the 3 columns*
#linear-algebra
tardy eagle
wintry steppe
do linear programming questions fall under this channel?
dusky epoch
i guess they would yeah
warm flicker
What if the first variable (x1) is a free variable? What will the rref look like?
gloomy arrow
I cant load my professors lectures so I am not sure how to do these problems
quartz compass
did you email your professor
gloomy arrow
Yeah it seems to be some technical issue with blackboard
She has doesnt want to upload it on youtube*
quartz compass
maybe try suggesting get something set up on google drive or drop box maybe, idk just an idea
well that sucks
gloomy arrow
But I got other youtube videos to help me thankfully
But how would I do this problem
Would I write H as a matrix then rref?
half ice
H can also be defined as:
[3 2][a]
[1 -3][b]
And that left matrix is A.
@gloomy arrow
I suggest actually carrying out that multiplication to see why it works
wintry steppe
hello
can someone point me to somewhere that talks about orthogonal matrices?
i don't understand why it has the properties that it has
potent totem
Can I just pick W as $F^4$, since every vector space is a subspace of itself. Then proving U+W=$F^4$ should be very easy.
stoic pythonBOT
Darius:
gloomy arrow
@half ice Ye I am getting that. Do I just reduce the matrix and find the column space?
Oh wait that is just straight up A that I needed
Okay okay I see I see
pallid rampart
@potent totem you cannot use W=F^4
Direct sum means that every vector is uniquely written as the sum of vector from U and W
sonic osprey
$+$ is different from $\oplus$
stoic pythonBOT
Zopherus:
potent totem
I completely overlooked that, ty.
maiden silo
$T: \mathbb R^3 \to \mathbb P_2$ can just be
stoic pythonBOT
KS42:
pallid rampart
*isomorphic
maiden silo
ye ye
stoic pythonBOT
KS42:
gloomy arrow
I just said that the zero vector is not included, that would not make it subspace
Or is there something formal I should do
sonic osprey
no thats right
maiden silo
yea thats sufficient
gloomy arrow
Alright thanks
maiden silo
though its also not closed over addition
gloomy arrow
How is it not also closed under addition?
maiden silo
like given x, y
and w, z
H(x, y) + H(w, z) does not equal H(x+w, y+z)
which is a condition
somebdoy correct me if im wrong
gloomy arrow
Oh its because of that 1 right?
maiden silo
yea
gloomy arrow
I said for my explanation "H is not a subspace because the zero vector is not an element of H"
maiden silo
thats fine
gloomy arrow
Alright thank you
maiden silo
its sufficient im pretty sure
maiden silo
the set of whatever spans the space
gloomy arrow
Makes sense sorry have to review some more
maiden silo
ye dont worry its confusing im taking it too
lokey can somebody help
theres 2 conditions
and i cant do it
straight it makes no sense i'm getting two matrices
gloomy arrow
Would I just turn this into a matrix and then the columns become the spanning set?
slow scroll
you can just write it as xv1 + yv2 + zv3 and then {v1, v2, v3} will be the spanning set (by definition)
gloomy arrow
I see alright
Would this be
[1 -1 0]
[0 2 1]
I just turned the right part into a matrix dont know if that works and why
slow scroll
it works because the solutions to Av = 0 are the same as the solutions to
x - y = 0
2y + z = 0
when taking v = (x,y,z)^T.
maybe that sounds a bit tautological... because it is. You can go between the systems of equations representation and the matrix representation and the variables all remain the same.
and yea, the matrix you got for A is right
gloomy arrow
The nullspace is just all the vectors that solve Av=0
Okay yeah that make sense that they would be the same as that system
Would I need to do row operations? Or would the size of the column space just be the number of rows because thats how many entries are in one column
Or do I need to see how many pivots there are
gloomy arrow
And for the nul space would it be the number of rows?
slow scroll
um, that was for Col(A)
gloomy arrow
Sorry I mean to say number of columns for the nullspace
slow scroll
ah ok. yep that would be correct
gloomy arrow
So the nullspace the and row space would have the same dimensions?
Because wouldnt the row space also have the same number of entries as the number of columns
slow scroll
nope, the dimension of the row space equals the dimension of the column space.
gloomy arrow
But when looking for like the vectors that span the rowspace wouldnt you reduce the matrix then look at the rows and then the rows become vectors
slow scroll
you do row reduction on the transpose of A and count the column pivots of rref(A^T) to compute the row rank. the proof that row rank = column rank isn't exactly trivial though
gloomy arrow
I see
I have not been doing rref on the transpose which makes sense why I was getting thinks wrong
Okay thank you
slow scroll
np
wispy perch
how do i go about proving that the eigenvalues of an orthogonal matrix is +/- 1
can someone give me a hint how to start
slow scroll
Hint: you should probably use the fact that any matrix A has the same eigenvalues as A^T.
wispy perch
oh that's a new fact that i don't know
quartz compass
I don't think that's necessary to know for this problem
slow scroll
Its just the first way I thought of doing it.
quartz compass
I'd just start from the eigenvalue equation Ax=lambda x and try to compute the inner product
wispy perch
the question includes a hint
"consider the norm of Av"
v is the eigenvector i guess
i do Av = lambda v and idk what to do yet
do i norm both sides?
quartz compass
yeah multiply it by its transpose on the left
wispy perch
v = A-1 lambda v
stoic pythonBOT
Merosity:
quartz compass
this is what I had in mind
quartz compass
yep now keep simplifying
quartz compass
yup it's just a scalar
wispy perch
thanks
quartz compass
you're welcome
don't forget to reason out that the reason you can divide by v^Tv is because eigenvectors are not 0
slow scroll
hm, I would have just done $Q^T Q v = Q^T \lambda v = \lambda^2 v$ and since $Q^T Q = I$ and the only eigenvalue of $I$ is 1, we have $\lambda^2 = 1$.
stoic pythonBOT
kxrider:
wispy perch
oooh interesting solution but how did you get lambda^2 v?
slow scroll
Q^T and Q necessarily have the same eigenvalues
wispy perch
oooh
quartz compass
the reason I mention that bit about eigenvectors not being 0 is because a very similar proof can be done to show that different eigenvalues of an orthogonal matrix necessarily have orthogonal eigenvectors
just work it through with Av=lambda v and Au = mu u instead
mu != lambda
slow scroll
hmm, actually, what I did might not be good. I assumed that the eigenspaces of Q are the same as the eigenspaces of Q^T. I'm not sure that is easy to prove... Its false, disregard the stupidity above
wispy perch
is I and -I "diagonal orthogonal matrices"?
quartz compass
well what conditions do they need to satisfy to be that?
wispy perch
i guess so
quartz compass
they're obviously diagonal, but orthogonal?
