#linear-algebra

2 messages · Page 321 of 1

slender yarrow
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As long as it's not worth 97% of your grade of something

winter pond
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Ah. No not really. Might get something for method at least.

slender yarrow
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Alala cliqpedia

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J'aime pas trop son contenu perso, il explique pas de manière ouf et il fait des slides blindées : la combinaison des deux je trouve que ça embrouille à fond

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Mais bon l'idée c'est que I_(L) c'est un vecteur qui représente la direction dans laquelle tu prends ta dérivée directionnelle

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Le problème c'est que t'as plein de vecteurs qui représentent une même direction (si je me prends un vecteur x, kx pour n'importe quel k !=0 représente la même direction que x)

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Pour se fixer les idées, par convention on prend I_(L) unitaire (I.e. de longueur 1)

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Un vecteur de longueur 1 c'est un truc qui tape dans le cercle unité

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C'est pour ça que tu peux écrire I_(L) comme cos(a) Ix + sin(a) Iy pour un certain a

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@hard verge

wintry steppe
slender yarrow
slender yarrow
hard verge
slender yarrow
hard verge
hard verge
hard verge
slender yarrow
slender yarrow
slender yarrow
rough pilot
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Guys, if I have a polynomial, say P(f)= f + f^2, where f is a linear transformation V->V

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I'm not getting how to find P(P(f))

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Please help 😢

wintry steppe
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if you mean to say P is the polynomial P(x) = x + x^2, then P(P(f)) = P(f) + (P(f))^2 = ... (you can finish it from here)

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polynomials aren't defined in terms of how they act on linear transformations, they're defined as sums a_0 + a_1 x + ... + a_n x^n

dusky epoch
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over what ring

amber crow
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Anyone on?

dusky epoch
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@amber crow no, this server has been abandoned for years if not decades. (/sarcasm)

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if you have a linalg question to ask then just ask it

amber crow
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Determine the value of A and B if the solution of this linear system is (4,6). Ax + By =-20. Bx + Ay = 15

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dis confsing me

limber sierra
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the solution being (4, 6) means that the system is solved when (x, y) = (4, 6)

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i.e. when x = 4, y = 6

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so substituting this in, we have that A * 4 + B * 6 = -20, B * 4 + A * 6 = 15

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solve.

dusky epoch
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well it is vaguely linear-algebraic in that it deals with a linear system

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so there's that

amber crow
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its linear equations

limber sierra
amber crow
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srry bbg

dim epoch
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exam is over so idk if you care anymore but no, when looking at it as a Z Module which we always did

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its cyclic but the generator is never linearly independent

clever totem
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🥲

dim epoch
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dont worry too much about it

clever totem
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all in all i have a good feeling i think

dim epoch
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the Signatur Aufgabe was absolutely gottlos

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but other than that it was fine

clever totem
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yeah everyone i knew just took random numbers

dim epoch
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so did i

glacial terrace
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So, T is invertible, then we can use Cayley-Hamilton Theorem to get a polynomial p(x) in F[x] such that p(A) = 0, which indeed has a non-zero constant term.

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Question: How can I prove that this polynomial is indeed the minimal polynomial? I know that the minimal divides any polynomial that has A as a root

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Another way to think of: How do I know if the polynomial given by the Cayley Hamilton is minimal?

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If it is

haughty berry
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Cayley Hamilton just says that p(A) (where p(x) is the char polynomial of A) is 0

dim epoch
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since every kernel of a ring homomorphism is an ideal in the domain

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so to generate these ideals the degree has to be minimal

glacial terrace
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That was to much, I think?

dim epoch
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This was how my course introduced the minimal polynomial

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if you're not familiar with some of these terms just ignore what i typed

haughty berry
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You can show this because T is invertible if and only if its representation matrix is invertible, and a matrix is invertible if and only if its determinant is non-zero, and since the constant term in the characteristic polynomial is (-1)^n det(A)*I, this is iff the constant term in the characteristic polynomial is non-zero.
And you know the minimal polynomial divides the the characteristic polynomial…

And you can then manipulate p(T) to get some polynomial = I, and that polynomial will have a zero constant term, so you can simplify it to T(…)=I

glacial terrace
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I am acquainted but not familiar

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hummm

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but what about having T(...) = I? I did that part but the minimal polynomial should have 0 as it image

haughty berry
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I don’t understand your question

glacial terrace
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How does getting T(...) = I helps

haughty berry
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You have Tq(T)=I (and also q(T)T=I) so you have that q(T) is the inverse of T (q(T) is that …)

glacial terrace
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q(T) is indeed a polynomial

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I got $\frac{1}{det(A)(-1)^{-(n+1)}}T(T^{n-1}+T^{n-2}c_{n-1}+\dots+c_1) = I$

haughty berry
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It doesn’t really matter what the polynomial is, as long as it exists

stoic pythonBOT
haughty berry
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So you have the polynomial

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What more do you need?

glacial terrace
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I am sorry but I don't see how this answer the question

haughty berry
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Because T*q(T)=I and q(T)*T=I. What’s the definition of an inverse?

