i cant figure out why my variance covariance matrix wont work
#linear-algebra
ocean sequoia
soft burrow
@tall surge you know about the cofactor expansion right? https://en.wikipedia.org/wiki/Laplace_expansion
(I wrote something previously but I was being dumb, anyway you should be able to calculate det(J) using that)
tall surge
you know about the cofactor expansion right? https://en.wikipedia.org/wiki/Laplace_expansion
@soft burrow yep I know this but this is way too bashy
I'm looking for some type of way to manipulate the block matrices to calculate the determinant
well the goal is actually to find the eigenvalues
so it'd be more helpful if someone could lmk about that
warm niche
Oh I finally get basis
I've always thought it was very abritrary that basis had to be defined based on a (euclidan) standard basis
but then when I learned change of basis
It makes sense that u cant change /define a basis unless u have some sort of standard
Unless I'm stretching my point?
prisma cairn
Yes you can define a basis without any sort of standard. All the basis is a set of linearly independent vectors that span the vector space. To confirm that this hold for a set of vectors you don't really need to be aware of the standard basis
dusky epoch
a basis is usually called standard when it has some properties that set it apart (intuitively or otherwise) from other bases, and most of the time this designation is somewhat arbitrary
i prefer to think of it as "the basis your space comes with, if any"
warm niche
@dusky epoch Ah I think that encapsulates what I was thinking
So in reality it is "abritrary"
But it is it intuitive
dusky epoch
like if it's the space of all polynomials of degree at most n
then you get the monomial basis {1, x, x^2, ..., x^n}
which is how you usually think of polynomials anyway
but the monomial basis isn't always the best to work with, depending on what exactly you're doing
warm niche
@prisma cairn When I thinking about this: I was thinking about vector space in Rn
I was thinking in particular not necessarily the entirety of the vector space but the very elements in the vectors themselves
Those numbers don't come out of thin air, but I think Ann perfectly explains it
prisma cairn
I don't really get what you mean, but ok yeah
limber sierra
are you wondering why the entries of vectors in R^n are real numbers? thats kind of... by definition
halcyon pollen
wanna learn linear algebra man
wintry steppe
didn't you f in math though
native rampart
f
mental pawn
Hello Peeps I need help
Is anyone available?
I dont want to come in in the middle of a problem
native rampart
Just ask
mental pawn
I have an augmented matrix
I am unsure what operation I need to perform to find the distribution of grassland, shrubland, forest at the end of the year
or how you recognise which operation to perform
I can paste the matrix
mental pawn
Yeah
dawn remnant
so just write down the recurrence equations - the distributions of grassland, shrubland, forest at year (n+1) in terms of them at year n.
Then rewrite that as a multiplication of a matrix by a vector:
$$
v_{n+1} = A v_n
$$
stoic pythonBOT
ConfusedReptile:
dawn remnant
and then a) is just finding the inverse of the matrix and applying it , and b) is finding eigenvectors of the matrix.
dawn remnant
it's a cool task by the way, I wish our algebra teacher gave us more such tasks and fewer hardcore proofs 😅
tall surge
hi i still need help on how to find the eigenvalues of this matrix https://images-ext-2.discordapp.net/external/oGvwtPNjHul4iAk6KpM8jJili3iRbdzpfBs5onEGPdA/https/media.discordapp.net/attachments/540211747613704221/759927441958895666/image0.png
mental pawn
I just wish I understood it... this is our final assignment and I am still trying to get my head around it
tall surge
think of each grid box as a block matrix
dawn remnant
@tall surge what's in the fourth box?
dawn remnant
only zeros?
dawn remnant
I just wish I understood it... this is our final assignment and I am still trying to get my head around it
@mental pawn What part do you not understand?
mental pawn
So I haven't even practiced one of these
it's brand new
I need repetition lol
so writing the equation formally
as you've done
I'm just thinking about what that would be with the numbers I have
okay so just to clarify ... part a is just the inverse of the square matrix
dawn remnant
well, if A transforms $v_n$ into $v_{n+1}$, $A^{-1}$ transforms $v_n$ back to $v_{n-1}$
stoic pythonBOT
ConfusedReptile:
mental pawn
@tall surge Just because people are asking questions ... doesn't mean they are stupid
dawn remnant
@tall surge that's not it, though, it's very much linear algebra...
so yeah, to solve a) the simplest way is just to calculate the inverse and apply it to the vector provided
mental pawn
Yeah but I'm unsure from this problem what the vector is
tall surge
Just because people are asking questions ... doesn't mean they are stupid
@mental pawn what????? when did i say you were stupid
mental pawn
I suppose it's the ifnal values of the shrubbery
tall surge
lol
mental pawn
sorry I misinterpreted your question
lol
er your statement
I misinterpreted you Jay007
lolol
tall surge
yeah i got u lol i just said that bc I had a question 😂
mental pawn
Okay well I wrote all that donw
down
ANd now I am going to experiment with it
thank you @dawn remnant
dawn remnant
Yeah but I'm unsure from this problem what the vector is
@mental pawn
Just $\vec{v_n} = (D_n,S_n,A_n)^T$
mental pawn
So I did something where I solved it by rref and came out with some numbers ...
stoic pythonBOT
ConfusedReptile:
dawn remnant
(it's a column vector, but I don't know how to write those in LaTex, so I wrote it here as a transposed row vector 🙂 )
dawn remnant
not sure what you mean
ah
See, that's like saying "from the question it doesn't seem clear what equation do they want me to solve" 🙂
mental pawn
🙂
dawn remnant
The question doesn't mention at all what to do. It wants you to solve a task.
