of all the nested surfaces inside it, if you think of it that way
#linear-algebra
quartz compass
quartz compass
yeah, like x^2 + y^2 + z^2 = 2 is contained inside it
(0,0,0) works, to give a specific point as well
it might help to think about x^2+y^2<=3 first if the 3D case is not as comfortable to you, but that's all it means
wintry steppe
When I was able to graph the sphere it clicked
quartz compass
ok quick question, what's the radius of the sphere? 😛
wintry steppe
well 3 is r^2, so it'd be sqrt(3) I'd say ?
quartz compass
perfect 👌
gloomy arrow
3 because thats the number of rows?
dusky epoch
no
gloomy arrow
oh 1
Because there is only multiplicity 1
Looking at my notes
Where as this one its 3 because the eigenvalue is 3 with multiplicity of 3
gloomy arrow
Ah so I would need to calculate and see how many eigenvectors
There were 2
So it would be 2
severe magnet
If I restrict the domain of a diagonalizable linear map F:V -> V in a way that it is still an endomorphism, and I know that $\text{dim Eig}(F, \lambda) = \mu(P_F, \lambda)$, shouldnt the amount of zero's of the eigenvalues (denoted by $\mu(P_F, \lambda)$) be the same for the eigenvalues of the restricted linear map?
stoic pythonBOT
Nabil:
severe magnet
if the eigenvalues under the restricted map are a subset of the eigenvalues of F ofc
eager burrow
The amount of zeroes? You mean the multiplicity of the eigenvalue zero? Because no, that might change. E.g. restrict the 2x2 zero matrix to a one-dimensional subspace, suddenly you only have one zero left.
severe magnet
hm maybe I should clarify what the mu means
I mean the multiplicity of any eigenvalue of F restricted
not necessarely 0
was my bad since idk the terms in english
severe magnet
ah perfekt
eager burrow
aber die vielfachheit des eigenwerts kann sich trotzdem verringern
da geht mein beispiel trotzdem noch
severe magnet
naja ums konkreter zu machen: Ich hab zwei Abbildungen $F,G: V \to V$ die diagonalisierbar und die Eigenschaft $F \circ G = G \circ F$ haben. Ich habe schon gezeigt dass $F(\text{Eig}(G,\mu_j)) \subseteq \text{Eig}(G,\mu_j)$ deshalb hab ich $F$ auf $\text{Eig}(G,\mu_j)$ eingeschränkt (da es trotz allem ein endomorphismus bleibt) und wollte zeigen dass die auf diese bestimmte menge eingeschränkte lineare abbildung weiterhin diagonalisierbar ist
stoic pythonBOT
Nabil:
eager burrow
hm, ich glaube auf der einschränkung solltest du immer noch eine basis von eigenvektoren haben
severe magnet
ja da der Eigenraum von G ja selbst ein Unterraum von V ist
das war mein Gedanke
zumindest
eager burrow
hmmm da muss man wirklich ein bisschen denken irgendwie 😄
ich denk mal ein bisschen
eager burrow
okay, ich war zu dumm und musste googlen, irgendwie ist das ein bisschen nichtstandard, aber der beweis ganz unten ist ganz nett: https://math.stackexchange.com/questions/62338/diagonalizable-transformation-restricted-to-an-invariant-subspace-is-diagonaliza
placid oracle
Can someone explain adjacency matrices to me? Can anyone explain adjacency matrices? heres a picture for example, and i think i set it up right but i just want to be sure:
Like for example, I got the adjacency matrix for picture a to be:
[0 1 0 1]
[1 0 1 1]
[0 1 0 1]
[1 1 1 0]
this is becauyse vertex 1 is connected to 2 and 4, v2 is connected to 1 3 and 4, v3 is connected to 2 and 4, and v4 is connected to 1 2 3
I think that makes sense, but now im asked to find which picture can be drawn without lifting a pencil or drawing on top of an already drawn line. Its obviously a.), but im not sure how to show that using the idea of adjacency matrices
subtle walrus
are you sure that the questions are supposed to be related to adjacency matrices
wintry steppe
@placid oracle is that the right diagram for adjacency matrices?
quartz compass
if it was a closed loop I could give an answer
in that case you can nearly raise the adjacency matrix to the number of edges it has and take the trace to see how many paths there are, but this doesn't quite work always
you could also have cycles that divide the number of edges so you have to remove them by inclusion/exclusion
and this is immediately given by mobius inversion
like for instance, let's say it has 6 edges, like c.) it has two loops of 3 so raising it to the 6th power would give a false positive since there are paths that end where they started in 6 steps
that's what the mobius function would account for and remove
$\Tr(A^n)=\sum_{d|n} c(d)$
stoic pythonBOT
Merosity:
quartz compass
$c(n)=\sum_{d|n}\mu(n/d)\Tr(A^d)$
stoic pythonBOT
Merosity:
quartz compass
too bad this doesn't answer your question
wintry steppe
I'm trying to add two vectors and find the magnitude of the resultant vector. I already know how to do this and I've been using the cosine rule. However, this calculator that's specifically designed for this returns a different result than the standard formula and I've been trying to figure out why.