wispy perch
i guess all the column and row vectors have to be orthonormal and every pair of vectors is orthogonal
yeah i guess they are
quartz compass
I think they need to be unit length too
quartz compass
yeah
wispy perch
i see
quartz compass
$P=\frac{vv^T}{v^Tv}$
stoic pythonBOT
Merosity:
honest notch
Anyone able to help with this
half ice
@honest notch
Still looking for it?
honest notch
yup
half ice
,w row reduce {{-6,-3,24},{8,5,-34},{-3,0,9}}
,w row reduce {{-6,-3,24,a},{8,5,-34,b},{-3,0,9,c}}
stoic pythonBOT
Requested by Kaynex#7520
half ice
There, Wolfram has it under the assumption
honest notch
damn that is not what i got
honest notch
im not sure how they assumed c either
half ice
5x + 3y - 2z = 0, sry
half ice
The bottom right most entry was
(5x + 3y - 2z) / some denominator
Meaning the bottom row reads
0 = (5x + 3y - 2z) / some denominator
Wolfram realized the form I was using and made it an assumption
olive coral
Can someone verify the answer for the row space? My row-echelon form matrix is
1 4 5 6 9
0 1 1 1 2
0 0 0 0 0
0 0 0 0 0
Shouldn't the basis for the row space of A be {(1 4 5 6 9), (0 1 1 1 2)}?
Oh wait, is the basis for row space not unique?
olive coral
Ohhh so both are correct?
dusky epoch
i can't say without checking explicitly
olive coral
Ah OK. Thanks anyway!
dusky epoch
,w (3,-2,1,4,-1) - 3 * (1,4,5,6,9)
stoic pythonBOT
Requested by Ann#0413
dusky epoch
yeah you're all set
honest notch
,w row reduce {{6,11,-4,a},{4,9,-1,b},{-18,-38,7,c}}
stoic pythonBOT
Requested by ๐๐ฒ๐ฅ#6098
fringe pasture
I'm supposed to analyze gausselimination in matlab using backslash/inverse(a)*b for 4 different files (with different sizes of N). Using this code:
and then fill in the times in a table like this
I've done that, but then they ask When N increases with a factor k, how is the runtime expected to grow according to the theory? Does it always line up with the theory? why not?
And I'm like.. what theory.. we havent read any theory about that i have no idea what theory Im supposed to compare it to
forest trail
Whats the difference between decomposing a vector and projecting it? Assume we are projecting the vector onto the basis vectors/unit vectors.
I read somewhere that the components of a vector are just scalar multiples of the unit vectors. Whereas a projection of a vector onto a second vector is the component of the original vector in the direction of the second vector. Is this correct?
potent moss
whats a good method to learn the cross product
forest trail
pallid rampart
Decomposition refers to writing a vector as the linear combination of other vectors, typically a basis
The projection that you described is an orthogonal projection
forest trail
each term including the magnitude of that component and the unit vector?
pallid rampart
?
You can decompose your vector such that each term is some multiple of the unit vectors
For example $\begin{bmatrix}3\2\1\end{bmatrix}= \begin{bmatrix}3\0\0\end{bmatrix}+ \begin{bmatrix}0\2\0\end{bmatrix}+ \begin{bmatrix}0\0\1\end{bmatrix}$, then the first term is the $x$ component, second term is the $y$ component, and third term is $z$ component
stoic pythonBOT
Whoever:
forest trail
so the components are scalars multiplied by basis vectors?
pallid rampart
Yes
But there are many basis for R^3, so you can choose any other set of basis vectors
And yes it can be subtraction when the magnitude is less than 0
forest trail
But there are many basis for R^3, so you can choose any other set of basis vectors
@pallid rampart you are referring the unit vectors in different coordinate systems?
pallid rampart
No
A basis is a set that is linearly independent and spans the space
But (1, 0, 0), (0,1,0), (0,0,1) is called the standard basis
pallid rampart
You can write each vector as the linear combination of basis vectors
Uniquely
Well if you havenโt seen this donโt worry about it
But your original understanding of projection is good
forest trail
oh ok
well i have seen vectors written as a combination of basis vectors (and i use them)
ok so whats the diff. b/w scalar projections and vector projection? And also b/w scalar projections and decomposition
pallid rampart
So here
We are given vectors a and b
Scalar projection of a onto b refers to the length a_1. The orthogonal projection (sometimes just projection) of a onto b is the vector a_1. The decomposition of a refers to writing a=a_1+a_2
forest trail
so we could think of y-component of the vector as the orthogonal projection onto the y-unit vector?
forest trail
sorry for the late response, but is the scalar projection a vector?
cursive narwhal
@forest trail Would it not be the case that the scalar projection be a scalar, as opposed to being a vector?
pallid rampart
Ok lemme say it again: orthogonal projection of a onto b is the vector a_1; the scalar projection is the length of vector a_1.
desert lava
no im aware of how it works
im asking for the operations that need to be applied in order to solve
gray dust
there are MANY ways to row reduce that matrix. they will all lead to the same rref
gray dust
paul has examples w/ solns
desert lava
not this specific one
gray dust
gray dust
the 2nd line immediately gives part of the solution. you can use that to backsolve the rest
desert lava
6:40 PM] dlp: it's not even in rref but it is solvable
i know
but i wanted to see it in rref
gray dust
i linked you an rref calculator
desert lava
also, this isnt rref?
desert lava
that is the rref calculator
gray dust
click the option "Reduced"
desert lava
thank you
ocean sequoia
how are you suppose to know what the latent dimensions are we compute the SVD
like how do we know what the singular values are attributed too
forest trail
Are the components of a vector {the part in bold => [45i + -56j]} a scalar or a signed number? From what I've read, magnitude are always supposed to be positive yet we have things like -2i or -9j. Does the negative here correspond to the magnitude or the negative of the unit vector (the negative of the unit vector would thus indicate opposite of the regular direction of the unit vector).
Are the things stated here true (https://farside.ph.utexas.edu/teaching/336k/Newton/node153.html) ? Especially the last paragraph.
dusky epoch
a scalar or a signed number?
"scalar" is just a fancy word for "real number"
scalars can be negative just fine
limber sierra
and yes, the entries of a vector need not be scalars
we can consider, for example, a vector space where vectors are elements of the set ${\begin{bmatrix}x\end{bmatrix} \mid x \in \bR}$ but the scalars are all from $\bQ$
stoic pythonBOT
Namington:
limber sierra
this would be called "the vector space R over the scalar field Q"
and we could do the same thing for, say, R^2 or whatever
where the vectors are from R^2 but we only consider scalars from Q
this is formally a different structure from "standard" R^2
broken hawk
tbf Iโd say then that the โentriesโ of the vector would have to be wrt a basis over โ (which of course is kinda hard to write down)
(for some very large value of โkindaโ)
violet comet
Could someone explain exactly how one arrives at the basic matrix for rotation about the y-axis? I wanted to use the matrices to help with me rotating an object in a program I'm working on. I've developed a basic understanding of vectors and matrices with the help of 3Blue1Brown's video on Linear Algebra.