How to do so is up to you. It just happens that linear algebra is one of the tools that can help, and this is why this question is in a linear algebra course despite not strictly speaking mentioning any matrices.
dawn remnant
you can choose the vector to be (D,S,F) and get one form of A, or to be (F,S,D) and get a different form (with swapped rows, I think)
whatever way you decide to do it, you should obtain the same answer in the end, because ultimately this task is about certain (linear) recurrence relations
mental pawn
Ohh
I really appreciate your input thanks
is it like a transition vector?
wait no
transition matrix
that's it
This is a Markov process?
dawn remnant
basically:
- Write down the recurrence equations - that is, write down $D_n$, $S_n$ and $F_n$ in terms of $D_{n-1}$, $S_{n-1}$ and $F_{n-1}$. This is the "essence" of the task - a) ask you to find $D_{n-1}$, $S_{n-1}$ and $F_{n-1}$ for a given $D_n$, $S_n$ and $F_n$, and b) asks about the fixed points of this system.
To solve that, start actually introducing linear algebra stuff:
2) Convert that system of equations into a matrix
3) Your system now looks just like $v_{n} = A v_{n-1}$, which is something you can do all sorts of cool linear algebra stuff with.
stoic pythonBOT
ConfusedReptile:
dawn remnant
This is a Markov process?
hmm, I don't think so - IIRC there you have discrete states and a matrix that transforms a state vector (representing a probability distribution over states).
mental pawn
Okay I'll disregard that
dawn remnant
also, note that you can technically solve this without using any LA at all
mental pawn
Oh yeah I know
I feel like I am on the cusp of everything clicking but I'm just not there yet
dawn remnant
I mean, you'll essentially be reinventing LA, but you can technically just do:
a) just solve the system for (D,S,F)_{n-1}
b) set D_{n+1} = D_n and the same for S and F and solve the system (or prove it's unsolvable)
mental pawn
Oh hang on
nods
So these distribution vectors are they always in a binary form like 1 0
like unit vectors... or am I way off there?
dawn remnant
I'm not sure, but I think you can also have your state vector be a probability distribution
like, in the case where the transitions aren't guaranteed, but probabilistic
mental pawn
oh /me nods
Okay well you've given me a lot ot work with I'm going to give it a go
thanks a lot 🙂
warm niche
I don't really get what you mean, but ok yeah
@prisma cairn lmao tbh I dont either I was super tired, but I get it
And i dont wanna think it about too much or else imma just confuse myself even more
mental pawn
@dawn remnant I get it now
@dawn remnant The vector for the end of the year is [0.36, 0.37, 0.27]
dawn remnant
for example, yeah
mental pawn
Well for this problem
I mean
I mean for this problem.
Oh my god
I think I get it
dawn remnant
Because the transformation is linear, scaling a vector by a constant will just scale the result too
so you could have also went with [36,37,27], for example
mental pawn
Sure sure
dawn remnant
but yeah, [0.36, 0.37, 0.27] would be my choice, for them to sum to 1
mental pawn
And I discussed that with my tutors
thank you for the massive dopamine rush ... bye now
halcyon pollen
easiest way to learn linear algevra?
native rampart
Learn linear algebra
halcyon pollen
youtube?
native rampart
Yes
native rampart
rocky wharf
.
dusky epoch
write out the equation AB = BA as a linear system of 4 equations in 4 unknowns, where the unknowns are the entries of B
split heart
Hi. If I'm given a matrix Amxn the maximum matrix rank is the lower value between m and n, right?
no problem
If you need i can rewrite
split heart
Nice, thank you
ocean sequoia
because then you are "changing" the columns here we are just changing the order
half ice
AB = BA
Implies A and B commute
split heart
When M1M2 does not = M2M1
@rugged basalt That's not associative property, you're commutating
rugged basalt
I don't understand
half ice
Or "are commutative"
split heart
Associative is (A.B).C = A.(B.C)
ocean sequoia
its because when you try to commute you are flipping rows and columns
you dont in associative
try some examples with 2x2 matricies
ocean sequoia
do some examples
seriously that will help i dont mean to be glib
like very easy numbers
its a good question
rugged basalt
ok
ocean sequoia
shit you dont even have to do a 2x2 matrix just do a 2x1 vectors
ifyou want 🙂
keep your numbers simple
rugged basalt
This is a rlly basic question but
Does the 2 in 2x1 represent the amount of horizontal rows or vertical columns?
It's rows right?
rugged basalt
ohhh
i think i get it now
I was mixing up vectors and matrices
Writing it down did help
ocean sequoia
it happens with matricies too
rugged basalt
what does?
ocean sequoia
associativity but not commutivity
rugged basalt
oh like (m1m2)m3 = (m1m3)m2
ocean sequoia
one is associate one is commutive
rugged basalt
but are they not the exact same things
they m1 m2 and m3 are all unknown
On both sides you multiply in the brackets first, and then multiply by the one on the outside
ocean sequoia
yea but m1 m3 m2 is not the same as m1 m2 m3
rugged basalt
im so lost
ocean sequoia
where are youlost
here
lets use a b c
a b c != bca
but
(a b)c = a (b c)
being on the left or right is important for matricies
rugged basalt
but doesn't abc = bca
rugged basalt
oh
ocean sequoia
should help
rugged basalt
im just thinking using numbers
ocean sequoia
🙂 gotta get use to matricies
rugged basalt
ripp
ocean sequoia
no worries
rugged basalt
(a b)c = a (b c) = a b c = bca
ocean sequoia
yes
rugged basalt
does it not
ocean sequoia
no
rugged basalt
both of those equations are the same things
ocean sequoia
wait why did you edit it
they are not for matrix multiplication
do some examples man
seriously itll help you to see it
ocean sequoia
yes it does
rugged basalt
and a(bc) = bca
ocean sequoia
no
rugged basalt
how
rugged basalt
because it matters which side the matrix is on
ocean sequoia
yes
rugged basalt
when doing the multiplication
ocean sequoia
wintry steppe
how is this false
void relic
the union of 2 lines
limber sierra
to be a bit more explicit
stoic pythonBOT
Namington:
limber sierra
then both of these vectors are in the union
stoic pythonBOT
Namington:
limber sierra
hence this is not closed under addition, so it is not a vector space (ie not a subspace)
this is just an example of what plurmorant was saying
void relic
hope you like solving cubics
icy blade
im not sure if this fits here but i have a line and I don't know how to interpret the curve notation
its just a linear line
like what would be the y=mx
mighty saffron
@rugged basalt Getting an equation for the determinant we get. a[(ba+b)(a)-(3a)(-1)] - 0[whatever] + a^2[(2)(-1) - (a)(a)] = a[ba^2+ba+3a] + a^2[-2-a^2] = a^2[ba+b+3]+a^2[-2-a^2] = a^2[ba+b+1-a^2]. so we get a = 0 works. also ba+b+1-a^2 works. or b(a+1) = a^2-1 or assuming a!=-1 b = a-1 since a^2-1 = (a-1)(a+1). If a = -1 then b can be whatever you want
i think so anyways
might have made arithmetic error
rugged basalt
Ok so I set the entries of B to a b c and d
Then solved for AB and BA and then made each element in the AB matrix equal to the corresponding element in the BA matrix
Which gave me 4 equations
1). -2a + b = -2a
2). b = -2b
3). -2c + d = a + c
4). d = b + d
Am I even on the right track? If so, what do i do now
dawn remnant
well, solve these four equations for a,b,c,d. Also see the hint - you'll get at least a linear space of solutions.