The vectors are: A = {108°, 63}, B = {190°, 17}.
From cosine law I got 62.92 for the magnitude of the resultant vector, but the calculator says it's 58.34.
The calculator's a flight computer called CX3 and this is actually a wind/course question.
Does anyone know how 58.34 was derived?
plain fjord
Uh, nasty, degrees.... not radians
wintry steppe
HELP
I've been trying to figure this out for the past 9 hours
Is there like a numerical discrepancy between sine law and cosine law?
plain fjord
Wait, what is the actual problem?
limber sierra
when you say 108 degrees and 190 degrees, are you measuring from the same axis
i'd assume "yes" but just making sure
because im getting neither 62.92 nor 58.34
plain fjord
Where do you even use vectors in polar coordinates with degrees like that?
plain fjord
Instead of radians I mean
plain fjord
Okay
limber sierra
no one measures navigational angles in radians
plain fjord
@wintry steppe can you clarify what you mean by cosine law
limber sierra
i mean... presumably they just mean the generalization of the pythagorean theorem
i'm not sure why it applies here though
oh, i think i see how you got 62.92
vector addition is head-to-tail, but you assumed it was head-to-head (or tail-to-tail) when calculating the angle
in other words, your angle is off by 180 degrees
im still not sure where the computer's 58.34 comes from though
aside: because of this 180 degrees "error" when we try to apply the cosine law, and because cos(180 - x) = -cos(x), we can actually get a formula for magnitude of vector addition based off the cosine law
$\abs{b-a} = \sqrt{a^2+b^2+2ab\cos \theta}$ where $\theta$ is the angle the two vectors enclose
stoic pythonBOT
Namington:
limber sierra
note the + instead of -
again, though, i have no clue where 58.34 comes from
maybe you inputted it wrong, or with funky settings
(radians? one vector in metres, another in feet?)
(angles measured from a different axis?)
(etc)
if this picture helps clarify where your confusion is
wintry steppe
@limber sierra yes, they're all measured from the true north pole.
@limber sierra I don't understand what you mean by 180 deg?
it's assumed that the wind is coming from 190 deg.
sorry, should've clarified that.
limber sierra
yes, and my point is that your calculation of the angle of the two vectors is off by 180 degrees
wintry steppe
how do I fix it?
limber sierra
add 180 degrees
again though, as i said
this doesnt give the computer's answer
i have no clue where the computer's answer is coming from
which makes me suspicious that something fucky is going on with the settings
like accidentally in radians, or unit mismatch (one in km/h, one in mph?), or whatever
or it's accounting for factors like air resistance?
limber sierra
uh
when you calculate the angle for the cosine rule
your cosine rule calculation should be
one sec
wintry steppe
This is what i did
limber sierra
$\sqrt{63^2 + 17^2 - 2(63)(17)\cos(190^{\circ} - 108^{\circ} + 180^{\circ})}$
wintry steppe
wait the arrow on 190 deg vector is wrong. it should point the other way
stoic pythonBOT
Namington:
wintry steppe
why am i supposed to add 180?
limber sierra
draw the picture and you'll see?
wintry steppe
I don't see anything
limber sierra
one sec
wintry steppe
what's wrong with the way i did?
aside from 190deg vector pointing in the wrong way
stoic pythonBOT
Couldn't find an attached image in the last 10 messages
limber sierra
wintry steppe
, rotate
stoic pythonBOT
stoic pythonBOT
limber sierra
er
wintry steppe
1 sec
limber sierra
this is better
note that we're measuring the angle of the yellow vector from true north
which means the angle needs to go in the same "direction"
also uh, i did it counterclockwise but it seems you did it clockwise
whatever
wintry steppe
and the white arrow is true north?
wintry steppe
your 108 deg is like 180+ deg
wintry steppe
yeah yours and mine are fundamentally the same.
limber sierra
ok let me try that again
wintry steppe
notice in my diagram im not using the true north as a reference but Vector A?
plan was to use the parallelogram technique to solve everything, but apparently the answers from CX3 is completely different from mine.
limber sierra
so the ylelow angle here
is what you found
but you'll note that you cant actually apply the cosine law to that
er
FUCKING
limber sierra
bleh
also tha tshould be
190-108
but you know what i mean
anyway, yeah, we cant apply the cosine law here to find the measure of a+b
not directly, at least
i really cant think today
wintry steppe
let me think
wintry steppe
so u wanna calculate the red angle?
holup
@limber sierra you're right about taking the reciprocal of the wind vector since wind is coming from 190, thus going to 010.