I've worked out/understood exactly how one arrives at the basic matrix for rotation around the z-axis and the x-axis (and verified it with Wikipedia) but for some reason that I don't understand yet I keep getting the matrix for rotation about the y-axis as the image I attached below whereas Wikipedia shows the same thing apart from the location of the negative sign.. I just have no clue how I keep getting a different answer for rotation about the y-axis but I get it correctly for the rotation about the x and z axis.
(I might be doing something blatantly incompetent but I would greatly appreciate it if someone points it out to me)
limber sierra
note that, when rotating about the y axis, a 0 degree rotation represents being aligned with the z axis
and as that increases, you mvoe counterclockwise towards the x axis
this might screw with your intuition a bit, since if you try and visualize this as polar coordinates/using trigonometric intuition, this means that the x axis is the "vertical" one
dry spear
I need help with b
I made a matrix with columns for a, b, c, d, e, but the RREF doesn't make an orthonormal basis for the rows or columns
gloomy arrow
For part c can I just treat the columns as vectors and add them
slow scroll
sure
meager tinsel
some help with Linear Programming?
here's the problem
so here's some system
Ax <= b
x >= 0
typical right?
Given that A is integer matrix and also for each integer vector b the solution (must be some polytope union right) is also integer prove that A is totally unimodular
I know what totally unimodualr is
aaaand its close
must be very easy
buut
my thoughts
maybe I can look at some specific b's such that Ax = b (solution of that) will be vertex of that thing...
basically i should prove that A is a bijections from Z_n to Z_n right?
civic tiger
idk if this is the appropriate place to ask but I have a question about Legrange polynomials
dry spear
if I have a (7,6) matrix, augmented with vector v, can there be one unique solution?
like if I have a leading 1 in 6 of the rows with a row of 0's at the bottom, does that still have one unique solution?
half ice
No, there can't be. A non-square matrix can't have an inverse @dry spear
Oh wait, unless you're counting the augmented part as another column
meager tinsel
any suggestions?(
gloomy arrow
The transformation is just taking a derivative so any constant would be in the kernel right?
pallid rampart
Yes
gloomy arrow
What about the spanning set, wouldnt be all real numbers or how would I denote that
pallid rampart
The question asks to find a set of polynomials that spans the kernel
Technically the set of real numbers does span the kernel, but there are better answers
gloomy arrow
How would you find the set of polynomials that span the kernel?
pallid rampart
There is no the set
Any constant polynomial will span the kernel
Well
Non zero constant polynomial, that is
gloomy arrow
Right I get that
Im not sure how to write that though
Or can I just word it like that
pallid rampart
You would just say
gloomy arrow
I mean is there a proper notation I can use
pallid rampart
ughhh
idk
but from what I've seen you can write
The set $(1)$ spans the kernel of $T$
stoic pythonBOT
Whoever:
stoic pythonBOT
Whoever:
maiden silo
Can anybody help me understand this?
I think it's true
because the orthogonal component of the column space
is the null space of the transpose
and using A^T A x = A^T b
it would be 0
forest trail
@limber sierra i read that scalars are invariable for coordinate system transformations whereas the vector components are not? ( <---- this seems to be true since for certain coordinate system transformations, the magnitude and direction of the vector must be preserved. But since the coordinate system has changed, the components will also have to change to preserve the original magnitude).
limber sierra
Sure, changing coordinates is just changing the basis
forest trail
So vector components cannot be scalars then?
limber sierra
er
the set of scalars has to be a subset of the set of valid entries within a vector, assuming youre defining scalar multiplication conventionally
since $\lambda\begin{pmatrix}1\1\1\\vdots\1\end{pmatrix} = \begin{pmatrix}\lambda\\lambda\\lambda\\vdots\\lambda\end{pmatrix}$
stoic pythonBOT
Namington:
limber sierra
i.e. if something can be a scalar, then it can also be a component of your vector
the relationship doesnt necessarily hold the other way around, but for "common" vector spaces like Q^n, R^n, and C^n it does
quartz compass
well there's a slightly different kind of convention taking place that I think he's reading about
" scalars are invariable for coordinate system transformations whereas the vector components are not"
dusky epoch
Q^n
common
quartz compass
here scalars meaning rank 0 tensors and vector components being rank 1 tensors
limber sierra
this is Qpression
also mero i have never seen that definition beforoe
does that even work as a definition
quartz compass
that's why I chimed in to help out
limber sierra
well apparently you know more about bad lin alg conventions than i do
in that case
quartz compass
haha it's more of a physics thing
like for example if you're standing at a point of space you can measure temperature (scalar) and look at wind velocity (vector) and look at the components of this vector in some chosen coordinate system
your temperature doesn't depend on the coordinate system, so changing coordinate system will leave temperature unchanged while vector components will change
which is what @forest trail is talking about
limber sierra
ah okay, i getcha now
i mean, i think my statements still hold then, but admittedly theyre less helpful
since the scalar field then is just R
quartz compass
yeah I think the ambiguity should be cleared up in whatever they're reading at some point if they read more closely at their text or whatever anyways lol
limber sierra
the reluctance of elementary computational linear algebra courses to talk about scalar fields other than R forever confuses me
like, thats one of the more useful/applicable takeaways from the abstraction that linear algebra provides
being able to consider what happens when we restrict scaling behaviour to a subset is very useful in, say, computing
probably far more useful than the adj formula for the inverse or whatever
/rant
limber sierra
freudian slip
quartz compass
but yeah inverses are bad computationally, it's cheaper to do row reduce the augmented matrix every time I believe
limber sierra
real chads define the determinant as the only alternating multilinear map satisfying blahblah
and do all their proofs based
quartz compass
idk I don't implement this kind of stuff so I wouldn't know lol
limber sierra
in a similar vein, i do all my calculus by considering dedekind cuts
limber sierra
quartz compass
lol excellent
forest trail
Thank guys, tbh with you, i'm not reading a linear algebra text specifically rather i'm doing physics (on my own, for pleasure and olympiads, mainly the International Physics Olympiad (IPhO)).
The doubts I post here just come from the questions I do in my physics book. So should I read a maths text side-by-side?
(I know Calc. I, i've read that we should go single var. Calc ---> lin. Alg. ---> multivar calc.) Or should i just post questions like this/learn things when I need them physics and not necessarily learn all of math (but I love maths!)
quartz compass
yeah what you're doing seems fine, just keep in mind there's some slight differences in words sometimes
doing linear algebra and multivar calc will be inevitable, the sooner the better
forest trail
the set of scalars has to be a subset of the set of valid entries within a vector, assuming youre defining scalar multiplication conventionally
@limber sierra What would be the valid entries within a vector? And why would we need to bring scalar multiplication here? Is it because the only way (at least what i can think of) that scalars directly relate (by directly i mean there are no "in-between operators" like dot product) to vectors by multiplication?