rugged basalt
What does "a linear space of solutions" mean
Also, I solved for b = 0. I don't see how I could solve for anything else
dawn remnant
well, for example, the system:
$$x + y = 1$$
$$x + y = 1$$
has not one solution, but rather a linear space of solutions: $x = 1-y, y \in \bR$
stoic pythonBOT
ConfusedReptile:
rugged basalt
wait would the matrix just be:
dawn remnant
Also, I solved for b = 0. I don't see how I could solve for anything else
@rugged basalt Seriously?
b = 0 from second equation. First equation is just 0 = 0, always true. So is the fourth one since b=0. We are left with one equation:
$$ -2c + d = a + c \implies c = \frac{d-a}{3}$$
stoic pythonBOT
ConfusedReptile:
dawn remnant
so $b=0$, $c = \frac{d-a}{3}$, and d and a are arbitrary real numbers.
stoic pythonBOT
ConfusedReptile:
dawn remnant
That's actually a two-dimensional space of solutions, even.
rugged basalt
how do u know they are abritrary real numbers?
dawn remnant
for any a and d, (a,0,(d-a)/3, d) is a solution.
that's how the answer normally ends up looking when you have fewer equations than variables
rugged basalt
Could:
1 0
0 1
be a solution?
I remembered we learned about something called the identity matrix
Sorry to stray away from what we were talking about for a sec
dawn remnant
as you can see, it indeed matches the formula for a = 1, d=1
and yes, it's a good idea to check that
since the identity matrix commutes with everything (not surprising, since it does nothing) so it would be weird if we solved for "all matrixes that commute with A" and didn't get the identity matrix among the solutions.
so the answer is:
$$
B \in {\begin{pmatrix} a & 0 \ \frac{d-a}{3} & d \end{pmatrix} | a,b \in \bR}
$$
stoic pythonBOT
ConfusedReptile:
dawn remnant
ah, yeah, that's a typo
$$
B \in {\begin{pmatrix} a & 0 \ \frac{d-a}{3} & d \end{pmatrix} | a,d \in \bR}
$$
stoic pythonBOT
ConfusedReptile:
rugged basalt
wait i set the top right corner to c
so wouldn't the bottom left corner be 0?
and the top left = d-a/3
dawn remnant
then they'll just be in a different order, doesn't matter much.
rugged basalt
so can i just switch them and it'll be fine?
dreamy iron
what is meant by derivatives being linear?
@slim gyro
it means they are linear i.e the function for calculating derivatives is a subspace
@ocean sequoia
Re: https://discordapp.com/channels/268882317391429632/540211747613704221/759916088946589707
This is Axler 1.C. You're "not yet expected" to know what linearity is (that comes at chapter 3). It's not essential, but it is helpful.
You are only asked to show that such a subset of functions is a subspace.....
for this reason, just use the rules of differentiation you'd have learned in a first class in calculus: https://en.wikipedia.org/wiki/Linearity_of_differentiation
if this is confusing you, just assume the rule are true (they are) and prove the subspace req.
twin loom
Is there a special way to calculate the magnitude of a column vector with complex entries?
Like I need to choose a to normalize this vector
$$\begin{pmatrix} a \ ai\end{pmatrix}$$
stoic pythonBOT
deadpan2297:
twin loom
but when calculating the absolute value in the normal sense gives 0
vague cedar
$|v|^2 = (v^*)^\intercal v$
stoic pythonBOT
jacob:
vernal pebble
rugged basalt
What does it mean by solve the three systems of equations
Like what am I trying to do?
void relic
it's just solving for those 9 variables
those will be the elements in the inverse matrix
rugged basalt
ohhh, gotcha
thank u
What does it mean by "check your work by calculating AA^-1" though?
what is AA^-1?
limber sierra
this is A
A^-1 is its inverse
what do you get when you multiply A and A^-1 (the inverse you found) together?
(you SHOULD get an identity matrix if you did it correctly)
slim gyro
@dreamy iron thank you, that's very helpful!
hidden ember
but here, v1 could also be a zero vector and x1 could be non zero?
how is this correct
dusky epoch
is this meant to be the solution to a problem
dusky epoch
please do
hidden ember
sry i should hv sent before
for the second one, how can it be linearly independent since x1 could be non zero and v1 could be 0
marble lance
They explicitly say v1 and v2 are not zero.