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Where q(T)=(-1)^(n+1)/det(A) (T^(n-1)+c_(n-1)T^(n-2)+…)

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Whatever you wrote up there

glacial terrace
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q(T) = T^-1

haughty berry
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Yes

glacial terrace
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I see this answer the second question but it has two!! I want the former xd

haughty berry
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I already gave the former

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In my first message

haughty berry
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Also showing that q(T)=T^-1 is sufficient anyway

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You don’t even need the first paragraph I sent

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You can totally ignore it actually

glacial terrace
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As you said, Cayley-Hamilton provides us the char polynomial which indeed has a non-zero constant term. How do we go from here to the minimal polynomial?

haughty berry
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You don’t need Cayley Hamilton

plucky lichen
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I'm looking through a solution to a particular problem here
https://math.stackexchange.com/a/1939462

And I'm trying to code an answer to this problem, but I've noticed in one of my test cases I am given a U of 0,0,0 which is the zero vector?
Part of my code indiscriminately converts this zero vector to a "unit vector" by which it tries to take 0,0,0/sqrt(0*0 + 0*0 + 0*0)
which then comes back as NaN basically...
And I'm really not sure how to deal with this...

In my specific test case, I'm basically positioned at 3,2,1 a laser is fired with vector 0,0,0 and the position of what I'm trying to hit is an object at 3,2,1 with a radius of 0.
Basically I am directly on top of an object with a radius of zero firing with a vector of 0,0,0 and I'm not hitting that object currently.

The test case seems to think I should be hitting this object... and mathematically I want to know why

More specifically I'm kind of interested in, what exactly is the zero vector? Does it have a magnitude? And mathematically how should this situation be playing out or what does it all actually mean?

haughty berry
# stoic python **mns**

You have this, just instead of dividing by the determinant times (-1)^n, you’re dividing by the constant term of the minimal polynomial

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Basically you have a minimal polynomial
[ m_T(x) = x^n + c_{n-1}x^{n-1} + \cdots + c_0 ]
You know that $m_T(T)=0$ so
[ T^n + c_{n-1}T^{n-1} + \cdots + c_1 T = -c_0 I ]
Then you get that
[ T\parens{-\frac1{c_0}\parens{T^{n-1} + c_{n-1} T^{n-2} + \cdots + c_1}} = I ]
So if you define the polynomial
[ p(x) = -\frac1{c_0}\parens{x^{n-1} + c_{n-1}x^{n-2}+\cdots+c_1} ]
You have that $T\cdot p(T)=I$ and $p(T)\cdot T=I$, so $p(T)=T^{-1}$

stoic pythonBOT
haughty berry
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So you have shown that T has an inverse and it is a polynomial

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You’ve answered both questions at once

glacial terrace
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oh

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Now I see how it relates with the minimal polynomial!!

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I was completely not using its definition

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Thank you very much for your time!

haughty berry
glacial terrace
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"Let V be a finite-dimensional vector space over a field F and Ω an algebraically closed field containing F ."

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So, for any polynomial p(x) in Ω, there's alpha in Ω such that p(alpha) = 0. I wonder what Ω containing F ( F subset Ω ) tell us

haughty berry
wet stratus
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well it's just convenient to be able to work over a field where every polynomial has a zero

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it should hopefully at some point become clear why your text does this

glacial terrace
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hum ok thanks

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just to make sure, T = [0 & 1 // -1 & 0] is semisimple. As F = R and Ω = C so that m_T(x) = (x - i)(x + i)

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for context: T is semisimple if mT(x) is a product of distinct linear factors over Ω[x];

zinc timber
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semisimple if the min poly is product of irreducible elements

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in F[x]

glacial terrace
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F or sigma?

zinc timber
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over F

wintry steppe
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algebraically closed fields are good for the soul

zinc timber
ivory shore
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the set of all polynomials of degree 3
does this include something crazy like x^8 - 3x^3, because technically this polynomial has a third degree term

dusky epoch
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no, x^8 - 3x^3 is a polynomial of degree 8, not 3.

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the degree of a polynomial is the highest degree of all its terms.

mental atlas
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guys, if If A is m×n matrix and n>m
how can i prove that A^T*A is a singular matrix.

