is this true for every matrix btw or only symmetrical ones?
#linear-algebra
reef prism
dusky epoch
it's true for matrices which have a single largest eigenvalue i think
by largest here i mean having maximal absolute value
dusky epoch
no i mean like if you have ±10 as eigenvalues and every other eigenvalue is strictly between -10 and 10 you may not get the kind of convergence you asked for
repeated eigenvalues don't mess with it afaict
reef prism
create systems of linear equations, no?
wary lily
yeah, I think I need to setup a 2 x 2 matrix for X with each entry a variable
then multiply out and system of equations
lavish jewel
sounds about right
reef prism
@lavish jewel didnt you try to tell me why vectors converge to the span of eigenvectors in a linear transformation
i still dont get it
lmao
lavish jewel
bruh
reef prism
3Head
reef prism
i mean, i do get it
feels like matrix multiplication over and over again seems to collapse every point onto an eigenvector
digital bough
I was given an old book about linalg, a book which is from the 80's and it was given by my father's acquintance and the book was written by the acquintance's lecturer in linalg. Either way, the author during the 90s posted a blogpost about teaching linear algebra where he also mentions Axler. Apparently he challenged Axler to prove a theorem without using determinants.
lavish jewel
feels like matrix multiplication over and over again seems to collapse every poi...
in the limit, that's about right
unless all the eigenvalues are identical, in which case your matrix is a projection matrix which is idempotent
reef prism
true
okay i understand it better now and can visualize
shame i couldnt understand it by reading through the mathematical proof
@digital bough who is axler
digital bough
Sheldon Axler, a mathematician and who is also an author of
Linear Algebra Done Right
lavish jewel
determinants are a mistake
digital bough
I wonder if that change of basis theorem is just a pain in the ass to prove without determinants or if it is just impossible(unknown) and therefore the challenge.
If it is the former then axler is still a chad
lavish jewel
which theorem do you mean?
native rampart
I was given an old book about linalg, a book which is from the 80's and it was g...
This
digital bough
I circled it in red in pic above
lavish jewel
the SVD is unique up to rotations
that should be enough, i think
that change of basis is a similarity transformation or eigenvalue preserving transformation
then again, those properties might require determinants to show
oh well
wintry steppe
Question: do you think having 50 min to answer around 13 questions almost all requiring proofs and rref calculations to be fair
Cause I got a 51/100 on my test, and she said I was technically cheating using a rref calculator so I should have gotten a 20/100. LIke BRUH I didn't even have time to finish the test WITH the rref calculator.
13 questions
50 mins
digital bough
How did she know u used rref calc?
wintry steppe
just compute fast lol
lavish jewel
and furious
wintry steppe
How did she know u used rref calc?
Cause I had a Augmented matrix , and then just the rref next to it with no work shown lol
wintry steppe
>13 questions
>50 mins
<:sully:651816820122189834>
exactly
3 min to do a proof????????????????????
bruh I BETTER GET THEM FAST HANDS
wintry steppe
it sounds like germany
She is from Russia lol
digital bough
Cause I had a Augmented matrix , and then just the rref next to it with no work ...
Weird, we dont have to show meaningless computations here unless explicitly asked for it. We just do them on scratch
wintry steppe
Weird, we dont have to show meaningless computations here unless explicitly aske...
Exactly, Im shocked she wanted to see the whole thing
That would have been like 3 pages long with everything else
Show each step? Are you kidding me
R2 + r1 -> r1 [ whole lotta bullshit ]
Like is this reasonable or is she right
digital bough
It doesnt look that bad to do by hand, i think it takes more time plugging that into a calc than just computing it but ye
I mean if there was no moment where you were supposed to show that you know how to compute then i guess she expected to see it here
And then again
50 min
13 w
wintry steppe
Thats at least two more steps I'd have to show.... at least 2 fully more drawn out matricies
x2 for the 2nd rref
so 4 more matricies to draw
yeah in 50 min
lol\
digital bough
I wouldve failed
wintry steppe
LOL
and it was like every problem
I really hope there's another LA teacher
im so done
digital bough
I think i wouldve failed, i mean i am a really slow thinker and reader i would have passed the computations but no way i would ever be able to prove anything lol
wintry steppe
Same here, maybe I should look into getting the dissability thing so I get extra time on the test lol
disability*
Anyway Ik my answers are probably not perfect and I deserve to get the ones wrong I didn't know but I'm glad yall think the time constraint is ridiculous too. Thanks guys haha.
digital bough
Well we still havent seen the entirety of the exam, it is pretty much just me agreeing.
wintry steppe
Lollll true
Its ok Its just like when I failed Calc 1 I just gotta retake it and get better
solid flower
i get two answers when i do this
depending on how i order the eigenvalues in D and consequently the eigenvectors in P
my $y_1$: $$ -2c_1e^{-x}+c_2e^{5x}$$ other $$y_1 = c_1e^{5x}+2c_2e^{-x}$$
stoic pythonBOT
ahmad_11
wintry steppe
If question as if a set of vectors are linearly independent or span in R^2, but only have 2 pivots and not the third, the answer is yes right ?
strong verge
steel moon
can you row reduce a linear transformation?
wintry steppe
can you row reduce a linear transformation?
if you choose bases of the domain and codomain and express it as a matrix, sure
lavish jewel
you test the definition of a subspace
properties of scalar multiplication and superposition
steel moon
wintry steppe
is that nlab
anyways
literally just check if (i) and (ii) hold
there isn't anything more to it
steel moon
i get a little intimidated by the matrices and the expression in general
lavish jewel
lol
as a hint
you never even have to multiply that out, you could replace the matrix by some letter A and everything will work the same
steel moon
i mean how do i visualize that set ?
lavish jewel
that depends very strongly on how you picture what matrices do
you can see it as the set of vectors that don't change direction when transformed via that matrix
they are only scaled up by 2
that doesn't really let you "visualize" it if you don't know the direction beforehand, though
steel moon
o
is it possible to tell the rank of a linear transformation just by looking at its image?
like rank = dimension(range) right, so if you see its 2d then its rank 2?
grand imp
learning lin alg from nlab? 
lavish jewel
sure thing
steel moon
thank you 🙂
wintry steppe
learning lin alg from nlab? <:monkagiga:651816911964995584>
given a vector space V and a subspace W, a quotient space of V by W is a vector space S together with a surjective linear map p: V -> S through which every linear map out of V, which is constant on the fibers of p, factors uniquely 
what's a coset??????
crystal totem
if i were to prove that a given linear transformation is invertible, how would i go about it?
all i know is that invertibility is true if and only if bijectivity exists, so would i have to show surjectivity (every out has input) and injectivity (every input is unique) to prove invertibility?
or is there another way?
bitter shuttle
Lets say I have 1 vector b.