bold python
If $U= [\bold{u}_1\dots\bold{u}_n]$ is a square matrix. How can I prove that $U$ is semi orthogonal iff $\bold{u}_j\neq 0$ for $j=1,\dots, n$ where $\bold{u}_j$ is orthogonal on eachother
stoic pythonBOT
homomorphism:
bold python
I just need a place to start
I was thinking of $U$ semi orthogonal $U^TU = I \Rightarrow \bold{u}_j \neq 0$
stoic pythonBOT
homomorphism:
dusky epoch
find three linearly indepedent polynomials in W
dim venture
I think I'm confused by the coefficients c0 ... c3. I thought T has a domain with rank 3
from my understanding, I have to find three sets of c0, c1, c2, c3 that satisfy f(-4) = 0
dusky epoch
"domain with rank 3"
dim(W) is 3 if that's what you were trying to say
also what's c4
dim venture
then my isomorphism is the linearly combination of a(first lin indep poly) + b(second lin indep poly) + c(third lin indep poly)
ah my bad c4 is typo
0+16*(a-c)x+4(b+c)*x^2+(a+b)*x^3
edit: it doesn't
does this work?
wanton spoke
@ocean sequoia
ocean sequoia
So I can do the math behind PCA its the eigendecomposition of the covariance matrix cool but I feel like a computer. I don't get the intuition behind why that tells us anything could someone help link the eigendecomposition and dimensionality reduction? I get how we uncorrelated the data because the eigenvector will be orthogonal to each other but i keep seeing the term projection I dont see where there is any projection going on here
wanton spoke
I'll explain it to you
wanton spoke
So basically what you want when you do a PCA is to dind the axes or plans (rarely higher dim spaces) that most explain the variation in your data
The eigendecomposition helps you do that
ocean sequoia
yea because the axis is basically the eigenvectors right
which are all orthogonal to each other
edited
wanton spoke
The variance is explained by the eigenvectors
ocean sequoia
in the eigenbasis arent the eigenvectors the axises?
wanton spoke
They are
Wait
this sentence is confusing
Your eigenvectors form a new basis
and each vector is an axis
ocean sequoia
yea sorry thats what i meant
im still working on making sure my language is exact
i tend to struggle with that because I dont come from a pure math background so I sometimes handwave terms
wanton spoke
So when you've done that work, you know that these axes explain the variance, but what does each one explain (wrt to the original data)?
The way to figure that out is to look at where your original vectors are wrt to the new basis
and if they are "close" to an axis or plan, then when you project your vectors you aren't losing much info
ocean sequoia
and if they are "close" to an axis or plan, then when you project your vectors you aren't losing much info
@wanton spoke are we projecting the covariance matrix? or the original matrix?
wanton spoke
The columns of the original matrix
Each one may be better explained by a different axis
ocean sequoia
ahhh
ok
so then when we get that result the columns that are closer by projection which will be larger vectors are the columns we want to keep
because if we project a onto b the size of b doesnt matter right?
wanton spoke
it's not really "closer by projection" (since by def of a projection you'll be in the space), you're looking at whether the original vectors similar to their projection (e.g you write your vector as sum of projection + an orthogonal vector, you'll want the orthogonal vector to be small)
The columns you'll want to kezp are those that don't lie in (or close to) the same axis
Because they basically tell the same thing
ocean sequoia
so the second part makes sense there
you're looking at whether the original vectors similar to their projection (e.g you write your vector as sum of projection + an orthogonal vector, you'll want the orthogonal vector to be small)
@wanton spoke
why are we writing the new vector as a sum of projection and orthogonal vectors?
wanton spoke
The projection you do is an orthogonal one
so by pythagor the vector is the sum of it and an orthogonal vector
ocean sequoia
ohhh
wanton spoke
Let me draw sthg
ocean sequoia
ok
wanton spoke
so let's say uk is your eigenvector, and Delta uk is the axis generated by this vexcto
I called Psi the factors
ocean sequoia
oh wait i think i get it the smaller the orthogonal projection the closer yi is to delta uk which should mean yi is a better indicator of variance
wanton spoke
Basically each component of Psi-k is the orthogonal projection of the i-th column onto the k-th axis
ocean sequoia
i think
wanton spoke
some columns are useless as they tell the same thing or almost as other columns (the columns are heavily correlated)
So you'll want to trim that down and keep only a few columns in order to avoid redundancy
oh wait i think i get it the smaller the orthogonal projection the closer yi is to delta uk which should mean yi is a better indicator of variance
@ocean sequoia it's not the orthogonal projection that need to be small but the component orthogonal to that projection
ocean sequoia
yes sorry
the smaller the sum of the orthogonal competent the better the indicator of variance that column is
is that correct?
wanton spoke
If it's small, then it means the axis is a good representative of this column
For dim reduction, you'll look at how many axes are good representatives
ocean sequoia
and so certain axis can be better representations of multiple columns
wanton spoke
Ans that number will be your new dimnension
Yep, if multiple columns are well represented by one axis, then you can eliminate some of them
And keep just one
ocean sequoia
ok so we can drop certain columns from the variance covariance matrix?
ocean sequoia
so if multiple columns are expressed by one axis we can just keep the vector that is closest to the axis?
wanton spoke
Yes
ocean sequoia
because that explains most of the variance
wanton spoke
Yep
ocean sequoia
if thats correct i get it
thats one of the best explainations i have ever heard
thank you so much
if we were on reddit id guild you
hot damn that was so well explained
wanton spoke
np
Tbh now that I think about I had found good explanations (way better than what I just did) on PCA on stackexchange
Lemme see if I have saved them
ocean sequoia
just to make sure you'll look at how many axes are good representatives and if one axis is a good representation of multiple columns that means the column with the smallest orthogonal projection is good representation of the axis so its the best representation of the variance in that data
wanton spoke
I think it was that
Tho I just skimmed through rn
just to make sure you'll look at how many axes are good representatives and if one axis is a good representation of multiple columns that means the column with the smallest orthogonal projection is good representation of the axis so its the best representation of the variance in that data
@ocean sequoia yep almost, just be careful, orthogonal projection isn't the orthogonal component of the sum of the two vects
wanton spoke
ocean sequoia
wanton spoke
Wait I think I said something not really true, because one way to do the dim red would be to remive columns from the original matrice, but a better way would just be to keep few factors that explaon most of it
Since in one case you lose info
wanton spoke
Tbh I've mostly used PCA for descriptive stats, not so much for dimensionality reduction
ocean sequoia
oh ive heard it used for dim red
thats why i was going through it hahaha
i do know you lose info
gaunt field
https://i.imgur.com/3uK7cTG.png can someone help translate this problem for me? I have problems understanding set builder notation
wanton spoke
Yeah it can be used for dim red
I've just not used ut luxh in that context
and yeah you do lose info
Vur I think you lose less info if you keep the factors
ocean sequoia
what do you mean by factors
wanton spoke
The problem with the principal components is that they're harder to interpret
Yep
it depends on your goal
wanton spoke
The thing with removing the initial columns is that there may complicated relationships between variables
ocean sequoia
that also makes sense tho because those principle compents will literally be the new basis for the covariance matrix
wanton spoke
If it's like we sais above and columns are well explained by a single axis, then yeah you just do the dim red on your initial data
But what if it's a combination of axes that explain it well?