The thing that confuses me is that some people the components of a vector are scalar multiples of the unit vectors, some say they are real numbers but not scalars (Check the last few paragraphs of this link-https://farside.ph.utexas.edu/teaching/336k/Newton/node153.html) and you say they may or may not be a scalar.
meager tinsel
Do I repeat my question if it wasn't answered and drowned or ...?
quasi vale
Yeah
meager tinsel
some help with Linear Programming?
here's the problem
so here's some system
Ax <= b
x >= 0
typical right?
Given that A is integer matrix and also for each integer vector b the solution (must be some polytope union right) is also integer prove that A is totally unimodular
I know what totally unimodualr is
aaaand its close
must be very easy
buut
my thoughts
maybe I can look at some specific b's such that Ax = b (solution of that) will be vertex of that thing...
basically i should prove that A is a bijections from Z_n to Z_n right?
hollow heath
Anyone know how to go about answering this question?
From what I learned the combinations of u,v,w will create a new vector b which lies in the same plane as u,v,w. And I can solve for b if b lies in the same plane as u,v,w. To check if columns lie in the same plane you have to find the combo of the columns that add up to zero. I am unable to find any combo that will add up to b being vector [0,0,0]
Maybe the planes are parallel and do not intersect. But in 3D planes can be in trouble without being parallel. You may need to move it up one axis so it aligns at an intesection. But uh idk how to integrate this knowledge to this triangular system
It seems the columns are independent as the first 2 columns on the left do not add up to the columns on the right.
Doing row and column picture in 3D seems to tedious a task and there should be an easier way?
I'm really lost as to how to answer this question. Its from Gilbert Strang Linear algebra and its applications 4th edition (Unfortanely can't find teachers solution so I only have odd numbered solutions)
Can someone steer me in the right direction?
storm python
i'm not going to read your essay above, but you should start from the bottom in a triangular system
you're trying to find the column [u v w] such that this is true,
you already know w = b_3
now go back up and solve for v, then back up again for u
agile wolf
Hi could someone give help me out with these two problems? I'm assuming i'm meant to use the projection equations but I'm not sure how to go about doing that
limber sierra
@forest trail that link is assuming the vector space in question is R^n
which is probably because its audience is introductory physics students
vectors from R^n can indeed only be composed of real numbers
in actuality, the structure of a "vector space" is far more versatile than this
but thats a side point
in any case, the point is: scalars and vector-entries are fundamentally different "things", but in the context you're currently working in, they're compatible
meager tinsel
I found there's a theore sorta of to the opposite side of what I'm trying to prove
but it seems easier too
doesn't give me a lot of hints
hollow finch
Assuming W is a subspace of V, why is it that $\dim(W)+\dim(W^{\perp})=\dim(V)$
stoic pythonBOT
nix:
wintry steppe
Well, you can show it by taking basis of W and basis of W^perp and showing its a basis for V
knowing that each vector v in V can be written as a sum of vectors v_1 and v_2 from W and W^perp respecitvely you can see that 'basis' spans V
Need to show linear independence
hollow finch
Right the projection theorem. The problem is that I'm trying to explain the 'why' to a student I'm tutoring.
Is there some intuitive reason? Or is it just how it be
wintry steppe
I always thought about W^perp as like kinda complementary vectors idk, I dont remember it that well but there was an intuition for it for sure
Or like, knowing that if two vectors are perpendicular then they are linearly independent, which is kinda 'visible'
forest trail
in any case, the point is: scalars and vector-entries are fundamentally different "things", but in the context you're currently working in, they're compatible
@limber sierra what are vector entries? and by "compatible" do you mean interchangeable?
limber sierra
vector entries are... numbers in a vector
i.e. each individual coordinate
and not quite "interchangeable" since they behave differently
but they can all be real numbers.
civic tiger
does anyone know how i can find, p, the polynomial coefficients of Lagrange by using linear algebra
i thought doing x\y_dot might work
low plank
mine or mortex's ?
low plank
yeah
half ice
Where both Ds are different matricies
dusky epoch
does this not make you thonk
low plank
....
half ice
Anyway have you got question 8?
low plank
my answer was that it wasn't diagonalizable
half ice
,w eigenvalues and eigenvectors of {{1,0,0},{4,1,-4},{4,0,-3}}
stoic pythonBOT
Requested by Kaynex#7520
low plank
its not diagonalizable is it
half ice
P =
[0 1 0]
[1 0 1]
[1 1 0]
D =
[-3 0 0]
[0 1 0]
[0 0 1]
half ice
You can kinda read the details off the eigensystem there
low plank
P^-1 would just be P with a -1 symbol above right?
half ice
P^-1 is P's inverse
half ice
Oh boy you're too late in the course to not know what an inverse is. Get that one fast
,w inverse of {{0,1,0},{1,0,1},{1,1,0}}
stoic pythonBOT
Requested by Kaynex#7520
low plank
oh you meant this
south wadi
hollow finch
We can always write
$\vec{v}=[\vec{v}]_{E}$
where E is the standard basis
stoic pythonBOT
nix:
hollow finch
We want a matrix which translates this coordinate vector into in terms of the basis B.
$P_{E \to B}[\vec{v}]{E}=[\vec{v}]{B}$
stoic pythonBOT
nix:
hollow finch
Use the fact that $P_{E\to B}^{-1}=P_{B\to E}$
stoic pythonBOT
nix:
hollow finch
Because finding $P_{B\to E}$ is really easy
stoic pythonBOT
nix:
low plank
@half ice i have two eigenvalues that are the same for question 9, would that make it not diagonalizable?
my eigenvalues are 1,1, and 3
hollow finch
What is the nullity of $A-1I$?
stoic pythonBOT
nix:
low plank
ive only found the eigenvalues so far
hollow finch
Because that will determine whether or not 1 has a geometric multiplicity equal to it's algebraic multiplicity.
It's possible to be diagonalizable with repeated eigenvalues if they are not defective.
low plank
okay
so i would make sure that the AM and GM are equal to each other
if they are i'd diagonalize and if not then i'd stop
meager tinsel
so nobody has any clues?
hollow finch
@low plank A will be diagonalizable if A-I has a nullity 2
Because then the eigenvalue 1 has two eigenvectors
If you have fewer than n eigenvectors for an nxn matrix, it is not diagonalizable
maiden silo
<@&286206848099549185>
wispy perch
this needs proving?
sonic osprey
do you think its obvious?
wispy perch
it does to me
sonic osprey
Can you explain?
wispy perch
suppose there is a vector v in the solution set, Av = b, multiply A^T both sides
still holds
sonic osprey
Sure, that shows that every solution to Ax = b is also a solution to A^TAx = A^Tb
But what if there are some solutions to A^TAx = A^Tb that aren't solutions to Ax = b?
wispy perch
OOOH
what i showed is just the solution to Ax = b is a subset of a solution to A^T Ax = A^T b
can i prove that the inverse of A^T exists
nvm
dusky epoch
who said A^T was square
novel ether
does anyone know why u is not [ 3, 1] ?
