S is just a subset
#linear-algebra
gray dust
show me where the book says a subset must be independent
gray dust
S is just a subset
wintry steppe
Yeah, but if S is a subspace of V, it has to be equal to L(S)?
gray dust
yes
wintry steppe
Why?
gray dust
use ur def of subspace
wintry steppe
A subspace S of V is a subset of V which contains 0, closed under addition, and scalar multiplication
gray dust
hence closed under finite linear combos right?
wintry steppe
Huh I guess it must equal the span then, since the span is just all linear combinations...
So we only really use span for a subset that isn't a subspace? Because otherwise we could just say S
wintry steppe
Yes I do
gray dust
so u can take the span of a subspace but indeed theres 'redundancy' in that u get the same set
vast needle
hey, I am not sure if this falls into this section but I'm trying to prove that given these two d-dimensional vectors, I can prove these given distance functions are a metric
I tried going to my TA's but they're all confused, and me and my peers are also completely lost
any form of guidance/resources would be greatly appreciated!
keen sierra
I tried going to my TA's but they're all confused, and me and my peers are also ...
Your TA's are confused? They don't know the answers? That seems hard to believe.
vast needle
Your TA's are confused? They don't know the answers? That seems hard to believe.
we have a new professor teaching this course and the way he is teaching isn't like the old professor. So the TA's have not really had experience with these types of questions
keen sierra
For (c), the triangle inequality fails to hold. There's a simple counterexample with d=1. Consider x=0 and y=2.
The first two are indeed metrics, so instead of seeking a counterexample, you will need to prove that the conditions for a metric are satisfied.
vast needle
bungo, you are a literal life saver
keen sierra
(b) is the harder one to prove. You will need the Cauchy-Schwarz inequality.
vast needle
do you know any good resources I could consult to see how to prove?
teal grotto
(b) is the harder one to prove. You will need the Cauchy-Schwarz inequality.
need a bit more than that
keen sierra
need a bit more than that
Yeah, I didn't mean to imply that it's all you need. 😁
keen sierra
do you know any good resources I could consult to see how to prove?
The first one is called the L1 (or taxicab) metric and the second is the L2 (or euclidean) metric, searching for these terms might help.
wintry steppe
http://mathb.in/69878 Can someone review my work?
vast needle
The first one is called the L1 (or taxicab) metric and the second is the L2 (or ...
I'll look into that, I'm also just confused what the L22 means, my prof didn't really explain it. I'm sorry if this is a basic concept
keen sierra
I'll look into that, I'm also just confused what the L22 means, my prof didn't r...
A general class of metrics is of the form $$\left(\sum_{i=1}^{d}|x_i - y_i|^{p}\right)^{1/p}.$$ where $p \geq 1$. These are called the $L^p$ metrics. Your (a) is the special case where $p=1$ and your (b) is the special case where $p=2$.
vast needle
A general class of metrics is of the form $$\left(\sum_{i=1}^{d}|x_i - y_i|^{p}\...
Appreciate the help! I'll look around online and do some of my own research. If you still happen to be around when I finish this problem do you think I could show you my work?
stoic pythonBOT
OurBelovedBungo
keen sierra
Appreciate the help! I'll look around online and do some of my own research. If ...
Sure, feel free to ping me. If I'm not around, I'm sure others can help.
vast needle
great thank you!
wintry steppe
hey all quick question
what is the second part of this asking for
i got the whole 'multiply AB and take the inverse'
but what is it saying by computing A^-1 and B^-1 and using ur previous answer
lean spoke
you know (AB)^-1 = (B^-1)A^-1
second part thinks you knew that and commands you to get the same answer from this
@wintry steppe
lean spoke
my pleasure
ripe birch
So I got these arbitrary matrices. It's given that they meet the criteria of the first line relating to the delta function. It's given that they are Hermitian and "i" runs from 1-4. The other lines were me starting to try to solve the problem but I'm at a bit of a loss as it's been 3 years since LA.
Question is trying to show that all eigenvalues of each gamma matrix is +/- 1. How is this possible given what we have here? Seems too arbitrary
teal grotto
i would emphasize that u are assuming that U is a subspace iff h is in U iff int h = b = 2b iff b = 0
the chain of implications needs to start off somewhere
hardy inlet
so like im looking at some stuff online and its a mess of proof by contradiction
it makes sense that its true in the reverse direction
but the forward direction is where im losing my marbes with the contradiction
distant schooner
i would emphasize that u are assuming that U is a subspace iff h is in U iff int...
ah okay that makes sense thanks
teal grotto
so like im looking at some stuff online and its a mess of proof by contradiction
if U is not a subspace of W and W is not a subspace of U, then there are vectors u in U and w in W with u not in W and w not in W.
if u + w is in the union of U and W, then u + w is in U or W.
if u + w is in U, then u + w + (-u) = w is in U, contradiction
if u + w is in W, then u + w + (-w) = u is in W, contradiction
teal grotto
fax indeed
teal grotto
looks good
hardy inlet
this one smells 🐟 , like can I take the cardinality of a vector space?
teal grotto
V is non-trivial, so there is some non-zero vector v in V
{0_V} can't possibly be equal to V
hardy inlet
i miss my real analysis...
distant schooner
i just said that the additive inverse of a in one subspace must exist by definition, but if it does then b must exist because (a + b) should be a member of the union and with (a + b) + (-a), b must also be a member of that subspace (and vise versa for b and the other subspace)
distant schooner
i miss my real analysis...
lmao, what course is this?
hardy inlet
advanced LA
distant schooner
ah okay
distant schooner
yeah it is
proper cradle
How to right finite dimensional vector space over field in union of proper subspaces of V? Or it is true only for finite fields?
hardy inlet
if U is not a subspace of W and W is not a subspace of U, then there are vectors...
I need to do some tweaking and understanding here. like i sorta get it
we start by assuming that they are disjoint sets
and then the contradiction shows that the element has to be in both sets, thus a contradiction
it just feels fuzzy and like im missing some details. Like is this all I need for the forward direction?
teal grotto
we start by assuming that they are disjoint sets
no
hardy inlet
well in the contradiciton we're assuming that right
teal grotto
not disjoint, just that they are not contained in each other
ex) take the intervals (-2,1) and (-1,2). neither is contained in each other, but they are not disjoint
hardy inlet
oh
right
different thigns
sigh
so there exists at least one element from each that aren't in the other, but our contradiction says that it must be in the other
teal grotto
yea, we are really proving the contrapositive
hardy inlet
like are we "proving the contrapositive"
teal grotto
no need for contradiction
hardy inlet
proving the contrapositive by contradiction
hardy inlet
am i 'allowed' to simplify the words "U is not in the subspace of W" to "U \not\subseteq W"?
im hesitant bc subset and subspace not same
teal grotto
U and W are both subspaces of V so its fine here
hardy inlet
ok thats what I was hoping for but didn't know if it broke math
and do we need it to be a proper subset?
a lot of the proofs i was looking at are proper
teal grotto
nah, doesnt matter here
hardy inlet
also are the logical and and ors acceptable in non-discrete math?