Is this true?
|2b|^2 = 4(|b|^2)
Can I assume this?
wintry steppe
|2b|^2 = (|2||b|)^2 = (2|b|)^2 = 4|b|^2
of course you can assume this, because it's true
bitter shuttle
also can i assume 2|b|= |2b|
bitter shuttle
yay thanks
nocturne oracle
how do u know |2| isnt -2
bitter shuttle
abs value cant be negative????
lavish jewel
steel moon
is the basis of a translated plane that doesnt cross through the origin doesnt exist since $\vec{0} \not\in V$, where $V \subseteq \mathbb{R^3}$ ?
stoic pythonBOT
add
limber sierra
right; a plane that doesnt pass through the origin cant be a vector space and hence cant have a basis.
steel moon
R^3**
steel moon
does a linear transformation thats rank 0 mean that everything gets sent to the zero vector?
bitter shuttle
also is 2|-b| = -2|b|?
limber sierra
does a linear transformation thats rank 0 mean that everything gets sent to the ...
yes.
these are equivalent conditions, in fact
limber sierra
also is 2|-b| = -2|b|?
no; again, try computing an example
|2||-b| = |-2||b|
but this is not the same thing as -2|b|.
steel moon
does it make sense to row reduce a linear transformation to determine column space
is that the same as row reducing the matrix that induces it?
nocturne jewel
row reducing only makes sense on a matrix that i'm aware
limber sierra
well you can write a linear transformation as a matrix
and then row reduce
and then the pivot columns of the row reduction will correspond to basis vectors from the original matrix
this is a key distinction; the pivot columns of the row reduction will NOT be a basis themselves
they'll just have the same POSITION as the basis vectors
alternatively, you could instead write the linear transformation as a matrix
then compute the transpose
then row reduce the transpose instead
and then the row space of the transpose is the column space of the original matrix
this is generally the approach I'd advise.
steel moon
😮
oh yeah i was also wondering what T(T(x)) = x for all x in R would look like if t is a transformation
stable kindle
just its own inverse
steel moon
like the graph flipped in y = x?
stable kindle
180 degree rotations around any axis, any reflection
nocturne jewel
reflect the triangle along any line and then do it again gives the same thing as doing nothing
steel moon
so the transformation returns itself?
bitter shuttle
say you have 2 vector - u and v
if these 2 vectors are equal in magnitude but not necessarily in direction, can you assume their inner product is equal?
or is there a geometric interpretation of 2 equal length vectors and its inner product
how about if the vectors span a complex space?
steel moon
is rank and nullity the same thing as a type of span for linear trnasformation? rank is like dimension of the range and nullity is dimension of the null space if i recall
nocturne jewel
$||u||=||v||\implies \sqrt{\langle u,u\rangle}=\sqrt{\langle v,v\rangle}$
stoic pythonBOT
moshill1
bitter shuttle
does this work if the vectors span R3?
nocturne jewel
yes
steel moon
T: R^3 -> R^3 as a rank 2 transformation implies that the range is a plane right
wintry steppe
ignore, misinterpreted
wintry steppe
T: R^3 -> R^3 as a rank 2 transformation implies that the range is a plane right
yes
rank 2 means the image is a 2 dimensional subspace
ie
plane
bitter shuttle
**moshill1**
is there is derivation for that assumption?
wintry steppe
the definition of the norm of u is sqrt <u, u> 
nocturne jewel
is there is derivation for that assumption?
Like Terra said, I derive it by reading the definition
If only proofs were that easy
wintry steppe
proof by definition
nocturne jewel
please 🥺
wintry steppe
if you want to prove something just redefine whatever you're working with to satisfy that
steel moon
If we have something like $\vec{x} = t\vec{v}, s\vec{u} + \vec{k}$ can you still say every vector is a linear combination of v, u, k? i feel like its not because the coefficient of k can never change
stoic pythonBOT
add
nocturne jewel
every vector x is a linear combination of v,u,k.
steel moon
even if the coefficient of k is 1?
nocturne jewel
It's still a linear combination of the 3 vectors
like how u+v+k is a linear combination
wintry steppe
if you want to prove something just redefine whatever you're working with to sat...
one of my analysis hw questions was to prove that the definition of measurable the book used was equivalent to another common one
very tempted to just write "follows trivially from the definition in this other book"
which vectors span R^3?
bitter shuttle
they are variable vectors
so its not explicitly defined
nothing is also said about if they are basis vectors
wintry steppe
i wasn't aware that R^3 could be spanned by two vectors
wintry steppe
good lol
bitter shuttle
@wintry steppe can you assume |4u - 8v|^2 = |(4u - 8v)^2|
wintry steppe
what does it mean to multiply vectors
bitter shuttle
i think it doesnt work
stable kindle
(a1, a2, ...) * (b1, b2...) = (a1b1, a2b2, ...) 
wintry steppe
do you know the properties of norms?
bitter shuttle
normalized vectors?
wintry steppe
norms
bitter shuttle
nope
wintry steppe
you've asked similar questions a few times now so i just want to make sure you're aware of the definition of a norm
well
you should try to justify this yourself
using the/a definition of a norm
bitter shuttle
i dont have any idea what a norm is @wintry steppe
bitter shuttle
oh
limber sierra
wait
hold on
when you write |a|
are you referring to the absolute value?