gray dust
fully solve the system (A-I)v2=v1 instead of picking out one solution @novel ether
let v2=(a,b). from the 2nd row of the augmented matrix of (A-I)v2=v1 we have a-2b=1, a=2b+1, v2=(2b+1,b). you chose to let b=1. the book chose b=0
fringe pasture
Hey guys, does anyone here know why making a matrix sparse makes it a lot faster to process compared to LU factorization? The bigger the N the more time you win with Sparse matrix
Got a presentation about my assignment in one hour and I still dont get it
wispy perch
how do i start working on this one?
i was thinking suppose v4 can be expressed as c1v1 + c2v2 + c3v3, then i did (v4 . v4) = (c1v1 + c2v2 + c3v3) . (c1v1 + c2v2 + c3v3). tried to show that right hand side can't equal to 1 but everything got messy and in the end idk what i'm doing
subtle walrus
@fringe pasture what do you mean by processed?
fringe pasture
I was supposed to make a table of runtime solving gauss elimination (backslash on matlab) on a big system matrix, I did, and now I can see that making the system sparse is faster than using LU
Gles means sparse
the bigger the N (eiffel) the more time it saves compared to LU
cant find any way to explain why
subtle walrus
so you solve a system of linear equations in matlab with sparse vs non-sparse matrices?
fringe pasture
yes
one sparse, and one just using LU factorization
and the left is just solving without either (Naiv)
regular gauss
subtle walrus
well, gauss algorithm will always take essentially the same amount of time
no matter the matrix
and if given a sparse matrix, the pivoting and whatnot would in general create a non sparse matrix
not sure which algorithm you are using to solve with sparse matrices
but in general there are better available algos for them
than just gauss
fringe pasture
x = A\b its builtin matlab function
subtle walrus
then i would assume it does something smarter
like e.g. performing a QR decomposition via Givens rotations performs better on sparse matrices
fringe pasture
First one is x = A\b, second one is [L.U] = lu(A) first and then backslash, and then last is making A = sparse(A)
subtle walrus
because you only have to get rid of the nonzero entries
well ok, not sure how the builtin matlab functions work
or if there even is a way to find out
but like in general when you do a decomposition, you want to keep sparseness
because that keeps down computation time obviously
and gauss algo does not do that
so use better algos
fringe pasture
But its so weird
Because the question is, which modell saves the most time and why
So i have to answer why either LU or sparse saves more time with backslash
and in the assignment it also says that Matlab uses gauss when you do backslash
potent totem
so since every element in U is also in V, cannot I literally take the same mapping function that maps U to W and use it to map V to W. Thus for every element u in U T(u) = S(u).
storm python
what if you have a proper subspace U of V? then you wouldn't know how some elements of v in V, v not in U would map to W
U can be smaller than V, so you need to define the mapping for elements in V-U such that linearity is maintained
storm python
sounds like it'll work
potent totem
ok tyvm
dusky epoch
let V = R^2, U be the x-axis, and S be any nonzero map on U
storm python
i apologize
dusky epoch
your construction would have T mapping everything off the x-axis to zero
storm python
will i get ban from being wrong so many times
dusky epoch
but then T(1, 1) + T(1, -1) would not equal T(2,0)
i don't think so
you'll just get me saying "big wrong"
storm python
every big wrong you say to me i'm learning somethin new
i'm somewhat a mister positive myself you see
shy atlas
thats the spirit pub 
slender yarrow
everytime she says big wrong you're closer to getting the big wrong role
storm python
tis the price to pay
slim laurel
Hey guys, can I get a hint on where to start with this?
I've tried googling but only the derivation for a matrix with full column rank shows up
dusky epoch
transpose and then apply that derivation
or in other words
apply the derivation you found for A^T
slim laurel
thank you I'll give it a go
trim galleon
i like matrices its too ez
pallid rampart
๐ค
Write V=U \oplus K for some K (K will be the orthogonal complement I think), then every v can be written as v=u+w, where u is in U and w is in K. Then define Tv=u
@storm python @potent totem
Oh W exist because V is finite dimensional
i.e. extending a basis of U to a basis of V, and the extra vectors will be a basis of W
pallid rampart
Yeah
If an element is in K, then it gets mapped to 0
Oh shoot W is already used
I donโt think the function will be convex
forest trail
@limber sierra could you accept my friend request on discord so we can continue this discussion in DMs? I would like to have all your responses collated so I can reference them later. Here in this channel, i have to constantly scroll up.
forest trail
Maybe write them down yourself as notes
@pallid rampart yeah thats what i want to do.
pallid rampart
What's preventing you from doing so?
mild tiger
det(kA) = k^n det(A) where A is n by n
but why
pallid rampart
Many ways to explain this
quartz compass
write kA = k*I*A and take the determinant of kI which is an nxn matrix with k on the diagonal
idk if that answers "why" necessarily
pallid rampart
Intuitively since the determinant is the volume formed by the unit vectors under the transformation represented by the matrix. If each unit vector is multiplied by k, then the volume in n dimension will be multiplied by k^n
Ig thatโs intuitive justification
kindred oxide
how do I find the inverse ?
half ice
I'll let A' represent the inverse. Multiply both sides by A':
A - 6I + 5A' = 0
A - 6I = -5A'
A' = 6/5 I - 1/5 A
@kindred oxide
civic tiger
why do i keep getting cucked
civic tiger
to get the matrix of A coefficients a0... a_m-1 would i have to x\y where x is the m-1 x m-1 matrix and y is the column vector from y0 to y_m-1
sorry if i wasnt clear when asking lol
civic tiger
ok no one cool..
grand ingot
vital swallow
I don't think there's enough information to answer this?
vital swallow
I think the interpretation is that e_1 and e_2 generate the nullspace,
but then T(e_3) would be the answer. but that could be arbitrary
half ice
Oh I see, that is the null space just in this case
vital swallow
heh. I think they mean : the null space of a known but undisclosed linear map T is ...
half ice
Then:
T[x, y, z]
= T[x, y, 0] + T[0, 0, z]
= [0, 0, 0] + [0, 0, c]
= [0, 0, c]
Oh wait I can't justify line 2 to line 3
vital swallow
what about $ T = \begin{bmatrix} 0&0&2\0&0&-3\0&0&4\end{bmatrix} $?
stoic pythonBOT
Alek:
vital swallow
the last column is totally unknown
half ice
Nice counter example, there's not enough info to say anything about [a,b,c]
daring solstice
Trying to understand this Gaussian_Elim.py I found from https://gist.github.com/num3ric/1357315
I don't quite understand this double for loop where it loops through the first equation and...
