swapping rows flips its sign
scaling a row scales the determinant accordingly
adding a multiple of a row to another row does nothing
#linear-algebra
dusky epoch
summer dagger
ty had a follow up question but lost it in these mess of tabs
Arow reduces to the matrixEA
AandEAhave the same determinan
So this would not hold true?
dusky epoch
...bad paste?
summer dagger
A row reduces to the matrix EA
claim: A and EA have the same determinant
or this would be true if you kept track of what you said above
dusky epoch
anyway, yeah no that claim is false as written
summer dagger
how would you prove that?
dusky epoch
...by giving a counterexample lol
[ 1 0, 0 2 ] row-reduces to the identity, yet its det is 2
not 1
summer dagger
ty
honest marlin
If a dot product is equal to the projection of one vector onto another, why isn’t it just equal to the larger vector?
earnest vessel
If you project (0,1) onto (1,0) you don't get "the larger vector", it is just 0
slow scroll
my immediate instinct is to use row expansion of the first row
but doesn't that give $a_0 (t-1)^{n-1}$ and not some nice expression involving all of the terms?
stoic pythonBOT
kxrider:
slow scroll
oh wait i did something wrong i see
dusky epoch
wintry steppe
thynk
summer dagger
@dusky epoch comming in clutch with that counter example you gave me came up on the exam
dusky epoch
?
rare barn
@half ice #4375 ok, ik that, but how does that become something thats not a matrix? thats three matrixes multiplied together
half ice
@rare barn
You're solving for an, not for Aⁿ. I'm sure they're related, but not the same. I'm not sure what an is
The previous question is important here
half ice
If we have a vector x of two values of the sequence, then Ax gives a new value of the sequence
Let's say x = [5]
[3]
Then Ax = [8]
[5]
So we can just get any value we want by multiplying by A a bunch of times
... Or, we could get the nth value by multiplying Aⁿ once
Namely, if we multiply [1]
[1]
By Aⁿ¯¹
Then the bottom number of that vector is the nth term of the Fibonacci sequence
dusky epoch
oh god please write [5, 3]^T instead of that spacing fuckery
slow scroll
n
o
half ice
Does it look right on your screen?
slender yarrow
rare barn
im still confused. Your multiplying matrices, how do you get the equation there then... isnt it supposed to be another matrix
slow scroll
What was the "previous question" asking?
half ice
@rare barn
Yes Aⁿ¯¹ is a matrix.
Aⁿ¯¹x is a vector.
The bottom value of Aⁿ¯¹x is a scalar
rare barn
the previous question asked to write A in the form of PDP^-1
why the bottom number
and
also
half ice
Can you see why A is useful to get values of the Fibonacci sequence?
half ice
Do you see how Aⁿ can be used to jump n values of the Fibonacci sequence?
So A² lets us skip one
A²[1, 1]^T = [3, 2]^T
Likewise A⁴[1, 1]ᵀ = [13, 8]ᵀ
rare barn
yes
rare barn
wow... ill delete that but look at the link. i used that here and it gave smth completely different
half ice
Yes I know how you diagonalized A
The question is asking you for the general formula for the Fibonacci sequence. You seem to understand why we can use Aⁿ¯¹[1, 1]ᵀ to find it, but are blanking on actually doing this
rare barn
i am, cant we just use the diagonilzation and use the power on D
i still have no idea how we are getting a formula that doesnt have a matrix...
half ice
A⁴[1, 1]ᵀ = [13,8]
Therefore a5 = 8
Note that 8 isn't a matrix
Without going too far, can you multiply Aⁿ[1, 1]ᵀ? What is the result?
OHMFG your page has an even better idea. Instead, do Aⁿ[1, 0]ᵀ
rare barn
half ice
That bottom value look familiar?
rare barn
OH!
Why was this so hard for me to understand......,
lmfao.. thank u for being patient
half ice
Lel np. Glad it clicked
lyric sphinx
You can use the following $\det(A^{-1}) = 1/det(A)$ $ \det(AB) = \det(A)\det(B),$ $\det(A^{T}) = \det(A)$ and $\det(xA) = x^n \det(A)$
dusky epoch
bad tex
lyric sphinx
Yeah I'm on my phone
dusky epoch
You can use the following: \ $\det(A^{-1}) = 1/\det(A) \ \det(AB) = \det(A)\det(B) \ \det(A^{T}) = \det(A) \ \det(xA) = x^n \det(A)$
stoic pythonBOT
Ann:
You can use the following: \\ $\det(A^{-1}) = 1/\det(A) \\ \det(AB) = \det(A)\det(B) \\ \det(A^{T}) = \det(A) \\ \det(xA) = x^n \det(A)$
oumid:
half ice
The size of the matrix
wicked trellis
Oh right, thanks
half ice
n = 4 for both cases
broken hawk
hint: distributive law
for future similar problems, #prealg-and-algebra is more suitable
or even better one of the questions channels at the bottom of the list
linear equations ≠ linear algebra
hexed swift
Okay thanks
vale arrow
Isn't this like 1×1 linear system Sascha Baer?
broken hawk
I mean it is in the scope of linalg technically
except by the time you actually do linalg you’d be supposed to be able to do this in your sleep
wintry steppe
I solved one side easily
if F is alternating it is easy to show F(u,v)=-F(v,u)
however im having hard time with if F(u,v)=-F(v,u) then F is alternating
does anyone see the magic trick ?
wintry steppe
Wait is that in response to me?
quaint heart
Yeah, it was lol
wintry steppe
i dont know how =3 helps
but yea the trick was to set v=u
=3 confused me lol
i can see that eigenvalues are ±1
how do i find eigenvectors to prove its diagonalizable though?
slow scroll
im completely lost on what to do here.
slow scroll
Its easy to see that $D_n = C_{1,1} + C_{1,2}$ where $C_{j,k}$ is the determinant of the $(n-1)\cross (n-1)$ matrix formed by deleting the $j^{th}$ row and $k^{th}$ column. Here, $C_{1,1}$ is just $D_{n-1}$ but $C_{1,2}$ is not obviously $D_{n-2}$ and I am struggling to figure out how to prove that $C_{1,2} = D_{n-2}$.
stoic pythonBOT
kxrider:
dusky epoch
take the matrix for C_1,2 and expand it along its first col
slow scroll
that is D_{n-2}, but you still have the matrix for C_1,2 expanded along its 2nd column to worry about, and so on....
dusky epoch
huh what no that's not what i mean
dusky epoch
the red crossing out is going from the original matrix to that for C_1,2
then the yellow is expanding C_1,2 along the first col
and what's left behind is the green block
slow scroll
what is that supposed to mean in terms of determinants tho? 
dusky epoch
that C_1,2 = D_n-2
slow scroll
do you agree that the matrix i wrote down in the picture above is the matrix for C_1,2?
oh wait a second! I don't really understand your diagram, but the determinant of the transpose of C_1,2 is obviously D_{n-2} i just realized 🤦
wintry steppe
@dusky epoch Are there any 2x2 matrices A such that AM = [ [1, k], [0, 1] ] where k ∈ R for any M?
i tried to multiply with entries
dusky epoch
hh what
dusky epoch
are you asking whether there exists a matrix A such that for all M, AM = that?
wintry steppe
yep
dusky epoch
no, because if M = 0 then AM = 0
wintry steppe
good counterexample!