@bitter shuttle
i assumed you meant vector norm since this is #linear-algebra
my prior statements still apply but like, now the arguments are even easier lmao
limber sierra
then what does |(4u - 8v)^2| even mean?
bitter shuttle
I think its just taking its inner product
wintry steppe
what does it mean to multiply vectors
.
gritty swift
it should be if and only if in (1) <=> (2) i think @wintry steppe
not sure how to prove it tho it makes sense intuitively if you think of R^n, i guess this is abstract tho
wintry steppe
isn't 2 implies 1 trivial?
like
if you have an inverse
you're one-to-one
lmao
tame mural
#1 and #2 are often considered equivalent definitions
Like if someone asked me what "one to one" meant for linear maps, I would've given definition 2
gritty swift
oh it even says "these are equivalent statements" at the top right
its pretty intuitive if you understand what the terms mean, and at least in R^n you can visualize
like, if the nullspace has more then {0} it can't be invertible since multiple vectors map to 0 we don't know where to map back to
if rank < n that means the same thing, vectors in the nullspace => not invertible
i got tripped up forgetting we mean invertible in the strict sense, ie. an inverse must exist for every vector in the range
@wintry steppe
orchid harbor
I have a question my teacher said there is one exception when u take a span there is a not infinite amount when is that?
i knew it lol
thnaks
tame mural
If you take the span of some vectors from a finite vector space.
That's another exception
orchid harbor
Lin Alg is so bad
I have trouble in this definitions
i understand it imagine all in head yes but when it comes to solving problems
i am bad
wary lily
I’m bad, and that’s good. I will never be good, and that’s not bad
wary lily
I was reading this and thought would it make sense to use this to find solutions of a system, considering the amount of computation one needs to do compared to simple row reduction techniques?
dusky epoch
lmao
wary lily
If I'm not mistaken, row reduction algorithms increase in computation linearly with the number of equations, but determinants not much so
dusky epoch
computation-wise, gaussian elimination takes O(n^3) time while this is O( (n+1)! ) i think
determinants literally take factorial time to compute
unless youre computing them by row reduction
wary lily
are there other benefits to Cramer's Rule?
dusky epoch
cant think of any
raw sand
gaussian is O(n^3)
raw sand
expansion by cofactors moment
dire thunder
best determinant algo is O(n^3.2) iirc
wary lily
row reduction to echelon form is less than O(n^3) tho, right?
dire thunder
rref is O(n^3)
raw sand
LU Decomp is O(n^3) and that is poggers champion
wary lily
O(n^3) is reduced echelon form, no?
lavish jewel
i think the fastest, when possible, should be with FFTs... sparsity aside
dire thunder
just use cofactor expansion
and that integers are closed under multiplication and addition
dusky epoch
||the determinant is a sum of products of the entries of your matrix, and Z is closed under both||
toxic escarp
I'm not really sure on how to approach Q4. I started off by trying to find the equation of the plane made by the points where the legs touch the floor, but it really hairy very quickly and I only just got started. Any suggestions?
tulip glacier
just to ask but since null spaces are the unique why isnt the ans neither?
lavish jewel
what does this have to do with null spaces?
basis 2 is presumably equal to basis 1, but the third basis vector was multiplied by -1
while basis 3 is presumably a permutation of basis 1
zealous junco
in general you can represent a non-zero vector in terms of any coordinate vector, you just need the right ordered basis
wary lily
I thought I can row reduce a row to zero, since they are multiples of each other
having a zero row gives me a zero determinant
fleet sun
Being colinear only implies they're multiples of each other if their line passes through the origin
wary lily
with multiples I'm referring to (x, y) tuples only, not the third column
do they necessarily need to pass through the origin to for collinearity to imply being multiples of each other?
fleet sun
yeah, like (1,2) (2,3) and (3,4) are collinear but they don't scale to each other
for their line misses (0,0)
leaden epoch
I read a book Hall Lie Groups, Lie Algebras
I have problems with 1.2.8 The Compact Symplectic Group
The author introduce a conjugate-linear map J: C^2n => C^2n by
J(α,β)=(-\bar β, \bar α)
α, β are in C^n
Then he says it's easy to check for all z,w є C^2n
ω(z,w)=<Jz,w>
ω - bilinear form, ω(x,y)=Σx_j•y_j
<,> - standard inner product on C^n
How he shows ω(z,w)=<Jz,w> ?
anyone knows good boo about conjugate-linear map
fleet sun
I thought I can row reduce a row to zero, since they are multiples of each other
Try writing the column (y1, y2, y3) as (mx1 + b, mx2 + b, mx3 + b)
Then C2 - mC1 = (b, b, b)
or 1/b (C2) - m/b (C1) = C3
So it's not full rank
native rampart
I read a book Hall Lie Groups, Lie Algebras
I have problems with 1.2.8 The Compa...
I don't think that's correct
leaden epoch
i think i posted wrong ω(x,y)=Σx_j•y_j
ω(z,w) is bilinear form as it's defined Symplectic Groups
z_jw_n+j-z_n+jw_j
native rampart
Take n=1 for example
Then,
$J(a,b)=(-\bar{b}, \bar{a})\
w((a,b),(c,d))=ac+bd\
\langle J(a,b) ,(c,d) \rangle= \langle (-\bar{b},\bar {a}),(c,d) \rangle=ad-bc$
Assuming sesquilinear in the first argument
stoic pythonBOT
DrunkenDrake
tulip glacier
basis 2 is presumably equal to basis 1, but the third basis vector was multiplie...
Ooh I always tot that was the null space, thks @lavish jewel
lavish jewel
there was no null space to speak about
no linear transformation was being spoken about
leaden epoch
@native rampart I'm not sure you wrote bilinear form right, for C^1 ω(z,w)=z1w2-z2w1
it's skew-symmetric bilinear form
I don't know how J acts on z, Jz, z is C^2
native rampart
Are you trying to prove the result for Sp(n,c)?
B[x,y]=(x,Jy) is not true over C^2n
leaden epoch
for Sp(n):=Sp(n,C) intersect U(2n)
Compact Symplectic Group
Sp(n) is the group of 2n 2n matrices that preserve both the inner
product and the bilinear form !.
wintry steppe
Help
dire thunder
do not multipost
wintry steppe
Sorry
dusky epoch
(and also see pins bc this isnt linear algebra)
wintry steppe
No english bro :))
dusky epoch
don't call me bro.
stone bear
How do I find minimum and maximum of such cases ?
Are there some methods or should I use hit and trial ?
wary lily
$$a + b + c = 44$$
$$a + b - c < 16$$
stoic pythonBOT
az
wary lily
I don't know of a method but you can be methodical about it.
dusky epoch
do you want the solutions of this over positive integers?
wary lily
You have 2 equations and 3 unknowns.