ocean sequoia
shit yea this is the first time im using it in practice so
hm yea so i guess i can train the model on the principle components?
wanton spoke
Well I don't know what's your end goal
ocean sequoia
ah its a competition so i cant share anymore without cheating
wanton spoke
ah
ocean sequoia
could i drop the lower principle components then move back to the old basis
wanton spoke
each component is a combination the original columns
ocean sequoia
like we perform the covariance matrix is in the standard basis(probably) so could i just drop the respective entries and move back?
wanton spoke
So yeah you can keep them but they'll also be harder to interpret
That's usually the trade-off of PCA
hmm moving back to the older basis seems like a bad idea to me (and can't really be done anyway)
ocean sequoia
wouldnt the new basis be in Q Lamda Q^-1?
wait no it would just be Q
right
those are the eigenvectors
ocean sequoia
does what im asking make sense
wanton spoke
since you're in lower dim
ocean sequoia
like idk how you would perform analysis on something thats Q Lamda Q^-1
which is the eigendecomposition which holds the princple factors right
wanton spoke
what is Q and lamda
noble crest
if this good place for graduate level linear algebra? Trying to figure out when ||A||_=||A^||, where * represents dual-norm/adjoint
ocean sequoia
noble crest
||A||=||A||
wanton spoke
Yeah well if you remove columns then your matrix isn't square anymore
ocean sequoia
true idk how to perform analysis on Q Lamda Q^-1 
maybe just drop columns and hope for the best
ocean sequoia
isnt that how we do the eigendecomposition of the covariance matrix
isnt pca the egiendecomposition of the covariance matrix
wanton spoke
So lambda is your covariance matrix?
and what do you want to do with that
I'm confused, are you doing that after the PCA?
Or are you referring to how you get the PCA
wanton spoke
oh
Well the axes are in Q
and the principal components are YMuk with uk an axis, M your metric and Y your data
ocean sequoia
yea
ok
hm im just not sure how we use it for dim red
I cant just go around dropping axises
im assuming
wanton spoke
look up inertia contributions
rugged basalt
can anyone help me do elimination
wanton spoke
anyway np
earnest vessel
Have you solved the systems of equations?
wintry steppe
does anyone know what the cross product of orthogonal vectors is
dawn remnant
does anyone know what the cross product of orthogonal vectors is
@wintry steppe nothing more special than the cross product of any two nonparallel vectors, I don't think.
rugged basalt
I get how to find the determinant
I got a - b^2 = 0
But what does that tell me about the solutions?
dawn remnant
if the determinant is zero, then there can't be a unique solution, because there exists a nonzero vector v such that $A v = 0$
stoic pythonBOT
ConfusedReptile:
dawn remnant
and therefore you can add a multiple of v to any solution to obtain another one - hence infinity or 0 solutions.
you'll have to check whether there is a solution, though. I think.
rugged basalt
What is a unique solution @dawn remnant
dawn remnant
The case where there's exactly 1 solution.
rugged basalt
And when would that occur?
Does a solution refer to a different vector that can be created with a given vector and matrix?
Like Idk what a solution even refers to @dawn remnant
dawn remnant
@rugged basalt I mean, this task is about the equation A x = k, where A and k are given, and the task is about when there exists solutions for x (that is, vectors x that match this equation).
wintry steppe
The lines are parallel, how can I calculate the shortest distance between the two
Can I just get the difference of the two poitns
then get the norm
azure trout
hollow finch
hey does anyone have any examples of vector spaces with strange addition/scalar multiplication rules? i need an example for a tutoring session tomorrow and the professor already used the u+v=uv, ku=u^k in class which is the only one i know.
edgy folio
does a spanning set always have to have atleast one solution?
soft burrow
hey does anyone have any examples of vector spaces with strange addition/scalar multiplication rules? i need an example for a tutoring session tomorrow and the professor already used the u+v=uv, ku=u^k in class which is the only one i know.
@hollow finch Define a"+"b by a+b+1 and c"⋅"x by cx+c+x. (These are actually ring operations over the reals.)
more generally you can use this result to construct more examples: https://math.stackexchange.com/a/1911327
it also might be good to just present some examples that are not numbers, e.g. the vector spaces of polynomials, sequences, functions
you could spice things up by presenting those typical examples + an additional condition; e.g. define the set of "Fibonacci-like" sequences by
$$\mathrm{Fib}:={(a_n)\in \mathbb{R}^\mathbb{N} ~:~ a_{n+2}=a_{n+1}+a_n\text{ for $n\geq 0$}}.$$
Then $\mathrm{Fib}$ is a real vector space.
stoic pythonBOT
bastian.uwu:
hollow finch
@soft burrow I was not expecting such a thorough answer, thank you!