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i proved that A is a singular matrix

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because its rref cannot be an Identity(equivalence thrm)

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but how to prove the A transpose times A

tranquil steeple
stoic pythonBOT
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Sven-Erik

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split parcel
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,w Bunny is considering joining a tennis club. The downtown club charges a $200 membership fee and $25 per hour to play. The uptown club charges a $300 membership fee and $20 per hour to play

stoic pythonBOT
quiet salmon
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hey, quick and stupid question:

Determine the dimension of the subspace $W \subseteq V = \bR^{4}$, $W = {(a, b, c, d): d = 0}$

stoic pythonBOT
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texaspb :rainbow:

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quiet salmon
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the answer for this is dimW=3, I don't get why

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a vector from my subspace W has 4 components

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why is the dim not 4?

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I mean intuitively it's clear that it's because of my 4th component being 0, but why does this imply that the dimension is 3

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in this case

gray dust
quiet salmon
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ok

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so

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hm

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ok I got it

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you cant find a basis of 4 vectors of 4 components right

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is that what youre trying to say

gray dust
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im just saying to find a basis

quiet salmon
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ok my first attempt would be to try this "base" B={(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}

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which doesnt work for W

gray dust
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ofc this requires knowing what a basis is

quiet salmon
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yeah

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I tried using the standard basis for R^4 but it cant span W

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so that was my first guess

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ok I dont get what you're trying to say

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I know what a basis is

gray dust
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B is a basis of W if B is linearly independent and spanB=W

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no wait, wording is off

quiet salmon
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I know what a basis is

gray dust
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B={(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}
here, spanB doesnt equal W

quiet salmon
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yes

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that's why I said it was my first attempt and then realized it wouldn't work

gray dust
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i suggest a more systematic way of finding a basis instead of guessing

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write an arbitrary vector in W as a linear combo of some independent vectors

quiet salmon
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ok

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that works

gray dust
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those independent vectors form a basis, dimW is its size

slim gyro
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hey friends, im having trouble grokking this paragraph in my book that comes after a proof. the statement being proven is that if you have a vector space V with basis B, then given any vector space W and any function from B to W, you can extend that function to a linear transformation from V to W

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g bar is the linear transformation from V to W and g is the function from B to W

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what im confused about is that after the proof he says that being able to extend g to a linear transformation depends on the fact that the basis vectors are linearly independent, and then he gives an example saying that a function g from the set {v, -v} to W can't be extended unless g(-v) = -g(v)

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im just confused as to why we can't always turn g into a linear transformation when it's defined on linearly dependent elements

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like in the example, i don't get why having g(-v) = -g(v) is required

wet stratus
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Well if g is linear then g(cv)=cg(v) for all scalars c. Using c=-1 we get g(-v)=-g(v)

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So this absolutely has to hold, otherwise g cannot be linear

dry vessel
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Hi I need help with this

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Could anyone please help me understand why
w_0 + 4w_1 - 2w_2 = 0
turns into
w_0 + 4w_1 = 0
and w_0 - 2w_2 = 0

gray dust
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context?

dry vessel
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So I have a linear graph of the equation

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w0 + w1x + w2y = 0

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i have values for x and y

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and am to solve for w0 w1 and w2

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I want to know what math rule is used to make

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w_0 + 4w_1 - 2w_2 = 0
into
w_0 + 4w_1 = 0
and w_0 - 2w_2 = 0

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do I need to give more context

gray dust
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yes can u post the full problem?

dry vessel
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let me post it here first then you can confirm

gray dust
dry vessel
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oh sorry I thought it was implied

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does that change the understanding?

gray dust
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these are assumptions on specific points on the line, not implied by the equation itself

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take the point (4,0), ie the equation holds for x=4, y=0

dry vessel
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Hmm how is it an assumption?

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its stated to us

gray dust
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assumption in that its an extra bit of info given by the problem

dry vessel
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hm ok

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so then why is the forumla split?

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when we substitute in the w012 equation

gray dust
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one known point on the line is (4,0), ie the equation holds for x=4, y=0

dry vessel
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out of interest this is the solution im showing you, I just havent a clue how we get there

gray dust
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so we can plug x=4, y=0 into the equation

dry vessel
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Ok

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and it gives y = -2

gray dust
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no, we plug x=4, y=0

dry vessel
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stop

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hold up

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that would mean

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that there are 2 insertions

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x = 4, y = 0

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and x = 0, y = -2

gray dust
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we're just focusing on the fact (4,0) is a point on the line for now

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we focus on the other point later

dry vessel
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you just helped me understand that the points are treated seperatly

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thats why we get
w_0 + 4w_1 = 0
and w_0 - 2w_2 = 0