No wait A is a matrix of equations it seems
I found a more beginner friendly code here but I don't quite understand how eliminating columns work here
https://github.com/ievans/GaussianElimination/blob/master/gaussianelimination.py
I think I'd need to write my own GaussianElimination script based off
I feel like I am missing some major concepts when looking at example scripts online
I found this vid on Youtube haven't seen it yet tho https://www.youtube.com/watch?v=ZDxONtacA_4
limber sierra
so are you trying to understand gaussian elimination programs when you don't know how to do it by hand?
daring solstice
I know how to do it by hand but I only know the method right now that involves using the elementary row operations
limber sierra
i'd recommend getting some practice row reducing actual matrices first
thats exactly whats going on here
like lets use your second code for example, the key thing here is the bounds
note that, due to how the bounds are changing
it's progressively making the matrix upper triangular
for the first column, when col = 0, it iterates through range(0, len(m))
i..e through ALL rows
but then for the next column col = 1
it only iterates through range(1, len(m))
and then for the nex, through range(2, len(m))
and so on
observe that it's essentially starting by
reducing the first column
and then the second
and then the third
and so on
since once a column is fully reduced, it can basically be ignored.
it'd probably be helpful to trace the execution of the algorithm on pen-and-paper
draw a 3 by 3 matrix and try and follow along with it
daring solstice
I agree. I wanted to do that as you were explaining the steps and I realize that may be extremely helpful way to break it down. I think doing that will help me visualize it better too. Thanks for the suggestion
I think I shall do that first and then come back with further questions if I have any after that
vital swallow
I can be beneficial to draw the "shape" of the matrix as you expect at each step for a generic example
daring solstice
Currently my understanding of Gaussian Elimination feels too 'aribtary' on what elementary row operations to choose in addition to what pivot to choose
limber sierra
well, the key insight is that it doesn't really matter
daring solstice
Probably becasue its a beginner level's introduction but it feels like I just 'eyeball' that the first row can be multipled to be subtracted from second row for example
I am baby
limber sierra
like some orders for gaussian elimination end up executing "faster"
but
any valid order will work, as long as it's eventually reducing rows and columns to the desired form
the algorithms you gave do a very "naive" approach
they dont try and optimize at all, just go through the simplest order
which is fine
[since after all, trying to optimize takes more computational time, so for small matrices its often not worth it]
daring solstice
I chose to find a naive approaches on purpose as I am still a beginner to the concept
As my own personal homework wanted to write my own script with numpy that visualises the naive Gaussian elimination so I can get better understanding
vital swallow
I usually recommend ignoring pivots when first learning the algorithm
daring solstice
That's what I am doing yes 
vital swallow
Your operation above looks good.
daring solstice
I should probably solve a few more by hand too to better familiarise myself with the algorithm but I noticed I take a fairly long time to do so. I guess more practice would lead to better accuracy and speed
vital swallow
I wouldn't call it "eyeball"ing. It's division: what number t satisfies t*(1) = 2 where 1 and 2 were the numbers leading your rows
daring solstice
Yeah I was about to say that
kindred oxide
can someone help me with this
vital swallow
I really recommend learning how to do this algo by hand, yeah. It's not the only important Linear Algebra algo, but almost any question in Lin Alg can be answered by some smart (but sometimes confusing or complicated) application of it
daring solstice
But occasionally it seems I can do R2 - (1)R1 sometimes if it satisfies the condition
And sometimes it seems I need to multiply both equations
Yeah I wanted to write a script in order to better understand how to solve it by hand but in hindsight I'm putting the cart before the horse there...
vital swallow
Multiplying both equations is always going to be optional. Preferred, often, if you want to avoid fractions in your hand calculations
Also, there are some extra considerations that come up with a (good) implementation of this algo for computers. Since computers represent decimals using a finite number of places, extra things can go wrong
That honestly gets pretty tricky in general
daring solstice
I read about how LU Factorisation is preferred to Gaussian for as computer due to floating point errors. I didn't fully understand it tho
vital swallow
LU facto.. is harder. It also is a bit weird to do by hand, since it leaves some of the info implicit because it is trying to do something more efficient for computers
Hey Retro, what do you know about determinants? You're starting with $UU^T=I$ and you have to get the determinant involved
stoic pythonBOT
Alek:
daring solstice
Thanks btw for explaining stuff in an accessible and beginner friendly approach btw @vital swallow
vital swallow
No prob @daring solstice I tried ๐
daring solstice
I'm trying to self teach myself linear algebra to prepare myself for a CS degree in a year so I'm a fresh newbie to post highschool mathematics...
It's kinda fun tho in its own way
I quite like Math but never really was good at it in the past but hoping that can change with more practice and effort
vital swallow
That's a great time to get into Linear Algebra, actually! Some examples may not be accessible yet, but the general ideas are
And Linear Algebra's applications are astoundingly wide-reaching
daring solstice
I heard linear algebra doesn't have a prereq beyond high school algebra and it's a good starting point before heading into Calculus too
vital swallow
most of it won't come up until multivariable calc (Calc 3 at many schools) or differential equations. but the ideas are working in the background everywhere you look, once you know what you're looking for
daring solstice
I am using a mix of 'Linear Algebra: Step by Step (Kuldeep Singh)' and Jim Hefferson's Linear Algebra textbook for my self teaching materials. I like the former's accessibility but I do understand I'd need to get used to the latter's rigor eventually. I guess I am gonna use the latter book as a sort of 'New Game+' to use an analogy
vital swallow
I don't know Singh's. I think the other I tried using to teach a previous semester.
yeah. Hefferon, you meant? I like it, but I'm not sure it's super easy to read.
I still love Sheldon Axler's Linear Algebra Done Right, but that may be better as a New Game+ than a real intro
it's better if you need to know the theoretical stuff really well (which you do for a lot of areas of comp sci, and definitely for math) but it's still written at a fairly low level
but in your position, a more direct computational book is probably better
daring solstice
Singh's is the screenshot I posted earlier
looks something like this
I'd add that book to the list tho :D
vital swallow
Looks like a good choice, from that pic
Axler is for after this book, then. since you haven't taken more math, you'll want to know better what sort of things he is talking about
daring solstice
It's really accessible to me. The same extract from Jim Jefferson's on Gaussian was a bit trickier for me to parse
vital swallow
Yeah .. that's the book I tried to use
daring solstice
I think I am not familiar with this type of rigor yet and I do feel that I am missing some sort of pre-req in terms of experience here
I figured using a more accessible book for linear algebra and then going back to this one can help expose me to the rigor more and get used to this style of writing while already having a familiarity with the concepts
rather than tackling two things at once
vital swallow
Yeah. Hefferon is technical and "correct" all the time, at the expense of readability
limber sierra
read hoffman-kunze coward
vital swallow
Namington, that seems like a good compromise book between Hefferon and (apparently) Singh. is there anything that makes it especially good?
limber sierra
oh im joking, it's far more rigorous than both of the aforementioned texts
it's a mathematician's reference
i mean its not exactly a category theoretic approach to linear algebra but
vital swallow
oh. maybe my view is too skewed
limber sierra
hefferson is rigorous but its not particularly so i dont think
daring solstice
Singh is comfortable it's kinda comrpomises a bit of rigor but it makes the transition from high school to undergrad math really painless and if used as a stepping stone I think would be fine
vital swallow
I felt like Hefferon was a bit ..stuffy maybe.
daring solstice
There is an argument to be made that starting in the deep end may save time in the long run in exchange for extra effort at the beginning but I started my preparation early and have time
willing to spend more time to make the transition easier
vital swallow
Of course [lol], Manin's book is the gold-standard
I actually disagree a bit, Clementine
limber sierra
like
this is how hoffman kunze defnes the determinant
this is certainly the way an algebraist defines the determinant
it's also totally unhelpful if you dont want
full hardcore raw abstraction
vital swallow
hah. indeed I didn't scroll that far
limber sierra
its like defining C as the closure of R
[which is indeed what my intro analysis course did]
vital swallow
I mean .. it's the only good definition of the determinant ... but still.
limber sierra
yeah again, it's a reference text
this is the "correct" way to define the determinant
but maybe not the most helpfufl way to think about it
at least at an introductory level
vital swallow
I'll have to glance over it at some point before I teach the undergrad class again.. just to laugh about not using it, of course
limber sierra
i mean its a well-written text for
what it's trying to do
and i think it's good to have a sort of "baby rudin equivalent" for intro algebra
but yeah, i certainly wouldnt use this for a typical lin algebra course
or any linear algebra course not supplemented by extensive lecture notes, for that matter
kindred oxide
Hey Retro, what do you know about determinants? You're starting with $UU^T=I$ and you have to get the determinant involved
@vital swallow so I tried to take the determinants of both sides and I got det(UU^T)=1 I don't know what to do next
stoic pythonBOT
Retro:
Compile Error! Click the 
reaction for details. (You may edit your message)
vital swallow
So Retro, do you know anything about det and how it treats products or transposes? (since you have a product of U and U^T)
kindred oxide
Tbh not too much
vital swallow
So the determinant has some awesome (and maybe tricky to prove) properties, like $\det(AB) = \det(A) \det(B)$
stoic pythonBOT
Alek:
vital swallow
I don't see how you could have possibly concluded (in a correct way) that det(U^2) = 1
vital swallow
This may be a simple question: if I know that the matrix Q satisfies
$ v^\dagger Q v = 0 $
stoic pythonBOT
Alek:
vital swallow
for all vectors v in C^n ... what do I know about Q?
by the dagger, I mean the conjugate transpose
must we have Q = 0 ?
hollow finch
I think it must be skew symmetric iirc
quartz compass
it's not necessarily 0, could be like a 90 degree rotation so that v^T and Qv are orthogonal
you can take the dagger of it to get that it's also satisfied by Q^T
vital swallow
ah of course. thanks. (bilinear forms are a weakness of mine)
quartz compass
I'm writing T to mean dagger
wait I must be
I am just wrong
for every vector v
obviously rotation can't work because it has eigenvectors with nonzero eigenvalues
if this has any eigenvectors, it necessarily must have 0 eigenvalues
otherwise you end up with that not being 0
@vital swallow
vital swallow
hm
quartz compass
to prove it's skew symmetric we can take the conjugate transpose, add it to itself
$v^\dagger(Q+Q^\dagger)v = 0$
stoic pythonBOT
Merosity:
quartz compass
for all v
but Q+Q^dagger is hermitian so in this case we know it's the 0 matrix
$Q+Q^\dagger = 0$
stoic pythonBOT
Merosity:
quartz compass
so it's $Q^\dagger = - Q$
stoic pythonBOT
Merosity:
quartz compass
past that I don't think there's much to be said
vital swallow
how does $v^\dagger (Q+Q^\dagger)v=0$ imply $Q+Q^\dagger=0$?
stoic pythonBOT
Alek:
quartz compass
because $A = Q+Q^\dagger = A^\dagger$ is hermitian
stoic pythonBOT
Merosity:
quartz compass
and earlier I said if it has eigenvalues, its eigenvalues are 0
vital swallow
ah yep
quartz compass
cool
vital swallow
thanks a bunch
quartz compass
yeah you're welcome
thorn robin
@vital swallow Q = 0
do you know the identity that allows you to write v* Q u in as a sum of terms of the form w* Q w?
cold topaz
if x1 = x2 = x3 = 0 in a matrix, what does that indicate?
broken hawk
what do these variables stand for
broken hawk
no Iโm genuinely asking that because those variables could mean anything
the notation isnโt standardized
I genuinely have no idea what you mean with them
cold topaz
oh
broken hawk
youโre gonna have to give me a definition
slow scroll
or a screenshot 
cold topaz
u know when we have a matrix of 3x3, we solve it. When we solve it, we find x1, x2, and x3.
broken hawk
and also โa matrixโ is extremely vague. what size does the matrix have for example
solve it as in find a solution to the equation $A \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}$?
stoic pythonBOT
Sascha Baer:
broken hawk
(where A is your matrix)
if the only solution to that equation is the 0-vector then that means the matrix has full rank
cold topaz
like this matrix, and set it equal to <0, 0, 0>.
and solve it.
I've already solved it and the results are x1=x2=x3=0
but what does that tell use about the vectors that we used in this matrix?
broken hawk
it means the matrix has full rank
but I feel like if I just told you everything else that implied you wouldnโt learn a whole lot from it
btw watch 3blue1brownโs videos
I have this gut feeling youโre not really grasping what is behind these calculations youโre doing (I may be wrong, itโs hard to tell)
cold topaz
please. it's 1:30 AM. lol
broken hawk
Iโm not saying right now
if itโs 1:30 am for you, go to sleep
https://www.youtube.com/watch?v=fNk_zzaMoSs&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab I feel like every linalg student should watch these videos
they give a great intuition for the basic notions of linear algebra
things like what a matrix really represents, what its rank means etc
cold topaz
i knw what vectors are and how they are formed under differnt circumstances lol
broken hawk
itโs a playlist
I watched these videos after taking a university level course in linear algebra and still found them insightful
cold topaz
oh
i didnt remember the word coefficient
they are the coefficients
please. just tell me. I'm prepared to take notes.
dusky epoch
Ax is a linear combination of the cols of A, with the coefficients being the components of x
if you solved Ax = 0 and found that x=0 is the only solution, then you have found that the only linear combination of the cols of A that sums to zero is the trivial linear combination, namely one where the coefficients are all zero
this is word for word the definition of "the cols of A are linearly independent"
@cold topaz
do you understand what i said in its entirety
warm wedge
Um
Orthoganility
Meaning when 2 vectors are perpendicular, right?
Meaning the angle between the vectors is 0
subtle walrus
we call two vectors orthogonal if their inner product is 0
subtle walrus
what's that?
warm wedge
Yes
subtle walrus
inner product is the scalar product or dot product
warm wedge
Meaning cos theta = u . v / llull llvll, right?
subtle walrus
sure, yes
subtle walrus
that is the standard dot product
broken hawk
inner product is the . there
warm wedge
I accidentally read it as linear product, not inner product
broken hawk
definition of what angles mean depends on the definition of the inner product
warm wedge
So if u.v = 0, the vectors are orthogonal
broken hawk
ye
subtle walrus
yes
broken hawk
perpendicular is a synonym to orthogonal
subtle walrus
orthogonality is a generalization of perpendicularity
warm wedge
Oh
subtle walrus
if your work in real vector spaces, it is synonymous
broken hawk
I suppose some people use โperpendicularโ to mean โorthogonal with respect to the standard inner productโ
whatโs your definition of perpendicular, loch?
subtle walrus
i'd define perpendicular in terms of basic geometry
warm wedge
Two*
subtle walrus
only in R^2
broken hawk
Iโve always used the words completely synonymously
btu I also just donโt use perpendicular
cause it takes longer to type
warm wedge
Oh
subtle walrus
i would say the better term is "orthogonal"
and i would in most cases reinterpret perpendicular to orthogonal
warm wedge
Okay
I am making a lot of typos today
Anyways, thank y'all so much!
Ok i get it now
So orthogonal is if one of the vectors or both are zero vectors
But if they are non zero but their dot product is zero, they are orthogonal and perpendicular
true oar
Hey guys can anyone help me with a vectors equation?
subtle walrus
you could try just asking
subtle walrus
ok, where is the problem?
subtle walrus
probably by applying definitions
what is the multiplication of vectors?
what is | |?
dusky epoch
ok, so sm ghosted me
subtle walrus
ok, what is the definition of | |, then?
true oar
It turns it to positive if its a negative
the question says its not the length of the vector
true oar
It determines its direction
subtle walrus
what does that mean
dusky epoch
the question says its not the length of the vector
can you show what the question says in its entirety
true oar
It's not in english
subtle walrus
post it anyway
dusky epoch
what language is it in
true oar
Hebrew
subtle walrus
ok, this is gonna be hard then
true oar
Do you want me to post it anyway?
dusky epoch
yes
dusky epoch
@limpid vine you are not helping
okay so can you translate what the first two lines say
@true oar
true oar
Alright
The question I already posted above
This says that the | | doesn't mean the length of the vector
dusky epoch
then what does it mean
true oar
absolute value
dusky epoch
no, what's $|\vec{u}|$ meant to be if not the length of $\vec{u}$?
stoic pythonBOT
Ann:
dusky epoch
what do you mean by "the scalar value of u"
true oar
I meant scalar
dusky epoch
Ok so if the scalar value of u is -5
what do you mean by "the scalar value of u"
what's the "absolute value" of a vector
dusky epoch
that's all fine
dusky epoch
I KNOW WHAT THE ABSOLUTE VALUE OF A SCALAR IS
But what does it mean in case of a vector
that's what loch and i have been trying to find out FROM YOU all this time
dusky epoch
well then nobody here can help you, unless they speak hebrew and can give an adequate translation of the problem
true oar
Alright then sorry to waste your time
storm python
sad
forest trail
"and not quite "interchangeable" since they behave differently" @limber sierra
how do they behave differently? Is it that coordinates depend on coordinate system whereas scalars do not? Do scalars not depend on the coordinate system because they are from fundamentally different things called fields (these fields obviously dont depend on coordinate system). Also can the set of numbers in vector "entries" differ from the scalar field over which the vector space is defined? Like you said "vector space r^n over scalar field q".
Also earlier you said the set of scalars has to be a subset of valid vector entries.
Then you say anything that can be a scalar and can also be a vector component but not the other way around.
Shouldnt components be a subset of scalars in the second case?
(Also i read why we use the word scalar instead of number as scalars can be any set of numbers)
restive flame
im having trouble understanding eigenvectors
steady fiber
what about them
restive flame
how to calculate them
steady fiber
so you have an eigenvalue $\lambda$ for matrix $\mathbf{A}$, and to find the eigenvectors, you find vectors so that $(\mathbf{A}-\lambda \mathbf{I})\mathbf{x} = \mathbf{0}$
stoic pythonBOT
PorosInMyAshe:
wintry steppe
@true oar I think you're mistaken. The |u| with the arrow on top means the length of the vector.
This is fairly standard notation; I have a professor from Israel who uses the same notation as everyone else.
I also don't speak any Hebrew, but it's written left-to-right, and I'm just going to give my best shot at deciphering it:
- Prove that if ||u|| ||v|| = ||u dot v||, then u = tv
- Prove the converse of 1
The rest idk
subtle walrus
actually i thought some more about this
and it might be that | | is just any norm
and not the euclidean norm
so it might just be "not necessarily the euclidean norm / length, but any norm"
true oar
I tried some more and I think that |a| means the length of vector a but if its | |a| x |b| | it means the absolute value of the lengths multiplied
subtle walrus
well, yes
pallid rampart
It's M doing a T pose
meager bolt
can someone help me with number 2?
quartz compass
any time I'm stuck I try to just combine whatever I'm given
they give you A=PDP^-1 and they give you B=2I+A^2
I'd just try plugging that equation into the other to get started to see if that gets you any ideas
meager bolt
B = 2I + (P D^2 P^-1)
So without adding 2I, B should have eigenvalues 1 and 1
But idk if adding 2I changes that
quartz compass
we can be trickier
and write I=PP^-1
and factor it out on either side
$B=P(2I+D^2)P^{-1}$
stoic pythonBOT
Merosity:
quartz compass
2I+D^2 is a diagonal matrix
that means we have technically diagonalized it and we can see what its eigenvalues are very simply @meager bolt
meager bolt
Hm so one itโs eigenvalues should be 3 right
meager bolt
@quartz compass would it just be 2,3?
quartz compass
yep
serene widget
hey all ๐ I think this is the right place to put this question, lemme know if not: if I have an absorbing markov chain with 5 states, is there anything special I need to do to find the foundational matrix once I've put it into standard form and found IOQR? the examples I've seen have been 4 states, so they end up making perfect 2x2 matrixes for all of IOQR and I'm uncertain if I can find F if I is 2x2 and Q is 2x3.
serene widget
upon further exploration, my question boils down to: how do I subtract Q from I if they are of differing sizes?
can I just simplify I to a NxN array matching the dimensions of Q, as long as I place the 100%s in the right spots (down the diagonal)?
maiden silo
$A^TA = QA^TP^TPAQ^T$
stoic pythonBOT
KS42:
maiden silo
does anyone know how i can prove that
given Q and P are orthogonal
i know that P^TP cancels out to I
covert skiff
Hey, how's everyone doing?
maiden silo
@gray dust do u know how to do it
gray dust
sry i'm too sick to do linalg
covert skiff
I'm doing good as well