I am a bit confused about the first example because 1x=2x+1≠x unless im misunderstanding
soft burrow
oh, the "1" is different in that example. what you should verify is that there exists an element, say $e\in \mathbb R$ such that for all $a\in \mathbb R$, $e\otimes a =a(=ea+e+a)$. Thus $e=0$.
stoic pythonBOT
bastian.uwu:
soft burrow
it can be slightly confusing but it helps to break our intuition of what "addition" and "scalar multiplication" should look like, for the sake of understanding the abstract definition. (that's why it's an example often presented in abstract algebra.) if that's your goal then it might be a good example, otherwise focus in the typical examples for the TA session
(btw the "0" in addition is different too! ||it's -1.||)
hollow finch
Hm that definition of scalar multiplication still seems to fail the axioms of a vector space :(
Is it perhaps over a different field than R?
the addition operation does work though! im definitely using it
i suppose i would just need a definition of scalar multiplication that goes with it such that all 10 axioms are satisfied
dim venture
the form should be in (-a+b) + (-b+2c)x + (-c+3d)x^2 + (-d)x^3 and then I'm stuck
is the matrix immediately a 4x1 with the coefficients in the polynomial above as its terms?
spiral star
the matrix is 4x4
it contains the coordinates of the images of your basis vectors under F1
so the first column is the B coordinates of F(1)
second column is B coordinates of F(x)
third column is B coordinates of F(x²) etc.
spiral star
F(x) = x' - x = 1 - x and [1-x]_B = [1, -1, 0, 0]^T so there might be a sign error in your second column
spiral star
i didnt check the others because im too lazy
but you get the idea
this is how you get the matrix for any transformation
you apply the linear map to your basis vectors and then use the canonical basis isomorphism
dim venture
this is how you get the matrix for any transformation
I'm applying it in the next part now
i didnt check the others because im too lazy
no worries, this explanation is very helpful already, appreciate it
spiral star
i dont think its a good idea to construct a matrix for the next part because the size is variable. you can argue at the level of linear maps
note that a linear map is fully defined by the images of basis vectors
dim venture
is the ker(F0) not the set of all constants?
and the range(F0) is a basis with n-1 vectors consisting of each column vector except the first
spiral star
you got the kernel right, what you say about the range might or might not be true because you didnt say which basis
make it concrete and write a proper proof
Choose a basis for $\mathbb{P}_n$ like $B = \qty{p_k = x^k \mid k \in \qty{0, \dots, n}}$.
is the bot offline? :(
dim venture
bot?
spiral star
ok guess i have to be my own bot now
and then you apply F_0 to the basis vectors
the range of F is just span{F(p_0), F(p_1), ... F(p_n)}
dim venture
F(p_0) is in the span?
spiral star
sure
the zero vector is in every vector space
i didnt say that this is a basis for the range
dim venture
ah right
spiral star
since F(p_k) = k p_(k-1) for k >= 1
you can see that the range is span{p_1, p_2, ..., p_(n-1)}
dim venture
the basis should be span{F(p_1), ... F(p_n)}
spiral star
yea exactly
dim venture
but I need to formally write it
spiral star
the set of the images is linearly independent if you take out the F(p_0)
because the basis was linearly independent
and k != 0
so they are a basis for the range
dim venture
just that simple? is that considered a formal proof since I found a basis
dim venture
so we first express the range of F0 as span{F(p_0), F(p_1), ... F(p_n)} but note that it is lin dep.
then in considering the removal of F(p_0), the set becomes lin indep and thus the span is a basis for the range
spiral star
yes, but you dont even need a basis
you just need to show what the range is
of course a basis is nice to have
range(F) = span{F(p_0), F(p_1), ... F(p_n)} = span{x, x², ..., x^(n-1)} would be sufficient
spiral star
matrix will take more effort because you cant even draw it.
since the dimension is n
dim venture
right
dim venture
linear maps as in the transformation of one basis to another?
or the literal transformation
not sure if that made sense
spiral star
you should always argue on the level of linear maps if you can. dont try to force them onto a matrix unless you have good reason to
like in problem we just solved
dim venture
okay I'll keep that in mind moving forward
that helps in part c as well
then I can determine isomorphism based on dimension instead of drawing a matrix with variable size
dim venture
and linearity
spiral star
sure
dim venture
I'm stuck on linearity
spiral star
differentiation is linear
dim venture
I don't think its sufficient just to state so
dim venture
oh its that trivial
spiral star
just as (c p)' = c p'
the exercise even states that F_k is a linear operator
i mean its obvious and i dont think you have to show it
you should be concerned with the isomorphism part
dim venture
linear, same dimension, injectivity proves isomorphism
spiral star
yes, and a linear map is injective if the kernel is trivial
dim venture
well the dimension definitely still holds because the polynomial still has a highest power of n
the last term of the transformation remains to be an*x^n
spiral star
you have lots of options here
for example you can take the image of basis vectors again
i think that is what i would do
and then use the rank nullity theorem
or just be done anyway
if you can show that the image of all basis vectors is linearly independent
then you have full rank
and then it must be injective
dim venture
okay this isn't so bad after all
we know the dim of range is n-1 from finding the basis just now (though not necessary). and the dim ker is 1 since its trivial and everything carries forward from there
spiral star
if the kernel was trivial it would have dimension 0
and i cant really follow your reasoning
dim venture
let me tidy my thoughts with the previous parts first and I'll write out something more clearly to solidify my understanding
dim venture
if you can show that the image of all basis vectors is linearly independent
@spiral star we know its linearly independent based on the assumption that k=/= 0, then the image is the span of all the basis vectors, thus rank = n
therefore by rank-nullity, n + 0 (since trivial ker) = n = dim(F_k)
spiral star
well yea but how do you know its linearly independent
dim venture
the basis vectors are 1, x, x^2, x^3, ... x^n
spiral star
so?
dim venture
isn't it just definition, they can't be represented by any linear combination of each other
spiral star
you need to show that the images are a basis
those are not the images
you need to show that F(1) F(x) F(x²) ... are a basis
it suffices to show that they are linearly independent
dim venture
range(F_k) = span{F_k(P_0), ... , F_k(P_k)}
spiral star
yes, but why is range(F_k) = P_n
thats where you need to show independence
if you choose to go that route
dim venture
theres something I'm missing or not understanding let me spend more time with this
spiral star
one (unfortunately a bit involved) way to show linear independence of the images is by literally applying the definition
take scalars lambda
such that the linear combination is 0
and then show that each lambda must have been 0
by using the basis you started with, after a bit of algebraic manipulation you get
and this implies already that all scalars were 0
of course there are other ways to show injectivity
unique fog
So this is my first ever post, so I'm sorry if I'm not too accurate. But I have this excersice in matrices where I would have to compute B^TAB, from a symmetric matrix A, and a invertible matrix B. How could this be done?