limber sierra
and the cosine law requires you to know the orange angle
but if you note
the green-blue here forms a parallelogram
(or at least it would if my drawing skills were better)
hence given the red angle
we can find teh value fo the orange angle
wintry steppe
isn't the angle between the yellow and blue line at the top equal to the angle created by the yellow line and the blue line at the bottom?
limber sierra
by subtracting it from 180 degrees
sure, i dont see why that matters @wintry steppe
limber sierra
ugh i dont wanna draw the diagrams again you should be able to figure it out
wintry steppe
I love you guys btw
limber sierra
the point ist hat you're looking at the wrong angle
wintry steppe
Thanks so much
limber sierra
you calculated the red angle in my diagram but should be looking at the orange one
limber sierra
(except adjusted to now have a 10 degree ray instead of 190)
wintry steppe
i love you more
limber sierra
fortunately you can GET the orange angle easily from the red one
by just subtracting it from 180
since its a parallelogram
wintry steppe
I meant i used the wrong angle
limber sierra
[i said "add 180" but]
[when we take the cosine this is equivalent]
[cos(x+180) = cos(x-180)]
wintry steppe
I subbed 180 from the answer and got 67 still not the same as cx3 but we'll see
I think i cans olve this
Let me try it on my own
can i try you on my own
i wanna piece of that coconut
anyways if u need help just tell us
lmao ok thanks
😍
nope im back to sqaure one.
ill show u why
forever be doomed lol
if we assume that the straight lines are parallel
then the angle 190 - 108 is same as the interior angle near hdg
even if i were to do 108 - 10 and find the reciprocal of that (so 180 - ans), I still get the same angle.
so, yes, technically you can apply cosine law to find the resultant vector in this case.
but i have no idea why my answer is wrong
jesus christ
what are you even supposed to find? i still don't know 🤣
see the thicc blue arrow that cuts across the middle of the box?
im supposed to find the length of that.
oh
well u have all 3 angles
so
cold topaz
is there a way to get the inverse of a 3x3 matrix without doing the whole standard matrix rref thing?
steady fiber
use adjugate
wintry steppe
^
placid oracle
@subtle walrus @wintry steppe super late answer, but yes
cold topaz
So I want to prove that A is invertible if and only if P^-1AP is invertible.Where do I begin?
limber sierra
suppose A is invertible
then $A^{-1}$ exists so you can consider the matrix $P^{-1}A^{-1}P$
stoic pythonBOT
Namington:
limber sierra
exploit matrix multiplication associativity to show that this is an inverse of PAP^{-1}
that proves one direction
can you see how i came up with that matrix?
i just did the naive "brute force" thing of trying to force the product to simplify to the identity matrix
anyway, once you've fleshed that out, try proving the other direction
cold topaz
i have this so far:
Let P^-1A^-1P exist. => P^-1 A^-1 P = (PA)^-1P = A^-1```
Then I'm stuck.
half ice
I'll use ' for inverse.
The inverse of P'AP is P'A'P, which you can prove by multiplying them together.
Note that this inverse is unique, and needs A to have an inverse, proving one direction.
odd kite
can't you also use determinant for this
limber sierra
P^-1 A^-1 P = (PA)^-1P
this is not true
just FYI
the argument isnt correct either way but
i wanted to correct this assumpton specifically
$(AB)^{-1} = B^{-1}A^{-1}$ which is not always equal to $A^{-1}B^{-1}$
stoic pythonBOT
Namington:
wintry steppe
ann laying down fax
pallid rampart
cold topaz
can we say λIx = Ax instead of λx = Ax?
pallid rampart
Well they're equivalent
wintry steppe
the latter is the proper way i think
pallid rampart
The latter is the more natural way
But we do use the idea in the first equation
Since
dusky epoch
these are the same thing
cold topaz
i like to keep the I, because it is present when A and λ are on the same side of the equation. It makes better sense for me to have it.
i just didn't want to make a rookie mistake.
pallid rampart
λx = Ax => λIx = Ax => λIx - Ax = 0 => (λI - A)x = 0 for some nonzero x => det(λI - A)=0
pallid rampart
Sure
wintry steppe
How do you convert a standard lpp back to canonical form
dusky epoch
lpp?
linear programming problem?
what are your definitions of standard form and canonical form?
wintry steppe
yeah
cold topaz
most of this list can be proven by cancelling the P and P^-1. No?
stoic pythonBOT
Ann:
dusky epoch
because they aren't
wintry steppe
This is my textbooks definition for canonical form
wintry steppe
for standard form my definition is
dusky epoch
alright
ok
well
the conversion from standard to canonical is done via slack variables
wintry steppe
I guess you would just add a slack variable
dusky epoch
you want your slack vars to be positive
well
≥0 rather
the left side falls short of the right side
so $a_{i1}x_1 + a_{i2}x_2 + \cdots + a_{in}x_n + s_i = b_i$
stoic pythonBOT
Ann:
wintry steppe
I was a little confused on that part but the part I was even more confused on was if you just let the slack variables be slack
i.e. not sure what to do to the slack variables after converting
dusky epoch
wdym
they just get added to your set of variables
their coefficients in the objective function are 0
wintry steppe
ah so then for example
if it was like z = x1 subject to x1 <= 500000
then canonical would just be z = x1 x1 <= 500000, x1,s1,>=0?
