P ? where it will work for n dimensions. In other words, I need to find a point on that plane that I can use as my tip, to go to the tail, which would be P, but it's going to be orthogonal to the plane.

[zinc timber]

since it's another quad form, you need to find the zeros. One option might be to EVD is

the col corresponding to 0 ev's will be the solution

[nocturne jewel]

Ill look into evd, only know svd from 1st year linal

[zinc timber]

i..e E_0

Eigen value decomposition

[nocturne jewel]

Yeah figured that was the acronym lol

[zinc timber]

yeah, but if you know svd, u must also know evd

[nocturne jewel]

scale the normal vector

[zinc timber]

because Quad forms are symmetric

[nocturne jewel]

Oh probably, I just dont think we ever called it that

[zinc timber]

yeah, idk what it's called either, I just call it evd

[quartz compass]

you could say diagonalization

[zinc timber]

evd feels fine honestly

[nocturne jewel]

Polar Decomp maybe?

[zinc timber]

nah polar is different

[quartz compass]

in this context you're literally diagonalizing the quadratic form, the eigenvectors making up the orthogonal matrix change of basis

[nocturne jewel]

My LinAl course never really went too much into decomps, just QR, Polar and SVD iirc

[nocturne jewel]

Oh that process is EVD?

[quartz compass]

sure

[fickle citrus]

@wide mortar Ask your things here.

And please don't ping me when I'm not around

[night wren]

does dim(V)=dim(W) iff V and W are isomorphic hold for infinite vector spaces?

[zinc timber]

yeah

though I think the spaces have to be banach

[slow scroll]

ye, and they don't have to be banach spaces. This is true from a purely algebraic perspective. a bijection between their bases induces a isomorphism between the whole spaces, and conversely an isomorphism restricts to a bijection of bases

[novel marsh]

Before going into linear algebra, what concepts should I have a good grasp on?

[zinc timber]

matrices, basic algebra is sufficient in my opinion

[novel marsh]

Thank you

[gray dust]

@novel marsh the following are not prerequisites but smooth the transition: systems of linear equations, matrix arithmetic, complex arithmetic, geometry intuition. what you DO need is mastery of high school algebra & some experience in proof reading/writing since most of LA is proof based

[zinc timber]

that's an interesting problem

[zinc timber]

though I don't see the solution

@quasi vale

[quasi vale]

we apply v on both sides of A^6 + .... I = 0, we get v + .... v = 0 which means v(the eigenvector) is 0 which is a contradiction

[zinc timber]

hmm but A^7-I is not the char polynomial

it's not specified,

only said that A^7-I = 0

$x^6+x^5+\cdots+1$ splits into factors. if the minimal polynomial is any one of these factor/prod of the factors then $m(x) | x^7-1$ i.e. $A^7-I = 0$

[quasi vale]

its pretty late but meant to say 'odd' dimensional real vector space

[zinc timber]

yeah I can understand that, that's not where my problem is

even I'm stuck, I feel like option B should be true

but I'm really not convinced

the problem with your argument is that given $x^6+\cdots + 1= 0$ means even $(x-k)(x^6+x^5+\cdots+1) = 0$ so you don't get that 1 is an eigen valuee

[zinc timber]

[wide mortar]

hi guyss please i need some help i didn't understand where that particular and general solution come from ?

i mean i understand he may get the particular solution form the pivot but how he get the general solution(2,47)

[quasi vale]

@wide mortar Since x2 and x5 are free variables, they let x2 = lambda1, x5 = lambda2

Now solve the equations in 2.45 backwards, you get x4 = 1 + 2x5 = 1 + 2(lambda2)

Now using this x4 and x5, find x3

and then using x3,x4,x5, find x2

and so on

If u don't understand, I can write it out for you

in full detail

ok 1 sec

[wide mortar]

thank you so much ❤️

[quasi vale]

u did it?

[zinc timber]

solved

[wide mortar]

i m trying to , since x2 and x5 are free variables so x2= lamda1 , x5= lamda 2
hmm okay than if i substitue x2 and x5 with their value in third equation for exple i got x4= 2x5+1 ( 2lamda2 + 1 )
hmm great x3 than = -2+ 2(lamda2 + 1 ) -3*lamda2

[quasi vale]

x_5 = lambda_2

x_4 = 1 + 2lambda_2

x_3 = -2 + 1 + 2lambda_2 -3lambda_2 = -1 - lambda_2

[wide mortar]

hmmm okayyy i understand you

then how he convert it in this form

[quasi vale]

Okay so..

In the end, u should be able to get x_5 = lambda_2, x_4 = 1 + 2lambda_2, x_3 = -1 - lambda_2, x_2 = lambda_1, x_1 = 2 + 2lambda_1 + 2lambda_2

Now you can write x_5 = lambda_2 as 0 + 0lambda_1 + (1)lambda_2

x_4 = 1 + 0lambda_1 + (2)lambda_2

x_3 = -1 + 0lambda_1 + (-1)lambda_2

x_2 = 0 + (1)lambda_1 + 0lambda_2

x_1 = 2 + 2lambda_1 + 2lambda_2

so the first column(containing no lambdas) becomes [2 0 -1 1 0](starting from x_1, ending with x_5)

second column(containing lambda_1) becomes lambda_1 [2 1 0 0 0] (factor out lambda_1)

third column(containing lambda_2) becomes lambda_2 [2 0 -1 2 1] (factor out lambda_2)

@wide mortar Do u get it?

[wide mortar]

well i understand this part 100% anyway x_5= 0lamda_1 + 1lamda_2

so x5 will be = 0 * lamda_1 + 1 * lamda2 and x2 = 1 * lamda_1 + 0 * lamda_2

??

[quasi vale]

Not sure what u mean

x_2 = (1)lambda_1 + (0)lambda_2, yes, x_5 = (0)lambda_1 + (1)lambda_2, yes

But if u look at x_1, x_3, x_4 they have a constant part of 2,-1,1

so that's why we add a 0 to x_2, x_5

so x_2 = 0 + (1)lambda_1 + (0)lambda_2 and x_5 = 0 + (0)lambda_1 + (1)lambda_2

[wide mortar]

aaaaaaaaaa okayyyy i thinkkkk i m tooo close

i will reread again your remarks ❤️

[wide mortar]

hmmmm great why you start each variable with the constant first x_4=1+.....
x_3=-1 +...
x2=0+...
x1= 2+...

[quasi vale]

just cause the answer given in the book starts with the constant

you can write the constant in the middle or in the end but it's better to write it in first

Have u seen the vector equation for a line in R^3?

it's something like r = (point) + k(direction vector)

so these constants make up the point

and the other columns make up the direction vector

[wide mortar]

actually i don't know what this type of demonstration are called (searching the particular solution and general solution ) since i v'seen it several time and i wanna explore more about this topic

[gray dust]

solving linear systems

[wide mortar]

@quasi vale thank you for you help i finally understand it ❤️ ❤️ ❤️

[quasi vale]

Np!

[wide mortar]

i appreacite a lot your effort thank you so muchhh ❤️

[quasi vale]

all good

[novel marsh]

Ah okay, do you need calculus at all?

[lavish jay]

,rotate

[lavish jay]

Is this right?

At first i turned the augmented matrix into equation, then I reorganize to get a pivot then cancel out X at the second equation