fossil rampart
If Anxn is diagonalizable and invertible , is A^-1 diagonalizable , would that also hold for A^T
spiral star
i'll just an alternative proof that leads to the same answer. you can also take a vector that is in the kernel and show that it must have been the zero vector
you will get the same answer
wintry steppe
I have a linear code where messages abc are encoded to abcde such that d = a+b and e = b+c, and I am trying to find a generator matrix for this. any help would be appreciated :D
nvm I think I got it
fossil rampart
Is an orthogonally diagonalizable matrix orthogonal ?
spiral star
not in general
you can find numerous counter examples
for example, hermitian matrices will have an orthonormal eigenbasis
but being hermitian doesnt imply that they are invertible
if a matrix is orthogonal it is necessarily invertible
so every hermitian matrix that isnt invertible would be a counter example
i guess if you want a concrete examples lets go with
[ \mqty[1 & 0 \ 0 & 0] ]
stoic pythonBOT
Flow:
dim venture
dim(U) <= dim(W) and dim(U) >= dim(W), but I don't know if the way I got here was right
spiral star
all you need to know is that there is an isomorphism between U and W
doesnt matter how it was constructed
dim venture
oh okay
spiral star
finite dimensional isomorphic spaces always have the same dimension
spiral star
T must be surjective
dim venture
I know a and c are true, and b is false. just not sure about d
spiral star
well just think it through
check what happens if S was either not injective or not surjective
dim venture
we know S must be injective
spiral star
yes
dim venture
so just check whether if S is not surjective and it still holds?
spiral star
well the trick is to assume dim(V) > dim(U)
you know S must be injective, but now it cant be surjective anymore because you lack dimensions
so S does not have to be surjective
dim venture
ohhh so that proves that S does not have to be surjective
the wording is so confusing
so do I tick the box or not?????????/
dim venture
then its injective and surjective
spiral star
yes
dim venture
then S is isomorphic
and ToS is isomorphic
so this just means it doesn't matter whether or not S is surjective
spiral star
well the wording is a bit off but you got the idea
S would be an isomorphism and the spaces U and V would be isomorphic
dim venture
well I didn't tick the box and it said it was right
spiral star
the statement is asking you whether it MUST be true that S is surjective
but it doesnt have to
we found examples for either case
so you cannot say anything about S being surjective or not
the only thing you know for sure is that S would be injective
spiral star
was it that bad? 😄
i think it was quite easy to think of counter examples for those statements that werent true
for example you could disprove b) and c) with the same example
take U = V= W and T = S = id
dim venture
what is id
spiral star
identity
dim venture
oh ok
spiral star
id(v) = v
since U = V, obviously dim U = dim V
and id composed with id is again id
and id is an isomorphism
so it is surjective in particular
that disproved both b and c
and to disprove d) you can take U = V and dim(V) < dim(W)
then S = id and T some injective map
TS is a composition of injective maps so it is also injective
but T cannot be surjective because dim V < dim W
so TS is not an isomorphism
dim venture
I'm glad that there exists such a smooth flowing solution
but realistically, I can't redo it on an exam
its my thought process that needs to improve (or experience)
spiral star
you get experience by writing proofs
you can try to formally justify my arguments for example
and then you will get more intuition for how and why it works
one example would be to argue that if dim(V) < dim(W) then there is no surjective linear map f: V -> W
dim venture
sorry, don't mean to change the topic, but for the previous question I asked about Ker(F_0) to be the set of all constants, can I write it as Ker(F_0) = Span{1}?
or Ker(F_0) = R
spiral star
if you identify R with a subspace of P_n then you can do this
span{1} works too
both is fine :)
dim venture
okay, I'll take that
if you identify R with a subspace of P_n then you can do this
could you elaborate on this, not quite sure what you mean
spiral star
i gotta go now, i can explain later unless someone else will do it in the meantime :)
dim venture
please come back later I still have more questions 😦 but appreciate your help flow, thank you very much
valid marsh
Hello guys 🙂 , I do have a statement to prove but I cannot find a way to do it
Here is the statement :
dusky epoch
$u \cdot v = u^Tv$
stoic pythonBOT
Ann:
spiral star
could you elaborate on this, not quite sure what you mean
@dim venture i dont think its worth going into this. anyway, there are different ways to define P_n, for example as a subspace of R^R or via formal polynomials. if we choose the subspace of R^R, then you could define P_n like so
and then you will find an injective linear map from R into P_n
which identifies a real number c with the constant function f(x) = c
so there is a subspace of P_n that is isomorphic to R
proper mauve
Can someone explain how to do this
https://gyazo.com/4aad8a91cab49959973ec7ca5f4d21f2
dim venture
^hoping to get some help with this question but C9 World Champ asked a question first
pallid rampart
Is the dimension of the null space of the conjugate transpose equal to the dimension of the null space of the original matrix
Nope
The answer is given in an exercise lol
hidden ember
if AX = 0 has a nontrivial soln, does that mean AX = y will not have a unique soln
pallid rampart
No
If AX=0 has nontrivial solution then for some y, the equation AX=y does not have unique solution, and for some y the equation AX=y has no solution
hidden ember
hmmm
i see
if i say that AX = 0 is consistent, i mean has a soln then A's columns are linearly independent right?
soft burrow
if "consistent" is "has no non-trivial solutions", yes.
dusky epoch
@soft burrow that's not what consistent means
@hidden ember AX = 0 is consistent always, regardless of what goes on with the columns of A
soft burrow
good to know, never heard of that term
hidden ember
i am talking about linear independence tho
dusky epoch
yes, and?
the linear independence or lack thereof of the columns of A says nothing about whether or not the system Ax = 0 is consistent
because Ax=0 is always consistent
ember wedge
wintry steppe
no
wet minnow
I don't understand lol
dreamy nimbus
i need help with the last 2 parts
wait
is the transpose of an orthonormal set
equal to its inverse
dawn remnant
Last two parts? Representing x as a linear combination of the basis?