dusky epoch
z = x1 subject to x1 + s1 = 500000, x1, s1 ≥ 0
wintry steppe
I thought the x1 + s1 constraint would always be >= 0, otherwise how would you do more than one contraint?
wintry steppe
our teacher gave us a solution to a problem and I'm confused by it
problem is
this is the first part of the solution I'm confused on
what is cT B
wintry steppe
where do the components of C come from
wintry steppe
<@&286206848099549185>
tired flint
for (b) is it better to use an arbitrary p = ax^3 + bx^2..... or p = ax^4 + bx^2..... or does it not matter?
eager burrow
Since your operator is restricted to polynomial of third degree, only polynomials with highest term x^3 are all you need to work with
gray dust
polynomial of third degree
degree less than or equal to 3
@tired flint you let p be an arbitrary vector in the domain of T which is P_3 not P_4, so the leading term of p(x) is definitely of the form ax^3
smoky lagoon
can you conclude a matrix has no free variables if the non trivial solution only has one possible solution. i.e the entire right side being 2 2 2 2
tired flint
thank you
hollow finch
@smoky lagoon Do you mean Ax=b has only one solution and b is not the zero vector?
idle echo
I have a 4x3 matrix representing my basis for a vector space
I have to find an approximation for a new vector, b
I don't know that I understand the question. Can I do matrix multiplication?
wintry steppe
How can I find the angle between vector a and b?
I know that i, j, and k are unit vectors, apart from that, I'm pretty lost.
We have a formula for angle in R^2 but I can't find the one for R^3.
i = <1, 0, 0>
j = <0, 1, 0>
k = <0, 0 ,1>
Wait so I just gotta ditch the letters and use the same formula as for R^2 🤔
cursive narwhal
For any Euclidean Vector Space, we have:
$$\cos(\alpha(x,y)) = \frac{\langle x,y \rangle}{|x| \cdot |y|}$$
where $x,y$ are vectors in that vector space. In this case, $\bR^3$ is a Euclidean Vector Space and you can give it a standard inner product. Basically, the formula that brzig stated above applies here.
stoic pythonBOT
Abhijeet Vats:
cursive narwhal
@wintry steppe
hollow finch
@cursive narwhal I might be wrong, but I think that definition is carried over to any real inner product space, Euclidean or not.
ocean sequoia
For another example of a linearly independent list, fix a non-negative integer m. Then (1,z,...,z^m) is linearly independent in P(F). To verify this, suppose that a0, a1,...,am, belonging to F are such that: a0 + a1z + ... + am*z^m = 0, for every z belonging in F.
If at least one of the coefficients a0, a1,...,am were nonzero, then the above equation could be satisfied by at most m distinct values of z; this contradiction shows that all the coefficients in the above equation equal 0. Hence (1,z,...,zm) is linearly independent, as claimed."
Im a bit confused as to what the contradiction is
hollow finch
I think I see it but someone should check this logic:
What you really have with that polynomialish thing is a homogeneous system of equations.
If (a0,...,am) is a solution to the homogeneous system, then it must be a solution no matter what z is
This is of course satisfied if all of them are zero because then you have the zero polynomial
If they arent all zero, though, then you have a regular polynomial of degree less than or equal to m.
For how many values of z could that polynomial then be zero? @ocean sequoia
For example, say a solution ended up giving you a cubic polynomial
a0+a1z+a2z^2+a3z^3=0
How many solutions could there be?
ocean sequoia
wait no an infinite number lol
hollow finch
Up to 3 if any of the coefficients are nonzero, but yes infinitely many if they arent
ocean sequoia
which in this case would be m
hollow finch
So if there are only 3 values of z such that that polynomial is zero, then that polynomial is not always zero which goes against that equality
Or up to m, yeah
ocean sequoia
thats alot for me im have to sit on this for a bit i think im starting to get it tho
thanks so much!
hollow finch
np. yeah its a bit abstract
balmy shuttle
Hi! i have a doubt about how can I get the solution of a vector.
I have the vector (2,3,1) and I have 2 of the 3 components (3,3,9) and (2,1,0) and I have to find the third one that it's (x,y,z)
How can I start that problem? 🙂
limber sierra
uh, im not sure what you're asking
do you mean you're soving $\begin{pmatrix}2\3\1\end{pmatrix} = \begin{pmatrix}3\3\9\end{pmatrix}+\begin{pmatrix}2\1\0\end{pmatrix} + \begin{pmatrix}x\y\z\end{pmatrix}$?
stoic pythonBOT
Namington:
balmy shuttle
The first one is the vector, and the other 3 are the components of the vector (v1=v11,v12,v13)
I study in catalan, so it's a bit difficult to explain it in english sorry 😒
That's the exercice I am doing 🙂
wintry steppe
slack variables get added to the maximization function right?
tight glade
half ice
Have you tried a few? Haha.
