\lambda \in \lambda(A) is a weird choice of notation I would say
#linear-algebra
brittle gyro
well it was good for Golub and Van Loan... who am I to disagree? haha
brittle gyro
I guess they didnt go for that to spare sigma for singular values only
Im familiar with \rho(A) to denote spectrum
haughty berry
\sigma(A) is a common notation, it denotes the spectrum of A
Wikipedia used their notation I think
zinc timber
Im familiar with \rho(A) to denote spectrum
it's one of those, I don't remember lol.
zinc timber
or "a is an eigen value of A" 
haughty berry
fr
brittle gyro
Ok, so Im thinking that if $A$ is symetric Schur decomposition guarantees that $\rho(A)$ (spectral radius) is indeed the 2-norm because you can write any $x$ as a linear combination of orthonormal eig vectors and things work our neat
stoic pythonBOT
Sydd
brittle gyro
Symmetric matrices are OP
zinc timber
A is self-adjoing or hermitian
zinc timber
self adjoint compact operators on a hilbert space are the sexiest object to work with
brittle gyro
I bet they are... isn't that what they use in quantum mechanics all the time?
but I guess they get unbounded there, maybe less sexy
not really sharp on functional analysis these days
zinc timber
for physicists, all operators are compact self adjoint anyway
brittle gyro
they cheat for sure lol
spare widget
for physicists, all operators are compact self adjoint anyway <:kekw:75202175665...
unfortunately transport equations involve skew-adjoint operators
zinc timber
then I don't care about transport equation
wintry steppe
Hi, I have this matrix $\begin{bmatrix} 3 & -2 & 5 & 1 & \bigm| & 1 \ 1 & 1 & -3 & 2 & \bigm| & 2 \ 6 & 1 & -4 & 3 & \bigm| & 7\end{bmatrix}$
stoic pythonBOT
n/c
wintry steppe
I have reduced it down to $\begin{bmatrix} 5 & 0 & -1 & \bigm| 5 \ 0 & 5 & -14 & \bigm| 1 \end{bmatrix}$
stoic pythonBOT
n/c
wintry steppe
So 5x = 5 + z, 5y = 1 + 14z
wintry steppe
Because if we view them in variables as x,y,z,u for each column
We can reduce down to get u = 0
So I just got rid of it completely
dusky epoch
so what you're ACTUALLY doing is solving a system of linear equations
wintry steppe
Yeah
dusky epoch
not working with a matrix stripped of context
but ok, now that that's cleared up: what did you want to ask?
wintry steppe
Okay so the solution is supposed to be (1,1,0,0) + t(1,14,5,0) according to the back of the book
But clearly since 5x = 5 + z and 5y = 1 + 14z, I don't see how we get that
dusky epoch
perhaps doublecheck your row reduction?
let's ask WA
,w row reduce {{3,-2,5,1,1},{1,1,-3,2,2},{6,1,-4,3,7}}
stoic pythonBOT
dusky epoch
okay so it sounds like this 1 should have been 5
very likely that you just fucked up the arithmetic somewhere
if you show all of your arithmetic i might be able to look over it
wintry steppe
I did $R_3 - 2R_1$ to get $0 \quad 5 \quad -14 \quad 1 \mid 5$
stoic pythonBOT
n/c
dusky epoch
can you show all of your work in one page please
wintry steppe
Then I did $3R_2 - R_1$ to get $0 \quad 5 \quad -14 \quad 5 \quad\mid 5$
stoic pythonBOT
n/c
dusky epoch
i don't want to read it piecemeal
wintry steppe
Okay
$R_3 - 2R_1 \longrightarrow 3R_2 - R_1 \longrightarrow \begin{bmatrix} 3 & -2 & 5 & 1 &\bigm| & 1 \ 0 & 5 & -14 & 1 & \bigm| & 5 \ 0 & 0 & 0 & 4 &\bigm| &0 \end{bmatrix}$
So I work on the matrix $\begin{bmatrix} 3 & -2 & 5 &\bigm| &1 \ 0 & 5 & -14 & \bigm| &5 \end{bmatrix}$ and $5R_1 - 2R2$ gives us $\begin{bmatrix} 5 & 0 & -1 & \bigm| &5 \ 0 & 5 & -14 & \bigm| &1 \end{bmatrix}$
stoic pythonBOT
n/c
wintry steppe
Wait 🤔
Did my 5 turn into a 1 for no reason
Okay so we get $5x = 5 + z$ and $5y = 5 + 14z$
stoic pythonBOT
n/c
wintry steppe
This should still give us $(1,1,0,0) + t(1,14,1,0)$ for the solutions, or am I crazy?
stoic pythonBOT
n/c
dusky epoch
Did my 5 turn into a 1 for no reason
appears so.
so you have x = 1 + z/5 and y = 1 + 14z/5
and z = 0 + z = 0 + 5z/5
to shoehorn z into it
this is unless you are working in a ring where 5 has no inverse
wintry steppe
Oh okay, so to make it easier on the eyes we let t = 5z...
dusky epoch
yes
er
no
not 5z
z/5
hopefully you're not the kind of person who despises the idea of 5 being a valid number to divide by
wintry steppe
hopefully you're not the kind of person who despises the idea of 5 being a valid...
What do you mean?
dusky epoch
**n/c**
you seemed reluctant to divide both sides by 5 here seemingly for no good reason
compact tartan
ay yall im just confused on some notation. does anybody know what the superscript T means in this context
halcyon spindle
It just want you to know it a column vector. T means transpose.
wintry steppe
I am trying to find all solutions for $x + y + 2z + 3u + 4v = 0, 2x + 2y + 7z + 11u + 14v = 0, 3x + 3y + 6z + 10u + 15v = 0$
stoic pythonBOT
n/c
wintry steppe
I found that this reduces to $x + y + 4v = 0, u + 3v = 0, z - 3v = 0$, but how do I write a solution with this information?
stoic pythonBOT
n/c
dusky epoch
.....you really should work on making your systems of equations look more presentable
hopefully you didn't make them hard to read on purpose
but anyway ok
so you have reduced the matrix of the system to this, right?
