ok, so we know that the norm of b is 1 cuz u did GM
#linear-algebra
novel hamlet
im not sure what you mean bby that question
lavish jewel
do you see that e1^T v = e1 dot v?
novel hamlet
that sounds reasonable, as e1^t*e1=1
lavish jewel
i don't follow your logic there
i just meant they are both equal to e11 v1 + e21 v2 + e13 v3 + e14 v4
regardless of what the norm of e1 is
novel hamlet
so bbasiically i will end up with (e1^tv)e1+(e2^tv)e2+(e3^tv)e3?
novel hamlet
i tought i can just getthat from the sum of projections?
lavish jewel
what is the task asking you for
novel hamlet
finding a matriix
lavish jewel
right
novel hamlet
for that [pi]col(A)
novel hamlet
find matrix for projection
lavish jewel
mhm
so what matrix does this? (e1^tv)e1+(e2^tv)e2+(e3^tv)e3?
i'll simplify it out for you
we already know that e1^T v is the scalar projection of v onto e1
so let's just call that p1
same for the other vectors ei
so we have p1e1 + p2e2 + p3e3
pi are scalars, ei are vectors
how can we express that operation, which is a linear combination, as a matrix-vector product?
novel hamlet
we eed to construct trnasformation matrix out of that and put p1 on correct slots on it?
lavish jewel
sure, that much is true
but this should be really straightforward
that is literally the definition of $\begin{bmatrix}e_1 ,, e_2 ,, e_3 \end{bmatrix} \begin{bmatrix}{p_1 \\ p_2 \\ p_3} \end{bmatrix}$
sigh
what is up with me today
stoic pythonBOT
Edd
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
novel hamlet
yes
lavish jewel
ok, that is p1e1 + p2e2 + p3e3
let us call $N = \begin{bmatrix}e_1 ,, e_2 ,, e_3 \end{bmatrix}$
stoic pythonBOT
Edd
novel hamlet
looks like dot product
lavish jewel
that just leaves the question of how to get the p_i
it looks like a dot product, but e_i are vectors
so it isn't a dot product, it's a linear combination
anyway
how do we get p1
well, p1 should be e1^T v, yeah?
novel hamlet
yes
lavish jewel
and in general, p_i = e_i^T v
how do we express that so that we get all the p_i in a vector, as we wrote before?
in other words
$W v = \begin{bmatrix} p_1 \\ p_2 \\ p_3 \end{bmatrix} = \begin{bmatrix} e_1^T v \\ e_2^T v \\ e_3^T v \end{bmatrix}$
what should W be?
novel hamlet
W should bbe colum of A? since it is a transformation?
stoic pythonBOT
Edd
lavish jewel
no
we're not using A anymore
we were done with A from the moment you did gram schmidt on its columns
novel hamlet
should it be just e1, e2, e3 ?
lavish jewel
arranged how?
signal latch
How can I find R and V from L and N?
novel hamlet
lavish jewel
no, that's the N we used before
that's size 4 x 3
you can't multiply that by v
since v is 4 x 1
novel hamlet
then im not really sure what is that W
lavish jewel
W is $\begin{bmatrix} e_1^T \ e_2^T \ e_3^T \end{bmatrix}$
stoic pythonBOT
Edd
lavish jewel
which is nice because
$W = \begin{bmatrix} e_1^T \\ e_2^T \\ e_3^T \end{bmatrix} = \begin{bmatrix} e_1 ,, e_2 ,, e_3 \end{bmatrix}^T = N^T $
stoic pythonBOT
Edd
novel hamlet
and N = (e1,e2,e3)
lavish jewel
so we know that our projection is Np, and that p = N^T v
so the ortho projection onto the span of the columns of A is given by the matrix N N^T
novel hamlet
okay, i go calculate this open now then
lavish jewel
this should also be equal to U U^T from the SVD you did before
ah
the matrix is rank 3
so U(:,1:3) * U(:, 1:3)^T
if you do it with the SVD
novel hamlet
i did....
this part works nicely there
{{sqrt(10)/5, -sqrt(10)/10,sqrt(2)/2}, {sqrt(10)/10, sqrt(10)/5,0},{sqrt(10)/10,sqrt(10)/5,0},{sqrt(10)/5, -sqrt(10)/10,sqrt(2)/2}}
but when i try to add multiplication with transpose it gets mentally challenged and starts to think sqrt is a word
lavish jewel
how did you come up with the notation Transpose[]
novel hamlet
copy pasted from wolfram
lavish jewel
what if you just say N = {whole matrix here}, N * transpose(N)
otherwise, try this out https://octave-online.net/
uses matlab notation and it's free
lavish jewel
like you would matlab
so exactly like in the image i shared above
up to where i do "version1"
the rest is with SVD, so just ignore that
lavish jewel
i didn't get ones on all corners
idk
i would expect the projection to be rank 3, not 2
novel hamlet
transpose should be correct tho?
lavish jewel
something is wrong, idk what
novel hamlet
yeah the top right and bottom left seem odd
lavish jewel
oh
ur third column is wrong
missing a minus on the fourth elem
same on the transpose
lavish jewel
aight. i hope you understood the procedure. i recommend you review your scalar and vector projections, as well as change of basis matrices
novel hamlet
yeah, i got this week a lot of do this
need to repeat this same process quite a few times
and then i need to figure out OLS method
and solve Ax=e4 with least squares method
novel hamlet
basically i need to solve for
where X is unklnown and y is matrix 4x1
(e4=0,0,0,1)?
lavish jewel
mhm
novel hamlet
so i need to calcylate A^te4 first and then do row reduce to calcuylate x?
so i would get A^te4 from this
novel hamlet
yeah i have Ax=e4 and i need to solve it so A^ty = e4
oh yes you are right why am i doing this
i should be doing this?
native rampart
f(l)=(l,x) is a linear function on l
Meaning f is completely determined by values of f(e_1),f(e_2)...f(e_n)
Basis vectors of X
How is (l,x) defined?