What about with [0,1]? [1,0]?
Oh wait, you almost have it. The negative is the only thing that's off
half ice
Have you actually tried those examples?
wintry steppe
multiply 4 with x and 8 with y and set that equal to 0
then multiply 2 with x and 4 with y and set that to 0
half ice
Just is log3(x) by definition
humble hazel
so its c?
half ice
Yaya
half ice
Convert them to exponentials
humble hazel
ok and then
half ice
And then they're each obvious haha
Well, some of them will be weird, but that's how you know they're wrong
humble hazel
tbh i dont know how to do this because this coronavirus online school is so hard
because i have no teacher
ok i turned them into exponentials and i found it thank you, also yeah the answer was C
limber sierra
this is not linear algebra
humble hazel
yeah i know i didnt know where to put it my bad, sorry
uneven crater
Let A be an NXN adjacency matrix of an undirected and unweighted network (a graph) without self-loops. Let 1 be a column vector of N elements all equal to 1. how can i use matrix formalism to write
uneven crater
the vector k whose elements are the degrees k, of all nodes i= 1, 2 = N
cold topaz
how do u find possible values of an unknown element of a square matrix, so the matrix can have eigenvalues?
slow scroll
you just compute the characteristic polynomial like you normally would, and find values for the unknown element that make it factor nicely or whatever.
cold topaz
yeah. i have hthe polynomial. but what then?
i mean, how do i know ho many values there might be?
slow scroll
what is the characteristic polynomial exactly?
cold topaz
lambda squared - lambda - lambdax + x + 9
slow scroll
ok, its quadratic in lambda so u can have at most 2 real eigenvalues. just use the quadratic formula to compute the values for lambda. The important thing is that you choose a value of x that makes the discriminant b^2 - 4ac >= 0.
when the discriminant is >0, you have two real eigenvalues. when its 0, you have one real eigenvalue (with algebraic multiplicity of 2), and when its less than 0, the eigenvalues are complex
cold topaz
so for the answer, we have write down the eigenvalues for all three scenarios?
slow scroll
well, no... what is the question asking? i thought you wanted to know the value of x that gives you eigenvalues (presumably real eigenvalues?).
cold topaz
yes
thats the question
since it is a determiant, we have to choose the one that is equal to 0?
slow scroll
Assuming ur familiar with the quadratic formula, this is exactly the same as finding the value of $x$ that gives the characteristic polynomial $$ \lambda^2 - (1+x)\lambda + (x+9) = 0$$ real solutions. A quadratic equation has real solutions whenever the discriminant $b^2 - 4ac \geq 0$ ( $>0$ if you want two distinct solutions).
stoic pythonBOT
kxrider:
slow scroll
i.e. a=1, b= -1-x and c = x+9.
cold topaz
since we are talking about eigenvalues, should the qeuation be equal to zero, instead of more or equal?
slow scroll
no. you set the characteristic polynomial equal to 0 to find the eigenvalues, but I am talking about $$\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$
when you have an equation $a\lambda^2 + b\lambda + c = 0$. The part inside the square root is called the "discriminant," and it needs to be non-negative to have real solutions, and strictly positive in order to have two \emph{distinct} real solutions.
wintry steppe
send a picture of your question
cold topaz
@wintry steppe this is all i have
river jasper
can someone explain to me the different between orthogonal and orthonormal?
wintry steppe
@golden garden Orthogonal means means that two things are 90 degrees from each other. Orthonormal means they are orthogonal and they have “Unit Length” or length 1. ... To get an orthonormal vector you must get the orthogonal vector and then divide each element by a weight so that the “magnitude” is equal to one.
shrewd slate
Each vector in an orthonormal list is orthogonal to all the other vectors and each vector has norm 1
wintry steppe
orthonormal is orthogonal scaled to unit 1
@cold topaz what were u supposed to do again?
cold topaz
find the eigenvalues
wintry steppe
oh
sharp merlin
wintry steppe
sos u have to subtract that matrix with (lambda)(Identity matrix)
sharp merlin
How do u do this
wintry steppe
do u know what that is
river jasper
wait can you explain orthogonal in terms of matrix
river jasper
oh LOL
cold topaz
@wintry steppe yes. I hve the polynomial
golden garden
my broother
wintry steppe
show me what u have @cold topaz
cold topaz
lambada squared - lambda - lambdax + x +9
slow scroll
what is the question?
slow scroll
it looks like you already have an answer?
slow scroll
oh ok. do you know how to compute the null space of a matrix?