[ 1 1 0 0 4 | 0 ]
[ 0 0 0 1 3 | 0 ]
[ 0 0 1 0 -3 | 0 ]
Yes
dusky epoch
okay
so you see that you have 3 pivot variables, yes?
they're a little out of order but still
the solution set can be parameterized in terms of y and v, with the other three variables (x, z, u) being expressed in terms of those
wintry steppe
Oh so I can do something like
[ 1 0 0 0 -3 | 0 ]
[ 0 1 0 0 3 | 0 ]
[ 0 0 1 1 4 | 0 ]
Wait no
Wait yes
Is that right? By just rearranging
dusky epoch
if you want to rearrange columns too, then sure...
that's a very good way to make yourself lose track of which variable is where though
fringe fjord
What do you mean by "solutions of a matrix"?
nocturne jewel
How would I find the general solutions of this matrix?
row reduce.
wintry steppe
ur putting in 4 inputs
wintry steppe
why noy a 3x3 matrix?
final lance
because the last column is the answer to the equation
wintry steppe
basically
you havea. 3x3 matrix
times a b c column matrix
to get 0 0 0 column matrix
let me get a pic
lavish jewel
this means your system has infinitely many solutions
wintry steppe
i cant draw the matrix now
lavish jewel
you can assign parameters to two of the variables, and write the third variable in terms of those 2 parameters
wintry steppe
lavish jewel
you know how when you row reduce a matrix, the goal is to make the entries per row correspond to the value of a variable you're solving for?
wintry steppe
sorry for horendous drawing
final lance
you know how when you row reduce a matrix, the goal is to make the entries per r...
yaeh i think so
wintry steppe
basically you want a,b, and c
lavish jewel
well, here you have arrived at cases where the row reduced matrix simply tells you
"0 = 0"
this is always true
regardless of what value your variables take
wintry steppe
ye
lavish jewel
so this matrix is telling you that 2 of the variables have no impact
since 0 = 0 regardless of what value these variables take
so one calls these "free variables"
final lance
yeah
lavish jewel
in some books, this is done by assigning an arbitrary parameter
e.g. y = t, z = s
and then you can write x in terms of s and t
or you simply state that x and y are free, and write x in terms of y and z
in your solution, this is what was done
final lance
ok thank you
fringe fjord
so this matrix is telling you that 2 of the variables have no impact
I think this is dangerously close to misleading. It just shows that two of the equations have no impact, but there wasn't, a priori, any particular variable associated with each equation.
lavish jewel
that's true, i got lazy in the explanation
do take note @final lance , that the real implication of those rows of zeros is just linear dependence of the equations
the parametrization you pick is a separate matter
final lance
I'm not quite sure what linear dependence of the equations or parametrization means can you eli5
lavish jewel
you haven't seen linear independence yet?
final lance
I mean I probably have but I don't know the name for it
My prof shows us how to do stuff but he doesn't really explain why it works or what its called
lavish jewel
ok. sadly i can't just teach you the whole thing either. but the key takeaway is that we could have just as easily taken, for example, x = s and y = t, then expressed z in terms of these two
there was no special reason to say y and z were the free variables
final lance
yeah that makes sense
we could have made x the free variable right
like x and y or x and z
lavish jewel
yep
final lance
ok
somber wigeon
Hello !
In one of my classes we always use a trick that I don't understand. I must be missing some linear algebra knowledge
In a finite $\mathbb{R}$-linear space, say we have a linear endomorphism $\phi$ with eigenvalues $\lambda_i $. We use a polynomial $P$ that sends $\lambda_i$ to $\mu_i$ and then say that $P(\phi)$ has $\mu_i$ as eigen-values. Why is that the case ?
stoic pythonBOT
plougue
fringe fjord
Are you clear about the meaning of P(phi)?
If you are, then you can just pick a v that is an eigenvector for lambda_i, and then compute that P(phi)(v) ends up being the same as P(lambda_i)·v = mu_i·v.
somber wigeon
for some reason I didn't think about putting an eigenvector into P(phi) 🤡 thanks !
paper ether
I already showed that L is linear
To show that L is injective, I need to show that if Lf = 0, for f(x) = a0 + a1 x + a2 x^2, then a0 = a1 = a2 = 0
I get that $$f(1)+f(0)+f(-1) = 3a_0 + 2a_2 = 0$$, so it seems that I have 2 free variables, so the kernel of L is not trivial
stoic pythonBOT
duchat
paper ether
did I do something wrong?
grim cliff
Given two endomorphism $t_1,t_2$
The generalized eigenspace of the pair $(t_1,t_2)$ is the w such that $t_1(w)=\lambda t_2(w)$
First off, does this seem familiar to any of you guys cause I haven't seen this def before
stoic pythonBOT
fajitas
gray dust
yes, this is sometimes called the generalized eigenvalue problem
grim cliff
yes, this is sometimes called the generalized eigenvalue problem
Is there a visualization that can help me think about this idea?
grim cliff
I'm just confused because it looks different from what I've seen before when looking up the generalized eigenvalue problem because I usually just see a single matrix
gray dust
if $A,B$ are hermitian with $B$ positive definite then the generalized eigenvalue problem
$$Ax=\lambda Bx$$
has real gen eigenvalues $\lambda$, and gen eigenvectors corresponding to distinct gen eigenvalues are orthogonal wrt the inner product induced by $B$
stoic pythonBOT
RokabeJintaro
gray dust
if a basis of gen eigenvectors exists then it reduces the quadratic forms induced by A & B, where some visual intuition lies
grim cliff
Do you know where I might be able to learn more about this? I.e. what to Google cause when I look up generalized eigenvector problem what comes up is generalized eigenvector a of a single matrix e.g. a sheer $[[1,1],[0,1]]$ which has a single eigenvector and a single generalized eigenvector
stoic pythonBOT
fajitas
gray dust
careful not to conflate gen eigenvector in jordan normal form & gen eigenvector in gen eigenvalue problem
see eg simultaneous reduction of quadratic forms
errant mist
If I have a positive symmetric nxn matrix in which all the entries A_ij > 0 is this matrix then irreducible? Looking at the definition of irreducible matrices I would think yes since we have that for every index i,j there exists an m s.t (A^M)_ij >0.
wintry steppe
i'm asked to describe a matrix for a transformation from $T_c : P^3 \to P^3$ where $T_c(P) = P(\alpha-c)$ with respect to $B_3 \equiv [1, \alpha-3,(\alpha-3)^2,(\alpha-3)^3]\newline$
I'm not sure where to start..
stoic pythonBOT
leadersheir
dusky epoch
is alpha the indeterminate for your polynomials?
let $P = c_0 + c_1(\alpha - 3) + c_2(\alpha - 3)^2 + c_3(\alpha-3)^3$ and write out $P(\alpha - c)$ in powers of $\alpha - 3$ i guess
stoic pythonBOT
Ann
wintry steppe
is alpha the indeterminate for your polynomials?
it's supposed to be treated like a variable
wintry steppe
let $P = c_0 + c_1(\alpha - 3) + c_2(\alpha - 3)^2 + c_3(\alpha-3)^3$ and write ...
i'm still not sure wht to do.. would u be able to show me matrix please?