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Thank you 😭

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But ok yeah lets continue with the one point

gray dust
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one known point on the line is (4,0), ie the equation holds for x=4, y=0
so we can plug x=4, y=0 into the equation
we get w0+4w1=0

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is this clear?

dry vessel
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so far yes because (0)w2 is 0

gray dust
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now (0,-2) is another point on the line, ie the equation holds for x=0, y=-2, so lets plug those in

dry vessel
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ok

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that gives w0 - 2w2 = 0

gray dust
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yes

dry vessel
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that means we have 2 equations :D

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awesome!

gray dust
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thats how we use two known points on the line to get these two equations from the original

dry vessel
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i see

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thank you man!

gray dust
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no prob

real bone
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whats the square root of an operator? our instructor never defined it for us but apparently we need to know what it is to solve a homework problem

wintry steppe
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the square root of an operator T is an operator S with S^2 = T

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what else could it possibly be?

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@real bone

real bone
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ok thats what i assumed thx

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i just wanted to make sure just in case i was wrong about it

plucky lichen
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I have a collision checking algorithm where I take a start point, a firing vector, and then I have a sphere at a location with a radius.
https://math.stackexchange.com/a/1939462

I'm working off of the example posted here...
In my current particular test case my
C = <-2,0,0>
r = 1
P = <0,0,0>
U = <1,0,0>

The math works out as follows

Q = <0, 0, 0> - <-2, 0, 0> = <2, 0, 0>
a = <1, 0, 0> * <1, 0, 0> = 1*1 + 0*0 + 0*0 = 1
b = 2 * <1, 0, 0> * Q = <2, 0, 0> * <2, 0, 0> = 2*2 + 0*0 + 0*0 = 4
c = <2,0,0>*<2,0,0> - 1*1 = 2*2+0*0+0*0 - 1 = 4 - 1 = 3
d = 4*4 - 4 * 1 * 3 = 16 - 12 = 4 > 0

Where d is the discriminant of my quadratic. I've noticed that in this particular case d is 4 which is greater than zero, and I'm trying to figure out why and I realise like.... indeed my sphere is on this line... but it's behind my starting point.... I'm basically collision checking a line not a ray, and in this particular collision I have hit an object behind me...
Is there any way to actually check for and rule out this case?

I'm currently kind of thinking maybe if I can get a unit vector from my start to my sphere location, and if it's opposite my given firing vector I know that I can't hit that object? Is there a fancy linear algebra way to do that?

Mathematics Stack Exchange
Calculate if vector intersects sphere

I have a 3D world and I have a gun who's position is defined by X,Y,Z which fires a bullet in a straight line with a direction defined by X, Y, Z towards a target who position is defined by X,Y,Z w...

ornate canopy
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can i get some help with this?

Desktop_Screenshot_2022.07.21_-_17.50.24.63_2.png
plucky lichen
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I'm currently kind of thinking maybe if I can get a unit vector from my start to my sphere location, and if it's opposite my given firing vector I know that I can't hit that object? Is there a fancy linear algebra way to do that?

It seems weird to me that this is necessary... when I feel like I'm already doing so much with these vectors... How can I already tell if my object lies on my line without knowing somewhere already that it lies on my line "backwards"

copper hinge
copper hinge
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Let p(t) = p_0 + td be a your line. If your sphere is determined by a center c and radius r we have all points x of the sphere satisfies |x-c| = r. Now plug in x = p(t). Square both side, left side turns into a dot product, expand. Get a quadratic polynomial, if the discriminant is negative the line doesn’t intersect the sphere. If not then either the line intersects the sphere once(one solution), or twice(two solution).

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If that what your trying to do.

worldly axle
wintry steppe
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Hi, I have a question on free variables for Ax=0 and Ax=b. In Gilbert Strang's video on Ax=b, the free variables are assigned 0, but for the null space in Ax=0, he uses free variables [0,1] and [1,0].

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why is it that for Ax=b, the x particular's free variable doesn't use the same set [0,1] and [1,0]?

wintry glen
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Hi, sorry for the interruption, but I need to show that for subspaces $V, W$ of $U$, $(V + W)/W \cong V/(V \cap W)$. I have access to the first isomorphism theorem that $V/$Ker$(T) \cong$ R$(T)$, so I could just find $T$ whose kernel and range are $V \cap W$ and $(V + W)/W$. Not sure how exactly to do that though.

stoic pythonBOT
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PhenomPlasma

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slender relic
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Could you try a map T:V+W -> V/VnW, where T(v+w)=v + VnW

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Have to show it’s well defined first, and then then kernel is W and it’s surjective hopefully

wintry steppe
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sheeesh high level stuff

plucky lichen
tulip cedar