or ig non boolean
and should these commas be ORs (nvm those are guaranteed at least one true since non containing)
teal grotto
it doesnt have to be all symbols, but if you want it to, it would be
$(\exists u\in U:u\not\in W)\wedge(\exists w\in W:w\not\in U)$
stoic pythonBOT
c squared
teal grotto
its fine as written now tho
hardy inlet
but if i get rid of the parenthesis it works
ok im done touching the forward direction. Is the final therefore safe?
aight bet. Now can we just say the reverse by definition or something? I know its supposed to be easy
teal grotto
if U subset W or W subset U, what is U union W?
hardy inlet
the larger
hardy inlet
is what i have atm
this is where we use the definitions of W and U right
doctor do we have the all clear? (after moving the forward/backwards arrows to the lhs)
teal grotto
all clear
hardy inlet
Epic. thank you c²
teal grotto
yw MattDog_222
cinder meteor
Just got to subspaces and vector spaces in my class, the logic looks sounds to me
Why does everything in linear algebra sounds like something out of a sci-fi movie
hardy inlet
i mean sci fi movies are made from linear algebra so
teal grotto
Why does everything in linear algebra sounds like something out of a sci-fi movi...
example?
hardy inlet
"Quick! Grab ur eigenvectors!"
teal grotto
alr, that was good
cinder meteor
“The alien morphed into its row echelon form”
cinder meteor
oblique prairie
not to minimod but the mods are gonna get mad if memes keep getting posted here
lavish jewel
indeed
oblique prairie
edd are you angry now
hardy inlet
aight math. for (a) i just need to show identity, closed under + and *?
lavish jewel
wdym by identity
cinder meteor
And that the 0 vector/matrix/whatever exists, but yeah
hardy inlet
additive identity 0
cinder meteor
Oh wait thats identity
hardy inlet
wdym by identity
group theory shaking intensifies
Not latexing matrices if its wrong, but it this good (albeit out of natural ordering)
lavish jewel
looks ok, maybe make a comment on the form of the sum
like "which is of tje form..."
hardy inlet
is \in W_1 not a sufficient claim?
hardy inlet
ok
I have literally no idea what part B means
like its asking for the intersection,but the heck
lavish jewel
what form does it have
hardy inlet
x -x
-x y?
lavish jewel
yeah
hardy inlet
so is (c) asking to show that W1+W2 spans M_{2,2}?
how is that possible
isn't it isomorphic to R^4?
or is it only a homomorphism
teal grotto
isomorphism
hardy inlet
so how can 2 vectors span R4
teal grotto
but its fine. you're just decomposing R^4 into the sum of two subspaces
W1 and W2 are subspaces of M_{2,2}
you just want to show that any vector M in M_{2,2} can be written as the sum of a vector in W1 and a vector in W2
hardy inlet
so i need <a,b,c,d> = xW_1 + yW_2?
teal grotto
idk what that means
hardy inlet
r4 vector = constant * W_1 + constant * W_2,
actually that probably isn't a constant
teal grotto
r4 vector = constant * W_1 + constant * W_2,
its ambiguous as stated rn, W1 and W2 are vector sub-spaces, not vectors
show that if m =
a b
c d
is a two by two matrix, then there is a matrix w_1 =
x -x
y z
in W1 and a matrix w_2 =
e f
-e g
in W2 such that
m = w_1 + w_2
hardy inlet
ok so with the constants the 4th coordinate is a given. so we just need to make the other 3 uniquely
since the 4th is independent of the other cells
hardy inlet
why not 0?
teal grotto
idk, just seems less trivial
lavish jewel
to come back to your question
the matrices are written with different coefficients
you can decompose them into the canonical basis
hardy inlet
lavish jewel
or some basis
hardy inlet
is what i made at the bottom of that image a basis
lavish jewel
e.g. if it helps you, [a, -a; b, c] = a[1, -1; 0, 0] + b[0, 0; 1, 0] + c[0, 0; 0, 1]
so that type of matrix forms a dim 3 subspace
hardy inlet
oh wait i can have more than 2 matrices 🤦
its been a while
we've not covered dimension and ranks yet in the course
lavish jewel
i see
hardy inlet
idk if my wxyz example answers the question, l ike is showing a particular arrangement enough to show W1+W2 is V
Does this get me anywhere
lavish jewel
yes, that's how i would've done it
hardy inlet
so i wrote
v_1 = a + x
v_2 = b - x
v_3 = -a + y
v_4 = c + z
which is similar to this in different orders
well different letters
do i need to zero out the duplicates, such as z and c being direct duplicates?
lavish jewel
i don't know what you mean by that
hardy inlet
like if we remove z[0,0,0,1] from the span we still span M_2,2
so we can 'throw it away'
lavish jewel
what you can do is substitute c + z by another variable
if you "throw it away" you're fixing a value of 0 in one of the matrices
this method is in general wrong
hardy inlet
so i just leave it as c+z
viral magnet
<@&286206848099549185>
you have to share the full question if you want help
hardy inlet
teal grotto
dont know where to proceed
i really hate that you are equating vectors with vector spaces
type mismatches 🤢
viral magnet
i really hate that you are equating vectors with vector spaces
also the variable names got swapped?
hardy inlet
🤦 yes. i've been up like 20 hours
viral magnet
dont know where to proceed
which part are you stuck on
hardy inlet
well i was stuck on making it some weird linear combination but i dont even think thats a valid approach since its a vector space
teal grotto
i thought we already solved this
teal grotto
up here. take e = 0 = g (might have gotten the letters mixed up)
teal grotto
why not?
hardy inlet
idk they said "this method is in general wrong"
hardy inlet
idk they never elaborated
teal grotto
um
well, you really just have to show that V is a subset of W1 + W2
or you can do a dimension count (or give an explicit isomorphism, there are a lot of ways to show that V = W1 + W2)
teal grotto
well, you really just have to show that V is a subset of W1 + W2
this is probably the most accessible
hardy inlet
like show V is a subset of w1+w2 and w1+w2 is a subset of V therefore V = w1+w2?
teal grotto
yes
teal grotto
but we already did that up here. take d = 0 = f
then just solve for a, b, c, e in terms of w, x, y, z
hardy inlet
variable names keep changing but anyways, i've reset to the original variables in the problem
hardy inlet
yeah it pains me that idk what that means
teal grotto
if W1 and W2 are subspaces of V, then W1 + W2 = {w1 + w2 : w1 in W1 and w2 in W2}
an element (called a vector) w of W1 + W2 can be written as w = w1 + w2 for some vectors w1 in W1 and w2 in W2
it makes no sense to write W1 = [a b c d] since W1 is not a vector, it is a vector space whose elements are vectors
lavish jewel
mattdog, this is why i rewrote (several messages back) the matrix as a sum of matrices that are 0 almost everywhere
viral magnet
variable names keep changing but anyways, i've reset to the original variables i...
You're typing this latex?
lavish jewel
to make it clearer to you that it refers to some element produced through a linear combination
viral magnet
$\begin{bmatrix}
x & -x \
y & z
\end{bmatrix}$
stoic pythonBOT
riemann
hardy inlet
yeah im using latex
im looking at my book and cant find jack
we never really cover anything useful in class
viral magnet
you're just solving a system of linear equations
hardy inlet
made up of what
viral magnet
an element (called a vector) w of W1 + W2 can be written as w = w1 + w2 for some...
this will be a system of linear equations very similar to yours
hardy inlet
ok but where are you reading this
teal grotto
variable names keep changing but anyways, i've reset to the original variables i...
idk what you mean. im just answering question (c)
hardy inlet
yeah but u said "as defined in your book" where do you see a V \subseteq W_1 + W_2
theres not even a subset symbol
teal grotto
W1 + W2 is defined in your book
definition 1.36
our goal was just to show that V is a subset of W1 + W2
teal grotto
**c squared**
and this does exactly that
hardy inlet
yeah but where are you even deriving this equation
teal grotto
what equation?
hardy inlet
v1 -v1
v3 v4
teal grotto
im just setting up one vector in W1 to add it to another vector in W2 to get back v
the process behind it is as follows
hardy inlet
where is this methodology from. I see nothing close in the pages
like i sorta see whats happening but its not described
stoic pythonBOT
c squared
teal grotto
just play around with the equations at the bottom now
hardy inlet
so is the only different that I had W1 not w1 and column vectors instead of matrix notation?
er well u didn't split
teal grotto
you seemed to for some reason not understand what i was doing. im just explaining my reasoning
hardy inlet
you seemed to for some reason not understand what i was doing. im just explainin...
this?
teal grotto
**c squared**
this was really all scratch work. i wouldn't include it in my proof without also including commentary around it
hardy inlet
I mean i had something like that with different variable names
hardy inlet
oof
teal grotto
there's no context/commentary supporting why you want those two things to be equal
hardy inlet
so i gotta show w1+w1 \sub V
teal grotto
other way
hardy inlet
but didn't we just show that $V \subseteq W_1 + W_2$
stoic pythonBOT
MattDog_222
zinc timber
are v_1, v_2, ... vectors here?
teal grotto
scalars
teal grotto
ye
teal grotto
**MattDog\_222**
my bad, didnt know you were trying to finish the proof
should be clear that W1 + W2 is a subset of V
hardy inlet
oh cause they're subspaces to begin with
therefore combining 2 subspaces is never larger than the original vectorspace?
teal grotto
Yeah just *verify* 4head
just from definition 1.36
teal grotto
alr, well forget it if you cant see it rn.
if w is in W1 + W2, then w = w1 + w2 for some w1 in W1 and w2 in W2
both w1 and w2 are in V
and because V is a vector space, then ...
hardy inlet
w in V? but why does it matter that V is a vector space
teal grotto
because addition is closed in V. the sum of two vectors in V is again a vector in V
hardy inlet
right, and those 2 vectors are w1 w2
teal grotto
right
teal grotto
W1 + W2 subset V
teal grotto
dude ur good. gts lol
hardy inlet
gts?
teal grotto
go to sleep
surreal otter
I´m hoping this goes here, because I have no idea where to put it. For the generators of a Lie algebra, are they one per continuous parameter? as in, are they proportional? or can a parameter have several generators? Was wondering.
nocturne jewel
yes
subtle gust
That can't be reduced to R REF
nocturne jewel
No, every matrix has an RREF
subtle gust
So you can always have leading 1s with 0s above and below them?
nocturne jewel
yes
wintry steppe
http://mathb.in/69878 Can someone review my work?
bump
heavy crown
Anyone?
For notation: A^(H) = Conjugate(transpose(A))
Given A is not the zero matrix, A is of order n x n and it is complex.
Assume A^(H) * Av=0 and prove Av=0.
--
If that helps i found that A*A^(H) is normal
but I got stuck
zinc timber
are you asking $A^*$ is non zero and $A^*Av = 0$ then show $Av=0$?
stoic pythonBOT
heavy crown
yes
zinc timber
because ^H notation is a little confusing to me
heavy crown
yea sorry I just didn't know how to put this as notation here
we usually put * in above the letter, so A* = Conjugate(transpose(A))
instead of the H
so assume A*Av=0 and prove Av=0
stoic pythonBOT
zinc timber
is this hint good enough?
heavy crown
The right side is equal to 0 because A*Av is zero right
zinc timber
yes
zinc timber
heavy crown
awesome
solemn lotus
anybody know any general interesting problems/theorems in LA to attempt to prove?
dusky epoch
try proving that the eigenvectors of a symmetric matrix are orthogonal
lavish jewel
how easy or hard?
zinc timber
ok so on a serious note, you can try 'Berkeley problem book', problems are good
lavish jewel
a very elementary one is that orthogonal vectors are linearly independent
solemn lotus
ive done that a few times on 5 different tests
lavish jewel
show that the pseudo inverse is an orthogonal projection onto the row space of a matrix?
solemn lotus
oof ty ill try that out
outer goblet
f(t)+g(t)=0 -> ln(f(t))+ln(g(t))=0 -> rt+st=0
wont it be true if r=-s
or does r=s also mean r=-s
zinc timber
what is F(R, R)?
zinc timber
you can try something similar to wronskian
wintry steppe
Let $V$ be a finite-dimensional linear space, and let $S$ be a subspace of $V$. Prove that $S$ is finite dimensional and $\dim S \le \dim V$.
$V$ is finite dimensional so it has a basis ${e_1,\dots,e_n}$. Similarly $S$ is a subspace of $V$ so it has a basis ${e_1,\dots,e_k}$. If $k>n$, then by a previous theorem which states that a set which has $n$ vectors that form a basis yet the cardinality of the set is $n+1$ is dependent, we conclude that ${e_1,\dots,e_k}$ would not be a basis, which implies that $k\le n$, and so $S$ is finite dimensional and $\dim V = n \le \dim S = k$
stoic pythonBOT
n/c
wintry steppe
Does this proof work?
outer goblet
its not for my question right?
zinc timber
Let $V$ be a finite-dimensional linear space, and let $S$ be a subspace of $V$. ...
check the last line
zinc timber
ig it's alright, unless someone finds a logical error which is invisible to me
subtle walrus
why is the basis of S finite
zinc timber
ah yes, nice
subtle walrus
considering thats part of the question and you present no argument, i think its incomplete
wintry steppe
@subtle walrus well because k ≤ n, and n is a natural number
(and k is natural too)
subtle walrus
what is k
wintry steppe
k is a natural number
subtle walrus
ok? how is it related to the argument
you just claim that S has a basis of k elements, but why
wintry steppe
Because it is a subspace of V
subtle walrus
thats circular
the first part of this proof is to show that a finite dimensional vector space cannot have a infinite dimensional subspace
so you kinda cant use that
(you can also ask for a hint if you cant think of an argument)
wintry steppe
Okay, let $B$ be a basis for $S$, so $B$ is independent and $L(B) = S$. Assume that $S$ is infinite-dimensional, which implies that $B$ is not a finite set of elements. However, $S$ is a subspace of $V$ which has basis ${e_1,\dots,e_n}$, and therefore $|B| > n$, yet a previous theorem states that given a set of $n$ independent elements of a vector space V, any n + 1 elements from the span of that set will be dependent. Therefore, $B$ is dependent which contradicts that it must be a basis for $S$, and so $S$ can't be infinite dimensional
stoic pythonBOT
n/c
wintry steppe
(you can also ask for a hint if you cant think of an argument)
does that work?
subtle walrus
L is span?
wintry steppe
Yes
subtle walrus
by independent you mean linearly independent?
wintry steppe
Yes
subtle walrus
the wording is a bit weird, specifically "from the span of that set" but i think you have the right idea
this also solves the exercise immediately
wintry steppe
Right basically I'm saying that |B| > n, so for example |B| is definitely at least as large as n + 1, (being infinite and all), so with B being a basis we should have n + 1 elements in the span of B that are independent, yet it contradicts the theorem
wintry steppe
Okay, it's just that apostol doesn't use linearly independent, just independent which is a bit strange I guess
subtle walrus
hm ok
wintry steppe
Axler does specifically use linearly independent
subtle walrus
there are other notions of independence so maybe i am nitpicky
they probably dont appear in a book on linear algebra
wintry steppe
Oh well, the subchapter is titled dependent and independent sets in a linear vector space
So that might limit it
subtle walrus
f(t)+g(t)=0 -> ln(f(t))+ln(g(t))=0 -> rt+st=0
how does the first -> follow?
outer goblet
how does the first -> follow?
cant i just do the logarithm
wont it count as scalar multiplication
subtle walrus
no
outer goblet
:(
subtle walrus
you have to show $\lambda_1e^{rt} + \lambda_2e^{st} = 0$ implies $\lambda_1 = 0 = \lambda 2$
stoic pythonBOT
Lochverstärker
outer goblet
ah
subtle walrus
in no world is "applying ln" scalar multiplication
scalars come from some field, and multiplication is well, generally multiplication
outer goblet
how would i show it then :8
subtle walrus
you try to work with $\lambda_1e^{rt} + \lambda_2e^{st} = 0$ or equivalently $\lambda_1e^{rt} = - \lambda_2e^{st}$ and see what kind of things this implies
stoic pythonBOT
Lochverstärker
outer goblet
yeah i tried that too but idk how that helps really
subtle walrus
hint: ||the exponential function has no roots in R||
outer goblet
yeah i thought about it, so e^st=-e^rt is never possible
but what if lambda_2 is just -1
subtle walrus
can you find $s \neq r$ such that $e^{st} = e^{rt}$?
stoic pythonBOT
Lochverstärker
subtle walrus
in fact you cannot
subtle walrus
hint: ||the exponential function has no roots in R||
why do you think i mention this hint?
gray dust
one can also use ||exp is injective & plug some t values||
gray dust
the way i went is ||plug t=0,1||
outer goblet
but isnt it supposed to apply for all t values
gray dust
yes so in particular it holds for ||0,1||
outer goblet
yeah but t=0.1 but what is r and s
outer goblet
i dont get what you mean for t=0 its just a=-b
ah
and the same a and b has to apply for all t
right?
outer goblet
yes
but like
a*e^rt+ b *e^st=0
do a and b have to be unchanged and fullfill it for all t
gray dust
a,b are constant. this eqn holds for all t
outer goblet
so
a e^rt = -b e^st , look at t=0 -> a=-b -> a e^rt = a e^st -> observe that only true for r=s
is this correct?
gray dust
justify ‘observe..’
gray dust
u divided by 0
wintry steppe
For an inner product, why is (z,z) ≥ 0 when defined axiomatically?
Where z = ax + by, for scalars a,b and nonzero vectors x, y in V
(z, z) >= 0 because that's part of the definition of an inner product
surely you're not asking why a definition is true
It just that it is used to prove the cauchy-schwarz inequality which states that |(x,y)|^2 ≤ (x,x)(y,y) however we haven't even defined | | yet
that means absolute value
I know
This is the proof they do, (z,z) = (ax + by, ax + by) = (ax, ax) + (ax, by) + (by, ax) + (by, by) = aa*(x, x) + ab*(x, y) + ba*(y, x) + bb*(y, y) ≥ 0
Where * represents the hermitian symmetry I guess?
Or companion relation
complex conjugation.
Yes
sick sandal
whats the difference between similar and congruent matrices? for me it seems the latter is identical to similar matrices just moved to bilinear forms (not implying transpose = inverse just saying they act the the same)
maybe a geometric view might help illustrate the idea better 
hardy inlet
this is false, correct? Im thinking for the counterexample let \
$U_1 = { (x, 0) }\
U_2 = {(0, y)}\
W = {(x, x)}\
V \equiv \mathbb{R}^2$
stoic pythonBOT
MattDog_222
lavish jewel
is \oplus here sum of subspaces or direct sum?
hardy inlet
Direct
lavish jewel
your example doesn't work then
lavish jewel
that much is true, but that sum is not direct
hardy inlet
isn't the intersection 0 tho?
lavish jewel
the dimension of a direct sum is equal to the sum of the dimensions of the subspaces involved
wait i'm a doofus, i was reading W as {(x,y)} this whole time
hardy inlet
yeah i thought saying {x,x} was a bad idea for r2
should I say like {z,z} for clarity
or like
a,0
0,b
c,c
hardy inlet
so then does that form a counterexample when reread
lavish jewel
mhm
hardy inlet
also when writing the sets should i use = or \equiv
lavish jewel
i think it's ok as it is
lavish jewel
i think this one is straightforward enough that you can just claim it. or if you really want, show it for U_1 + W and say it's similar for U_2 + W
should be simple enough to do it in 1 or 2 lines
hardy inlet
so i'll prob add the U_1 + W = V and similarly for U_2, but aside from that does this make sense?
lavish jewel
looks ok to me, idk if u want a separate symbol for the 0 vector, for clarity
hardy inlet
the book used {0} so i'll leave that alone, although i personally prefer \vec{0}
ok does this properly show what u were saying (changed conclusion)
lavish jewel
seems pretty clear to me. admittedly, i'm bad at this stuff cuz i haven't dealt with it much. maybe others will say it's too verbose for their taste, but looks good to me.
hardy inlet
verbose meaning too much?
is this pretty much what I literally just did for showing U1 + W = V? just an extra coordinate?
like just a raw computation
lavish jewel
yeah, looks like it
but if they're asking for it there, then they also probably wanted it in the previous one
hardy inlet
yeah i'll ask the guy if my 'similarly' is acceptable 😂
(i mean its literally the same thing with (0, y-y)
well no
uh
lavish jewel
(0, y-x) and (x,x)?
lavish jewel
yeah. i would say that's certainly within the realm of "similarly" 😛
hardy inlet
i used matrices augmented with I_3 from the prereq to prove something last HW and he said I cant do that bc we haven't gotten there so Imma just type it explicitly 😅
lavish jewel
hardy inlet
i mean i got partial credit cause it was right but the wrong method
he's a picky boi
lookin 🔥 ?
idk what to make of this. Like it implies v1:v4 are L.I. , so adding or subtracting from one another cannot create a zero vector or change the span. Like it makes perfect sense that its true just how to elaborate that
but wait it doesn't say they're linearly independent, just that they span... :/
lavish jewel
then just show you can obtain the original vectors from these
zinc timber
ye those replacement lemmas
lean spoke
hi, how can i do these 2
wintry steppe
to find the RSS its just XTX * XTY for beta hat right?
stoic pythonBOT
wintry steppe
wait sorry
im just a bit confused on how to visualize the 'solve' function on R
bc i have this for RFF
RSS
sorry
and i dont know how to understand what solve is doing
zinc timber
idk R so can't decode, sorry
wintry steppe
no worries
wintry steppe
did you mean $(X^TX)^{-1}X^Ty$?
ah ok so this is beta hat then
hardy inlet
then just show you can obtain the original vectors from these
so do I start with ${v_1,v_2,v_3,v_4}$ and arrive at ${v_1-v_2, v_1+v_2, v_3+v_4, v_4}$ or the other way
stoic pythonBOT
MattDog_222
lavish jewel
other way unless you wanna show the operation is invertible
hardy inlet
so is that a valid operation/step?
subtle walrus
i cannot parse this
dapper gorge
Let $T$ be a linear operator on a vector space $V$. Let $V_\lambda$ be the subspace of eigenvectors associated with $\lambda$. We show that for any $v\in V_\lambda\setminus {0}$, $(T-\lambda)^m(v)\neq 0$ for all $m\in\mathbf{N}$. Use strong induction on $m$. For the base case, take $m=0$, so that $(T-\lambda)^0(v)=v$ for all $v\in V_\lambda\setminus{0}$. For the inductive step, assume the statement is true for all $\leq m$. We have $(T-\lambda)^{m+1}(v)=(T-\lambda)^m((T-\lambda)(v))$, where $(T-\lambda)(v)\in V_\lambda$ is nonzero by the inductive hypothesis. And we can apply the inductive hypothesis again for $(T-\lambda)^m$.
stoic pythonBOT
Croqueta
dapper gorge
This is very silly, I know.
I can't see what is wrong with it to be honest, although it is of course complete nonsense, as $(T-\lambda)(v)=0$.
stoic pythonBOT
Croqueta
dapper gorge
This is a question about induction
dapper gorge
you use strong induction
so assume it's true for 0,...,m
btw, I don't really know how strong induction works (that's why I'm asking, so maybe the mistake is just very obvious)
subtle walrus
yes but (T-\lamba)^1 = (T-\lambda)^0 \circ (T-\lambda)^0 or what?
dapper gorge
no
subtle walrus
so your inductive step assumes m>1
so there is no way you get from m=0 to m=1
if you did manage to prove it for m=1, the induction works out
dapper gorge
the inductive hypothesis gives $(T-\lambda)(v)\neq 0$, since $0\leq 1\leq m$ (?)
stoic pythonBOT
Croqueta
dapper gorge
ok I think I se what you are saying
subtle walrus
this is the "all sheep are black proof"
you assume that you can write (T-\lambda)^m = (T-\lambda) \circ (T-\lambda)^{m-1} but this is only true for m > 1
dapper gorge
so there is no way you get from m=0 to m=1
if you did manage to prove it for m=1...
Yeah, I applied this same argument (starting at m=1) to show the intersection of generalized eigenspaces is zero, but as it seemed too easy, I tried non sense and I arrived at this
yeah I see
if m were 0, the inductive hypothesis wouldn't apply
Okay, thanks
hardy inlet
is this proper/correct?
subtle walrus
looks good but it should be \subset and not \in for your basis
(or remove the curly braces)
also i have to mention this only works in characterisitc not being 2 please ignore this
winter flume
Any sites that are good for self teaching linear algebra?
proud gate
Can someone please check my computation over here: 1/p + 1/q = 1.. I am literally brain farting and can not figure out which step of the computation went wrong..
The answer should've been |x|^2 instead of 1
ripe birch
@winter flume KhanAcademy i believe has a decent LA section. Or you can always go my favorite route and use /sci/ wiki's recommendations for books and get them from LibGen
Also, I've got a problem here. I have 4 matrices $A_i$ where i runs from 1-4, they meet the following criteria:
1: They anticommute in such a way that ${A_i, A_j} = 2\delta_{ij}$
2: $Tr(A_i) = 0$
I am to try to prove that these operators can only act on even-dimensional spaces. How in the world is that possible?
stoic pythonBOT
Ferret
ripe birch
ugh. i guess my LaTeX sucks. That's supposed to be a kronecker delta so it's 0 when i =/= j and 1 when i = j
if it helps, i've also shown that A_i's eigenvalues are equal to +/- 1
vast needle
do we need to use matrices to get eigenvectors??
because I just spent a ton of time doing it with equations, and don't wanna have to redo it again if I don't need to
limber sierra
an eigenvector is just a vector v that satisfies Tv = lambda v for an eigenvalue lambda
you can represent T as a matrix or a more conventional function
and either representation can let you compute the eigenvectors
vast needle
oh thank god
zinc timber
Also, I've got a problem here. I have 4 matrices $A_i$ where i runs from 1-4, th...
does {Ai, Aj} mean their commutator?
ripe birch
oh i ended up figuring it out @zinc timber. That's their anti-commutator though. Ended up being "A is hermitian, so it's diagonalizable. It's diagonaled Trace is a summation of N eigenvalues. W/ eigenvalues of +/- 1 we need even N entries (since Trace has to be 0) to get 0."
hearty rapids
so the way you check if a vector space is indeed a vector space is by checking if it's closed?
shit, prof dave might be even better than trev tutor
gonna do basis and dimension tomorrow
after that linear regression and orthogonalization
sick sandal
to check if a set is a vector space you need to check if the axioms that define a vector space are met
when you say closed i assume you meant closed under addition and scalar multiplication
should also mention you get most of the axioms for free when working over a field like R but thats not always the case
hearty rapids
to check if a **set** is a vector space you need to check if the axioms that def...
yes closed under addition and scalar multiplication
wdym by "most of the axioms for free"?
like they are automatically true?
sick sandal
pretty much
for example addition is commutative and associative naturally by definition of a field
well considering you do prove the set is closed under addition and scalar multiplication over R (or any field K)
then you get them
wintry steppe
hello
gray dust
like they are automatically true?
no, one technically must check all vector space rules, but if theyre satisfied its often due to the field structure
eg addition in R^n commutes bc addition in R commutes
pure tangle
Hi could someone please check my steps for a problem. I have a 2x5 augmented matrix and I’m asked to make the row 1 value 2 and row 2 value 4 variables pivots. Would I just make column 1 0’s and then get the values above and below x2 and x4 to be 0s?
trim galleon
show example
pure tangle
So I have this matrix and I’m asked to find a basic solution by pivoting x2 in 1 and x4 in 2
Sorry for some reason the image didn’t upload
3x1 - x2 + x3 -2x4 = -2
X1 - 2x2 -x3 +x4 = 1
So would I just get zeros in the first column
And then a zero below row 1 column 2
And a zero above row 2 column 4
hearty rapids
no, one technically must check all vector space rules, but if theyre satisfied i...
i see
halcyon spindle
So I have this matrix and I’m asked to find a basic solution by pivoting x2 in 1...
You just need to kill(make it zero) the entries below first entry of column 1 , then your basically done. Doing that will get you REF so you can do a back substitution. You will end up having 2 free variables x_3 and x_4.
pure tangle
Thanks for the response. So if I wanted my pivots to be x1 x2 would this be the right solution:
halcyon spindle
A little bit more work but yeah from here you should be able to find the solutions.
If you want to find x_1 and x_2 by reading it up the augmented matrix you need to scale be (1/3) for row 1.
Wait hmm.
halcyon spindle
x_3 and x_4 can be anything, I usually just set it to 0.
pure tangle
Mb I’m a little confused. If I want x2 and x4 to be pivots won’t they be 1
halcyon spindle
Ah wait sorry I misread you, hold on a sec.
I am a bit confused now, you want the column corresponding to x_2 and x_4 to be pivots that is not possible, since 3 in column 1 corresponding to x_1 is a pivot, then 1 in column 2 corresponding to x_2 would be our second pivot. We cannot have anymore pivots after that.
pure tangle
Here’s the wording:
Find a basic soluiton by pivoting x2 in equation (1) and x4 in equation (2).
halcyon spindle
Yeah I am not sure, I need to reread that section in my book again to see if I missed something.
pure tangle
Idk if this makes a difference but this is from an operations research class
heavy crown
is there any theorem or like something I can tell, about a matrix that is diagonalizable over R ?
If a matrix A is diagonalizable over R by a unitary matrix , so I can say that A is real and symmetric, heremetian, normal.
but in case it's just diagonalizable not by unitary matrix can I say anything about A?
fringe fjord
Not much. It's a sufficient condition for an n×n matrix to be diagonalizable that it has n different eigenvalues (and there are a lot of matrices with that property), but that is not necessary.
lavish jewel
yeah, only properties related to being diagonalizable, if you have no other info
like comment on the geometric and algebraic multiplicity
heavy crown
okay good to know, thank you two )
void meteor
I'm working on a gram schmidt orthogonalization algorithm rn. There is a line under my second for loop that is Q[:,k]=Qj@Q[:,k] which goes through each column of k and makes it orthogonal to the column qj. Does anyone know what happens if Qj@Q[:,k] = 0, and any tips on what the algorithm should do if that happens?
dusky epoch
nothing, Q[:,k] is just already orthogonal to qj in that case
er
mb
i meant that Q[:,k] is parallel to qj
so it would have zeroed out anyway
pure tangle
Could someone please help me understand how to “choose” a column to pivot on? I thought pivot columns just occurred. Here’s the wording of the question and it’s a 2x5 augmented matrix. This is in relation to operations research if that helps
Find a basic soluiton by pivoting x2 in equation (1) and x4 in equation (2).
<@&286206848099549185>
spare widget
I'm working on a gram schmidt orthogonalization algorithm rn. There is a line u...
If you get a 0 throw that vector away, it's linearly dependent on the others and continue with the remaining
tulip quiver
,rotate
proven lotus
What would be the fastest way to solve this set of 3 equation, knowing that the marked numbers are the same
stoic pythonBOT
proven lotus
Yup, that's also an option, but isn't there a shorter method whenver those marked numbers are the same?
I need to solve a lot of these without a calculator
And everytime, those numbers will be the same
spare widget
The matrix is symmetric, you can do cholesky if you want
But tbh for a 3x3 matrix even direct inversion is easy
i.e. cramer's rule is 3 cross products and 1 dot
spare widget
I need to solve a lot of these without a calculator
Invert the matrix then
Diagonal is not the same?
proven lotus
The main diagonal is never the same, but the others ( from left to right, bottom to top) are always the same
proven lotus
Q^T * D * Q decomposition
Sorry could you tell me how this would work exactly? I'm an engineering student and I haven't seen this method yet
spare widget
whatever just set variables a,b,c on the diagonal and compute the inverse
proven lotus
On the internet I found something with "LU dec"
spare widget
You know how to compute inverse using row reduction?
proven lotus
Transpose?
spare widget
Do you know how to compute the inverse using row reduction?
proven lotus
I know how to do rref, yes
spare widget
Then do it by substituting a,b,c on the diagonal
Or you can directly compute the adjoint terms since 2x2 dets are easy
Then you'll have the inverse in tetms of a,b,c
Then you can plug in any abc
proven lotus
Okay, I'll try this method rn, thanks for your help!
spare widget
As for the Q^T * D * Q thing
That would be the eigendecomposition
I doubt you want to solve potentially cubic polynomials though
Also once you get the determinant in terms of a,b,c
Check for what abc it's equal to 0
proven lotus
It sounds interesting but I'm in my first year of uni. We haven't seen any advanced mathematics yet
spare widget
Because those make the matrix singular
proven lotus
Okay, let me try
spare widget
for 3x3 matrices det = dot(r1, cross(r2,r3)) btw
The adjoint elements can also be formed by cross(r2,r3), cross(r3,r1), cross(r1,r2)
wintry steppe
How is this a definition? This seems like more of a theorem
(x,y) is the inner product
of V
spare widget
I guess it would work as a definition for inner product space
Because there cos(theta) will lose its original meaning
e.g. consider L_2(a,b)
But you can nevertheless define the meaning of "angle between two functions" to satisfy the above
I agree that for R^n it should be a theorem though
Also for other euclidean spaces
wintry steppe
Does L_k denote the subspace spanned by the first k elements in your list?
spare widget
fringe fjord
It could be a theorem in R^2 and R^3 where we have a preexisting concept of angle, but even in R^4 it begins to feel more like a decision than a deductive consequence of something we already know.
fringe fjord
Based on which definition?
fringe fjord
That seems to be very close to the inner-product definition anyway.
spare widget
that would be the point of the theorem
To show that u^T G v = cos(theta) * |u| * |v|
where G is the metric tensor
But it would be a definition for arbitrary inner product spaces, that I agree with
fringe fjord
I mean, the thing I feel is a definition in higher dimension is the decision to use the word "angle" about a number derived from what we think its cosine should be rather than something that more intuitively measures some quantity in a more or less additive way.
spare widget
But even in R^d
Two vectors (non-collinear) span a 2d subspace
So it's back to the usual definition of cos
fringe fjord
For example, in contrast to defining something like: Euclid taught us what an angle in 2D is; if we can embed Euclids plane isometrically in our space, then we'll declare by definition that the image of two lines have the angle in the ambient space that Euclid says they had in 2D.
spare widget
It's for stuff like functions that this definition cannot be derived anymore from your basic middle school geometry
Since students usually know cos(theta) = b/c before they know what an inner product space is, I understand where the question is coming from
fringe fjord
Can't we just as well embed the usual 2D plane in a function space, once we have an inner product?
spare widget
I can pick two curvy functions
The cosine between them has no meaning wrt the school geometry studied
fringe fjord
The functions might be individually curvy, but the subspace they span is isomorphic and isometric to the usual plane.
spare widget
what's the map between say exp and cos and 2d vectors?
fringe fjord
If you say b/c then how do you know what b and c should be without having an idea of right triangles? And you need the inner product to be able to speak about the triangle being right.
spare widget
Since students usually know cos(theta) = b/c before they know what an inner product space is, I understand where the question is coming from
If you were taught inner product spaces in middle school I agree
fringe fjord
Yes, that's what I don't understand.
How do you apply "b/c" in a space other than R^2 or R^3, if you don't have an inner product yet?
spare widget
The theorem the user is asking for would obviously define what the inner product is
And then it will ask to prove that cos(theta) = ... wrt that inner product
You can probably cook up an inner product for which the geometric meaning does not hold, idk
I guess the point is that one could prove this as a theorem given what they know from middle school
I wouldn't be surprised if it's an exercise to prove it for R^2 or R^3 in some lin alg textbook
Yup, they do prove it as a theorem in a number of places
Albeit typically for orthonormal
But the non-orthonormal version is just u^T G v
@wintry steppe here you go
wintry steppe
@spare widget I know about that
However that does not use a general inner product
That uses the dot product
An inner product of a vector space does not have to be the dot product
spare widget
I thought you wanted it as a theorem because you wanted the proof, but yes it's for dot product
wintry steppe
I know how to prove it for the dot product case, however it does not seem clear to me at all how it can be a definition for the (much more) general case
spare widget
The general euclidean inner product would require u^T G v = cos(theta) * |u| * |v|
For non-euclidean stuff, e.g. L_2 it becomes a definition
i.e. it loses its meaning of cos(theta) = b/c
It's just a definition at that point by analogy to the euclidean case
wintry steppe
Then what does it mean, if it loses its meaning?
spare widget
that is a generalization of the meaning of angle for functions for example
For functions it would still measure how different those are in some sense
wintry steppe
How do you define angle in that case then?
spare widget
$\langle f, g \rangle = \int_{a}^{b}f(x)g(x),dx$
stoic pythonBOT
criver
spare widget
And then you just choose theta to be the value that satisfies cos(theta) = <f,g> / (|f| * |g|)
Then this is the angle between f and g wrt this inner product
You could define a different inner product though
wintry steppe
This is only true for C(a,b) though, no?
spare widget
Then you'll get a different angle
For example
$\int_{a}^b\int_{a}^b f(x) k(x,y) g(y),dy ,dx$
stoic pythonBOT
criver
spare widget
And k chosen to satisfy the properties of inner product
Then you'll get a new definition of theta wrt this inner product
And you can have as many definitions of theta as you have inner products
They don't have to agree either
There was a good video with an example, let me try to find it
See 56:40 for an example: https://youtu.be/2c8XQlQApx8
spare widget
This is only true for C(a,b) though, no?
L_2 is not C
You can take any inner vector space and use its inner product to define an angle
wintry steppe
Yeah no offense I can only think of an angle geometrically
I don't get how you could do it any other way
spare widget
Did you check the video?
Just think of it as a similarity between two vectors (not necessarily geometric vectors)
wintry steppe
Then how can you use the word angle?
spare widget
By analogy/generalization
That's the point, one sees that cos(theta) = <u,v> / (|u| * |v|) is true for euclidean vector spaces
So to generalize one now uses this as a definition instead of a theorem for other vector spaces that have an inner product
It's a standard way to generalize things in mathematics
I can give you an example from numerics
It's a common problem there to have some function f (e.g. an image) that we want to compress
f is typically from an infinitely dimensional vector space
We choose instead a finite dimensional subspace with basis vectors some functions and want to find the best approximation of f there
This leads you to a Galerkin formulation
subtle gust
hey y'all
spare widget
JPEG for instance uses cosine functions as a basis
subtle gust
does anyone have a good summary sheet for the most imp theorems in this course?...
spare widget
While JPEG2000 uses wavelets
subtle gust
i'm 3 weeks in and it's already hard to keep track of all the theorems
subtle gust
does anyone have a good summary sheet for the most imp theorems in this course?....
or a cheat sheet or smthng
spare widget
The most important theorem I remember was about symmetric real matrices having real eigenvalues
subtle gust
The most important theorem I remember was about symmetric real matrices having r...
i don't think we reached this part...
wintry steppe
How does this follow from because (x_1,x_1) != 0? Because we haven't defined the inner product for V
In fact, why is $c_1(x_1,x_1) = 0$?
stoic pythonBOT
n/c
subtle gust
we're still in matrices and determinants and stuff...
spare widget
Then all of those are baby stuff I guess
the important ones are the ones to do with vector spaces
The row reduction and determinant theorems are just tedious
subtle gust
the important ones are the ones to do with vector spaces
yeah we won't cover those until like week 5
subtle gust
The row reduction and determinant theorems are just tedious
true
spare widget
Eitherway, for the end of your course: https://www.engineering.iastate.edu/~julied/classes/CE570/Notes/strangpaper.pdf
spare widget
yeah we won't cover those until like week 5
Axler starts with those
halcyon spindle
Linear algebra done right. He is the author.
subtle gust
yeah we're doing the howard anton's book
subtle gust
Axler starts with those
never heard of this guy tbh
halcyon spindle
As for keeping track of theorems, just do the exercises.
spare widget
Halmos, Axler, Hoffman Kunze, there was some other but I forgot the name, all standard
wintry steppe
How does this follow from because (x_1,x_1) != 0? Because we haven't defined the...
Bumping this, why is $c_1(x_1,x_1) = 0$?
stoic pythonBOT
n/c
spare widget
tbf I don't remember the row reduction and det th/proofs
halcyon spindle
Halmos, Axler, Hoffman Kunze, there was some other but I forgot the name, all st...
Also Sergei Treil linear algebra done wrong.
subtle gust
As for keeping track of theorems, just do the exercises.
yeah i do that, but would be useful to have all the theorems in one sheet
halcyon spindle
Why?
spare widget
Bumping this, why is $c_1(x_1,x_1) = 0$?
dot the sum with x_1
subtle gust
each theorem is a consequence of another one
wintry steppe
dot the sum with x_1
But we haven't defined the dot product to be the inner product of V
spare widget
And use dor(x_i,x_1) = 0 for i!= 1
Instead of dot use inner prpduct
the one that you defined
wintry steppe
There is no defined inner product though, that's the thing
We just have axioms for what it must satisfy
spare widget
And they also should have defined <x,y> = 0 if x,y orthogonal
wintry steppe
However, there is no reason why V can't have multiple inner products
spare widget
They should have defined that if x,y orthogonal then <x,y> = 0
Then by using this definition
wintry steppe
And because we haven't defined any inner products on V, why would I use the dot product?
That seems arbitrary
spare widget
because the theorem says orthogonal vectors
wintry steppe
Yes, which means the inner product is 0
spare widget
And if it says orthogonal vectors it assumes a definition of orthogonality
Yes
So use that
spare widget
But you know that orth => <x,y> = 0
wintry steppe
I do
spare widget
Then use it
wintry steppe
Use it how?
spare widget
$\sum_i c_i x_i = 0 \implies \langle x_1, \sum_i c_i x_i \rangle = 0$
stoic pythonBOT
criver
wintry steppe
Yes I can I think, thanks let me try
paper ether
how do I show that {1, cosx, ..., cos nx} are linearly independent on [0,1] without using matrix techniques?
my prof wants us to somehow turn
p(x) = a0 + a1cos(x) + ... + an cos(nx)
into a polynomial of degree n
and show that this polynomial only has finitely many zeros
wintry steppe
Okay, so, $\left(x_1, \sum_i c_ix_i\right) = (x_1, c_1x_1) + (x_1, c_2x_2) + \dots + (x_1, c_kx_k) = \sum_i (x_1, c_ix_i)$
fringe fjord
Do you know Chebyshev polynomials?
stoic pythonBOT
n/c
paper ether
no
fringe fjord
Shame; they're tailor-made for this.
paper ether
lemme look into it
wintry steppe
Then $(x_1,c_ix_i) = (c_ix_i,x_1) = c_i(x_i,x_1) = c_i(x_1,x_i) = c_i \cdot 0 = 0$
stoic pythonBOT
n/c
wintry steppe
So all of the other terms die for sure, which means that $\left(x_1, \sum_i c_ix_i\right) = (x_1, c_1x_1) = c_1(x_1,x_1)$
stoic pythonBOT
n/c
fringe fjord
You may need to derive them yourself using cosine addition formulas and sin²(x) = 1 - cos²(x).
wintry steppe
But actually, I don't know that this inner product is equal to 0 @spare widget
Why is the implication true?
Oh nevermind I am dumb
I guess it's because the inner product of 0 and any vector is 0
spare widget
you could use $\langle x_1, \vec{0}\rangle = \langle x_1, 0\cdot\vec{0} \rangle = 0\langle x_1, \vec{0}\rangle = 0$
stoic pythonBOT
criver
spare widget
If you need that <0, x> = 0
wintry steppe
Except $\sum_i c_ix_i \in \mathbb{R}$
spare widget
For $0\cdot x = \vec{0}$ you need a separate proof from the vector space axioms
wintry steppe
i.e. it's not a vector
halcyon spindle
A real number is a vector.
wintry steppe
Sorry you're right
spare widget
x+y is a vector, a * x is a vector
spare widget
There's no 0 dot x in the proof
halcyon spindle
First time seeing that type of indexing.
stoic pythonBOT
n/c
wintry steppe
I guess that's less ambiguous
spare widget
$\langle x_1, \sum_i c_i x_i \rangle = \langle x_1, \vec{0}\rangle = 0 \implies c_1 \langle x_1, x_1\rangle = 0$
stoic pythonBOT
criver
wintry steppe
Indeed
spare widget
The first equality is from sum..= 0
wintry steppe
No don't worry I got it now, thanks for the help
spare widget
My best advice is to pick several books up, it helps a lot with proofs
since some show the proofs in a different way
wintry steppe
I found that this book is vastly different from the book I used before (Axlers)
spare widget
Halmos has a whole book of solved problems
though his are rather interesting compared to today's books
I think it's called solved problem something something finite dimensional vector spaces
wintry steppe
$||x||^2 = \sum_{i=1}^n |(x,e_i)|^2$ if $x \in V$ and $V$ has orthonormal basis ${e_1,\dots,e_n}$
stoic pythonBOT
n/c
wintry steppe
Does this mean that this formula does not hold if V does not have an orthonormal basis?
Because it's supposed to be a generalization of the Pythagorean theorem for an arbitrary vector space
zinc timber
if V doesn't have an orthonormal basis then yes you are right, but again all finite dimensional inner product spaces have an orthonormal basis, so you are wrong as well (kinda)
(gram-schmidt)
spare widget
Because it's supposed to be a generalization of the Pythagorean theorem for an a...
And if parseval's identity holds, then the orthonormal basis e_i is termed closed
$\lim_{n\rightarrow\infty}|v - \sum_{i=1}^{n}\langle v, e_i \rangle e_i| = 0$
stoic pythonBOT
criver
spare widget
Parseval's identity and the basis being closed are equivalent
so if the parseval identity holds/the orthonormal basis is closed then the generalized parseval identity holds
What you know as dot product:
$\langle u, v\rangle = \sum_{i=1}^{\infty} \langle u, e_i \rangle \overline{\langle v, e_i\rangle}$
stoic pythonBOT
criver
spare widget
The bar on top is conjugation, if you're not working with the complex field it's redundant
The infinities are also redundant for finite dimensional vector spaces
You can prove the above from parseval's identity + the polarization identity
This is in fact quite an important result in practice
It lets you find the best finite number of terms approximation of some function
As an example
$|u-f|^2 = |\mathcal{F}[u] - \mathcal{F}[f]|^2$
where F[x] is the fourier transform of x
stoic pythonBOT
criver
spare widget
That means that in the discrete setting you can just sort F[f] to find the m largest coefficients and use them in F[u] while setting everything else to 0, and this will give you the best approximation with m coefficients in the L2 error
In practice one chooses the DCT basis for images for example
Basically orthonormal bases => allow you to find the best m-approximations
golden monolith
Hello, I would need some help at solving a differential equation Y'=AY, with A not diagonalisable
Is there anyone free to help ?
zinc timber
bold sun
hey can someone please help me out on this question:- im not really sure how to do it/what to do ?
i tried something like this but it didnt work ik its wrong
dusky epoch
can you explain the purpose of your calculations?
i think there is a much easier way to do this
proper cradle
you need to add extra elements from P_3 in a such a way that they are not in span(p_1,p_2) and that span P_3
dusky epoch
and i'm certain you don't actually need to go through all this legwork at all
wintry steppe
hey can someone please help me out on this question:- im not really sure how to ...
so what you need to do is complete to basis, of P3
what we know about P3 is that it constructed by 4 variables: ax^3+bx^2+cx+d
you found that p1 and p2 are linearly independent but you are missing two more polynomials to complete the basis.
if you take p3 = x and p4 = 1 you will have a spanning set which is also linearly independent(basis for P3).
dim(p1,p2,p3,p4) = 4 and dim(P3) = 4
P3 = span{p1,p2,p3,p4}
bold sun
can you explain the purpose of your calculations?
umm i wanned to try to make it into the identity matrix for coefcinets to see if it spans but i got lost like half way thru but yh if easier way yh lmk plz
dusky epoch
well elay_dadon kind of already gave away what i have in mind
proper cradle
umm i wanned to try to make it into the identity matrix for coefcinets to see if...
since dim P_3 is 4, they are two vectors from P_3 so it will not work
dusky epoch
you can take p_3 = x and p_4 = 1
you are missing two vectors anyway and these are inexpressible as linear combinations of yours (think about the high power coefficients in any such linear combination)
bold sun
hmm like how did u guys figure it out straight away tho is what i dont get lol ive been trying for ages
and why them specific ones how du know its p3 is x and p4 is 1 straight away
spare widget
Do you know how to find a solution to an underdetermined system?
Or find vectors spanning the complement of a subspace
dusky epoch
and why them specific ones how du know its p3 is x and p4 is 1 straight away
there are many other ways to complete your basis
this is just the simplest one
wintry steppe
hmm like how did u guys figure it out straight away tho is what i dont get lol i...
so the intuition for it is that since you have only two vectors, so you have two vectors that are linearly independent, and also we can see that their degree is 3 and 2.
let's look at the standard basis: {x^3,x^2,x,1}, we have two polynomials that have x^3 and x^2.
for the basis to be linearly independent you need to more polynomials that will be linearly independent with p1 and p2.
when we look at the basis what pops out is x and 1 because x^2 and x^3 already there and for them to be linearly independent, we need to more polynomials that will satisfy it and span P3.
bold sun
Do you know how to find a solution to an underdetermined system?
if you mean finding the general sol yh
spare widget
Well then do so
bold sun
but that was for matricies not polynomials...
spare widget
But the poly have coefficients
You can write a poly as p(x) = (x^n ... x 1)^T v
v is a vector from R^{n+1} there
stoic pythonBOT
Kaishin