I rewrote the previous question
since it wasn't readable
wary lily
agree
dusky epoch
i mean ok we can let d := a+b for simplicity
then we have d+c = 44, d-c < 16
so d - (44-d) < 16
which gives d < 30 i believe
so a+b < 30
so i guess b can range between 1 and 29
wary lily
they need to be positive integers all
quasi vale
@dusky epoch shouldn't the range of b b/w 1 and 28? a is a positive integer
dusky epoch
i was going to play that off as a typo and have plausible deniability about it
but good catch
quasi vale
ty 😄
wary lily
damn, it's less than 16, I was mentally taking it as less than and equal to 16
I have never seen column operations before
is this allowed only in the context of finding triangles for dets or...?
looking at the matrix as a system of equations, this doesn't make sense like row operations
wintry steppe
In this situation, we could row reduce the last row to have three pivots, which would make colA have 3 dimensions, right?
wintry steppe
Ah alright, we were taught to look for the amount of pivots, so explaining the dimensions based on the linear independence* of the columns didn't come up in my head.
wintry steppe
Thank you @dusky epoch
wary lily
why would I need to use combinations of row operations and cofactor expansion after reducing the matrix to row echelon form?
In row echelon form, isn't the matrix upper triangular and the det just the product of the main diagonal?
blissful orchid
how do i rotate the matrix [-1 0 ] and [0 1] counter clockwise by pi/2 ?
wary lily
left multiply by $\begin{bmatrix}
-1 & 0 \
0 & -1
\end{bmatrix}
$ I guess. But I'm a beginner.
stoic pythonBOT
az
blissful orchid
**az**
hmm i thought you do -cos(pi/2) and sin(pi/2) and then for the second matrix cos(pi/2) and then sin(pi/2)?
wintry steppe
are you trying to rotate the two points (-1, 0) and (0, 1)
or actually rotating a matrix 
wary lily
now that I gave that dumb answer, I'm starting to understand how this work
blissful orchid
ooops sorry i meant vector
rotate the vectors pi/2 radians counter clockwise
to find the standard matrix
wintry steppe
left multiply by
$$
\begin{pmatrix}
\cos\theta & -\sin\theta \
\sin\theta & \cos\theta
\end{pmatrix}
$$
and put $\theta=\pi/2$
wary lily
standard matrix for this rotation, IG
stoic pythonBOT
(T*Terra, dqⁱ ∧ dpᵢ)
blissful orchid
ahhh so i can use that formula despite previous transformations done to the vectors?
wintry steppe
huh
blissful orchid
I had to reflect the vectors about the y axis and then rotate it by pi/2
and the vectors were [1 0] and [0 1]
wintry steppe
im pretty sure with vectors that simple you dont need to actually go through the hassle of setting up a rotation matrix
just
draw the transformations lol
blissful orchid
im more of a plug and chug type of guy
wintry steppe
wary lily
wintry steppe
When he (the lecturer) says "non-zero column vectors which are mutually orthogonal" he means an orthogonal set, which are also the columns of R(A). Right?
The way to solve this would be to just apply the orthogonal decomposition theorem, and the answer would be y-hat.
Am I interpreting this correctly?
amber sierra
Can anyone tell me what I'm doing wrong / if my final matrix in row echelon form is wrong?
The starting matrix is:
1 2 1 0
-1 -1 1 0
3 4 a 0
Im trying to find what value(s) of 'a' will the system have nontrivial solutions. My final matrix in row echelon form is:
1 2 1 0
0 1 2 0
0 0 a+1 0
Which would mean a = -1
However, my textbook says the answer is a = 1.
also i covered someone elses question so theres also someone above me who asked a question ^^^
wintry steppe
Can anyone tell me what I'm doing wrong / if my final matrix in row echelon form...
1 2 1 0
-1 -1 1 0
3 4 a 0
R2 + R1
1 2 1 0
0 1 2 0
3 4 a 0
R3 - (3 * R1)
1 2 1 0
0 1 2 0
0 -2 a-3 0
R3 + (2 * R2)
1 2 1 0
0 1 2 0
0 0 a+1 0
I get the same answer
amber sierra
oh what the heck
eh okay ill just ask my professor if the textbook is wrong then thank you!
wintry steppe
Let's both wait for someone more clever to chime in : ~)
amber sierra
yea 🤣
nocturne jewel
what's d in the transformation?
severe remnant
Let's just say 8
severe remnant
Yh 8
nocturne jewel
so write 8 if it's 8
severe remnant
Does that matter for the proof
nocturne jewel
well there's a symbol that isn't defined
severe remnant
Ah ok
nocturne jewel
for all I know d is a function itself
and showing linearity means showing
0 maps to 0
any linear combination of inputs gives a linear combination of outputs
severe remnant
I'm kind of struggling with what polynomial to pick
nocturne jewel
T[0]=0
T[cu+dv]=cT[u]+dT[v] for u,v in V and c,d in K
or you can split the combination test into scaling and addition
severe remnant
Yeah I get that but for r^2 we're mapping from a,b to P1
nocturne jewel
yeah
nocturne jewel
what's the 0 vector of R^2
severe remnant
0
severe remnant
0,0?
nocturne jewel
yes, what happens to (0,0) in the transformation?
severe remnant
That I don't understand tbh
nocturne jewel
in the point (0,0), what does a=?
severe remnant
0
nocturne jewel
and what does b=?
severe remnant
Also 0
nocturne jewel
so plug a=b=0 into the polynomial, what do you get?