slow scroll
ah ok :p
wintry steppe
@cold topaz so that simplies to lambda ^ 2 - 2(lambda) + x + 9
x+ 9 is suppposed to be kept together in this question iirc
wintry steppe
u typed this: lambada squared - lambda - lambdax + x +9
u had a " + " infront of the x
cold topaz
u typed this: lambada squared - lambda - lambdax + x +9
@wintry steppe notice I have lambdaX + x 😄
wintry steppe
oh haha
sharp merlin
@slow scroll wait i confused myself
wintry steppe
true
cold topaz
that's where ive stopped.
slow scroll
@sharp merlin yea
wintry steppe
it would be a lot easier if i saw a visual
sharp merlin
so then u do x1 = 4x 2
wintry steppe
like a legit visual of your work
slow scroll
yes, that is correct. If you look at the matrix, if x3 was anything else, there would be a -5 that you can't get rid of
sharp merlin
do u know why its saying this is wrong
nvm found the mistake
isnt the 0 vector here if u plug in s and t
make s and t 0
slow scroll
when s=t=0, you get (0,2,0,0)
cold topaz
@wintry steppe yeah. sorry. i was away
wintry steppe
no worries.
i was also away for dinner lolo
anyaways u should have lambda ^ 2 - lambda(x) + 9
then u gotta set the equation to 0 and solve for lambda
cold topaz
quadradic. right? @wintry steppe
wintry steppe
yes
cold topaz
so are we gonna have two distinct answers or our answer will be from a range?
wintry steppe
2 different answers i think
cold topaz
i mean will we have x=y,z. or x is going to be from a range of numbers?
wintry steppe
well you don't have a middle number so you wont get actual values
sharp merlin
@slow scroll
for this one
its just 3 times 1st vector -2 times 3rd vector right
wintry steppe
are u supposed to multiply B with [x]B?
sharp merlin
oh
wintry steppe
augmented matrix is taking the numbers from a system of linear equations and putting them into matrix form
so u have to find the inverse of B
then multiply it with [x]B
wintry steppe
u have to put ba and b2 into augmented matrix
sharp merlin
i used a calculator
wintry steppe
show me what you got for your inverse
wintry steppe
ugh
wintry steppe
yah thats what i got too
so now multiply this matrix with the cooridnate vector [x]B
to get x
sharp merlin
the final answer in in fractions
but the correct answers is whole numbers
so its def wrong
@wintry steppe
im pretty sure this isnt inverse
wintry steppe
it is
slow scroll
that is right
wintry steppe
can u reduce [4 4 8] ^ T into [ 1 1 2] ^ T ?
hasty violet
can you repost the original question please? @wintry steppe
wintry steppe
thats not from the question, its from the last picture hydra posted
hasty violet
I don't think you can. In the picture the question asked to find [x]b in terms of standard coordinates.
so it would be [4,4,8]
wintry steppe
ok
pale shell
Hi
hasty violet
Yeah, [1,1,2] is a vector in the same direction. But it is not the correct one. Not always correct to "simplify" in math
lol
hi earlten
waddup?
hollow finch
@pale shell No
pale shell
Mk
pale shell
o
hasty violet
XD
could be that I know what they are but they are called differently
but I am willing to learn
regardless
pale shell
Basically R^n is just a space so like R^2 is two dimensional space
hasty violet
oh ok
I know it then
n-1 is a vector that is 1 less
than the dimension of the space
pale shell
Oh so when you think about span
It is where you have a set of vectors
Do you know what a linear combination is?
hasty violet
yes
pale shell
Mk
hasty violet
ax + by
pale shell
So it is all possible linear combinations of those vectors
An example
What do you think the span of the î and j hat are
hasty violet
R2
pale shell
Yea
hasty violet
but i , j , k are R3 right?
pale shell
Mhm
hasty violet
and i, j ,k are the basis vectors usually right?
pale shell
For R3 they are
hasty violet
I see, in that case the answer is no. You cant span RN with n-1
for R3 you would need 3 or more vectors
however each vector can have less than 3 dimensions i think.
pale shell
Wot
hasty violet
like ijk
pale shell
Wait but in R3 vectors are defined as having three entries
hasty violet
oh. Sorry I misunderstood then.
When you asked whether the space R^n can be spanned with n-1 vectors
I thought you were referring to the amount of vectors when you said "n-1"
pale shell
How
hasty violet
Think about it this way
you are in R3
there are 2 vectors
imagine any 2 vectors you want
there is 1 plane that contains those vectors
pale shell
I can visualize that
hasty violet
there is a vector that is normal to that plane
that vector cant be expressed using the 2 vectors
pale shell
But I mean a proof generalizing that to R^n
pale shell
Wot
hasty violet
you know what a dot product is?
hasty violet
it should work because you need to prove that all the vectors lie on a "plane"
and has a vector normal to it
"plane" doesnt have to be 2d plane
hasty violet
yeah
hmmm.....
Let me first start with R2
n -1 vectors. So 1 vector
V = [x,y]
there is a vector perpendicular to the 1d "plane" of V.