dusky epoch
$P(α-c) = c_0 + c_1(α-3 - c) + c_2(α-3 - c)^2 + c_3(α-3 - c)^3$
stoic pythonBOT
Ann
dusky epoch
$= c_0 + c_1(α-3) - cc_1 + c_2(α-3)^2 - 2cc_2(α-3) + c^2c_2 + c_3(α-3)^3 - 3cc_3(α-3)^2 + 3c^2c_3 (α-3) - c^3c_3$
stoic pythonBOT
Ann
dusky epoch
i guess it's just this
collect like terms (carefully)
idk what you were meant to do here other than algebra hell
wintry steppe
i was supposed to figure out a matrix.. i think i know wht to do from here.. thank u 🙌
ripe compass
what happens when you multiply two unit vectors?
outer goblet
can someone explain what happens here
why does T(vj) =-b/a U(vj) lead to it beaing a part of the intersection
dusky epoch
T(v_j) ∈ R(T) by definition
T(v_j) ∈ R(U) by virtue of -β/α U(v_j) = U( (-β/α)v_j ) ∈ R(U)
outer goblet
ah right scalar multiples of a vector means its still in the subspace
spare widget
what happens when you multiply two unit vectors?
If you dot them you get the cosine between them, if you cross them in 3D you get an orthogonal vector to both with length equal to the sine between them.
If you multiply componentwise, then the resulting vector has length <=1.
If they are 4D vectors and the multiplication is the quaternion multiplication, then their product is another unit length quaternion (versor) that represents a rotation equivalent to applying the rotations of the two consecutively.
If you multiply them in the sense of geometric algebra then you get: uv = u dot v + u \wedge v, where u \wedge v is a 2-vector
spare widget
u dot v = cos(theta), and there's probably something one can say about the magnitudes involved in the u \wedge v components, my guess is something involving the sine between the two
zinc timber
proper cradle
<:surprisedpikachu:556851991763091456>
how's the NET?😁
spare widget
u dot v = cos(theta), and there's probably something one can say about the magni...
Ah I think it will be that the determinant formed by the u \wedge v components would come out to +-sin(theta)
half robin
Not sure if this is the right place for this, but does anyone know how eigenvalues and eigenvectors are calculated computationally? I can find a few things like the shifted power method and QR method online but both look terribly slow, they can't possibly be the standard way
lavish jewel
there's also stuff like givens rotations
but generally it's very computationally heavy and iterative
spare widget
lanczos?
lavish jewel
lanczos is also a classic, yeah
lavish jewel
you basically pick an algorithm based on what you know about your matrix, since some methods are faster than others
half robin
I am trying to implement PCA from scratch, so it's the eigenvalues of a covariance matrix that I need.
lavish jewel
then lanczos is a good call
half robin
Okay cool, will look into that and the book. Thanks!
lavish jewel
but tbh this isn't something that makes much sense to do yourself unless it's to learn the algorithms
you're probably going to make a slower version of something that already exists
half robin
Yeah I realize that, it is purely for the sake of curiosity. I'm probably better off just using numpy since PCA is my focus
lavish jewel
aight, all good
you can also just write matrix diagonalization, or eigendecomposition, or singular value decomposition in google scholar
and you will get thousands of hits on small variations for special types of matrices
especially if you type PCA, because machine learning has made (robust) PCA a hot topic and people swarming to it
tired jungle
what's the point of affine space, affine maps, duals, and the different types of mappings, and like orthonormal basis stuff? What is it building up to
wintry steppe
they're fundamental topics in linear algebra
spare widget
what's the point of affine space, affine maps, duals, and the different types of...
you can represent rigid-body motion through affine maps
orthonormal bases you can use for example to extract the best m-component approximation of some data
for example JPEG uses the DCT basis, and keeps only more "perceptually significant" data of images
it's basic tools/notions that are used in a lot of applications
zinc timber
how's the NET?😁
solemn lotus
what would be the basis of the polar coordinate system on R^2? for some reason im struggling to understand how this works
fringe fjord
Polar coordinates are not linear; only linear coordinate systems have bases.
spare widget
Doesn't it have a basis at each point?
In cartesian coordinates that would be (cos(t), sin(t)) and (-r sin(t), r cos(t)) afaik
solemn lotus
ngl i didnt ever consider the concept of a linearity for a coordinate system
how exactly do we define that? im only familiar with the concept of linearity for transformations
severe mica
Hi all,
I wrote an intuitive explanation of eigenvectors and eigenvalues. Thought I'd share it here in case if anybody finds it valuable. It is basically repackaged information by 3Blue1Brown
Constructive feedback is HIGHLY appreciated.
spare widget
how exactly do we define that? im only familiar with the concept of *linearity* ...
Look into manifolds and specifically charts
here's an illustration: https://youtu.be/OMCguyCnTQk?list=PLJHszsWbB6hpk5h8lSfBkVrpjsqvUGTCx&t=233
slow scroll
ngl i didnt ever consider the concept of a linearity for a coordinate system
a coordinate system is linear there exists a linear transformation T such that if x is a point in space, then T(x) is its coordinate representation
oblique prairie
tensor calculus 
wintry steppe
whats the difference between the absolute error loss and the conditional mean ?
besides the fact that absolute error loss uses the median
oh is it just that
conditional median minimizes absolute error loss
and squared error loss is minimized by conditional mean
lavish jay
In this exercise i have to determine the values of a that makes the system has: no solution, 1 solution and infinity solutions
But, in which cases i have to work on
?
i do the gauss matrix
then if i have 2 lines diferrent infinity solution
3 diferent lines one solution
but how about no solution?
dim lotus
put the system in augmented form and row reduce it
then check value of a that make the last row [0, 0, 0, b] where b is nonzero; this means the system is inconsistent
when your pivots are nonzero, the system has a unique solution
when one of the pivots is zero, the system has infinitely many solutions
wintry steppe
if we have a matrix A with a row of zeros, why is detA = 0?
nocturne jewel
if we have a matrix A with a row of zeros, why is detA = 0?
Laplace along that row
you'll have 0det(C_1j)+0det(C_2j)+...+0det(C_nj)
wintry steppe
thank you, Mosh. can you explain it without Laplace by using only elementary transformations (my book mentions that fact before mentioning Laplace and says its obvious, lol).
the rows will be linearly dependent, making the determinant zero
there are a few ways to look at it
ohh i see, thank youu!!
versed rock
lavish jay
,rotate
stoic pythonBOT
tired jungle
What is a frenet serret frame?
Having trouble understanding what it actually is, went over it super quickly in class or maybe I’m just slow in the mornings
I tried looking it up but it seems out of reach on Wikipedia at least for my current level
dim lotus
what i should do next?
can you tell me the value of entry (3,3)?
halcyon spindle
Let look to see when a^2-22 = 0. It happens when a = sqrt(22), -sqrt(22).
slow scroll
What is a frenet serret frame?
idk but if you do #get-advanced-access and ask in #diff-geo-diff-top someone might have an answer
halcyon spindle
So we have a-4 is not equal to 0 at does values.
dim lotus
Plegasus is helping you
lavish jay
can you tell me the value of entry (3,3)?
a^2 -14
dim lotus
do as Plegasus suggested
lavish jay
So we have a-4 is not equal to 0 at does values.
so if a=/= + -sqrt(22) and a= 4 no solution
?
and i continue do that
in the cases for each entry of a
halcyon spindle
At a = sqrt(22) we have the given equation in 0 = a^2 -4 = 22-4. Which is obviously false. So the system is inconsistent for that given a.
wintry steppe
hello, is anyone able to help me with linear
I am just getting started and I'm already confused
halcyon spindle
At a = sqrt(22) we have the given equation in 0 = a^2 -4 = 22-4. Which is obviou...
Also for a = -sqrt(22).
A system Ax = b is inconsistent iff there exist a pivot in the last column of the echelon form of the augmented matrix.
wintry steppe
** is anyone able to VC? I have very basic questions
halcyon spindle
Also for a = -sqrt(22).
If a was any other value we have a^2 - 22 is our last pivot.
lavish jay
which result would lead to each case
halcyon spindle
You have a pivot in every row, and a pivot in every column, so unique solution exist.
Looking at the above screenshot you gave.
wintry steppe
How would you solve a?
nocturne jewel
Put it in an augmented matrix and row reduce
wintry steppe
Aug -> rref or just ref
halcyon spindle
Either case works.
wintry steppe
I am confused about free variable vs. pivot variables
I solved it that way and was still incorrect
halcyon spindle
The pivot variables would either have constant values or be expressed in terms of the free variables. Free variables can take any values.
nocturne jewel
Yeah, you have multiple ways to write a plane/line/etc
nocturne jewel
so your answer might be right, just different initial point or a scaled direction vector
wintry steppe
Why are they using that initial point?
dim lotus
yea you can say z=z since it is a free variable aka not zero in a pivot position
wintry steppe
z isn't in a pivot position :( ??
why are pivots so special and hard to alter ?
Dear Eve: Basic var = 1 assigned value, free variable can be anything
x1, y1, z(1,1), w(1,-1) ? but z has 2 options
but they are thesame
dim lotus
pivots are entries where i = j and we like them to be nonzero. if they are zero, they correspond to a free variable
find the equation of the line. you have the slope and the point (-2,2) and these two are enough to find the equation of the line. once you have the equation of the line, plug 5 for x and u is the output
wintry steppe
i am an undergraduate linear algebra
lunar sorrel
what are the rules, generally, of matrix manipulation?
so if i want to turn something into row echelon, i can do whatever i want, so long as i change the entire row?
wintry steppe
it's an nice exercise to find those by yourself
teal grotto
so if i want to turn something into row echelon, i can do whatever i want, so lo...
search up elementary matrices
teal grotto
so if i want to turn something into row echelon, i can do whatever i want, so lo...
but essentially, yes. if you do the same operation to each entry in the row
or add one row to another
wintry steppe
but essentially, yes. if you do the same operation to each entry in the row
if by operation you mean multiply both sides by a nonzero element yea
half ice
the bottom row is z=z ?
The bottom row is 0 = 0. z is free since there's no pivot in the third column.
halcyon spindle
Yes.
lunar sorrel
ty ty
spiral osprey
I need some help/explanation for this project I need to do
Create a 4x5 augmented matrix and draw a network corresponding to your augmented matrix. Remember, your system needs to be consistent (include your row reduction into the project with all the work shown).
What does she mean when she says "Network corresponding to your augmented matrix"
zinc timber
Hey can I have a hint here
are you aware of operator norms and spectral radius?
blissful thicket
unfortunately not
this exercise is presented right after introducing inner products and their properties
zinc timber
ok say $T-\sqrt{2}I$ is not invertible then there is one non-zero vector $v$ s.t $(T-\sqrt{2}I)v = 0$
stoic pythonBOT
zinc timber
show that this gives norm |Tv| > |v| contradicting the assumption
spiral osprey
K so she means a network with flow and analysis
she said if we watched 1.5 we would understand what a network is, but we went over networks in 1.6 so thats where the confusion came from
Um ok so next question
So If I make something like this
Doing the augmented matrix first then the network would be stupid right?
I should just make the network and then solve
because to be fair I have no clue how I would do it backwards
stoic pythonBOT
Potato
outer goblet
can someone reccomend me a youtube plaaylist where i can study how to solve lin alg proofs for linalg 1/2 , my lecturer is completely useless and talks about completely unreleated things
outer goblet
every time i tried to search for something in google all i get is stupid videos on the most basic things, not the stupid proofs that dont make sence
like book ir yt video on how to do linear algebra, but with basicailly never actually writing down the matricies, cuz this is how 90% of my homework is
tender dragon
find the coordinates of the points of intersection of the curve 2/x + 3/y =13 and the line 2x + 3y = 2
brittle gyro
stuff around Perron's Theorem is driving me nuts... how can I show that $\rho(A)<1 \implies \lim_{k \to \infty} A^k =0$?
stoic pythonBOT
Sydd
lavish jewel
rho(A) as in spectral radius?
stoic pythonBOT
Sydd
lavish jewel
the usual proof is to pick an eigenvector of A
and study A^k v
then since v is an eigenvector, this is equivalent to lambda^k v
brittle gyro
But what if I can't get a basis of eigenvectors?
lavish jewel
as abs(lambda) < 0, lambda ^k -> 0
we don't need a basis, just one eigenvector associated to the largest eigenvalue
but if you want you can consider an upper triangular form of A
brittle gyro
we don't need a basis, just one eigenvector associated to the largest eigenvalue
I dont get this. To show A^k -> 0 I would need to show A^k v-> 0 for any and all v, right?
lavish jewel
ah, this is exactly how it's done in wikipedia
if A is diagonalizable, what i wrote suffices. otherwise, you argue using the JCF
brittle gyro
if A is diagonalizable, what i wrote suffices. otherwise, you argue using the JC...
is this Jordan form?
brittle gyro
Does it make any difference if I take Jordan form or Schur? As in upper triangular?
lavish jewel
hmmm
i've never seen it done with schur decomp, you can try
it still holds that A^k = QU^kQ^-1 for the schur decomp, but i don't think U has any special structure other than being upper triang
whereas the jordan form has the eigenvalues (and ones) in blocks
brittle gyro
Yeah I guess it's unnecessary, Schur helps if you need that the similarity matrix to be orthogonal
but in this case I guess whatever
anyways, I just missed the entire exposition on wikipedia, will take a look now, thanks!
zinc timber
If I'm correct schur only works on C right?
tho spectral radius doesn't make sense for real matrices so ig it's ok
or does it?
wintry steppe
I need a hint for this problem: If a square matrix has a row or column of zeros, prove that it is singular.
Can't use determinants.
Nonsingular matrix A is defined as exists B such that AB = BA = I, and singular is defined as (not) nonsingular
zinc timber
you can also try to show that there is a non-zero vector v with Av=0
wintry steppe
Oh ok
zinc timber
lavish jewel
let's see
we wanna take a transformation f: V -> V
V is R^3, and f is the identity transformation
you can chain so-called "change of basis" transformations
for example, we know the identity transformation in the canonical basis is the identity matrix I_3 here
so what one can do is take a vector in the basis B, convert it to the canonical basis, apply the transformation I_3, and convert it to the basis B'
if we associate a matrix M to the basis B, and a matrix M' to the basis B', the overall matrix is M'^-1 I_3 M
and the columns of M and M' consist of the basis vectors in B and B', respectively
hopefully someone will double check i didn't mess up the order, but that's the basic idea
grave garden
( im here and processing about it )
lavish jewel
the intuition behind these "change of basis" matrices is the interpretation of matrix multiplication as a linear combination of the columns of a matrix
so if a vector v is given in some basis B, and we express this basis as a set of vectors that are themselves given w.r.t. the canonical basis, as was the case here, then we can go from the coordinates in the basis B to the canonical basis by simply taking Bv = w, where w is the vector now in the canonical basis
from this same expression, one can see that if you are given a vector w in some basis (here, the canonical), and we want to find its coordinate vector v in some new basis B, then this is done by solving for v in Bv = w, so that v = B^-1 w
note that there's no need to interpret it this way, someone else might have a different/better explanation
this is equivalent to the first equation in the first image you shared, btw
they write it as v = sum x_i v_i, where the v_i are vectors in V that form a basis. this can be rewritten as v = Mx, for a matrix M whose columns are the v_i
v is the vector expressed in the canonical basis, and x is its coordinate vector in the basis of the v_i (the columns of M)
i'll go ahead and double check in octave while you digest this
grave garden
Yes yes
lavish jewel
do you prefer reading matlab or python code
grave garden
I know some python code
lavish jewel
ok
here's the setup of the matrices we're dealing with, using the same names i wrote above
and here's the result, which matches what the solution in your image shows
for practical means and purposes, -5.55e-17 is close enough to zero
lavish jewel
maybe someone can fill the gaps while i eat :x
grave garden
Is there a name for that notation ?
grave garden
maybe someone can fill the gaps while i eat :x
It's okay, enjoy it mate ! I might go to sleep soon too as it is night here
wintry steppe
computational linear algebra is not linear algebra
karmic dagger
How should I go about the proof on this?
I was thinkingof maybe using the lemma that 2 subspaces' union is only a union if one of them cntains the other
to reach a conclusion that all the subspaces are contained in one subspace
which would then be equal to V
nocturne jewel
Not seeing anything blantantly wrong w/ that
lavish jewel
is only a subspace*
wintry steppe
How does this show that it's true for every matrix A? They just showed it's true for A = I? What
hardy inlet
neither me or my my few classmates I know; know what this means/is asking
what are ABCDXY
teal grotto
linear maps from V to V
hardy inlet
so like $A,B,C,D \in \mathcal{L}(V,V)$
stoic pythonBOT
MattDog_222
teal grotto
also linear maps
teal grotto
uh
idk what u have in class sometimes
i think ik what u mean tho
like, ST is shorthand for S o T
same idea here
teal grotto
ye
nocturne jewel
They have to be for the matrix multiplication / transformation composition to make sense
hardy inlet
i honestly dont know where to start. I assume its a "direct" proof
teal grotto
add the two equations
maybe subtract them
see what you get
try and solve for X and Y
hardy inlet
oh like AX + BX + AY + BY = C + D or the other way
hardy inlet
is this useful or even legal
teal grotto
Yes. now your next line should be (A+B)(X+Y)=C+D
A+B is invertible
so you can solve for X+Y
next try to solve for X - Y
hardy inlet
why is it (A+B)(X+Y) and not (X+Y)(A+B)
nocturne jewel
matrix * isn't commutative
hardy inlet
i dont think i did it right
oops i forgot the inverse on the RHS but idk what good it does
😬
thats the 
mean
I haven't prooven anything idk what to do once i get X+Y X-Y
teal grotto
you have expressions for X + Y and X - Y
add them together to get an expression for 2X
now you have solved for X
so you can solve for Y
then in your proof, you can choose to show none of this work and make it look trivial by saying, here: this is the X that works and this is the Y that works. qed
hardy inlet
but 2X != X 
teal grotto
divide by 2 dawg
teal grotto
assuming you're not working over a field where 1 + 1 = 0, then yea
hardy inlet
is Z_2 a field?
teal grotto
ye
teal grotto
you can solve for Y now yes
hardy inlet
this feels messy
teal grotto
yea but u can simplify that
hardy inlet
how tho
teal grotto
how would u simplify 7 + (x+10)/2
hardy inlet
12 + x/2
hardy inlet
(24+x)/2?
teal grotto
ye
teal grotto
this feels messy
now just do that same thing here
teal grotto
ye
teal grotto
no they're different expressions
hardy inlet
theres a negative
so is it just done. Are 'we' trying to show that there exists a "combination" / "composition" of A B C and D
teal grotto
well, for a formal write up, i wouldnt show any of this scratch work
i would just show that those specific choices of X and Y work, since you need to give me the X and Y first
instead of assuming they exist, then reverse engineering what they should be
hardy inlet
so I need to put X and Y back into the original?? 🤮
teal grotto
yea. and just check that it works
however, this scratch work shows that if such X and Y exist, they must take on those specific forms, so they are unique
teal grotto
neither me or my my few classmates I know; know what this means/is asking
no. you have to show there exists X and Y satisfying these equations.
- Write down what X and Y should be.
- Show that they satisfy the equations
hardy inlet
is that not what this is doing?
teal grotto
okay, that is
teal grotto
i cant even fit it in the margin but like this?
idk what this was doing
hardy inlet
it was the same thing but I accidentally wrote Y=
hardy inlet
no clue how this simplifies to C
teal grotto
combine like terms
the first and the third terms look similar
second and fourth also look similar
teal grotto
is that not what this is doing?
terms of this last fraction
hardy inlet
did I ever say I hate math?
teal grotto
lol
hardy inlet
Im not spending a fortune printing in color so this is staying a disgusting black and white mess
teal grotto
i hate having to print my hw out
like, im typing it online. can it just stay online?
hardy inlet
ikr
but with this snow they might have us turn it in digitally idk
its sorta helpful to see physical error corrections
teal grotto
yea its snowing crazy where i live too
hardy inlet
did I ever say I hate math?
now we're not really supposed to start with "C = ... = C" right
its supposed to be AX + BY = ... = C
teal grotto
you dont have to. the way you have it written now, starting from top to bottom, i wouldnt
but
if you wanna look super cool
hardy inlet
👀
teal grotto
you could start
from C = ((C+D)+(C-D))/2
and work your way backwards
to end up at C = ... = AX + BY
hardy inlet
teal grotto
what a chad
teal grotto
i have the flammable maths hoodie that just has $\lfloor 10^2\ln2\rfloor$
stoic pythonBOT
c squared
teal grotto
printed on it
teal grotto
nice
hardy inlet
next is a 1 direction (no not the band) proof. not sure if this is supposed to be some quadratic formula stuff
pretty certain that showing its surjective or injective works
tired jungle
What’s the dual of a dual? What would it look like?
teal grotto
solve for I. factor
in the finite dimensional case, its isomorphic to the original space
hardy inlet
I = A - A²?
teal grotto
should be I, not 1
hardy inlet
hardy inlet
(whats next )
teal grotto
thats it
hardy inlet
is that some formula or something?
teal grotto
somebody else asked a question about double dual spaces
hardy inlet
sowa it is my answer sufficient or was "double dual spaces" an unrelated topic
nocturne jewel
You haven't shown (I-A)A=I
hoary void
(not part of anything academic but i was just wondering) where did you get that? i would love to buy one
hoary void
nocturne jewel
probably
hoary void
alright ty
nocturne jewel
teal grotto
You haven't shown (I-A)A=I
doesnt really matter since hes working in a finite dimensional vector space and the maps are from a vector space to itself
hardy inlet
the short one (the one in the pic) has been out of stock for a long time; but yes 3B1B shop
nocturne jewel
Oh Is commutativity guaranteed in fin-dims?
teal grotto
if you know that AB = I
then it can be shown that BA = I as well in the finite dimensional case
nocturne jewel
yeah not commutativity that
hardy inlet
so do i only need to show AB?
teal grotto
it depends
do you know, or have you shown in your class, that if AB = I, then BA = I?
if you have, then just make sure to cite the result (if you can do that)
otherwise, just show that (I - A)A = I (which is trivial)
hardy inlet
idk if we've shown it or not but I guess i'll show the trivial proof
idk if that is correct sicne i used the given condition that says it equals I
quartz compass
idk if that is correct sicne i used the given condition that says it equals I
it's probably worth knowing that you can prove that the left inverse is equal to the right inverse always, so you don't have to prove it for a special case
it's easy to prove too, it's possible to do it in one line, I suggest giving it a try and I'll give you some hints or show you after a bit if you're still stuck
nocturne jewel
idk if that is correct sicne i used the given condition that says it equals I
If we're being technical: You assumed A was invertible half way through
So the 2nd half is null
little frigate
Any suggestions regarding how to quickly find whether that matrix is invertible or not?
I found that it was singular using wolframlpha
nocturne jewel
Quickest way (That isn't trial and error/Gauss-Jordan) is determinant
little frigate
Yeah I was initially planning to do Gauss-Jordan for something else related to the inverse,
hardy inlet
cant u also augment with I_5 and row reduce to see if the LHS = I_5?
little frigate
I'll look into it, thanks
oh I'm stupid
yeah it has to do with the rank, so rank(A) = n
nocturne jewel
cant u also augment with `I_5` and row reduce to see if the LHS = `I_5`?
That's Gauss-Jordan
hardy inlet
oh ok. idk terminology :(
hardy inlet
If we're being technical: You assumed A was invertible half way through
is the first half good then?
hardy inlet
this part?
nocturne jewel
wdym
I said 1st half
but you just do that backwards basically
then you haven't assumed invertibility
hardy inlet
so this is wrong?
teal grotto
first column plus third column is second column
cant be invertible (because the rank - the number of linearly independent columns - is at most 4, since one column is a linear combination of two others. invertible matrices always have full rank)
nocturne jewel
so this is wrong?
Please read what I'm saying
I said the 2nd half is wrong
The 1st half is fine and correct
outer scaffold
Hi sorry to hijack the channel but can someone tell me how did the solution arrive at the conclusion that "a + b = -3 and a.b = 3"?
nocturne jewel
Vieta's Results
outer scaffold
OHHHH
TIL
Damn thank you that was the quickest help I've ever received lol
Also now I remember studying it a few years back
worldly bear
can’t you just use quadratic formula on it to find a and b
nocturne jewel
Unneeded computation
and 1/a+1/b follows directly from Vieta's
However that goes more in #prealg-and-algebra / #precalculus, as it's quadratics
hardy inlet
This better?
teal grotto
before you say "Therefore ...."
just write <==> I = (I - A)A
its easy enough to see
cuts down on 50% of the other stuff you've written
and if you want you could format this to fit on one line ✅
hardy inlet
idk if we can use I = AA^-1 <=> A^-1A = I
teal grotto
thats true because thats the definition of an inverse element
and ur not explicitly writing that. literally just put this
$$A^2-A+I=0\Longleftrightarrow I=A-A^2\Longleftrightarrow I =A(I-A)\Longleftrightarrow I=(I-A)A $$
stoic pythonBOT
c squared
teal grotto
one liner
wintry steppe
What exactly is a vector?
teal grotto
an element of a vector space
wintry steppe
Okay, what does a vector look like?
teal grotto
it depends on how your vector space is defined
wintry steppe
Also is there a name for that weird looking R we can notate dimensions with?
sleek sundial
R^n?
wintry steppe
R^2 is a regular Euclidean planes dimensions
sleek sundial
r is just the reals
wintry steppe
I know it’s a tuple but is there a specific name?
teal grotto
the reals
wintry steppe
r is just the reals
The reals?
teal grotto
the real numbers
fringe fjord
The $\mathbb{R}$ symbol in particular?
stoic pythonBOT
Troposphere
teal grotto
thats black board bold font
wintry steppe
**Troposphere**
Yes that
nocturne jewel
Also is there a name for that weird looking R we can notate dimensions with?
Yes, it's the Reals
sleek sundial
yes
nocturne jewel
Yes
teal grotto
yes
nocturne jewel
As opposed to the not real numbers being called the reals
fringe fjord
Opposed to imaginary and complex numbers.
wintry steppe
I need to go back over algebra, teachers did it poorly last year
nocturne jewel
So the tuple R^n is meant to represent dimensions correct?
R^n is n-dimensional space
wintry steppe
Okay
hardy inlet
**c squared**
didn't do the one liner since the teacher doesn't really like a long string like that
nocturne jewel
Yeah that's better
fringe fjord
R^n is not itself a tuple, but is the set of all n-tuples of real numbers.
teal grotto
do you know how to use the align environment?
wintry steppe
Wikipedia does a better job at explaining then my teacher for algebra 1 did last year
teal grotto
because 
nocturne jewel
$\mathbb{R}^n:={[x_1,...,x_n]^T|x_i\in\mathbb{R}, 1\leq i\leq n}$
stoic pythonBOT
Mosh
hardy inlet
R² = <x, y>
R^3 = <x, y, z>
R^4 = <w,x,y,z>
...
R^n = <v_1, v_2, ... v_n>
wintry steppe
R^n is not itself a tuple, but is the _set_ of all n-tuples of real numbers.
Okay that makes sense as to why it’s not in parentheses
So would this also be valid: R^(n)?
Probably not just a random thought
fringe fjord
That would be confusing at best.
nocturne jewel
I've never seen anyone write R^(n)
hardy inlet
do you know how to use the align environment?
wintry steppe
Would it still be valid? Confusing or not.
nocturne jewel
<:sadcat:547124024585486359>
$\iff$
stoic pythonBOT
Mosh
teal grotto
\begin{align*}
A^2-A+I=0&\Longleftrightarrow
I=A-A^2\&\Longleftrightarrow
I=A(A-I)
\end{align*}
nocturne jewel
iff
wintry steppe
iff
Right
I’m going to go watch an algebra lecture, I’ve forgotten most of it already.
fringe fjord
There are some other contexts where a parenthesized superscript has a different meaning an one without parentheses, so R^(n) will make people wonder if you're sing such a notation and what you mean by it.
wintry steppe
I see I see
stoic pythonBOT
c squared
nocturne jewel
commonly $f^{(n)}(x):=\dv[n]{x}f(x)$
stoic pythonBOT
Mosh
teal grotto
ive seen (R^n)^* for the dual of R^n
hardy inlet
$Y_{(n)}$ order statistic
stoic pythonBOT
MattDog_222
wintry steppe
ive seen (R^n)^* for the dual of R^n
Interesting
hardy inlet
**c squared**
hardy inlet
ty
typo
i feel like im gonna have to multiply by something but I cant multiply by T^-1 bc thats what we wanna find right. i feel so lost
zinc timber
is S invertible?
teal grotto
it's probably worth knowing that you can prove that the left inverse is equal to...
this one
teal grotto
do you know, or have you shown in your class, that if AB = I, then BA = I?
or, this one really
hardy inlet
so like STU = I = UTS?
teal grotto
yes, but here parentheses imo are important
zinc timber
think of it as $S(TU)$ then reverse the order
stoic pythonBOT
hardy inlet
S(TU) = I = (TU)S = T(US) <--> US = T^{-1}
zinc timber
that's why u need the invertibility of S, which I have been asking
hardy inlet
nothing is stated as invertible
hardy inlet
oh i gues so
its just asking for 1 of 3 inverses i see
since S, T, U, all have inverses
S^-1 = TU
U ^-1 = ST
zinc timber
nothing is stated as invertible
teal grotto
ryu, got any interesting problems?
hardy inlet
oof
teal grotto
one of those problems where you just gotta brute force it
hardy inlet
sufficient?
zinc timber
one of those problems where you just gotta brute force it
ig you can try it, A be a real matrix with positive (non-m
negative) entries, then A has a positive (non-negative) eigen value
teal grotto
hurb
hardy inlet
u guys do math for fun?! 
teal grotto
is this a true or false? or just like, prove the result
hardy inlet
(i guess im the same way with computer science)
zinc timber
do is a bold assumption
teal grotto
u guys do math for fun?! <:sippy:920835242980442113>
im procrastinating so hard rn
zinc timber
is this a true or false? or just like, prove the result
lol prove or disprove for added difficulity
hardy inlet
im procrastinating so hard rn
I do the same thing in this programming server. Thank you for your service
zinc timber
im procrastinating so hard rn
not more than me
teal grotto
what are u procrastinating?
hardy inlet
doing this weekly linear algebra assignment until thursday night
teal grotto
i feel that
zinc timber
also project work
teal grotto
yea, thats more than me. i just have two hw assignments that i half way finished. i dont feel like doing all the typing on latex to finish them
due tmrw
hardy inlet
20?
teal grotto
eff word
zinc timber
2 assignments actually
teal grotto
10 pm his time
stoic pythonBOT
The current time for Ryuzaki is 08:57 AM (IST) on Fri, 18/02/2022.
teal grotto
bruh lol
hardy inlet
,ti
stoic pythonBOT
You haven't set your timezone! Set it using the interactive timezone picker with ,ti --set.
hardy inlet
,ti --set cst
teal grotto
,ti
stoic pythonBOT
Your timezone has been set to CST6CDT!
Your current time is 09:27 PM (CST) on Thu, 17/02/2022.
You haven't set your timezone! Set it using the interactive timezone picker with ,ti --set.
teal grotto
Your timezone has been set to `CST6CDT`!
Your current time is **09:27 PM (CST)**...
this is incorrect
stoic pythonBOT
The current time for MattDog_222 is 09:28 PM (CST) on Thu, 17/02/2022.
teal grotto
,ti --set est
stoic pythonBOT
Multiple matching timezones found, please select one!
1. America/Creston 08:28 PM
2. Australia/West 11:28 AM
3. Brazil/West 11:28 PM
4. EST 10:28 PM
5. EST5EDT 10:28 PM
6. Europe/Bucharest 05:28 AM
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Please type the number corresponding to your selection, or type `c` now to cancel.,ti --set EST
stoic pythonBOT
Your timezone has been set to EST!