severe remnant
0 and 0*x
nocturne jewel
which collapses to just 0, the 0 polynomial to be specific
severe remnant
Oh yh
nocturne jewel
so T[(0,0)] --> 0(x), so 0 vector of R^2 maps to 0 vector of P1
severe remnant
Lmao I don't know what I was thinking there for a sec
pseudo prawn
I have no idea what y’all are talking about but it seems fun lol
nocturne jewel
so you can either test scaling/addition seperately or do 1 linear combination of R^2 vectors, whichever you've learned/more comfortable w/
if u,v are in R^2 and c,d are in R, want to see what happens to T[u+v] and T[cu] (OR) what happens to T[cu+dv]
nocturne jewel
First you need to see if the map is bijective
severe remnant
Would it be the same still in this case because it's 0
Oh
So see if it's one to one and onto first
nocturne jewel
Yes, see if Ker(T)={0} and Im(T)=R^2
severe remnant
And if it is then what's the next step
nocturne jewel
Tbh I'm not 100% sure on if there's more than one way, but I was taught to find the matrix representation then use Gauss-Jordan to find the inverse of the matrix
wintry steppe
Yes, see if Ker(T)={0} and Im(T)=R^2
(you only have to show the first one of these)
second one, or rather, their equivalence, follows from the rank-nullity formula
nocturne jewel
Isnt that just for operators or is it for all transformations?
wintry steppe
as long as its a linear trasnformation between two spaces with the same dimension
it's fine here since R^2 and P_1 have the same dimension
severe remnant
I'm struggling to find the inverse map for this
I know how to do it from p1 -> r2 but not the other way
severe remnant
@nocturne jewel
nocturne jewel
what?
severe remnant
Do you know how to do this
I tried making a system of equations and setting them to equal 0 but I got stuck at this part so idk what to do
nocturne jewel
Ker(T)={f in P2|T[f]=0}
so yeah, set each entry of the matrix equal to 0 then solve the system
severe remnant
Yeah that's what I did but I'm not sure how I'm supposed to solve it in terms of one variable
I tried to make b the free variable
nocturne jewel
what do you mean? there are 3 variables
severe remnant
A+b =0, b-dc = 0, and a+dc=0
nocturne jewel
solving the system should yield:
a+b=0
b-8c=0
c=t means we can write the vector as [-8,8,1]^T which corresponds to f=-8+8x+x^2
and you can replace 8 with d, but you defined d=8 so idk why you wrote d's
severe remnant
Why is c = t
nocturne jewel
it's the free variable
severe remnant
Why can't it be b
nocturne jewel
it can be, you just get a fraction entry 
severe remnant
O
nocturne jewel
(divide what i wrote by 8 and you get that)
nocturne jewel
I mean that will give dim(Im(T)), which is just the number of basis vectors
severe remnant
So how do we find the basis for the image
nocturne jewel
What's the canonical basis for M2x2
severe remnant
1,0 0,1
nocturne jewel
no clue what that is
nocturne jewel
2x2 matrices w/ 1 in 1 entry and 0 else
severe remnant
I thought m2x2 and r2 were the same spaces though
nocturne jewel
No..
severe remnant
So [1,0,0,1] ?
severe remnant
Ok so we got 4
nocturne jewel
yes, but one of them clearly cant be in the basis we want
since the (1,2)-entry of the output is always 0
severe remnant
Oh yeah
nocturne jewel
so the basis for the image is the "reduced" canonical basis
severe remnant
So just the other 3 ok, but is there a way I can show how I got to that conclusion
nocturne jewel
$Im(T)=span{E_{1,1},E_{2,1},E_{2,2}}$
stoic pythonBOT
moshill1
nocturne jewel
I mean.. just go through the logic I said
Start with the canonical basis, then argue why you can remove E_12
severe remnant
Alright and is there a quick way to look and tell if it will be an isomorphism or do u have to go through all the calculations to show it
nocturne jewel
$Im(T)\subset M^{2\times 2}=span{E \text{whatever}}$
stoic pythonBOT
moshill1
severe remnant
Do we know if it is 1-1
nocturne jewel
Alright and is there a quick way to look and tell if it will be an isomorphism o...
I personally cant answer / give a meaningful answer
severe remnant
Sorry not 1-1
I mean onto
Because from the kernel we proved it's 1-1
No worries, thanks for the help though!
gray dust
@haughty dust this is calc material, not linear algebra, and don't multipost
solid flower
solid flower
looks like it is to me
yea but
why
i thought the non-zero first entry had to be 1
i.e pivot
quartz compass
no, that's reduced row echelon form
solid flower
oh
quartz compass
yeah, just a slight difference between REF and RREF
quartz compass
probably they're just being lazy and saying they're the same for your class
quartz compass
no
solid flower
so then why
quartz compass
I already said twice
dusky epoch
why what
stable kindle
why what who
solid flower
why its in row echelon form
stable kindle
dusky epoch
do you agree that $\bmqty{2 & 1 \ 0 & 0}$ is in row-echelon form?
stoic pythonBOT
Ann
dusky epoch
Y/N
dusky epoch
solid flower
wait one second
dusky epoch
i was going to say that -1+i is a nonzero number just like 2
sonic osprey
what are these things
zinc tapir
p and r are vectors where p is the optimal search direction and r is the gradient descent direction
west orchid
solid flower
$Im(ix-y^2) = x$ right?
stoic pythonBOT
ahmad_11
dusky epoch
are x and y real?
dusky epoch
where did ix - y^2 come from then 
dusky epoch
how did $-i^2y$ become $-y^2$
stoic pythonBOT
Ann
solid flower
oh
it would be just y^2 right
hmm
so to simplify its actually $Re(ix-i^2y) = Re(ix+y)$?
stoic pythonBOT
ahmad_11
solid flower
the answer key says its -y tho :(( ,for a)
dusky epoch
wyhy ar eyouia\h erok sdfgjskagjhga
lavish jewel
you started with x + iy?
dusky epoch
why are you so hellbent on squaring y
solid flower
sorry thats a typo
solid flower
you started with x + iy?
yea as z
lavish jewel
so, what is i(x+iy)?
lavish jewel
so far so good
solid flower
ok yea
lavish jewel
simplify it a little
lavish jewel
so what's the real part
solid flower
-y
lavish jewel
congratulations
solid flower
how is the im part -x tho
lavish jewel
it isn't?
lavish jewel
,w imaginary part of i(x+iy)
stoic pythonBOT
lavish jewel
wow that's cursed
dusky epoch
QGKJFGDFKHSDGDSFG
lavish jewel
to be fair that's more Wirtinger calculus
dusky epoch
Re'(x)...
lavish jewel
lol
also, depends on what you wanna do really, but for distances and angles and the like, the dual space would be the complex conjugates, it's a sesquilinear form
sure
you could do it as (x + iy + x - iy)/2
so also in the complex conjugates
orchid harbor
Guys quick question
When i have this kind of thing tbh never seen like this
If there are all zeros at the first row and column
Where should i start then?