A = [a,b]
V dot A = xa + yb = 0
this linear equation can be solved.
there is a solution. therefore for R2 n-1 vectors are not enough
now for R3
n-1 vectors. So 2 vectors
V = [x,y,z]
W = [g,h,i]
pale shell
Wait
hasty violet
Now we have the perpendicular vector
pale shell
Wot
hasty violet
oh nvm
mistake
yeah I forgot to add 3 entries to the vectors, did 2 instead
now we have perpendicular vector
A = [a,b,c]
xa + yb + zc = 0
ga + hb + ic = 0
This system should be able to be solved I think.
so for Rn
there will be n-1 equations in the system of linear equations
each with n terms
I think if we can prove something like "if there are n terms in each equation, n-1 linear equations can be solved"
then we are done
pale shell
Yeah
Well wait
That means there is a free variable right
If you have more variables than equations
hasty violet
oh yeah
a,b,c are the variables
the xyzghi are the constants
hmmm
Wait!
Wait!
I know!
pale shell
?
hasty violet
the system of linear equations only needs at least 1 solution
doesnt have to be exactly one solution
if there is a free variable
it means there are infinite solutions
hasty violet
There we go
so for R^n, if you pick ANY n-1 vectors
there will be infinite amount of vectors that are perpendicular to all n-1
👌
pale shell
But why did you set the dot product equal to zeor
hasty violet
because there is a theorem in mathematics
if the dot product of 2 vectors is 0
they are perpendicular.
hasty violet
or are you referring to
a1b1 + a2b2?
pale shell
When you dot two vectors
You are essentially projecting one onto the other, then multiplying the length of the projected vector by the length of the other vector
hasty violet
Yeah hahaha.
The math way to say that is ABcos(theta)
so if the dot product is 0
it means the projection is 0
which means perpendicular.
pale shell
Yeah
I guess that is how it would have to work out because
That is the only factor which could possibly go to aero
hasty violet
Yeah so for Rn there are n-1 linear equations which equal 0. Each has n terms
since there is always a free variable
there are always infinite vectors.
that cant be spanned using the n-1 vectors
pale shell
Very interesting
hasty violet
yeah linear algebra is so powerful. I never realized how powerful it can be until today.
pale shell
Mhm
hasty violet
When I tried this question.
hidden sable
Can somebody explain what a companion matrix is?
dusky epoch
no
no, rent free, that is not true
it's only true for the companion matrix of the polynomial x^n if anything
please do not spread misinformation
@hidden sable the companion matrix of a monic polynomial p is a specially constructed matrix for which p is the charpoly
autumn kraken
Hi, is archetype the correct English term for f^-1 (N) ?
autumn kraken
ah ok thanks
stoic pythonBOT
madmike:
autumn kraken
Is the above sufficient to prove the following:
stoic pythonBOT
madmike:
autumn kraken
?
stoic pythonBOT
madmike:
autumn kraken
my argument is that if f(x) is in C or in D, the x will be in both sets f(C u D) and f(C) u f(D)
is this wrong?
dusky epoch
"if P then (Q and R)" does not exactly prove "Q iff R"
dusky epoch
wrong approach
you have a statement saying two sets - f^-1(C \cup D) and f^-1(C) \cup f^-1(D) - are equal
so to prove that, you need to prove that every element of f^-1(C \cup D) is also an element of f^-1(C) \cup f^-1(D), and that every element of f^-1(C) \cup f^-1(D) is also an element of f^-1(C \cup D)
autumn kraken
I see, then I'll try starting with proving subset
stoic pythonBOT
madmike:
autumn kraken
Can I not go like this? 🤔
subtle walrus
sure, but this is not linear algebra
solar osprey
for the first two matrices I keep getting that the matrix is diagonalizable for all values of a
But I am having doubts because the question says "THE" value
subtle walrus
it also says THE matrix, but lists 3
solar osprey
guess my prof is just dumb
or unattentive to details
😆
regardless maybe ive answered incorrectly anyone can review my method ?
eager burrow
nah that's fine, your result should be correct
hi-speed-quick way: for upper triagonal matrices, you can read off their eigenvalues from the diagonal, so you see that, for example, A1 has the eigenvalues 1, 1 and -2
non-diagonalizability can only happen if you have an eigenvalue which shows up more than once, but less eigenvectors than eigenvalues; however, for A1 there's no problem, because you can even write down explicitly two linearly independent eigenvectors to the eigenvalue 1, namely just (1,0,0), (0,1,0)
and A2 has three different eigenvalues 1, -1 and -2, so this will always be diagonalizable
okay, this doesn't look like the hi-speed-quick way if you write it all out, but it's a way to see diagonalizability without having to do any calculation
solar osprey
Can we discuss these?
for part c , I showed that if C is an eigenvalue of A then C^2 is an eigen value of A^2. And since A^2 is the zero matrix, it's only possible eigenvalue is 0 and hence the only possible eigenvalue of A is 0.