Let {e_1,e_2...e_n} be a basis of X,think of functionals f_i,such that f_i(e_j)=0 if i is not j and 1 if i=j
These functionals will be the basis for your dual space
spiral sparrow
Hi guys I’d love some help on question 1 part 3.
Any help Is appreciate
Is that all that needs to be said for the answer?
Or do I need to do some calculations of some sort
I’m sorry I’m so bad at matrices
Great thank you so much 😊
Yes we do haha
Once again thank you
native rampart
I think the point of invoking Th1 is to show that if L is an element of double dual space,then L(f)=f(x) for some fixed x,for all f
Yes
The iso from a vector space to dual is not natural iso,for example
stoic pythonBOT
mirzathecutiepie
native rampart
idk how that affects natural isomorphism
I mean,There could be some natural isomorphism we don't know of
That's not simple to prove
Response I got for the same question
Th1 shows l(x) is completely determined by l((1,0,0...))=a_1,
l((0,1,0,...))=a_2...
l((0,0,0...n))=a_n
(l,x) is a linear function in l, which means it's completely determined by values of $(f_1,x),(f_2,x)...$ Where {$f_1,(f_2)...(f_n)$} is a basis of dual space.
$(l,x)=c_1(f_1,x)+c_2(f_2,x)+...c_n(f_n,x)$, where $l=c_1f_1+c_2f_2...c_nf_n$
stoic pythonBOT
DrunkenDrake
native rampart
Write (f_1,x)=f_1(x),(f_2,x)=f_2(x)...
So that ends up being ($c_1f_1+c_2f_2...+c_nf_n$)(x)=l(x)
native rampart
T(x)=$c_1T(e_1)+c_2T(e_2)...c_nT(e_n)$ ,where x is $c_1e_1+c_2e_2...c_ne_n$
stoic pythonBOT
DrunkenDrake
stoic pythonBOT
DrunkenDrake
native rampart
Th 1 applied here
Now (f_1,x)=f_1(x)
(f_1,x) is T(f_1)
You are defining T: T(f)=(f,x)
native rampart
You are defining T: T(f)=(f,x)
Applying this,RHS simplifies to
$c_1f_1(x)+c_2f_2(x)...c_nf_n(x)$
stoic pythonBOT
DrunkenDrake
native rampart
=($c_1f_1+c_2f_2...c_nf_n$)(x)
=l(x)
stoic pythonBOT
DrunkenDrake
dusky epoch
bruh the bad tex
native rampart
Ok,I think I get how Theorem 1 is useful here.
you can show $L(f_i)=f_i(x)$ where $x=\sum{L(f_i)e_i}$
stoic pythonBOT
DrunkenDrake
native rampart
I don't get why you care about the bilinear form at all,tho
$f=\sum{c_i f_i} \implies L(f)=\sum c_i L(f_i)=\sum c_i f_i(x)=f(x)$
stoic pythonBOT
DrunkenDrake
native rampart
Which book?
stoic pythonBOT
mirzathecutiepie
wintry steppe
Can someone link me to a tutorial on how to solve matrixes raised to a power like 2n
A^2n
A^2n+1
stoic pythonBOT
mirzathecutiepie
wintry steppe
Make it into 1’s on diagonal and 0 the rest ?
wintry steppe
Shit
I need to know that to raise to powers?
Currently we just doing A^2
Real numbers
I’m dumb ok I’ll go learn
Yeah it’s a hw problem
wintry steppe
It’s filled with 1/2
frosty vapor
i always want to ask before i drop my questions
so i did 17 and so when working on 20 i don’t really understand where to get more leverage
wintry steppe
4x4
frosty vapor
i got to showing nullity(L_A) + rank(L_A) = nullity(T) + rank(T) but from there i need something else to get to the result we want
spiral sparrow
Second time in here 😭 but if anyone can help me with this I’d greatly appreciate it
stoic pythonBOT
mirzathecutiepie
spiral sparrow
thank you so much !!
gritty swift
$A = S\Lambda S^{-1} = A^T = (S^{-1})^T \Lambda S^T \implies S^{-1} = S^T$
is this a valid proof for a symmetric matrix having orthogonal eigenvectors? assuming you can diagonalize
stoic pythonBOT
uli
gritty swift
I do? where
I think you misunderstood the steps lemme rewrite it
$A = A^T$ so if $A = S \Lambda S^{-1}$ then $(A)^T = (S \Lambda S^{-1})^T = (S^{-1})^T \Lambda^T S^T \implies S^T = S^{-1}$
gritty swift
no I'm just substituting $A = S \Lambda S^{-1}$ then "distributing" the transpose
stoic pythonBOT
uli
gritty swift
oh, its just that $\Lambda^T = \Lambda$ since its diagonal so for that to equal the other guy I thought $S^T = S^{-1}$ must be true
stoic pythonBOT
uli
lavish jewel
the thing is you could put rotation matrices between S and lambda and it would still work
so you have to show there is no rotation in the middle when you transpose
gritty swift
oki
oh what you mean is $S(S^{-1})^T \Lambda S^T S^{-1} = \Lambda$ does not imply $S^TS^{-1} = I$ and $S(S^{-1})^T = I$ i think i understand now
stoic pythonBOT
uli
lavish jewel
it's easier if you move s's to each side, isn't it?
lavish jewel
$S \Lambda S^{-1} = S^{-1T} \Lambda S$
stoic pythonBOT
Edd
gritty swift
$\Lambda = S^{-1}S^{-1T} \Lambda SS$
stoic pythonBOT
uli
lavish jewel
$(S^T S) \Lambda (S^T S )^{-1} = \Lambda$
stoic pythonBOT
Edd
lavish jewel
idk if you can come up w something more easily from this side, now that i think about it lol
gritty swift
wait does $(S^{-1})^T = (S^T)^{-1}$? I'm confused how you moved $S^{-1T}$ to the lhs
stoic pythonBOT
uli
gritty swift
oh cool
lavish jewel
project the solution onto the row space
novel hamlet
i did it the problem to A^tA part and then i tried to invert it but it is not invertible
lavish jewel
doing the classic (A^T A)^-1 A^T is the same as projecting the solution onto the row space and forgetting about whatever was lost to the null space
novel hamlet
lavish jewel
right, that's only invertible for matrices A with full column rank
novel hamlet
yeah My a is 3 colums 4 rows
lavish jewel
and are the 3 columns lin indep?
novel hamlet
lavish jewel
so no
the inverse indeed does not exist. you'd do a pseudo inverse
or in other words, replace A^T A by the orthogonal projection matrix onto its column space
novel hamlet
so pseudo inverse and then continue normally?
lavish jewel
pseudo inverse and you are done, there is nothing else to do
novel hamlet
i dont multiply it by A^ty anymore?
lavish jewel
(A^T A)^-1_economy_sized A^T is the pseudo inverse
keep track of your matrix dimensions and what is getting mapped to which space
A maps R3 to R4
A^T maps R4 to R3
lavish jewel
(only subspaces thereof, to be fair, but you have to at least make sure you can multiply the vectors)
novel hamlet
that is for A^tA
lavish jewel
do you wanna understand what that does or do you just wanna solve it?
you're jumping the gun
novel hamlet
to be honest this is first time i have ever needed the pseudo inverse to do anything
lavish jewel
you need the pseudo inverse of A
novel hamlet
currently googling what it does
gritty swift
coulden't you just throw away dependent guys then invert? (I also haven't done pseudoinverse)
lavish jewel
that inverts from the wrong side
novel hamlet
ah yues, i keep forgetting that multiplication is no longer commutative
lavish jewel
i'm unfamiliar with all of the notation and nomenclature :x
what is (5) here?
it looks like it calls the transpose the dual space, and it's talking about what i know as conjugate inner product
the transpose turns a vector into a function that you can apply to a vector to map to the field
i might be way off anyway tho, i'm sleepy and i don't know squat
it looks to my untrained eye like you have $x \in X$, with dual space of the form $x^T$
stoic pythonBOT
Edd
lavish jewel
and the dual space of the set with elements $x^T$ has elements of the form $x^{TT} = x$
stoic pythonBOT
Edd
slow scroll
do you understand the identification in theorem 3?
lavish jewel
please do correct me if i'm mistaken, actual math isn't my thing
wintry steppe
(i deleted the explanation because i think someone else will give a better one)
slow scroll
well, i mean, what the actual map is
right, so you just send $x$ to the map $l \mapsto (l, x)$.
stoic pythonBOT
kxrider
slow scroll
right, so you just send $x$ to the map $l \mapsto (l, x)$.
stoic pythonBOT
kxrider
slow scroll
sry bout that 
lavish jewel
it looks to me like fixing x, l are in X'. fixing l, x are in X''
wintry steppe
the notation in this book is terrible lmao, (l, x) should be the number l(x) but it seems as though the author is also writing it for the function l \mapsto (l, x)
slow scroll
let $f_x : X^{'} \to \bR$ be the map $f_x(l) = (l,x)$. Then, the claim is that $x \mapsto f_x(l)$ is an isomorphism.
stoic pythonBOT
kxrider
slow scroll
alright, since X'' and X have the same dimension, it suffices to show that the kernel of this map is trivial.
stoic pythonBOT
mirzathecutiepie
slow scroll
agh oops, i meant to write $x \mapsto f_x$.
stoic pythonBOT
kxrider
lavish jewel
that looks about right to me
slow scroll
now, if x is in the kernel, f_x = 0, i.e. (f, x) = 0 for any f in X', and you just need to show that x = 0.
coarse rain
rank-nullity theorem
wintry steppe
showing the kernel is trivial shows injectivity
and if the sets are the same size yes surjectivity also follows
slow scroll
its like how showing determinant of a square matrix is nonzero suffices to prove invertibility
coarse rain
Rank-Nullity Theorem states that the dimension of the kernel + dimension of the image is equal to the dimension of the domain.
Therefore, if the dimension of the kernel is 0, then the dimension of the image is the dimension of the domain. Because the codomain has the same dimension as the domain, this implies that the image is the codomain.
ah
slow scroll
i thought i saw you in this channel talking about rank nullity the other day 
wintry steppe
"all such linear functions are of this form" just prove surjectivity manually
slow scroll
its like first isomorphism theorem for vector spaces
coarse rain
:deenothink:
slow scroll
i was probably just seeing things
but yea manual proof of surjectivity works to i think
wintry steppe
think of it this way: lets say A(u+x) = Au. Then, because A is linear Au + Ax = Au and Ax = 0. This means that x is from the kernel. This shows injectivity iff ker={0}
coarse rain
rank-nullity ain't that hard to prove by yourself actually. Get a basis of the null space, extend that basis into a basis of the domain, show that those other basis vectors have linearly independent images.
I definitely shoulda looked at context
lavish jewel
but somewhere down the line it turned into seeing what the fuck
coarse rain
so you're confused on the natural isomorphism between a vector space and its double dual?
wintry steppe
the slicker solution is to read another book
any decent linear algebra book should have this
coarse rain
this book has exactly the same layout/font as baby rudin
basically it's saying that the isomorphism $v \mapsto f_v$ such that $f_v (\varphi) = \varphi(v)$ is a natural isomorphism from $V \to V''$.
stoic pythonBOT
Alphyte
coarse rain
and to prove surjectivity you take any element f in V'', compute the image of a basis of V' under f to find v in V that maps to f under that isomorphism.
misty storm
I started out with this, which is a one sheet hyperboloid
and I want to end up with the closest thing to its general equation
I managed to turn it into this
kinda need help turning it into this
stoic pythonBOT
mirzathecutiepie
coarse rain
I'm not too sure what the notation of the book is either
wintry steppe
this is the same thing as writing f(x) for both a function and its value at a point
(i don't meant to imply that writing f(x) is bad notation, but i think it can be misleading or confusing here)
the idea is really that the elements of X are acting on the functionals on X
lavish jewel
do they ever use cdot for functions? like to be able to say f_x(\cdot) instead of f_x(l)
wintry steppe
each $x \in X$ determines a function $X' \to \mathbb F$ taking a linear functional $l$ to $l(x)$. that is, $x$ "acts" on $X'$
stoic pythonBOT
TTerra
wintry steppe
that's what i mean
lavish jewel
i mean to denote a function instead of a function evaluated at a particular argument
stoic pythonBOT
mirzathecutiepie
lavish jewel
right, that's what i was thinking too. you have x in X, l in X', and l(x). they wanna say that there is an isomorphism from x to x(\cdot) that acts on l
and this x(\cdot) is in X''
jaunty sage
can anybody help me with this i am very lost, like i have 0 clue as to what i should do
jaunty sage
can anybody help me with this i am very lost, like i have 0 clue as to what i sh...
. please someone help with this one
lavish jewel
the value of m has to be 3
lavish jewel
that the first scenario they give you has no sol. implies the matrix is rank defficient
jaunty sage
by relation they mean this right?
lavish jewel
so it has rank at most 2
jaunty sage
so it has rank at most 2
either 0,1,2 yes?
lavish jewel
mhm
now in the second scenario they give you an example with a unique solution
this would mean the null space of the matrix has only the 0 vector in it, and the matrix has full column rank
so r = n
jaunty sage
what does full column rank mean? do u mind explaining?
lavish jewel
otherwise, any equality that has a solution would have infinitely many of them
the number of independent columns is equal to the rank of the matrix
jaunty sage
ah so according to the image the relation is number 2?
lavish jewel
remember the rank can be smaller than both m and n
jaunty sage
the number of independent columns is equal to the rank of the matrix
AH alright
lavish jewel
D: good luck
jaunty sage
goodluck
lavish jewel
and yes, the relation would be number 2 in the second image you shared
lavish jewel
0 would be a pretty lame matrix as it would be empty
if n = 1, you would have a single column
i think that works
2 should also work
it can't be 3
lavish jewel
we still would get 3 x 1
lavish jewel
but then either all equations would have a solution, or the ones that have a solution would have infinitely many
jaunty sage
do you mean either no solution or infinitely many?
im not sure i follow that statement
lavish jewel
no
a 3x3 matrix can have rank 1,2 or 3
if it is rank 1 or 2, then problems would have either no sol or infinitely many
if it is rank 3, all equations have exactly one sol
which would be case 1 in your second image
but we already know that is not the case
jaunty sage
ok i see
lavish jewel
so either 1 or 2
lavish jewel
well, both scenarios are with the same matrix
jaunty sage
oh i see
lavish jewel
you're supposed to put all the info together
jaunty sage
so when i asnwer the question i take both into consideration
lavish jewel
yes
lavish jewel
1 or 2, i think only the zero vector has rank 0
jaunty sage
1 or 2 alright
then we established r=n
because of the second scenario
and values of n would be 1 or 2
right?
what about c
also in this question what all methods am i supposed to do
lavish jewel
it has infinitely many solutions, so you have to find the particular solutions and all the components in the null space
so it'll be some x_c that is constant + a*x_n for any values of a
where x_n is in the null space of A
this is also the approach you would use for part c
jaunty sage
this is also the approach you would use for part c
i didnt follow
lavish jewel
also for part c, recall that A has linearly independent columns because r = n. the transpose has lin indep rows
spiral sparrow
But with the previous question I sent in here, I’m still confused on it a bit is it alright if I could go into dms with you?
If not that’s alright
jaunty sage
also in this question what all methods am i supposed to do
@lavish jewel i think i kinda understand what you mean, moving on could you let me know what all i have to do for this one?
lavish jewel
read what i said again and look at number 3 in the summary pic you sent
stoic pythonBOT
mirzathecutiepie
lavish jewel
remember we had m x n, n = r, n < m
jaunty sage
remember we had m x n, n = r, n < m
ah gotcha that sums up the other qquestion the one with abc
lavish jewel
to be completely fair, there was an F_L(x) in the middle, too but that looks ok
i think my example of that with transposes was a good way of looking at it, too
jaunty sage
ok last question how should i do the part e
coarse rain
finding the vector or verifying that they are orthogonal?
jaunty sage
finding the vector or verifying that they are orthogonal?
both
is row space the transpose of column space?
jaunty sage
column space of transpose
original matrix A transposed and the column space of that is that what u mean?
coarse rain
yes
jaunty sage
and also how would i verify the second part of that question
lavish jewel
you find a basis for the null space and find the projection of x onto it
it should be 0
jaunty sage
you find a basis for the null space and find the projection of x onto it
i have 2 values for the basis of null space
i just dot product those vectors and then verify?
lavish jewel
sure. if the dot product is 0, the 2 vects are ortho to each other
i disappear into the night now. GG
jaunty sage
gn
undone sky
Does it make sense to talk about $V/W$ when $W\cong B\subset V$ is the only relationship between V and W?
stoic pythonBOT
Uchigawa
undone sky
Gotcha ty
nocturne jewel
$\begin{bmatrix} 2&0&0\0&1&-1\0&1&-1 \end{bmatrix}$
stoic pythonBOT
moshill1
nocturne jewel
I have this matrix which comes from plugging in an eigenvalue to find its eigenvectors, how do I go about finding the subspace which is spanned by the eigenvectors?
I got ${\vec{x}\in\mathbb{R}^3|\vec{x}=t[0,1,1]^T,t\in\mathbb{R}}$
stoic pythonBOT
moshill1
wintry steppe
it would be the kernel of that matrix
where presumably by "plugging in an eigenvalue" you mean you subtracted off an eigenvalue from the diagonal entries of whatever matrix whose eigenvectors you're trying to find
if that's not what you meant, please clarify
nocturne jewel
where presumably by "plugging in an eigenvalue" you mean you subtracted off an e...
Yeah that's the matrix I got from tI-A
wintry steppe
not exactly sure what that has to do with finding the matrix's kernel but you got the right answer so it seems like you understand it
nocturne jewel
jaunty sage
what do i do in this question
if rank is 1 then there is one linearly independent column right
wintry steppe
yes
wintry steppe
maybe write out what it means for the columns to be linearly dependent
and note that they can't both be zero, for rank to be 1
for example one of the columns should be a scalar multiple of the other
jaunty sage
if c and d are 0 isnt rank 1?
wintry steppe
that will be true but you're not allowed to set the values of the matrix's entries
jaunty sage
for example one of the columns should be a scalar multiple of the other
isnt that linear combination in a sense
wintry steppe
it is
jaunty sage
i would perform rref on this matrix or ?
wintry steppe
write out what it means for the columns to be linearly dependent
wintry steppe
d will be a scalar multiple of c, and you can try to find that scalar
for the columns to be linearly dependent means there is a nontrivial linear combination of them equaling zero
jaunty sage
so like x(c1) + y(c2) =
is this correct tho or am i thinking wrong?
wintry steppe
im not sure what you're saying there
you haven't said what the right hand side of this equation is
wintry steppe
yeah, and x and y are not both zero
wintry steppe
yeah
jaunty sage
what do i do after that then
wintry steppe
well you wanna solve for d right
to do that you need to make sure y is nonzero. why is that the case?
jaunty sage
cuz if d is 0 then rank is not 1?
jaunty sage
im sorry if i dont follow fast enough but this subject is really confusing for me
wintry steppe
take your time
jaunty sage
no
wintry steppe
right
jaunty sage
because that would not satisfy a>0 right?
wintry steppe
yeah, if xa = 0 and a > 0, then x = 0. but remember you can't have both x and y equal to zero, so having y = 0 is impossible
so now you can solve your second equation for d
after that you just have to do some funny algebraic manipulation to simplify whatever expression you get to be in terms of only the entries of the matrix
jaunty sage
ah sounds like a hassle
wintry steppe
jaunty sage
so would the solution to d be like -xc/y?
jaunty sage
ah sucks
wintry steppe
you should be able to solve it now
frosty vapor
so i did 17 and so when working on 20 i don’t really understand where to get mor...
similarly can anyone help with this 😭
restive wraith
so for solving recursive functions our class has us do it with geometric equations, however, we have to do this for a 3d vector instead of 2d and i am completely lost on what this step would look like
this is the 2d one, any idea what it would look like in 3d
obtuse kraken
proud lava
is there any way u can find matrix A if u already have the eigenvalues?
proud lava
but is there a method to finding one of those matrices?
wintry sphinx
if you take any triangular matrix, the eigenvalues are just the elements on the diagonal
proud lava
i dont understand
wintry sphinx
what part don't you understand
proud lava
lemme think about it a little more
ok so could we go through an example?
say u have eigenvalues 5,7, and 9
and there are no 0's in the matrix
@wintry sphinx sorry for ping but coulld u show me the process?
wintry sphinx
There are no zeroes in the matrix?
would've been helpful for you to state this beforehand
proud lava
sorry
i tried making a 2x2 first that worked for 5 and 7
im pretty sure u can use a pattern from that 2x2 or something to find a 3x3 for 5,7, and 9
but im sorta confused
wintry sphinx
are there or are there not zeroes in the matrix
proud lava
no zeros @wintry sphinx
wintry sphinx
that's a little harder, but all you have to do is pick a linearly independent set of vectors that span the space
and then use those as the eigenvectors
proud lava
hmm ok
wintry steppe
so i did 17 and so when working on 20 i don’t really understand where to get mor...
what happens if you restrict phi_gamma to T(V)? you should get an isomorphism whose image is the same as the composition (1). now you can apply 17
and if you know the ranks are equal...
oh that was at like 3 pm LOL
wintry steppe
so the rank of T is the dimension of T(V) and the rank of L_A is the dimension of L_A(F^n), yeah?
so if you want to apply exercise 17 you're gonna need an isomorphism of V and F^n taking T(V) to L_A(F^n)
frosty vapor
hi
im moving back in here since chill got busy
but it seems i last deduced that L_A = phi_gamma Im(T) or i guess in these terms = phi_gamma T(V)
er
ya
native rampart
You mean range(L_A)=phi_gamma(Im(T))?
frosty vapor
umm wait now im confused
ok so first he told me to write L_A as phi_gamma T phi_beta^-1
okay
then
oh
i took the images of both sides right
and thus yes what you said
yes
man i am so bad at this
this problem ruined my day
i should have been studying for my exam on tuesday
h
native rampart
You mean range(L_A)=phi_gamma(Im(T))?
The problem is done with this step
native rampart
There is an bijection from range(L_A) to Im(T)
native rampart
Which is to say range(L_A)=Im(T)
native rampart
Range is image
frosty vapor
yeah range uses domain the image might just come from some other set
oh
i guess here though since both started from F^n
then
in the end its the same
?
since we went from F^n to V
which is the domain for T anyway
then we get range(L_A) = phi_gamma range(T)
native rampart
range and image are synonyms
frosty vapor
oh
okey i think
i think this makes sense
since there is a bijection between the ranges
the ranks must be equal
and so the other equality also falls out
woah...
okay time to type this at hyperspeed
native rampart
A=B as vector spaces only when the bijection is also a isomorphism
native rampart
Yes
frosty vapor
thank you
i submitted my hw on time
but i will try to understand this problem so i am going to do it again but slowly
sonic osprey
the dot product is an example of an inner product
brisk fable
yea I just found out
I just don't know what else would be considered an inner product
jaunty sage
what do i do here
native rampart
what's the difference between inner product and dot product?
Every inner product is a dot product wrt some basis
(Assuming a basis exists)
jaunty sage
what do i do here
dim N(A)=1 since there is 1 free variable
do i find the rank using dimN(A) = n - rank(A)?
sturdy portal
Each row from an augmented matrix in row echelon form can be perceived as a vector from the basis in the solution set, if the latter is viewed as a vector space.
Is this statement correct?
restive wraith
does anyone know how to make a all non 0 matrix given eigenvalues
heavy coral
im a bit confused how to approach this problem
lavish jewel
this belongs to complex variables, but we can take a look
do you know how to find roots of complex numbers?
orchid harbor
If we have Ax=b
b= 0
Then we have unique solution for homogeneous system?
Always?
orchid harbor
In what cases its not unique?
The w are the coeeficients of chemical reactions coefficients
Lemme show u
b
(b)
Is it reasonable to expect that any chemical reaction has always a unique solution? Always has many solutions? Why?
dusky epoch
In what cases its not unique?
when the matrix has rank less than the number of unknowns
magic light
If I want to prove that ImT is all the symmetrical matrices from M_2x2 and I show they have the same dimension, does that prove it?
T is from M_2x2 to itself over R
dusky epoch
and I show they have the same dimension
"they"?
magic light
ImT and the symmetrical matrices
magic light
you still haven't said what T actually is
I’m asking in general,
wintry steppe
what's T.
magic light
If they have the same dimensions why does it matter what T is?
wintry steppe
ty for clarifying
dusky epoch
Why is it not sufficient?
because the space of all symmetric matrices is not the only dim 3 subspace of R^2x2
there exist others
magic light
I thought if two spaces have the same dimension they are the same
dusky epoch
they're isomorphic but they are not the same
you can't handwave your way out like that in linalg
magic light
Thanks
gritty swift
when do we use $\det(A - \lambda I)$ vs $\det(\lambda I - A)$?
they seem similar since $\det(\lambda I - A) = \det(-1(A - \lambda I)) = (-1)^m \det(A - \lambda I)$ for m rows, if I factored out correctly
does the sign ever make a difference?
stoic pythonBOT
uli
dusky epoch
not rly
youre most likely considering this determinant for the purpose of finding eigenvalues
which are not affected by this
gritty swift
ok cool
lavish jewel
btw @wintry steppe
that is a much nicer way to look at it
as two maps that behave the same way
ah, aight. GG
yeah
it makes more sense tbh
cuz a function is not the same as a function evaluated at a point
not really tbh 😛 i believe in you
at some point later
warped cape
the proof does use a basis, but the definition of the mapping does not
if that helps
and the proof is for an arbitrary basis
yeah i think it's fine
it proves that (l, x) is an isomorphism
yes it does not prove naturality
does your book define what "natural identification" means?
warped cape
if you want to prove it's a natural isomorphism you can work through the CT definition of a natural transformation, with ((-)^{**}) being a functor
stoic pythonBOT
Muf
warped cape
graceful vortex
@obtuse kraken Where are you having problems ?
stoic pythonBOT
mirzathecutiepie
warped cape
well they write (l, x) to mean the same as (-, -)
and yeah, their proof doesn't cover naturality
they might just mean natural in the "this it the obvious one" sense
but also know it is natural, so they leave it in
stoic pythonBOT
mirzathecutiepie
mirzathecutiepie
mirzathecutiepie
mirzathecutiepie
mirzathecutiepie
restive wraith
i am so so lost on what this is asking to do, is anyone able to rephrase this
warped cape
@wintry steppe yeah, that's correct, except you don't construct F such that F{x} = l(x), F is defined as taking {x} to l(x)
same with G i think
but the intuition works for either, just a technical detail
uhh
ohhh i see
yeah, then you're correct
go ahead :)
stoic pythonBOT
mirzathecutiepie
warped cape
wait how is Y^\perp defined?
ohh, it says that if you take (l, x) as the bilinear form from the definition, then (Y^{\perp \perp} = Y)
stoic pythonBOT
Muf
faint lintel
I'm trying to prove that the number of non-pivot columns in a RREF matrix corresponds to the nullity of that matrix
Not sure how to go about it
I know that a non-pivot column represents a free variable that can be expressed as a linear combination of pivot columns
but going from there idk
warped cape
kinda, the wording in the definition is probably something like "Y^\perp is the orthogonal complement of Y under the bilinear form ..."
ivory sleet
how would i prove that two graphs with identical Laplacian matrices are isomorphic
little citrus
@main charm
Clown how is Harvard ?
Damn you used to be my student here, you forgot me?
main charm
Damn you used to be my student here, you forgot me?
?????
faint lintel
Why is this bottom statement true
hollow finch
I'm trying to prove that the number of non-pivot columns in a RREF matrix corres...
Rather than show non-pivot columns correspond to nullity, use the fact that pivot columns correspond to rank. Then use the relationship between the rank and nullity of of a matrix to prove that what you can say about the rank implies what you want to say about the nullity.
faint lintel
oh dope
Uh
I mean I'm trying to prove that pivot columns correspond to rank
and I want to use the dimension theorem
So like either direction
I don't get
I don't get why rank = # of pivot columns or why nullity = # of free variables
both are confusing to me
So like if I understand one I can get the other through the dimension theorem
but the issue is I don't get why either are true
hollow finch
I see
How comfortable are you with change of basis? I can see that being one way to see it
faint lintel
pretty comfy with the matrix representation and stuff
hollow finch
So first, keep in mind that rank is the dimension of the column space. Which means rank tells you the number of independent columns.
Take this matrix in rref for example,
$$\begin{bmatrix}1&0&2\0&1&1\0&0&0\0&0&0\end{bmatrix}$$
stoic pythonBOT
nix
hollow finch
how many independent columns are there?
faint lintel
2
hollow finch
right
not coincidentally the number of pivot columns
All non-pivot columns in a rref matrix are dependent on the pivot columns. Does that make sense?
faint lintel
that's because non-pivot columns correspond to free variables and as such you can write the free variables in terms of the independant variables or vise versa
?
hollow finch
That is true
faint lintel
so how does that linear dependance imply null space?
hollow finch
Because row operations do not change column relationships/column dependence.
faint lintel
I don't see the connection
I know that's true
but how does that imply that the free variables correspond to linear dependance?
hollow finch
**nix**
If you do a bunch of row operations on this matrix, the third column will always still be 2 times the first column plus the second column. I recommend testing this out at some point. So then if I multiply by (2,1,-1) I will always get zero. The null space is preserved by row operations. That's why when we look for the null space of a matrix we put it in rref, because then the column relationships become much more obvious.
I like to think this corresponds to the rank of a matrix. the number of pivot columns in the rref still gives me the number of independent columns of A. i.e. the dim of the column space so rank
But if you want to think about nullity, you could imagine finding a vector to cancel out each non-pivot column. So there are as many independent vectors in the null space as there are nonpivot columns.
That's how I think of it anyway
faint lintel
The idea that row operations don't change linear dependence makes sense. I just am struggling to see how "this column has free variable" connects to "this column is a basis vector for the nullspace"
hollow finch
that's because non-pivot columns correspond to free variables and as such you ca...
Like you said here, we can write free variables in terms of independent ones. That actually gives us a vector for the null space since we can use that relationship to cancel out the columns. If its a null space vector of the rref matrix, its a null space vector of the original matrix since that relationship is unchanged by row operations.
faint lintel
since we can use that relationship to cancel out the columns
in this sentence the columns refers to cancelling out the dependent columns?
hollow finch
yeah
hollow finch
**nix**
like how i got (2,1,-1) is in the null space from the relationship of the columns here
because 2*col1+1*col2=1*col3 -> 2*col1+1*col2-1*col3=0
faint lintel
I see I see
hollow finch
it's a great question. good for you making sure you understand it conceptually. it will serve you well.
faint lintel
this whole worksheet is a conceptual nightmare
I know it's for the better
but man it's hard
gritty frigate
I will make a stupid question but.. Provided that A = mxq matrix and B = qxn matrix the is it true that:
-AB = A(-B) ?
Or if u is an scalar is
(uA)B = A(uB)?
hollow finch
I will make a stupid question but.. Provided that A = mxq matrix and B = qxn mat...
yes to both. by linearity
gritty frigate
Thanks
bitter shuttle
quick question
im considering taking an intro linear algebra course in high school
are the concepts of unitary matrices + hilbert spaces covered in intro linear algebra course
nocturne jewel
are the concepts of unitary matrices + hilbert spaces covered in intro linear al...
I dont think my 1st year LinAl course covers them.. but it's probably school-to-school type deal
hollow finch
are the concepts of unitary matrices + hilbert spaces covered in intro linear al...
Probably not. My intro course in college didn't even cover them.
wintry steppe
unitary operators and matrices were discussed thoroughly in my second linear algebra course, but no hilbert spaces
so not quite "intro"
but still a first year course
wintry steppe
can someone explain why (iii) is a subspace of r4
do i just need to check if the columns span r4?
hollow finch
do i just need to check if the columns span r4?
Span is not a requirement for a subspace. What are the two requirement for a subset to be a subspace?
wintry steppe
0 vector and close under addition & scalar mult?
hollow finch
Yeah
Luckily with this one, we can see right away that one of these is definitely not satisfied
wintry steppe
i know (iii) is consistent so it must have the 0 vector but how do i go about finding if it’s close under addition and scalar mult?
does consistent mean it is also close under addition and scalar mult?
hollow finch
Oh i see you said iii not ii. My b. I meant ii does not have the zero vector.
Consistent definitely does not show either of those. All it tells us is that the subset is nonempty.
To check for addition and scalar mult, just check whether or not the sum of homogeneous solutions is itself a homogeneous solution and wether or not a scaled homogeneous solution is still a homogeneous solution
All subspace tests go that way. Add them (that is, two arbitrary elements) and scale them and see if they still satisfy the original definition (in this case, being a homogeneous solution)
nocturne jewel
@hollow finch is your pfp James Grime?
hollow finch
<@!196803594576592896> is your pfp James Grime?
lol ye
nocturne jewel
noice
hollow finch
i like e
restive wraith
would anyone be able to rephrase this, i am not sure how to do it and ive been completely lost all day
hollow finch
would anyone be able to rephrase this, i am not sure how to do it and ive been c...
I think I see it. w1=Aw0 right? How then can you express the transformation of multiplying by A in terms of its eigenvectors/eigenvalues? Since we know how A transforms the eigenvectors, perhaps it would be helpful to get whatever we're multiplying "in terms of" them...
restive wraith
im not quite sure how the eigenvectors/eigenvalues will help me get to the answer in this
like ik they are used
w1 would be something along the lines of (u+v) * w0 right?
bronze beacon
Hey guys is anyone familiar with the theorem every sub space of V is a part of a direct sum equal to V
I need it to answer a question but I’m not sure how to apply it
It’s this question
bitter shuttle
does anyone know the difference between inner product and dot product
seems interchangable
and also why sometimes there is row/column vector
and sometimes vector is just <x, y, z>
warped cape
the dot product is just a special case of an inner product
nocturne jewel
dot product is the inner product of R^n
zealous junco
does anyone have or know a good proof that spectral radius is almost a norm?
stoic pythonBOT
Anticipation
ashen harbor
hello
how can I tell when 2 lines are clearly not parallel?
for example
the guy in the video says this is clearly not parallel
oh wait
is it if it stretches or contracts to meet the points
so if
v1 was <1,2,3>
and
v2 was <2,4,6>
would that be parallel?
warped cape
yes, they are parallel if the direction vectors are multiples of one another
ashen harbor
thank you
ashen harbor
what subject are you doing? Knewton sometimes can do proofs
sudden nacelle
me?
lavish jewel
which part is troubling you
pallid pumice
Can anyone give me a resource on how to calculate T(e1), T(e2), etc from vectors v1, v2, etc
professor completely skipped what he was doing and just gave us the answer
he also gives T(v1), T(v2) etc
lavish jewel
do a change of basis before applying T
pallid pumice
havent learned basis yet
lavish jewel
can you show what they did?
pallid pumice
idk if i should since its last weeks quiz assignment
but he basically just added or subtracted T(vn)
so like T(e1) = T(v1) - T(v2) + T(v3)
lavish jewel
i see, i get the idea
he basically asked you to teach yourself how to do a change of basis during the quiz, pretty clever
do you know gaussian elimination?
lavish jewel
mhm
the idea is as follows
if you make a matrix I = [e1 e2 e3]^T
actually, letm e start from the other side
make a matrix V = [v1 v2 v3]^T
you can augment this matrix as [V I]
you do gauss jordan until the V matrix now looks like an identity
this will turn the matrix I into the matrix you are looking for
since the row operations you do are the ones required to get e_i out of the v_i vectors
so those row operations tell you "you need 3v1 + 2v2 - ..." or something of the sort in order to get e1, for example
then you would do T(3v1 + 2v2 - ...) (or whatever it is you got) and use properties of linearity to instead do 3T(v1) + ...
pallid pumice
so id have a 3x3 matrix as an example from v1,v2,v3 and then the identity matrix
and the row operations i do to the matrix from vectors i do to the identity matrix?
and then the columns of that matrix are e1 e2 etc?
lavish jewel
do you know how to invert a matrix?
lavish jewel
yeah
sudden nacelle
@lavish jewel how they got to nn.T
lavish jewel
the dot product $n \cdot v$ is equal to $n^T v$
stoic pythonBOT
Edd
lavish jewel
you can expand both into a sum following the definition of dot product and vector multiplication, and you will see they are the same
after that, they just threw in a parenthesis because multiplication is associative
formal cargo
Can anyone help me with this problem from Trefethen&Bau?
I found these two posts but I'm having trouble really understanding the proofs: https://math.stackexchange.com/questions/1109755/a-projection-p-is-orthogonal-if-and-only-if-its-spectral-norm-is-1 and https://math.stackexchange.com/questions/491007/p-2-1-implies-p-p
lavish jewel
by two-norm there you mean the induced norm, right?
formal cargo
Yup
Showing inequality is pretty simple:
Then, following one of the links I shared, I think I can prove one half of the equality (although idk why 2-norm of x = 1)
Not really sure how to approach it from 2norm of P equal to 1 showing P is orthogonal
lavish jewel
i'd look for something related to the inner products and cauchy schwarz
in 2d, for example, having P be an orthogonal projection would mean the basis vectors are ortho to each other, and the matrix vector product (which yields a vector of inner products) would end up something like [P1 v; P2 v] = [v1 cos theta1; v2 cos theta2]
lavish jewel
and from the orthogonality of the vectors p1 and p2, we should get something like cos theta 2 = sin theta 1
maybe not the most straighforward way, but it should be pretty geometrically reasonable
@waxen flume well, how do you define the span
stoic pythonBOT
spitfire
formal cargo
@lavish jewel Thanks for your response. Any ideas on ^^?
Inner products and cauchy schwarz get me to the same place I'm stuck at
lavish jewel
you can arbitrarily set the 2-norm of x to 1
the denominator of the definition of the induced norm normalizes that anyway
WLOG you can just take the scalar factor out of x and put it in front of the expression and just forget about it
induced norms are submultiplicative
regardless of what the norm of x is, $\Vert P x \Vert_2 \leq \Vert P \Vert_2 \Vert x \Vert_2$
stoic pythonBOT
Edd
lavish jewel
but now that i look at it again that's not useful lol
anyway, if you already showed the inequality $\Vert P x \Vert_2 \leq \Vert x \Vert_2$, you can just say x has norm a, and then $\Vert P a \hat{x} \Vert_2 \leq \Vert a \hat{x} \Vert_2 = a \Vert \hat{x} \Vert_2 = a$ for an $\hat{x}$ with norm 1
lavish jewel
that sounds about right. GG
waxen flume
for this one
would I get
A = [x1 ; x1]
since the output is just x1 of the transformation
and its in R2
lavish jewel
not quite
stoic pythonBOT
Edd
waxen flume
what is A then?
lavish jewel
what happens to the y axis if you project onto the x axis?
waxen flume
turns into x axis
waxen flume
x component
lavish jewel
what if i give you a different vector, say [w1; w2]
the matrix has to work for any vector
not just [x1; x2]
lavish jewel
when the input is [x1; x2], yes
lavish jewel
right
waxen flume
how do i figure out the inside of the 2x2 matrix
lavish jewel
that's what i'm walking you through
the easiest method is usually to see what happens to the canonical basis vectors
[1;0] and [0;1]
what should happen to [1;0] when you multiply by A?
waxen flume
what is A in this case
lavish jewel
that is what you are looking for
waxen flume
1A + 0A?
lavish jewel
look
waxen flume
not sure
lavish jewel
the image is TELLING you what happens
they are telling you Ax = T(x) for any vector x
A is the matrix that performs the linear transformation
waxen flume
A = T(x)/x?
lavish jewel
what is T([1;0])
limber sierra
A = T(x)/x?
you cant divide by a vector
so this doesnt make sense
but you can solve T(x) = Ax through the process Edd is outlining.
waxen flume
yeah..
lavish jewel
how do you find the projection of a vector a onto a vector b?