Should i just start with -3 making it 1?
novel hamlet
how do i find basis for perpendicular complement for this vector space?
I know v1 and v2 are basis of V
lavish jewel
what is v3?
novel hamlet
linear combination of v1 and v2
if you row reduce them when finding basis you get
lavish jewel
all righty. then you need a vector perpendicular to both
novel hamlet
yeah, i forgot how to calculate this
lavish jewel
by inspection should be easy enough, i think
novel hamlet
how is it done?
lavish jewel
you could put a vector x, y, z as the third column in a matrix, while v1 and v2 are the first 2 columns
you rref it and the result should be an identity
so that will give you 3 equations with 3 variables
or if you can just "see it", then the procedure is going "oh, i see it!" and writing the solution down 😛
novel hamlet
guess i try that 3 equations approach
lavish jewel
1, -5, -4, for example, should work
i think, anyway
another approach is to write the conjugate transpose of v1 and v2 as rows of a 2 x 3 matrix and multiply by a vector x,y,z
the result should be 0. this one will have infinitely many solutions in terms of some parameter
novel hamlet
lavish jewel
you would have to try to rref it by hand
i guess that gets messy
the other approach i wrote should be kinder on your soul
novel hamlet
so basically (v1,v2)*(x,y,z)=(0,0,0)
lavish jewel
yeah
fervent gulch
could someone guide me with this.
wintry steppe
look up "orthogonal projection formula"
fervent gulch
I did and some things are complicated.
Let L = Span { u } be a line in R n and let x be a vector in R n . By the theorem, to find x L we must solve the matrix equation u T uc = u T x , where we regard u as an n × 1 matrix (the column space of this matrix is exactly L ! ). - > I don't understand like the u T uc= u T x thing
wary lily
$proj_v\mb{u} = \left( \frac{\mb{u} \cdot \mb{v}}{\abs{\mb{v}}^2} \right)\mb{v}$
stoic pythonBOT
az
wary lily
wouldn't that work?
dire thunder
hi az
wary lily
dire thunder
wotsup
wary lily
how you doin?
dire thunder
could be much better ig
wary lily
going through LA basics, I am
wary lily
no, I'm sorry, this is the vector projection of u onto v
not the orthogonal one
but that's the orthogonal one
the projection is orthogonal
dire thunder
Let L = Span { u } be a line in R n and let x be a vector in R n . By the theore...
please use proper notation because the stuff with equation is not really clear and clarify theorem you are referring to
wary lily
Vimes, my formula is orthogonal projection, right?
fervent gulch
please use proper notation because the stuff with equation is not really clear a...
this isn't mine, i looked that up and paste it here.
wintry steppe
i found the thing you copypasted
wary lily
haha
wintry steppe
you know the formula is right below that paragraph right?
fervent gulch
yes
wintry steppe
could someone guide me with this.
anyways, you can now do this computation by straightforward computation
the explanation here for why the formula works is a little too complicated imo
fervent gulch
hold on what was the straightforward computation called
fervent gulch
okay imma try it out.
wary lily
definition of c is confusing me
what is <b, a>/<b, b>?
aren't those vectors themselves? a and b?
wintry steppe
a, b are vectors. c is a scalar
we want to find a vector v parallel to b with the property that a - v is orthogonal to b
dire thunder
tterra come to russia
wintry steppe
vectors parallel to b = scalar multiples of b
dire thunder
fervent gulch
I solved it but I hope I got the solution right, I got 1 1 -3
wintry steppe
what is <b, a>/<b, b>?
inner product of a and b, divided by the inner product of b and b (i.e., norm squared of b)
wary lily
O
dire thunder
I solved it but I hope I got the solution right, I got 1 1 -3
so proj_v(u)=v you mean?
fervent gulch
yea
wary lily
thanks
wintry steppe
knowing how to do orthogonal projection without trigonometry is important
repeat it and you get gram schmidt
wary lily
yeah, I'm still not there
dire thunder
or just define gram-schmidt because you can
wintry steppe
also cosines and arccossines are annoying
fervent gulch
what is gram schmift
dire thunder
what is gram schmift
procedure for generating orthonormal list of vectors with the same span as given one
wintry steppe
what is gram schmift
a process that takes in linearly independent vectors and spits out orthogonal vectors with the same span
the idea is to keep doing orthogonal projection
wintry steppe
all u need
for example, if you performed gram-schmidt on {b, a} in this picture, then you'd get {b, a - c*b}
which are orthogonal
and with the same span as {b, a}
(the order matters)
fervent gulch
gotcha
wary lily
having orthogonal bases, makes computations easier, r?
wintry steppe
orthogonal bases make life easier
wary lily
yeah, I can imagine that
wintry steppe
if your basis is orthogonal then it's very easy to express vectors in terms of the basis
wary lily
yeah, finding suitable scalars made easy
fervent gulch
so did anyone else get 1 1 -3?
wintry steppe
if $(u_1,\dots,u_n)$ is an orthonormal basis of $(V,\langle\cdot,\cdot\rangle)$ then for any $v \in V$, $$v = \sum_i \frac{\langle v, u_i\rangle}{\langle u_i,u_i\rangle}u_i$$ (i.e., any vector is equal to the sum of its orthogonal projections along orthogonal basis vectors)
stoic pythonBOT
(T*Terra, dqⁱ ∧ dpᵢ)
wintry steppe
if you normalize the u_i's to an orthonormal basis it's even simpler
fervent gulch
I entered to see if i was right, i was wrong. the solution was 1/11 1/11 and -3/11.