If so, the "simple formula" is that the coordinate for a given basic vector in an ortonormal basis is just the scalar product of the vector with that basis vector.
so $\alpha_2 = x \cdot x_2$, for example
stoic pythonBOT
ConfusedReptile:
static dust
A rectangle has a perimeter of 60cm. The length of the rectangle is 3 inches less than twice the width. What is the length?
feel like smooth brain bc I can't solve this
half ice
I would let the width be x.
What's the length, in terms of x?
Also, not a linear algebra question haha. We can finish this here though real quick
tall thunder
Hey, if a set has a 0 vector in it ? Is that whole set “linearly dependant”
half ice
Yes
half ice
Do you know the definition of linear dependence? There's a nice equation for it
tall thunder
I think so, it’s dependent if there is a linear combination right ?
half ice
v,u,w are linearly dependent if
au + bv + cw = 0
Has a non-trivial solution
For a,b,c
Basically, "is there a way to combine these such that I can get the zero vector?"
limber sierra
given a set of vectors ${v_i\mid i\leq I}$, this set is called linearly dependent if $\sum_{k=1}^{I}a_iv_i = \mathbf{0}$ has a set of solution scalars ${a_i \mid i\leq I}$ such that not all $a_i$ are equal to $0$
stoic pythonBOT
Namington:
limber sierra
if 0 is in your set then
just set all the scalar coefficients on all the other vectors equal to 0
and the scalar coefficient of 0 equal to 1 or whatever
then you have $0v_1 + 0v_2 + \dots + 1 \cdot \mathbf{0} + \dots + 0v_I = \mathbf{0}+ \mathbf{0} + \dots + \mathbf{0} + \dots + \mathbf{0} = \mathbf{0}$
stoic pythonBOT
Namington:
limber sierra
but the coefficient of one of your vectors (0) is not 0
and yet this sum is equal to 0
so it must be linearly dependent.
tall thunder
I’m on part b, do you think my explanation is good for credit or naw ?
She’s just asking if the set is dependent or independent, and I understand that it’s dependent but I’m not sure if my explanation is good
limber sierra
I have no clue how detailed an explanation they want since I don't know the rigor standards of your course
half ice
I'll call those "vectors" u, 0, w
Then if we have the equation:
au + b0 + cw = 0
You can solve it with (a,b,c) = (0, 1, 0)
That's a non-trivial solution, and so this set is dependent
@tall thunder
You've got to at least know the def lol. I don't think even engineering courses will let you get away with just ignoring it
tall thunder
Gotchya ‘
strange dove
Hi can someone help me with this
The questions asks to "highlight the error" and i literally don't see one
limber sierra
how do you add a scalar to a vector?
strange dove
you can't add a scalar to a vector, but i dont see a scalar being added here
ocean sequoia
@strange dove what does the dot product give you
strange dove
a scalar quantity
ocean sequoia
whats up
ocean sequoia
you sure?
strange dove
yea, the x dot 2y is equal to a scalar, and then your left with + x - y which are vectors
ocean sequoia
shadow horizon
f(x)=100x+600. f(1)+f(2)+f(3)+...f(n)=1,300,000,000. Assuming x is greater than or equal to 0 and x can only be an integer, how would you solve this?
Yes
How would I apply that to a function though?
alright, thanks
pearl elm
is there a particular strategy for gauss elim where the number of var exceeds the number of equations? I am supposed to get a row to be all zeroes right?
ivory moon
Can you integrate an augmented matrix
pearl elm
can someone give me an example of doing this
im not sure I can always get a row to be all zeroes.
stoic pythonBOT
Namington:
limber sierra
the number of variables clearly exceeds the number of equations
pearl elm
ok thats what I figured
limber sierra
but you certainly cant get a 0 row
pearl elm
sometimes you can though?
limber sierra
perhaps you meant the other way around; the number of equations exceeds the number of variables?
limber sierra
sure, sometimes you can, e.g. the matrix $\begin{pmatrix}1&0&0&0\0&0&0&0\end{pmatrix}$
stoic pythonBOT
Namington:
limber sierra
@ivory moon what do you mean by "integrate a matrix"?
matrices form a vector space so componentwise integration kind of makes sense
and that would, of course, work with augmented matrices
but dont expect something nice and linear out of it
ivory moon
@limber sierra
limber sierra
in what context is this coming up?
again you can do componentwise integration but it wouldnt really be satisfying in that case
you'd just be adding an "x" to the end of every number
ivory moon
I was just curious
limber sierra
(and ignoring constant terms because yuck)
half storm
Is that standard notation?
half storm
I meant for component wise integration over a matrix, is there a standard notation.
limber sierra
not that im aware of thedon
but if you just wrote something like
Let $\int A(x)\dd{x} = \left(\int a_{i,j}(x)\dd{x}\right)$ for a matrix $A$ of variables $x$
stoic pythonBOT
Namington:
pearl elm
I am not sure if I can just do augmented matrix operations here
and just convert it back and work it out
limber sierra
@pearl elm yes, that would be best.
pearl elm
ok
half storm
looks way better
limber sierra
i mean
fundamentally augmented matirces are systems of equations
theyre just a convenient representation
where we dont write all the variables
or the + signs
and instead arrange them in a nice row pattern to convey the same information
every operation on an augmented matrix is an operation on an equation of a system of equations
multiplying a row by C corresponds to multiplying both sides of an equation by C
adding a row to another corresponds to adding an equation to another (think the "elimination" method of solving linear systems, if you're familiar)
swapping rows is literally just rewriting your system in a different order