I have T(A)=AM map
and im interested if T is always diagonalizable no matter what M is
or if there exists an M for which T is not diagonalizable
I thought if I could choose an A so that AM = that then that would be a counterexample
because that [1 v] [0 1] is not diagonalizable
wintry steppe
I see that eigenvalues for A are 1 and -1
but I dont see how exactly I should prove that A is diagonalizable
the hint says to factor A²-I₂
can I factor liek A²-I₂=(A+I₂)(A-I₂)?
half ice
Yes you can
wintry steppe
Yeah
the next step in hint is to consider possible ranks of the factors
it can be 0 1 or 2 right?
wintry steppe
does that work like that?
the rank of their product is 2 since its I₂
I can see that
but I dont remember a theorem
or a fact
that says factors have to be both the same rank
half ice
Yes that's true. The rank of AB ≤ the lower rank of A or B
Since AB is max rank, so are A and B
Another way to say this, the determinant of the product is not zero, so neither of the factors have a zero determinant
half ice
Of A? I'm not sure, how do you figure?
wintry steppe
leap of faith
that wouldnt help anyway
its not enough to show eigenvalues to prove A is diagonalizable right?
I'd have to at least show that dim(eigenspace) = multiplicities of lambda
half ice
If you have two different eigenvalues, then you definitely have a diagonalizable matrix
wintry steppe
is that a theorem too?
half ice
Since each has at least (and at most) a 1D space, and that adds to the matrix size
A matrix is diagonalizable ⇔ the dimensions of the eigenspaces sum to the matrix size
wintry steppe
the hint says to prove diagonalizability through ranks of the factors which we have said to be 2 and 2
but whats the next step after saying the rank of A-Id = 2 ?
means it has a nonzero determinant
half ice
I actually assumed A² - I had rank 2 because you didn't give any zero eigenvalues, but we don't have those, do we?
So I lied, I don't know what the rank of A² - I is
Actually, the rank has to be zero since it's the zero matrix right?
half ice
Means we don't know much about the ranks of A + I or A - I
The zero matrix is a matrix of all zeroes, the only matrix of rank zero
half ice
A² - I = 0
wintry steppe
ah
right
well Im sure eigenvalues for A are +1 and -1
so maybe if I could show that I could claim that A is diagonalizable
half ice
Well, A² = I and thus has rank 2
A then has rank 2 as well
Where are you getting those eigenvalues from?
half ice
You're probably right lol but we gotta justify it
wintry steppe
can I say that
since its a 2x2 matrix and rank of A is 2
then A is diagonalizable ?
half ice
No I don't believe that's enough
wintry steppe
it looks to be so easy
problem is 1 line
i cant tell what theorem to apply though
dusky epoch
They must both be 2, as the rank of their product is 2
the rank of the product is 0 since the product itself is 0
wintry steppe
absolute state
dusky epoch
A^2 - I = 0
wintry steppe
so rank of A is 0?
half ice
Yes we recognized the problem
wintry steppe
okay we got to the point that rank of A is 2
but that doesnt say anything aobut its diagabilty
maybe saying that rank = number of nonzero eigenvalues shows that it is diagonalizable?
so if 2 = number of nonzero eigenvalues then A is diagable
but then how do i find eigenvalues of A
I only know A²=ID
what do?
lyric sphinx
@wintry steppe A can only have for eigenvalues 1 or -1 thus null(A - I) and null(A + I) are the eigenspaces
A is diagonalisable iff the sum of the dimensions of these null spaces is 2 which is the same as saying A is diagonalisable iff rank(A-I) + rank(A+I) = 2 by the ranks theorem
If either one of these rank is 0 then A = +-I is clearly diagonalisable
So the only case remaining is when both these ranks are 1
But then they add up to 2 and thus A is diagonalizable again
astral junco
is the dimension of this vector space equal to 3?
swift plaza
Yes, since ${1, x, x^2 }$ is a basis
stoic pythonBOT
killer_memestar:
astral junco
thank you, i appreciate it.
delicate spire
hi, im trying to find a basis for the subset of polynomials p in P2 with the property that p(1) = 0. How would i go about doing this? i always get confused when we work with polynomials.
wicked trellis
Can anyone help me out with this question
dusky epoch
what've you tried and what's giving you trouble here
dusky epoch
uhhh what
you can't just assume a_2 = a_3 = a_4 = 0, and yet you just did
if you wanted to expand along the first row (or perhaps the first col) you would have four 3x3 determinants to untangle
swift plaza
Do you know how an elementary row operation affects the determinant?
swift plaza
- interchanging two rows means the determinant switches sign, 2) adding a multiple of some row to another doesn't change the det, 3) multiplying a row by a constant scales the determinant by that same constant
and then notice that B can be obtained from A by applying row operations
that should help you on your way
dusky epoch
ah well
yeah, that's doable
another thing that could be done is to notice that there must exist a matrix Q such that B = QA
dusky epoch
seems ok
wicked trellis
Thanks
honest marlin
Could someone help me understand what the cross product actually represents? I know it has something to do with projection but I don’t get it.
slow scroll
the result of the cross product is a vector perpendicular to plane spanned by the two vectors u are taking the cross product of
charred stirrup
"The most important thing about a basis for V is that every vector in V can be written uniquely as a linear combination of the elements of B. The uniqueness there is important; it means that there is a 1-1 correspondence between ordered lists of scalars of length n=|B|=dim(V) and V itself. ".
can someone define "unique"
does that mean for a certain given vector one and only one linear combination of the basis will give that vector?
dusky epoch
yup
that's what makes a basis a basis
minor nitpick though: you'd have done better to say "clarify" rather than "define" there
charred stirrup
thank you!
astral junco
Trying to find the rotation matrix that takes <0,5> to <3,4>
i'm not doing this right am i?
dusky epoch
i mean first off why are you using the empty set symbol $\varnothing$ for your angle
stoic pythonBOT
Ann:
dusky epoch
if you wanted a lowercase phi, that looks like this: $\varphi$
stoic pythonBOT
Ann:
astral junco
oh...yeah that's what i was going for
dusky epoch
but still, once you have -5sin(θ) = 3 and 5cos(θ) = 4
why complicate it more
you have cos(θ) = 4/5 and sin(θ) = -3/5
astral junco
good point
dusky epoch
so you know your angle is in quadrant IV, and so you can, for example, evaluate it as arcsin(-3/5)
astral junco
ah i gotcha. much appreciated, thanks.
charred stirrup
how are you supposed to say this?
Let T:U->V be a linear map
the colon in my head says " such that" but saying T such that U something V doesnt seem right
half ice
@charred stirrup
T is a function from domain U to codomain V
It's not really a "sayable" notation. It says a lot very quickly
You could interpret it literally. "T is U arrow V"
dusky epoch
U -> V = "from U to V"
charred stirrup
for matrix A , B , and C
ABC means to apply the transformation of C, then B, then A?
sharp lodge
Yes
patent orbit
slow scroll
err i think i got it now, but I'm a bit confused with the way it asks to do it. I plugged c_n into the formula from part c to get the determinant matching the original matrix, namely, the vandermonde determinant for n. Then I computed A_n by dividing the product at the top (the one we want to prove) by the terms that come after A_n in part c.
From there, you get that A_n = the vandermonde determinant for n-1.
So if we let $V_n$ be the vandermonde determinant for some $n$, then we can say that $$V_{n} = V_{n-1}\prod_{k=0}^{n-1} (c_n - c_k)$$
stoic pythonBOT
kxrider:
slow scroll
um soo uhh am i doing this right and what would i have to do next to complete the induction?
im a bit thrown off because the problem says to assume n-1 and prove n, but i think i assumed n, and need to prove n+1 with the way i have it set up
I think i had to assume n in order to compute A_n tho
gentle knoll
The equation you have there is right
So assume the formula holds for V_{n-1} and plug it in
That will prove it for n
patent glacier
is there a rule for finding out the dimension for a vector space of polynomials with multiple variables
like for example i know that the dimension of a euclidean space Rⁿ is n
and the dimension of a vector space of polynomials in x with degree n is n+1
but say it's a vector space of polynomials in x,y with degree n
is there a rule or some way to figure it out other than writing out {1, x, y, xy, x², y² ...}
half ice
This depends on a bunch of stuff. What's your basis? What polynomials are in your space?
If your basis is {x, x²} then you can represent x² + x in a 2D space
patent glacier
there's no given basis, it just says the vector space of all polynomials in x, y, z of total degree less than or equal to 3
and to find the dimension
i wrote it out and got 20 (didn't check if it's independent yet) but i was wondering if there was a general rule for it
presumably the vector i wrote out is a basis
half ice
Cool cool, then this is a 20D space!
I assume you're going to want to use combinatorics for similar questions
patent glacier
for more variables/degrees this would probably get tougher so i was wondering if there was some formula/rule for it, yeah
half ice
You just need the number of terms you could make by multiplying any three of x, y, z
Or any two, or any one, and then throw 1 in there
dusky epoch
"20D space" 🤢
half ice
This is a "stars and bars" problem, I believe
Unless there's an easy way to do it if you're summing all cases?
half ice
Do you know of the choose function? If so, have you seen a stars/bars problem before?
patent glacier
i know what choose is but no, this is my first time of hearing of stars/bars problems
dusky epoch
wait what are y'all talking about
find the dim of the space {f in R[x,y,z] | deg f <= n}?
half ice
Okay, let's say we have x, y, z. You can pick any one of them up, and keep them. You can even pick them up multiple times!
This is a "combinations with repetition" problem. We generally find the number of ways to do this with the "stars and bars method"
Let's say there's two different instructions. A star (×) means to pick up. A bar (|) means to move right.
So ×|××| means xy²
charred stirrup
why does the yellow vector get stretched by a factor of 2? doesnt the "2" in the matrix tell you how much we are scaling in the j direction?
half ice
There's always 5 stars/bars, three of them are stars. There's 5C3 ways to choose any 3 of x, y, z
Or 10 ways
dusky epoch
@charred stirrup do the matrix multiplication and you'll see
it's not just the 2 in the matrix
half ice
You don't need to move off z. The process ends when you are on z and have picked up 3.
You have to do that so the stars and bars are a bijection with the possible choices
But you have the right idea. ×|×|× is xyz
patent glacier
why do you have to move off y when getting xy²
half ice
You've got to at least "get to" the last one. You don't have to pick any given element up
charred stirrup
for two matrixes A and B,. AB is not the same as BA, except when matrix B or A is a column vector?
patent glacier
oh i see
half ice
I just used that idea to show there are 10 terms of degree 3.
Same question but degree 5: there are always 7 stars/bars, 5 of them are stars. 7C5 = 21
dusky epoch
@charred stirrup matrix*vector multiplication is only defined when the matrix is on the left.
charred stirrup
because in matrix multiplication we are transforming from "right to left"
patent glacier
there are 7 because, as you said previously, the stars and bars have to be a bijection with the possible choices?
half ice
Imagine n choices, and you pick up k of them. There are n - 1 bars and k stars
half ice
You've got to get to the last option, and have to pick up enough of them
half ice
You can also count the bars, rather than the stars. Same thing. That may be easier here.
5C2 ways to get degree three terms.
4C2 ways to get degree two terms.
3C2 ways to get degree one terms.
2C2 ways to get degree zero terms
half ice
Oh sick
Yeah that works
So 6C3 is your final answer
If you have k variables and want to get up to degree n, then it's just (n+k)Ck
patent glacier
yeah, 6C3 = 20
half ice
I think unless I'm doing it wrong
Cool cool, I learned today. Thx for going through this with me
patent glacier
np, i learned a lot too, thanks again
charred stirrup
when you dot product how do you know which vector is projecting onto the other?
ohh nvm
dot product is a scalar
i get to pick what i project on by using the unit vectors?
slow scroll
yea
charred stirrup
u the best
slow scroll
$a\cdot b = |a||b|\cos{\theta}$
stoic pythonBOT
kxrider:
slow scroll
b proj on to a is $\frac{a\cdot b}{|a|}$ basically c:
stoic pythonBOT
kxrider:
charred stirrup
I'm slightly confused by row operations- is it not possible to say row 3 - row 3? that's not possible right
but you can say things like row 3 > -row 3
nvm got it
slow scroll
no because thats the same as multiplying by zero, and thats not a elementary operation
charred stirrup
but things like multiplying by any scalar of the real numbers is allowed?
slow scroll
right, thats because multiplication by non-zero scalars are invertible
charred stirrup
god bless you
slow scroll
npnp
charred stirrup
the easiest way to prove that a matrix is not invertible is by manipulating it and getting a 0 row ?
slow scroll
yea
charred stirrup
love u
slow scroll
charred stirrup
for 2. shouldn't it say that A is the inverse of B? why does it keep the notation the same?
since we are transforming from right to left
slow scroll
@charred stirrup it just wants u to see that B is a left and right inverse at the same time
charred stirrup
ohh
doesnt really matter since if they are inverse, we will get the identity matrix regardless of how they are arranged
slow scroll
right, its contrasting the idea of "invertibility" with "left invertibility" and "right invertibility." Namely, an invertible matrix is a matrix that has a left and right inverse.
... and you can derive from that, that the left and right inverses must be equal to each other
broken hawk
I don't really understand what you're asking :/
broken hawk
what do you mean with "transform first by B -> A" or whatever it was you wrote
charred stirrup
AB is not equal to BA, cause we apply the transformation from right to left
broken hawk
that's not really the right explanation for why they're not the same; rather the reason you do them from right to left is because they're not the same
if they were the same you could apply them in any order
charred stirrup
your explanations always hype
broken hawk
like how in 1+2+3, you can add the numbers in any order you please in your head
anyway, with transposes I've found it easiest to not think too much about the meaning when it comes to linear functions, and actually just think about the matrices themselves
there is a concept related to the transpose called the adjoint map, but it's not as straightforward as e.g. inverse functions
so honestly, just think of (AB)^T = B^T A^T as a rule for matrix calculations
charred stirrup
does a symmetric matrix mean transposing it effectively doesnt do anything?
dusky epoch
a symmetric matrix is BY DEFINITION a matrix equal to its own transpose.
charred stirrup
what notation does the fancy w mean?
wintry steppe
in this case, it means $\mathbf{w} = (x,y,z)$ for some $x,y,z \in \bR$
stoic pythonBOT
brittle juniper
I think they were talking about the omega
wintry steppe
oh kek
charred stirrup
oh
dusky epoch
yeah $\omega$ isn't a w
stoic pythonBOT
Ann:
dusky epoch
it's just a greek letter
charred stirrup
in my head it says angular frequency but that's not right
dusky epoch
they're defining the $\omega$ function there
stoic pythonBOT
Ann:
charred stirrup
what should I think in terms of linear algebra?
wintry steppe
just a function
dusky epoch
it's just a letter tbh. don't read too much into it
charred stirrup
ohh
wintry steppe
that they defined there
dusky epoch
a given letter can stand for several different things
charred stirrup
why are r s and t bounded between 0 and 1? I thought they are just scalars that stretch the three vectors
charred stirrup
dusky epoch
alright so it's taking me too long to make this in mspaint so fuck it
wintry steppe
lol
dusky epoch
so like
charred stirrup
it's the same parallelepiped as the 3b1b video right?
dusky epoch
yes.
charred stirrup
yup ok im following
dusky epoch
consider $P(\mathbf e_1, \mathbf e_2, \mathbf e_3)$. that's the unit cube.
stoic pythonBOT
Ann:
charred stirrup
e1 e2 and e3 are unit vectors that make the unit cube?
charred stirrup
yup
dusky epoch
so the thing is
the points in the cube are precisely the points whose coordinates (again, in the standard basis) are between 0 and 1
and as such, each point inside that cube may be represented as $$c_1 \mathbf e_1 + c_2 \mathbf e_2 + c_3 \mathbf e_3$$ with $c_i \in [0,1]$
stoic pythonBOT
Ann:
dusky epoch
it's the same thing if you replace e_1, e_2 and e_3 with another LI set of three vectors
it just gets you a slanty-slanty cube
(in 3b1b terminology)
wintry steppe
lol
charred stirrup
should i represent the vectors in the matrix as columns or rows? or will it not matter?
dusky epoch
det(A) = det(A^T) so yeah it doesn't matter
charred stirrup
this is true, but if I wanted to show that they were in the same plane, wouldn't I just instead show they are linearly dependent?
charred stirrup
gonna take that as a no and just make a 0 row
dusky epoch
sorry, i ran out of steam there a bit
charred stirrup
dont worry
you're really good
if i got a plane equation
like 2x + 3y + 5z = 10
what is "10" ? the magnitude of x y and z?
dusky epoch
hhhhhh it's just a constant
it takes a bit of a stretch to like... give it an actual geometric interpretation
ime like... most of the time, what matters is if that constant is zero or not.
charred stirrup
bet
wintry steppe
bet
dusky epoch
$\beth$
charred stirrup
if i have a 2x3 matrix
stoic pythonBOT
Ann:
charred stirrup
i can say that the span of the columns is in R^2 but the span of the rows is in R^3
dusky epoch
yes, because the cols of a 2×3 matrix are two entries tall
and the rows are three entries long
charred stirrup
the rank of that matrix will correspond to the dimension of the space spanned by its columns
?
dusky epoch
it will be definitionally equal to said dimension.
dusky epoch
test coming up?
dusky epoch
let me quote 3b1b here
Eigenvectors and eigenvalues is one of those topics that a lot of students find particularly unintuitive. <...> I suspect that the reason for this is not so much that eigen-things are particularly complicated or poorly explained. In fact, it's comparatively straightforward, and I think most books do a fine job explaining it. The issue is that it only really makes sense if you have a solid visual understaning for many of the topics that precede it. Most important here is that you know how to think about matrices as linear transformations, but you also need to be comfortable with things like determinants, linear systems of equations and change of basis. Confusion about eigen-stuffs usually has more to do with a shaky foundation in one of these topics than it does with eigenvectors and eigenvalues themselves.
dusky epoch
hm?
charred stirrup
Ax=0
If I wanna find a vector x that the transformation A will SEND to 0
that will be then
i have to do the matrix and put it into rref right?
basic variables written in terms of free variables
broken hawk
I mean you don't have to
you can also just make an educated guess, but it may be hard
for 2x2 matrices I'd guess. 3x3 at least do a bit of reduction
broken hawk
and then of course, don't forget that when you do a row reduction, taht's a change of basis. so the vector you then find will be in terms of the basis of the reduced matrix
so you have to unchange the basis
(when I say guess I mean of course you still have to check it actually works :P)
and there may not be any solution to Ax=0 except for the 0 vector, of course
A needs to be noninvertible
honest marlin
I'm really having trouble understanding the dot product. If the dot product shows how much 1 vector is projected onto another, how is it convayed with just 1 number?
half ice
Length
viscid apex
in matrices does $A^{2} = A \rightarrow A = I$?
stoic pythonBOT
kakamaika:
dusky epoch
not necessarily
viscid apex
I know its not but i cant find a disproving example
brittle juniper
0
dusky epoch
if A is invertible then yes. but there are plenty of matrices which aren't invertible yet satisfy A^2 = A.
example: diag(1, 1, 1, 0)
viscid apex
?
stoic pythonBOT
Ann:
astral junco
trying to find a basis for span(S) that is also a subset of S.
dusky epoch
have you determined dim(span S) yet?
astral junco
no i'm kinda lost in lin alg. would that be the first step?
dusky epoch
that's the most logical first step
astral junco
ok and then what would i do with that?
dusky epoch
once you find dim(span(S)), you know that's how many vectors your basis will have to contain
astral junco
ok i'll start with that.
astral junco
how do i determine the span of S in a program like matlab?
i took all vectors and put them in a matrix and got it in rref, but i'm kinda as to what to do now
kinda lost*
slow scroll
if you still have the matrix and know how to find the column space of a matrix in matlab, then you can just do that to get the span of the column vectors in the matrix.
astral junco
ok i'll look up how to do that, thanks.
slow scroll
np
wintry steppe
you can't exactly represent the span of vectors without a basis, though.
Your question isn't too well defined.
slow scroll
@astral junco if you have the rref form of the matrix, the pivot columns of the rref form correspond to the column vectors of the original matrix that make up a basis for the column space i.e. the span of the column vectors.
astral junco
i think i found them. columns 1, 3, and 4?
so they are basis for the span of all the vectors in the matrix and a subset of the set of all the vectors in the matrix?
sry if these are dumb question
slow scroll
columns 1, 2, and 4, and 5 of the original matrix make up the basis for the span of those vectors. The basis vectors are elements of the subspace used to describe the other elements of the subspace. This space is a subspace of R4
astral junco
i was having a lot of trouble with this, thanks
slow scroll
np c:
slow scroll
i want to prove by induction that the number of multiplications needed to compute the determinant of a $n \cross n$ matrix by co factor expansion is $$\sum_{k=2}^n \frac{n!}{k!}$$ However, for $n=2$ I am getting only 1 multiplication, when it should actually be 2. Whats the deal here?
stoic pythonBOT
kxrider:
slow scroll
see wetf?
prime sand
dusky epoch
that sure is a geometry problem
prime sand
FFFF!!! xD someone told me this was linear
sand hedge
wym this is #probability-statistics
frosty vapor
rare barn
is this true or false? why? The span of three nonzero vectors in R
4 is a three-dimensional subspace of R4
frosty vapor
they can be any number of dimensions less than or equal to 4 tho
wait
ya three vectors right
hec
now im confusion
well for sure it can be less than three consider three colinear or coplanar vectors
rare barn
hmm confused now...
forest jay
how did they rewrite it like this?
and @rare barn that sttement is not necissarily true
if one or more of the vectors are linearly dependant
then it could be something like a plane or even a line
ahhh i dont get how they rewrote it like that
rare barn
@forest jay so like {1,2,3,4} and {2,4,6,8} and {1,2,5,9} wouldnt be a 3d subscape of R4
forest jay
yeah because if you reduce it, it's a 2d subspace
rare barn
got it!
one more q. how do i show that every square matrix is/is not similar to a diagonal matrix
forest jay
well not every matrix can be diagonalized right
rare barn
@forest jay its not asking if a square matrix can be diagonalized tho, its wehther they can be similaer
quaint heart
Diagonalizable means similar to a diagonal matrix lol
slow scroll
What about in the situation when an operator has no eigenvalues?
slow scroll
the null space of A - \lambda I is the space of eigenvectors, right? What if the operator has no eigenvectors? Then wouldn't the null space just be 0?
err i guess what they said makes sense. If you start with an known eigenvalue, then there should be a non trivial solution to the null space, my bad.
dusky epoch
...yes, if the operator has no eigenvalues, then the kernel of A - λI is going to be trivial for all λ in K
broken hawk
@forest jay This is an extremely common trick. You introduce new variables, which to avoid confusion ill write as x1, x2 and x3 instead. And you name them x1 = y, x2 = y' and x3 = y"
Now the idea is as follows:
x1' = (y)' = y' = x2
x2' = (y')' = y" = x3
x3' = (y")' = y"' = ??
and now you look at the original differential equation, put in the right expression for ??? and then simply replace the derivatives with x1,x2 and x3
This gives you a first order differential equation for x1,x2 and x3 in terms of only themselves
I like to represent the x as a vector, personally
forest jay
got it, thank you a ton!
prime rock
Stupid question-- what's the easiest approach in finding the inverse of a 3x3 matrix?
I would take the determinant * the transposed 3x3 matrix, right? But would the signs change?
winter reef
I'd build 3x6 matrix where its AI (next to each other, A is your matrix, I = 1s on diagonal, rest are zeroes.) and row echeleon until A=I. The matrix on the right is the inverse of A
rare barn
Two questions, I have to show why they are either true/false or sometimes true: A line in R2 is a subscape of R2
and If the eigenvalues of a 2×2 matrix A are 0.4 and 0.7, then all solutions of the dynamical system xk+1 = Axk tend toward 0 as k → ∞.
quaint heart
What if the line is not through the origin?
And for the second one, if it had 2 distinct eigenvalues we can diagonalize it so it is of the form P^-1 A P where A is a diagonal matrix whose entries are the eigenvalues
Then A^n approaches the 0 matrix as n goes to infinity
So the second one is true
rare barn
got it, and the first one only works if the line does not go through the origin
quaint heart
Yeah lines through the origin are 1d subspaces
rare barn
@quaint heart im having a little trouble understading ur explanation for the 2nd one
quaint heart
So we have a matrix A = P^-1 B P where B is the matrix of eigenvalues
The solutions to that dynamical system are (x_0, Ax_0, A^2 x_0,...)