So I got the numerators but had no clue how to get the denominators.
fervent gulch
okay, but i'll do another question instead
dire thunder
i mean
you could just do
u = kv+w
where <v,u> = 0
this would give you four linear eqs of four unknowns
and finding k you get kv as your projection
this if you are not ok with formula
dire thunder
are not you familiar with that if x,y are nonzero vectors
then <x,y> = ||x||||y|| cos (theta)
wherer theta is angle between vectors
fervent gulch
perhaps, i just forget things easily
dire thunder
i mean this basically shows us how we get information about angle between vectors from their inner product
you are right that cos theta is zero
since dot product is zero
but why theta is zero then?
fervent gulch
because theta is negative
fervent gulch
i was but now no
dire thunder
you are saying that theta is zero because theta is negative
which is quite meaningless
fervent gulch
okay so what were you trying to demonstrate when theta is 0
dire thunder
ok so another hint: what do you say about vectors if their dot product is zero?
fervent gulch
the vectors are 0
fervent gulch
so 2 nonzero vectors
dire thunder
yes
there is one very important definition
about vectors having zero dot product
fervent gulch
okay
digital bough
then cb is proj_b a
My book did it like in az’s example but during the lecture the lecturer did it your way.
Is there any reason at all why the book would take the hard way instead of the easy way?
Oh sorry for ping, forgot to disable
I assume it is perhaps more clear that it works for angles > pi/2 aswell when presenting it the hard way
umbral trout
Given 3 Complex numbers, z1,z2,z3
it's also given that z1*z2=z3 and that 0<d<-c and -f<c<0,
which of the following answers is true? prove it, if it's wrong give a counter example
can someone help me with this question? been trying to solve it for the past couple of hours and I can't seem to get to an answer
stable kindle
sorry was cleaning my keyboard
hazy sparrow
has anyone watched 3b1b’s series on linear algebra?
sonic osprey
Yes
nocturne oracle
no
plain ibex
Yes
tiny palm
hi sorry, both my L and U are correct but for some reason D was marked incorrect. I've checked my work over but i can't seem to find what's wrong. would my answer actually be correct? perhaps it was a computer grading error
potent wraith
@lost ermine ask here. Or perhaps #groups-rings-fields
lost ermine
oo thanks
potent wraith
Lets first help @tiny palm
lost ermine
oke yea
potent wraith
@tiny palm If you mark your matrices as
1 0 0 d1 0 0 1 u12 u13
l21 1 0 0 d2 0 0 1 u23
l31 l32 1 0 0 d3 0 0 1
Where l, d and u are unknowns
You can just multiply these to get
d1 0 0 1 u12 u13
l21*d1 d2 0 0 1 u23
l31*d1 l32*d2 d3 0 0 1
Mutiply these also to get:
d1 d1*u12 d1*u13
l21*d1 l21*d1*u12 + d2 l21*d1*u13 + d2*u23
l31*d1 l31*d1*u12 + l32*d2 l31*d1*u13 + l32*d2*u23 + d3
Now you just need to match the cells with A
d1=-5 d1*u12=-5 => u12 = 1 d1*u13=10 => u13=-2
l21*d1 = -25 => l21=5 l21*d1*u12 + d2 = -20 => 5*-5*1 + d2 = -20 => d2 = 5, l21*d1*u13 + d2*u23 = 70 => 5*-5*-2 + 5*u23=70 => u23=4
l31*d1 = 5 => l31=-1, l31*d1*u12 + l32*d2=25 => -1*-5*1 + l32*-20=25 => l32 = -1, l31*d1*u13 + l32*d2*u23 + d3 = 72 => -1*-5*-2 + -1*5*4 + d3 = 72 => -10 -20 + d3 = 72 => d3=102
@lost ermine OK post it here now, hopefully someone more experienced will see. And also #groups-rings-fields might be even better.
lost ermine
yea i am not allowed on that channel
potent wraith
(check #get-advanced-access )
lost ermine
I need help with this problem, i have a solution written down but i do not think its rigorous enough or even right, i need help with that. I will upload follow up pictures to clarify the definnitions
This defines the phi-subscript star function
and this defines the pi-subscript set function
wintry steppe
forget the linear structure of E and form C(E)
linear map on C(E)
can you post the full question (i.e., the question with every part included)? it's still not very clear to me what C(E) or C(X) or C(Y) are supposed to be
sonic osprey
so what's your solution?
lost ermine
lol
sonic osprey
Uh what? You said you had a solution right?
lost ermine
yea one sec
so first i defined PIe as a linear function induced by a set map, that maps the basis-functions into the element of the space E. Same as PIf
the element is gotten from the basis function
and then i showed that both sides of the equation equal the same express at the end when get their value at a linear combination of the basis
this is to show that a linear equation satifies it
but i am not sure how to show that if the function satisfies the equation then it is linear
sonic osprey
I'm not really sure what you're saying, can you write it out in more detail?
solid flower
how would i know how many roots there are of a complex number such as $(1+\sqrt3i)^{\frac{1}{2}}$?
stoic pythonBOT
ahmad_11
wintry steppe
When he (the lecturer) says "non-zero column vectors which are mutually orthogonal" I think he means an orthogonal set, which are also the columns of R(A).
The way to solve this would be to just apply the orthogonal decomposition theorem, and the answer would be y-hat.
Am I interpreting this correctly?
hazy fable
whats R(A) here
wintry steppe
With R(A) he means the Range of A, which he uses interchangeably with the column space
wintry steppe
No example question of this type is given
My lecturer writes in a very confusing manner, I just want to know if I am interpreting it correctly
This is by far the most stressful class I've had to attend, due to how poorly structured and phrased a lot of the lecturers' writings are.
lost ermine
Need help with this, anyone?
hazy fable
This is by far the most stressful class I've had to attend, due to how poorly st...
it seems like you understand it correctly, but its hard to say given what information i have
lost ermine
wdym alphas
hazy fable
well your P is defined by the alphas
lost ermine
i been trying to show P(f)(a+b) = P(f)(a) + P(f)(b)
u mean choose an n and see how to works?
hazy fable
thats not necessarily true
it depends on how P is defined
which is what you need to prove
hazy fable
i dont have the answers at hand but seems doable
you can certainly find some condition such that P is linear
fervent gulch
question: Find x so that the triangle with vertices A(−8, −9, −6), B(x, −17, −4), and C(−4, −19, 3) has a right angle at A.
hazy fable
this condition is fulfilled if the vector AB and AC are orthogonal
fervent gulch
ok so what steps do i take
lost ermine
do a dot product and solve for x
but first
figure out
the vectors]
like figure out AB and AC
lost ermine
no
AB is A-B
its the dot product between AB and AC
this has to equal 0
i think AB = A-B or B-A
same with Ac
AC
fervent gulch
could u demonstrate through visuals and like show me how its done. i'll do the calculation but formatting can be hard
fervent gulch
yea i just mess up quite often
lost ermine
so to get the vector that is AB u take A-B
which is one side of the triangle
then u at A-C which is AC the other side of the trianlg
then u take their dot product and set equal to 0
and solve for x
stoic pythonBOT
ahmad_11
bitter shuttle
that is equivalent to $(e^{i\frac{pi}{3}})^{\frac{1}{2}}$
stoic pythonBOT
τεnsors
bitter shuttle
its easier to see why its pi/6 if you use the euler's exponential form
solid flower
ooh
what i was doing was
i simplified it down to $\sqrt(1+\sqrt3i)$ and then took a and b as that
stoic pythonBOT
ahmad_11
bitter shuttle
yeah i not sure how you would go about square rooting it
the first method that instantly poped was euler's formula
solid flower
**τεnsors**
so this is the best way to find theta and for r ig the best way is just to plug in a and b into the pythagoras' formula
bitter shuttle
pythagorean theorem doesn give you theta...
phyathorean theorem describes sidelengths not the angle measures
solid flower
i meant for r
bitter shuttle
wait its not 1
sorry
its like 2 or something
but getting the angle measure doesnt require its length
just take tan(b/a) and use it eulers formula
solid flower
yup
thanks a lot
also would you happen to know how one would predit how many roots there would be to each of these?
bitter shuttle
all of these can use eulers formula......
solid flower
or i think you have to think along the lines of $z=(re^{i\theta}^{n})$ right
stoic pythonBOT
ahmad_11
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
solid flower
all of these can use eulers formula......
yea
bitter shuttle
question is the projection of a unit vector u on any vector v just their inner product?
vectors are in R2
wraith coral
Any three non-zero vectors in R3 form a basis of R3? is this true?
bitter shuttle
Any three non-zero vectors in R3 form a basis of R3? is this true?
i, j, k
any transformation of these 3 vectors that preserves their inner product are basis vectors
bitter shuttle
2i, 2j, 2k
bitter shuttle
didnt you want the basis vectors?
wraith coral
yes
wintry steppe
Any three non-zero vectors in R3 form a basis of R3? is this true?
no
take (1, 0, 0), (2, 0, 0), and (3, 0, 0)
potent pollen
Hello! I've been struggling on the following exercise for a little while now : "Let $u$ be linear operator from a finite dimensional inner-product space $E$ to itself, show that $u$ is non-expansive (that is, $| u(x) | \leq |x|$ for all $x\in E$) if and only if the eigenvalues of $u^* \circ u$ are in the interval $[0,1]$". I've shown the "easy" implication (the "only if" part) but I'm having difficulties with the other one. I've tried two different techniques. The first one is that since the eigenvalues of $u^$ are that of $u$ then maybe (I hope) we can show that the eigenvalues of $u$ are in $[0,1]$. If that's correct then I think I can conclude using the Schur decomposition, but it wouldn't be very satisfactory since we haven't seen it in class and so I'd have to provide a proof of that as well. My second technique has been less fruitful and consists in diagonalizing $u^ \circ u$ (we can because it is necessarily self-adjoint) with an orthogonal matrix. It quite simple to show that if $U$ is the matrix of $u$ in the standard basis then there exists an orthogonal matrix $P$ such that for all $x\in E$, $$| P^t, U^t, U, P, x | \leq | x |$$
but I've yet to find a link between $| P^t, U^t, U, P, x |$ and $| U x |$ (even though I know orthogonal matrices do not change the norm, that is $| P Y | = |Y|$)... Is any of these techniques doomed to fail? Am I even going in the right direction?
stoic pythonBOT
Canatime
potent pollen
(This is my first time asking a question here ^^, tell me If that's not where I should have asked this)
sonic osprey
its alright here but this is usually higher level than most of the questions that end up here
you might have more luck getting an answer in #advanced-analysis since this is functional analysis ish
potent pollen
Cool, should I just copy-paste my question there?
sonic osprey
yeah thats fine
potent pollen
Ok, thanks!
solid flower
1,1,0 is associated to eigenvalue 6
mortal juniper
solid flower
im not sure what this part means
mortal juniper
yea i thought it was a bit redundant
dusky epoch
it just lets you save time calculating things
the problem asks which of these vectors satisfy Av = 6v
mortal juniper
ah i see
we cant add eigenvectors together and call it that its associated to its other eigenvector right?
only multiplication right?
dusky epoch
we cant add eigenvectors together and call it that its associated to its other e...
that sure was a bunch of words you just said
you can add two eigenvectors w/ the same eigenvalue and get another eigenvector with the same eigenvalue
mortal juniper
if the eigen values have different values can we still add up the eigenvectors?
limber sierra
if two eigenvectors are associated with different eigenvalues, their sum isnt guaranteed to be an eigenvector.
mortal juniper
ah i see thanks
limber sierra
example: (3, 1, 1) is not an eigenvector of this matrix
mortal juniper
so if we add v1 and v2 which is = to (3,1,1) we cannot say that that it is associated 2, 1 or 0 am i correct?
limber sierra
indeed, note that the first entry got multiplied by 5/3, whereas the second and third entry got multiplied by 2
limber sierra
so if we add v1 and v2 which is = to (3,1,1) we cannot say that that it is assoc...
its not an eigenvector at all
so yes, its not associated to any of the eigenvalues
since its not an eigenvector
mortal juniper
however if i add by itself i can say it is right
since it is scalar multiple of itself
limber sierra
sure
if u, v are eigenvectors both associated with eigenvector s, then this means Au = su and Av = sv
and so A(u + v) = Au + Av = su + sv = s(u + v)
so u+v is an eigenvector as well.
the catch is that they need to have the same eigenvalue for this to hold in general.
mortal juniper
ah i see that was interesting thanks!
keen coyote
how is (4.2) a formula for projection onto the complement of Ker A?
lavish jewel
that's what you call a pseudo inverse
since b = Ax, you would have VS^-1W^* times W S V^* x
you can see W^* W is an identity, and so is S^-1 S
you end up with V V^*, which is an orthogonal projection matrix
it projects x onto the span of V, which is the row space of the matrix