Your current time is 10:28 PM (EST) on Thu, 17/02/2022.
teal grotto
how much snow did u get today matt
zinc timber
ig you can try it, A be a real matrix with positive (non-m
negative) entries, th...
this is true
hardy inlet
prob like 6 inches
teal grotto
think we got like 4 or 5 where im at
hardy inlet
got more when it hit like 2 weeks ago
teal grotto
supposed to snow tonight too. i have to be up early as well
hardy inlet
damn
hardy inlet
my statistics teacher cancelled irl class but I still have to go in to turn in the paper copy of this homework :(
teal grotto
eff
stoic pythonBOT
teal grotto
this is true
i actually dont know where to start
hardy inlet
yea, thats more than me. i just have two hw assignments that i half way finished...
i'll be ur scribe
hardy inlet
give me what u want typed and I'll type it while u help me think through the next problems 
teal grotto
but then i have to type what i want typed
or i have to write legibly, and just cant do that
hardy inlet
wait do u not even have it on paper?
teal grotto
no
i have outlines in my head for sketches of proofs that im hoping are right and easy to turn into full proofs
hardy inlet
dang
teal grotto
am lazy
hardy inlet
wanna do an easy one?
only thing idk is if its supposed to be L(R^4, V^4) or V
I "took" abstract algebra so I know why its iso
generally speaking
hardy inlet
4 = dim V4 = dim(V) + dim(V) + dim(V) + dim(V) = 4 dim(V) = 4
teal grotto
never said that V was finite dim
teal grotto
R^4 is finite dimensional tho
hardy inlet
i kinda dont know what it means for these to be iso
teal grotto
i kinda dont know what it means for these to be iso
there is a bijective, linear map between the two spaces
hardy inlet
ye i know the bijection nature but I more specifically dont know what L(R^4, V) is
teal grotto
that vector space is the set of all linear transformations from R^4 into V
hardy inlet
isn't that saying $V^4 \cong V$?
stoic pythonBOT
MattDog_222
teal grotto
no
so like, given a v = (v1, v2, v3, v4) in V^4, can you give me a linear transformation from R^4 to V?
hardy inlet
theres only 4 ways right
R4 → V
<a,b,c,d> → a
<a,b,c,d> → b
<a,b,c,d> → c
<a,b,c,d> → d
anything else isn't a bijection
well constant times
ca, cb, c"c", cd
teal grotto
R4 → V
<a,b,c,d> → a
<a,b,c,d> → b
<a,b,c,d> → c
<a,b,c,d> → d
im not really sure what this means
hardy inlet
fair point. like we take the 1st component and map that to the singular component of V, and do that for all 4
since u cant do a + b for example
because then its not a bijection
teal grotto
so like, given a v = (v1, v2, v3, v4) in V^4, can you give me a linear transform...
a linear transformation is defined by what happens to its basis elements. pick any basis of R^4 (use the standard one for now). try and relate this basis to the element v in V^4 defined up here
hardy inlet
a[1,0,0,0] + b[0,1,0,0] + c[0,0,1,0] + d[0,0,0,1] = [av1, bv2, cv3, dv4]
hardy inlet
but thats a transformation from R4 to V4 not R4 to V like L says
teal grotto
for each v in V^4, you have given me a unique linear transformation in L(R^4,V)
sounds like a map from V^4 to L(R^4,V) to me
hardy inlet
sorry some neurons arent connected rn
i gave an isomorphism that maybe looked like this
$$\phi : \mathbb{R}^4 \to V^4$$
$$\phi(a, b, c, d) = (av_1, bv_2, cv_3, dv_4 ) \in V^4$$
stoic pythonBOT
MattDog_222
teal grotto
no
you gave a map that looked like this:
$\Phi:V^4\to\mathcal{L}(\mathbb{R}^4,V)$
$$\Phi(v)=T_v:\mathbb{R}^4\to V$$
where $$T_v(x_1,x_2,x_3,x_4)=(x_1v_1,x_2v_2,x_3v_3,x_4v_4)$$ if $v=(v_1,v_2,v_3,v_4)\in V^4$.
stoic pythonBOT
c squared
teal grotto
ig you can try it, A be a real matrix with positive (non-m
negative) entries, th...
does this have to do with the spectral radius or something?
or is there an elementary approach?
zinc timber
you'll be surprised but it's more of a topology problem than a LA
teal grotto
hmmm
hardy inlet
this is what I dont understand
teal grotto
ive made this like, as clear as i can lol. what part specifically dont you get? the circles arent making clear what ur not getting lol
hardy inlet
T_v maps to V but when defining Tv u say its contained in V4
wouldn't that be $T_v : \mathbb{R}^4 \to V^4$
stoic pythonBOT
MattDog_222
teal grotto
i meant plussssss
eff
$\Phi:V^4\to\mathcal{L}(\mathbb{R}^4,V)$
$$\Phi(v)=T_v:\mathbb{R}^4\to V$$
where $$T_v(x_1,x_2,x_3,x_4)=x_1v_1+x_2v_2+x_3v_3+x_4v_4$$ if $v=(v_1,v_2,v_3,v_4)\in V^4$.
stoic pythonBOT
c squared
hardy inlet
right
teal grotto
like, \Phi(v) is a linear map. and i am relabeling it T_v (for some reason, you dont have to).
point is, \Phi(v) is the map which takes the vector (x1,x2,x3,x4) in R^4 to the vector x1v1 + x2v2 + x3v3 + x4v4 in V
it should be easy now to show that Phi is linear, i.e., Phi(av + w) = a Phi(v) + Phi(w). then you can either show that Phi is injective and surjective, or just give the inverse of Phi, and show the inverse is also linear
hardy inlet
i suppose its an isomorphism
aTv(x1,x2,x3,x4) = Tv(ax1,ax2,ax3,ax4) and same with addition
hardy inlet
to show injective we just show that phi(0,0,0,0) is the only thing that gives zero right
teal grotto
yea. Phi(v) = 0 if and only if v = 0
hardy inlet
i dont see how thats true. we cant uncollapse a dim1 to a dim4
teal grotto
you'll be surprised but it's more of a topology problem than a LA
Phi(v) = 0 means that T_v(x) = 0 for all x in R^4
zinc timber
teal grotto
wrong thing
zinc timber
teal grotto
i have an outline for ur question ryu
zinc timber
teal grotto
let r be the largest magnitude of an eigen value of A.
then the largest magnitude of an eigen value of A/r is 1, so there is some eigen value of A/r on the unit circle
i just need to show that it has to be 1
zinc timber
what if that eigen value is complex?
zinc timber
it does matter
teal grotto
no like, the eval of A/r should be 1, the one i specified
zinc timber
are you not trying to show that the ev corresponding to magnitude 1 will be the eigen value we were looking for?
say we have Eigen values 2, 4i, -4i then A/r will have ev ½, i, -i
how are we to know there's one eiten value with mag 0<ev≤1
teal grotto
shoot
zinc timber
🙈
hardy inlet
im gonna take a pause whilst I dont ponder
teal grotto
thats less or equal to one
zinc timber
x²+y²+z²=1
teal grotto
where u gettin that from dude lol
zinc timber
in R³
teal grotto
the map from R^n cut zero to R by |Ax|/|x|
teal grotto
im just not seeing this. thought A was a map from R^n to R^n. its a matrix with positive entries
zinc timber
oof