It's pretty easy to show that as k tends to infinity A^k tends towards the 0 matrix
So x_k = A^k x_0 will tend towards the 0 vector
broken hawk
what do you know of diagonalization?
rare barn
wdu mean
broken hawk
like, essentially the exercise is asking you to diagonalize A and write down the change of basis matrix
rare barn
is that exactyl what theyre asking though
i dont see what they mean by realtive to B is diagonal
broken hawk
not quite, they want the basis, not the matrix representing the basis
which is like, only a difference in notation, I guess
rare barn
how do i get there after diagonalizing though
broken hawk
that’s why I’m asking, what do you know about diagonalization
when is a matrix diagonalizable and what does that even mean?
rare barn
it has linearly independent vectors
broken hawk
no
that’s invertible
the zero matrix is diagonalizable (for it is diagonal itself)
rare barn
"the diagonalization theorem states that an n×n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors, i.e., if the matrix rank of the matrix formed by the eigenvectors is n."
or am i confusing this
broken hawk
no, that’s right
you said something else
you missed the eigen- bit
which is very important
now, basically you want to go into a basis where the original transformation becomes a diagonal matrix
which is precisely the eigenbasis, the basis of eigenvectors
so you have to find the eigenvectors
rare barn
ok, im going to try and figure this out and ill let u know my answer
rare barn
what kind of question i sthis, it looks similar to diagonalization but not really?
broken hawk
honeslty, solve that by brute force
write P = [[x, y], [z, w]] and solve the system of equations you get from doing PA = PC
broken hawk
I mean calculate it out
I don’t feel like verifying things you could just type into a calculator
please stop pinging me, too, I’m no longer available for help
rare barn
okay sorry!
broken hawk
there’s various algorithms for inverting matrices, probably the most straightforward being gaussian elimination
well, the actually most straightforward is plugging it into wolfram alpha / mathematica
but you can just do it by hand
google gaussian elimination invert matrix
broken hawk
that also works. that’s cramer’s rule
it’s however significantly more tedious
imo
adj A is the matrix where the ij-the entry is the determinant of the 2x2 matrix you get by removing the ith row and jth column of A
so you would have to compute nine of those determinants, plus the determinant of A itself
which is definitely more work than gaussian elim
dense holly
det(A^-1) = 1/det(A) right?
proper crescent
dense holly
How is that the more general version
astral junco
I'm trying to find an example that shows U is not a subspace of M_22. Can someone walk me through this or point me in the right direction?
half ice
@dense holly
Consider:
det(AA¯¹) = 1 = det(A)det(A¯¹)
So you can get any "power rules" from that formula
Namely, det(Aⁿ) = det(A)ⁿ
dense holly
I see. Thank you
glad pagoda
In general, the determinant is a multilinear operation, meaning it's pretty flexible with how you toy around with the algebra
proper moat
@astral junco So you just need to find a counter example that violates subspace properties right?
Obviously, the 2x2 matrix of all 0's is in U
So now just check for closure under addition and scalar multiplication
you can very easily come up with a counterexample for atleast one of those 2 properties
slow scroll
@astral junco consider the 2x2 matrices $ \begin{pmatrix} 1&0\0&0 \end{pmatrix} + \begin{pmatrix} 0&0\0&1 \end{pmatrix} $
stoic pythonBOT
kxrider:
slow scroll
I think u got the rest
proper moat
mmm you should have let him think about it lol
slow scroll
Oops I didn’t even see that you said something
astral junco
@proper moat i think that's enough for me to think about it. idk i'm completely lost in linear algebra and i can't really review what we are supposed to know because i don't have time.
@slow scroll thank you i will look at that
proper moat
@astral junco the example kxrider gave you is sufficient
to prove U is not a subspace
of M_22
astral junco
@proper moat ok thank you. i'll have to run through the axioms of subspaces to confirm so i understand it.
this is very difficult for me.
proper moat
what part, just finding a proof?
perhaps it might be beneficial to study some common proof techniques?
patent glacier
how do i show that a linear map isn't invertible?
quaint heart
Show it's not injective
dusky epoch
there are many ways to do that tbh
one of the most straightforward would be to show it fails to be injective
and in particular for linear maps it is possible to further reduce the effort required to show non-injectivity by finding a nonzero X such that L(X) = 0, following the problem's notation
quaint heart
I think in this case you need to show the existence of a nontrivial antisymmetric matrix
patent glacier
to show it's not injective, i would have to show that there exists two different X and Y such that L(X) = L(Y)?
dusky epoch
that would be a direct use of the definition of injectivity
or rather of the negation thereof
reread my suggestion; you might find it easier to carry out
patent glacier
okay thanks, i'll try it
dusky epoch
note that there may be many possible matrices (indeed, there will necessarily be infinitely many) which L sends to the zero matrix, so worry not about trying to find The One.
patent glacier
okay thanks, i found one
by the way, just wondering, is this definition of inverse somehow related to the definition of inverse for matrices?
patent glacier
oh right
so if a transformation matrix isn't invertible then i know that the linear map it represents isn't invertible and vice versa
quaint heart
Yep
patent glacier
okay thanks
scenic solar
If you have a finite dimensional inner prduct space V with basis {v_i} let A be a matrix with A_ij = v_i*v_j. Is this matrix always invertible? Assume that V is over the real numbers if you want.
dusky epoch
not only is it always invertible, it even has a name
it's called the Gram matrix of your basis
crimson oak
can someone help me make sense of this question
https://cdn.discordapp.com/attachments/229386752861798401/567886318961426493/innerprod.JPG
this is essentially saying that the dimension of vectors in V that are orthogonal to w are all the vectors except w since <w,w> cant be equal to 0 right
slow scroll
its saying that the space of vectors orthogonal to w form a subspace 1 dimension less than V
quaint heart
Do you know about the orthogonal complement?
quaint heart
There are lots of other elements with <v, w> =0
They are called orthogonal elements
Do you know about orthonormal basis?
@crimson oak
crimson oak
ye
quaint heart
So you can choose an orthonormal basis that includes a (scaled down version of) w
And from there it's easy to see that every other basis element besides for w is in the orthogonal complement of <w>
You just need an orthogonal basis, not an orthonormal one but whatever
crimson oak
Ok that makes sense
This is similar to the proof someone gave me in another server, that's a useful tool taking the orthonormall basis of V, sincs it must always exist
Thanks!
dense holly
whats the difference between a subset and a subspace when you re talking about a vector space
dusky epoch
a subspace is a special kind of subset which meets the two closure criteria
dense holly
so i see here that a subspace has to follow these three requirements to be considered a subspace
what is a subset then
is it nothing more than a set of vectors
dusky epoch
indeed
dense holly
the null space of a matrix is 0 iff it rref to identity matrix right
dusky epoch
not zero
half ice
Oop, yes that
dense holly
wait so if the determinant is non zero it's null space is 0? @dusky epoch
dusky epoch
yes bc then it's nonsingular and thus invertible
dense holly
oh right. my class has a long list of things on the invertible matrix thm and that was one we just added
whats the difference between basis and col space
gray glen
sinful hornet
Column space is the span of the column vectors. A vector basis is a subset of a vector space in which all the vectors are linearly independent and span that vector space
wintry steppe
ya'll need help
sinful hornet
iirc
dense holly
pls
isnt the column space of a matrix just the span of it's basis?
@sinful hornet is that kinda what u were saying
sinful hornet
Aight so, a column space doesnt always form a basis
Because sometimes the column space might not be linearly independent
dense holly
the col space is the span of vectors in that matrix but it can be simplified from a bunch of vectors to just as many as rows as there are
right
like if i have 3x5 matrix
span of 5 vectors gets reduced to span of 3 vectors bc it needs to span R3
sinful hornet
The column space is the span of the column vectors, so in other terms, its the set of all linear combinations of the column vectors.
It doesnt have to span R3
I havent had Linear Algebra in a minute tho so I might be wrong
dense holly
oh right bc it can be less i think
i see what us saying
bc like back to the 3x5 matrix
sinful hornet
Or more, it doesnt have to span R3 always
dense holly
u can have zeroes in the bottom entry for everything so it can end up spanning r2
if its 3x5
cant the max it span be R3?
bc theres only 3 rows
how can it span more
i think it would have to be R3 or less
sinful hornet
oh yeah yeah ur right because when you find the linear combination
you get a sum of constants times a 1x3 matrix
But I was speaking in general
rather than for the specific example of a 3x5
dense holly
i see
sinful hornet
and row space is the same thing but for rows btw
dense holly
okay so i understand col space now can we talk about null space
sinful hornet
yeah what about it
dense holly
if im asked to find a null space, i just set Ax=0 and find answer in parametric form
is that all there is to it
sinful hornet
yes
Basically you just find all the vectors,x, that are an elemtent of Rn, which that Ax=0
Then solve that homogenous equation
patent glacier
if i am row reducing a matrix to find its determinant and along the way (before completion) i end up with two identical rows or columns, i can conclude that the determinant is 0, right?
dusky epoch
yes
patent glacier
kk thanks
to solve this i just put the vectors into a matrix as columns and then find the absolute value of the determinant?
dusky epoch
yes
dense holly
will a matrix have the same determinant as its form in reduced row echelon form?
slow scroll
not necessarily. Remember, multiplication to rows/columns by a scalar scales the determinant by that scalar
dense holly
oh right
half ice
@dense holly
However a zero determinant will be preserved.
Zero det ⇔ Does not reduce to identity
slow scroll
nothing. different names
slow scroll
yea. What is the determinant of the identity?
astral junco
1?
slow scroll
ye.
slow scroll
err yea
astral junco
ah ok thanks. think i get it now
granite light
dusky epoch
warm sluice
If I have F = qv x B (where F, v and B are vectors), how do I isolate the B?
Especially in the x direction
lyric sphinx
You can use the rules e_x X e_y = e_z, e_y X e_z = e_x and e_z X e_x = e_y assuming e_x, e_y, e_z is a positive oriented basis
dense holly
What's the easiest way to describe what the heck a basis is
winter reef
least amount of vectors that span a space
brittle juniper
a linearly independent collection of vectors that spans the space 🤔
half ice
For example, I can describe ℝ² as the span of [1, 0], [0, 1]. We call that a basis of ℝ²
I can also describe it as the span of [0, 1], [1,0] [1,1] but that's "too much". That [1,1] doesn't add anything. Since that's larger than we need, it is not a basis. It IS still a spanning set.
@dense holly
brittle juniper
It's a matrix tho
brittle juniper
The columns vectors form a basis for a particular subspace of R³
it's a basis for neither R² nor R³
dense holly
But doesnt it span all of r2
half ice
It spans the vector space represented by [a, b, 0]
There's a bijection between that vector space, and ℝ² but they're not the same
They're also each a 2D space
winter sage
how can a basis for column space with 2 matricies with 3 variables each be dimension 2 but have 3 variables
half ice
Remember the definition of dimension?
The dimension is the number of elements in a space's basis
winter sage
o
wait
im stupid head
its a plane in 3d
i was only thinking about planes in 2d coordinate system and wondering how to get z in there lol
astral junco
i'm trying to find which values of t make the matrix invertible. is my logic here correct?
half ice
I can't follow line 4 → line 5
astral junco
when i was rearrainging the equation? maybe i made a mistake
half ice
You've circled a few terms then wrote a line that I can't follow
Anyway yes, you want to find which values of t cause the determinant to be zero. This will involve solving a quadratic
astral junco
sweet i'm on the right track then at least
yeah think maybe i made a mistake rearranging the equation
half ice
bt² - b²t - at² + a²t + ab² - a²b = 0
(b - a)t² + (a² - b²)t + ab² - a²b = 0
Then quadratic formula and you're done
astral junco
alright thanks. much appreciated
wintry steppe
can someone help me in the question-delta channel
winter sage
how do i calculate the left null space of a matrix?
ive scoured the internet but the best i could find is finding null space of A transposed
and to do that i have to get rref of A transposed which im pretty sure is super inefficent
is there a better way
slow scroll
nope no other way that i know of
quaint heart
@winter sage what do you mean left null space?
Oh nvmd I see
Why is that inefficient?
quaint heart
You can do it manually I guess
Why can't you do rref on the right?
You can definitely right multiply by a matrix
winter sage
whats right
ive only learned up to four fundamental subspaces and right isnt one of them
quaint heart
What do you mean fundamental subspaces?
winter sage
column space row space null space left null space
quaint heart
Oh lol I just mean that putting a matrix A in rref just means multiplying on the left by an invertible matrix.
You can also multiply on the right
The rref that you would wanna get would be the transpose of the rref that you would do on the left
I don't see why you wouldn't use the transpose actually
It seems like a good solution
dusky epoch
is [1, 0, 0, 0] even in that subspace
dusky epoch
okay wait
what are you referring to as [2,0,1,-1]
is it $\begin{bmatrix} 2 & 0 \ 1 & -1 \end{bmatrix}$
stoic pythonBOT
Ann:
wintry steppe
yeh
dusky epoch
i... i don't think that one's in the subspace either
wintry steppe
thanks
rocky hill
So my prof assigned a proof, if A is similar to B, show that A^2 is similar to B^2. It's literally just one line of algebra to show that A^2 = P B^2 P^(-1)
but that feels deceptively easy
like... way too easy
am I missing something?
broken hawk
proofs can be easy
winter reef
depends on what you did, but yeah its basically multplying A A I think
broken hawk
and there really is nothing more to it
upper jewel
Don't know quite if this is the correct chat for this, but I think it's close enough. In my high school trig/precalc classes I was never taught matrices, so I'd like to ask for some textbooks to read or online recommendations from people here so I can at least give linear algebra a shot.
broken hawk
well we did just open up a channel called #books-old and there’s at least one suitable recommendation there
slow scroll
khan academy
Linear Algebra Done Right (less matrices, more vector space theory)
Linear Algebra Done Wrong (the book I use to teach myself)
Linear Algebra Hoffman & Kunze (a classic people recommend)
broken hawk
3blue1brown’s Essence of Linear Algebra series is pretty much mandatory watching to give you the right intuitions
@slow scroll would you mind writing a short review of these (at least the ones you’ve used) for the books channel?
slow scroll
even if i haven't finished?
broken hawk
(doesn’t have to be several paragraphs but more than “I liked it”)
if you’ve seen enough to have a good picture of how the book’s like go for it
slow scroll
alright
upper jewel
Hmmm thanks for the recommendations and it being so quick
I was reading the strang book, and got confused by his terminology with this line saying "We multiply A by "elimination matrices" to reach an upper triangular matrix U. Those steps factor A into L times U, where L is lower triangular". But I hope the lectures help out with that.
I understand the math behind it when he showed the example
I just don't understand what he means by things like upper triangular matrices and elimination matrix.
broken hawk
did you skip ahead? surely he explained those terms somewhere
(I haven’t read the book myself, the recommendations there are crowdsourced)
upper jewel
It's in the second page of the book
upper jewel
sure want me to screenshot the pages then post it here?
broken hawk
sure, but start at the beginning
broken hawk
I think he’s just describing what he’ll cover in this chapter
you’re not supposed to understand it per se yet
upper jewel
oh XD
broken hawk
I mean he even puts “elimination matrices” in quotes
slow scroll
wow very different layout from my book lol
broken hawk
interesting that joel (friend of mine) would recommend a book that starts on matrices, which iirc we both agreed was a rather meh way to go about
(there’s two main approaches to linalg)
(the “matrices are useful!” and the “but they’re lame!” way)
(I’m in camp two)
slow scroll
yea i was gonna say my book starts at vector space axioms and constructs matrices only to solve to the problem of representing linear transformations between spaces.
broken hawk
I should read through friedberg/insel/spence just so I can leave a review though
it was “our textbook” for linalg I&II but I never opened it
just followed the lecture
which supposedly followed it pretty closely
slow scroll
is this linear algebra? [insert "is this" meme]
maiden lintel
@broken hawk Oh I forgot to mention linear algebra as a prereq for my recommendation
bitter urchin
rare barn
viscid ibex
Put a_0=2 and a_2=8 and see what you get
rare barn
im thinking its in both?
slow scroll
2+8t^2 is not in the kernel of T
rare barn
yeah i figured it out, my b i didnt say it here
is this always false? for every n x n matrix, Nul A = Col A^T
slow scroll
its true for the zero matrix ¯_(ツ)_/¯
rare barn
is that it though?
slow scroll
hmm no clue
slow scroll
wait im dumb. The null space of the zero matrix is K^n, not zero 🤦
I don't see why you couldn't come up with nontrivial examples of Col(A^T)=Null(A) with square, symmetric matrices where rank=nullity or something like that
slow scroll
does anyone know what lambda^{n-1} is supposed to be?
cloud tundra
lambda raised to the power of n-1
slow scroll
huh wouldn't the coefficient just be 1 then?
slow scroll
yea. (-1)^{n-1}?
cloud tundra
Why do you think that?
dusky epoch
slow scroll
wait a second oh. Nevermind what I said. I have no idea how i would find the coefficient of \lamda^{n-1} tho. Binomial theorem or somethin?
quaint heart
It's just plus or minus the sum of the \lambda_i
Basically when you multiply n binomials of the form x+a_i together you can get the x^m coefficient by looking at all of the ways you can choose m x's and n-m a_i's
This is strictly stronger than the binomial theorem
@slow scroll
If you let all the a_i be the same this gives you the binomial theorem
cloud tundra
you can think of it like this:
we have n terms, so if we want lambda^(n-1), then out of those n terms lambda should be chosen in exactly n-1 terms
so how can we do that? we can leave out one term, meaning that we won't select lambda from that
we can do that in n different ways (n different terms to omit lambda)
then if we leave out lambda from the first we get
lambda1 * (-1)^(n-1) lamda^(n-1)
similarly with the others
in the end, we get
lambda^(n-1) * (-1)^(n-1) * [ lambda1 + lambda2 + lambda3 + ... + lamdaN ]
slow scroll
Sorry for late reply, but lambda1, lambda2, ... don’t have their own coefficients?
If you multiply the first n-1 terms together, then lambda^{n-1} will have coefficient (-1)^{n-1}. Then when you multiply on the last term, (lambda_n - lambda), lambda^{n-1} should now have a lambda_n multiplied onto it. What am I missing?
cloud tundra
lambda1, lamda2, ... are the coefficients
the "variable" in that polynomial is lambda
lambda1, ... are just constants
slow scroll
Yea yea I know what you said, but I think the term with lambda^{n-1} should just have the coefficient (-1)^{n-1} lambda_n
cloud tundra
But what if you take the lambda from the last n-1 terms?
swift plaza
just try it with an example, i.e $(\lambda_1 - \lambda)(\lambda_2 - \lambda) = \lambda_1 \lambda_2 - (\lambda_1 + \lambda_2)\lambda + \lambda^2$
stoic pythonBOT
killer_memestar:
slow scroll
ok yea i see now
slow scroll
im super stumped on the last part. The first part looks like the characteristic equation for a triangular matrix, but A is not triangular
and its not obvious to me that you can just add some q(lambda) at the end to make things better lol
spring wolf
It's not obvious until you start trying to expand det(A-lambdaI)
(trying it on small examples should make it clear)
This channel is currently occupied @placid oracle
Although I probs should have tagged @slow scroll
slow scroll
im trying example with 3x3 rn
spring wolf
Okay, good luck 👍
placid oracle