pallid rampart
A^2 is not the zero matrix
pallid rampart
Oh yeah part (c)
The first one is part (c)
Sorry misread
yeah that seems about right
solar osprey
for part d
I followed the same procedure kinda
if C is an eigenvalue of A then C^2 is an eigen val of A^2
if v is the corresponding eigen vector
then it follows that v = C^2 v
which is true only if C = 1 or -1
which also shows that 1/C is an eigen value
I didnt really show 1/C is an eigen val explicitly
lol
so not really sure
solar osprey
lol
If it looks simple, I usually doubt it a lot
Doesnt that answer part d as well?
i mean e
elfin ingot
quickie basic
suppose a b and c are linearly independant in V ( V is a vector space )
prove that a+b , b+c , a+c are linearly independant
in V
dusky epoch
use the defn
elfin ingot
yea i am just checking solutions
1 sec
c_1(a+b)+c_2(b+a)+c_3(a+c) must equal 0 right?
dusky epoch
no
elfin ingot
c_is are scalars
dusky epoch
you start out assuming that what you wrote is 0
dusky epoch
and you wanna prove that c1 = c2 = c3 = 0
elfin ingot
yea
yea i got it tysm
1 mroe
is the basis of F^(mxn)
set of columns of ddimension m and rows of dimension n?
im p bad with matrices lol
dusky epoch
there's no such thing as "the" basis
elfin ingot
yea a*
elfin ingot
oh okay sorry
the space of all mxn matrices over a field F
thats F^(mxn) ig
and the set is
i dont know how to write but but basically the set has the column vectors that jsut keeps slidding 1 ( the standard basis ) and the rows the same
with dimensions m and n
1 0 0 0 , 0 1 0 0 ... etc those are the rwos ( m times )
and the same for columns
is this considered a basis
dusky epoch
it's not because none of these things are even in F^(m × n)
elfin ingot
oh fuck
yea
okay what if i just fill the rest with zeros?
so okay A_ij has zeros everywhere except at ij
is this now a basis?
yea ig am i right?
subtle walrus
the set of all these, yeah
subtle walrus
it's basically the canonical basis in F^(m*n)
subtle walrus
a mxn matrix has m*n entries
elfin ingot
yea yea
1 mroe stupid questilon what does F-vector space mean
vector space over F?
subtle walrus
yes
elfin ingot
👍
elfin ingot
so i just wanna recap and ask a few questions , the coordinate of a vector relative to a basis is the n-tuple of scalars that are in the linear combination basically
now does this theorem say
that there exists a matrix ssuch that you can change the coordinate of a vector relative to basis to another?
i am just getting confused with the concepts here ig
subtle walrus
essentially yes
elfin ingot
yea cool now can u prove this with me?
subtle walrus
a basis is just a way of "putting coordinates" on your vector space
subtle walrus
how does the book prove it?
elfin ingot
that u can change like coordinates
is this how like
in claculusu u change from polar to rectangular?
and stuff?
subtle walrus
well, the cool thing is that you can choose the best coordinates
subtle walrus
i mean kinda like in calculus
elfin ingot
can u prove that with me id get it
the text just confusus me with what he wants to show
i dont understand 1 bit of the proof
lol
subtle walrus
are we looking at theorem 7 or 8
elfin ingot
no lets look at 8
8 is the more general ig as he said
if we prove 8 we get 7 i think
so iw ant to show that the basis exist
i dont even get the 'it is clear that' lmao
subtle walrus
7 you construct the matrix between 2 bases
elfin ingot
yea
subtle walrus
8 you show that every matrix gives you a basis change
subtle walrus
ok, so
this is just a lot of unfolding definitions
if you know (i) is true you can also apply it to any basis vector in your second ordered basis, say
say $a_j'$
stoic pythonBOT
Lochverstärker:
subtle walrus
you don't know how to represent it in your basis $\mathcal{B}$
stoic pythonBOT
Lochverstärker:
subtle walrus
but you know the theorem is true, so $[\alpha_j']\mathcal{B} = P[\alpha_j']{\mathcal{B}'}$
stoic pythonBOT
Lochverstärker:
subtle walrus
but $[\alpha_j']_{\mathcal{B}'}$ is just the zero vector, except in the j'th position
stoic pythonBOT
Lochverstärker:
subtle walrus
so you can just compute this matrix vector product
elfin ingot
what do we get XD
subtle walrus
it becomes $P\begin{pmatrix}0 \ \hdots 1 \ \hdots \ 0 \end{pmatrix}$
stoic pythonBOT
Lochverstärker:
subtle walrus
hmm
elfin ingot
yea okay
subtle walrus
i dont know how to latex
elfin ingot
fuck that i get u
subtle walrus
well, you can just compute this
subtle walrus
you get the vector $(P_{1, j}, \dots, P_{n, j})$
stoic pythonBOT
Lochverstärker: