They are LI
#linear-algebra
wintry steppe
p1(t)=p2(t)+2p3(t)
quaint heart
You can select 2 of them which are linearly independent to get a basis
placid oracle
wait, apparently you are wrong though
wintry steppe
:o ?
placid oracle
answer choices are one of these
quaint heart
There are a lot of choices of basis
wintry steppe
Oooh lol
spring wolf
I mean there's always more than one basis
wintry steppe
The first
placid oracle
that is true
wintry steppe
There are many, but from the list I'd say no. 1
spring wolf
might have thrown people off
placid oracle
how is t2, 1 equivalent to that though?
spring wolf
your polynomial p_1(t) = 1+t**^2**?
wintry steppe
t2, 1 would be the canonic
So to speak
Since it toggles the entries of the polinomial at2+b for a, b constants.
Which is what you'd get from the first too
You can think of polynomials as a vector space
In here you're asked to find the span of (t2, 0, 1), (t2, 0, -1) and (0, 0, 1)
You can say that the span is accurately represented by (t2, 0, 1) and (t2, 0, -1) or (t2, 0, 0) and (0, 0, 1)
placid oracle
yeah
wintry steppe
👌
placid oracle
ty!
wintry steppe
Npnp!
slow scroll
im a bit stuck on 1.9. The best i can come up with is that geometric multiplicity equals algebraic multiplicity because the eigenspace of the block matrix in 1.8 has k vectors, and the characteristic equation would have a (z-lambda)^k term, algebraic multiplicity of k.
Not sure how to make the argument that geometric can be less than algebraic.
oh yea, and i know geometric definitely can't be greater because we can write the operator in this form where the first k basis vectors contribute to the algebraic multiplicity of the eigenvalue.
royal timber
can someone help me with a problem?
i suspect theres an easier method than computing the 3x3 determinants
as if i equate some 2x2 determinants to 0 it gives the right answer
but i don't know what 2x2 determinants to computwe
cause some 2x2 determinants gives the wrong answer when equating them to 0 and solving for a/b/c
slow scroll
maybe try putting the matrix in row echelon form, and then find values of a,b,c that eliminate a pivot column
or uhmm something like that 
royal timber
\begin{pmatrix}1&2&3&14&4&10\ 0&0&-3&-36&3&-24\ 0&0&0&0&-4&-18\end{pmatrix}
stoic pythonBOT
boilhats:
\begin{pmatrix}1&2&3&14&4&10\\ 0&0&-3&-36&3&-24\\ 0&0&0&0&-4&-18\end{pmatrix}
```Compile error! Output:
! Missing $ inserted.
<inserted text>
$
l.11 \begin{pmatrix}
1&2&3&14&4&10\ 0&0&-3&-36&3&-24\ 0&0&0&0&-4&-18\end{pm...
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.
LaTeX Font Info: Try loading font information for U+msa on input line 11.
(/usr/local/texlive/2018/texmf-dist/tex/latex/amsfonts/umsa.fd
File: umsa.fd 2013/01/14 v3.01 AMS symbols A
It gave me this matrix when I tried your method
slow scroll
wait but where is a, b,c?
royal timber
i replaced them
a is 2 b is -3
and c is -4
The solution from the manual is a=2,b=-3 c=-2
I tried to calculate the rank of the matrix with an online calculator
and in both cases the rank is 3
so there must be something wrong with the problem
i think
royal timber
There are 2 commutative square matrices with n rows and n columns(A and B). How do I show that if A^3=B^3=I(identitity matrix) than rank(A+B)=n
royal timber
Is there a matrix( let's call it A) such that A^n=n*A ,for n>=1, besides the zero matrix?
dusky epoch
A^n = nA for all n ≥ 1?
royal timber
yeah
wanton glade
If given a set called S, how do you find an orthonormal set with the same span as S?
dusky epoch
👏 gram 👏 schmidt 👏 orthogonalization 👏
wanton glade
Yeah I did that, Im just confused what you do after that
native lodge
You write down those vectors as your basis? I’m mean you said you did all the work so those vectors you got are the new basis vectors now
dusky epoch
normalize lol
once you're done gram-schmidting your set you're left with an orthogonal set of vectors
all that's needed to make it orthonormal is to normalize each vector
wanton glade
Which is just making length = 1?
native lodge
Yes
Take the magnitude of the vector and then divide each entry in the vector by that magnitude to normalize it
wanton glade
Ok so after gram-schmidt I got this
that looks messed up, hold on lol
\left{
\begin{bmatrix}1
\ 1
\ 1
\end{bmatrix}
\begin{bmatrix}5
\ -1
\ 2
\end{bmatrix}
\right}
stoic pythonBOT
EatRocks:
Compile Error! Click the 
reaction for details. (You may edit your message)
wanton glade
So i got that, I get the normalization on the first as 1 / root3 which is correect
native lodge
Then you take the magnitude of each vector and divide the entire vector by that magnitude
Yes, your first one is correct
wanton glade
the second is 1/ root 2 which is what is confusing me
wanton glade
yeah i got that
native lodge
So then divide the second vector by 1/sqrt(30)
wanton glade
im not getting 1/Sqrt(2)
the orthogonal basis is
\left{
\begin{bmatrix}1
\ 1
\ 1
\end{bmatrix}
\begin{bmatrix}3
\ -3
\ 0
\end{bmatrix}
\right}
stoic pythonBOT
EatRocks:
Compile Error! Click the reaction for details. (You may edit your message)
wanton glade
idk if that helps
native lodge
So you wanted to orthogonalize (5 -1 2) with ( 1 1 1)?
wanton glade
No, im supposed to use gram schmidt on (1 1 1) (5 -1 2)
and then get an orthonormal set with that basis
native lodge
I’m dumb lol, you get (1/sqrt(2) -1/sqrt(2) 0) when you normalize (3 -3 0)
wanton glade
No I was unclear lol, do you still get that by dividing 1 / root 30?
native lodge
Ok, so I’m getting a bit lost here. So what were the original vectors, and what were your results?
native lodge
So for (3 -3 0), you take the magnitude which is sqrt(18) which is equal to 1/3sqrt(2)
Now divide (3 -3 0) by 1/3sqrt(2), which will leave you with 1/sqrt(2) * (1 -1 0)
When we say take the magnitude of the vector, we are referring to your orthogonalized one, not the original one
native lodge
No problem
royal timber
Find X for $\begin{pmatrix}-7&-8\ 4&-7\end{pmatrix}$
royal timber
Find X
quaint heart
I can certainly find an X that works, I don't know if it's a unique X though
Just diagonalize, take 4th root, undiagonalize
junior relic
Well, I don't think there's a unique X
quaint heart
Yeah there never is actually
slow scroll
i'm having trouble seeing how they use $(A-\lambda_r I)\mathbf v_r = \mathbf 0$ to come up with the last equation
stoic pythonBOT
kxrider:
slow scroll
what happens to matrix A tho? 
quaint heart
Eigenvectors happen lol
You're applying A to it's eigenvectors, so it just scales them by the eigenvalues
slow scroll
i mean, what are they doing to $\sum_{k=1}^r c_k \mathbf v_k = \mathbf 0$? When you apply $A$ to it, you get
$$ \sum_{k=1}^r c_k \lambda_k \mathbf v_k = \mathbf 0$$
Then what? The only way you can get the last equation is by assuming $\sum_{k=1}^r c_k \lambda_r \mathbf v_k = \mathbf 0$.
stoic pythonBOT
kxrider:
slow scroll
er wait that last part isn't right
quaint heart
You have to apply A-lambda_r I to it
Not just A
So it kills the last term
(A-lambda_r) v_r = A v_r -lambda_r v_r =0
slow scroll
okay i got it now lol.
thank u C:
round wharf
So I have XYZ is a triangle, the vector XY is (4,-3) and the vector XZ is (-2,3), I'm then asked to calculate |ZY|
fringe cave
draw it out
dusky epoch
why are you unsure of your answer?
round wharf
Because I'm not sure if like the method I used is correct
native lodge
like darkrifts was saying, this problem has small enough numbers where you can just draw it if you wanted to
native lodge
but in this case, |ZY| will be connecting the two vectors that are originating from the origin
native lodge
yes
this is basically coordinate geometry for now, but that interpretation of vectors will only go so far lol
dusky epoch
mmmh... are you actually taking a linear algebra class
round wharf
No I'm uh not haha
I know a bit and yes I've watched 3 blue 1 brown videos
But they didn't help much
native lodge
well we just had a 2d vector, so just thinking of them in the xy plane is "good enough" and likewise with the 3d vectors, so then you run into higher dimensional vectors, which are impossible to visualize, so I mean you can think of vectors as a list of numbers and that works, though I admit it sounds like the most loose definition ever lol
dusky epoch
bc at some point, you're going to have to work in and with spaces of dimension above 3. that's when it'll start to be hard to draw everything
also
3b1b's vids aren't meant as a linalg course
round wharf
Oh :<
dusky epoch
but rather as a supplement to an actual course
native lodge
you can always look up gilbert strang's lectures on linear algebra, though the book is handy to have as well
round wharf
I see, ok thanks
native lodge
I think his fourth edition linear algebra book can be found online
round wharf
From earlier
native lodge
it would be a column
dusky epoch
i mean for that matter, all vectors students work with before LA are "column"
round wharf
I see
native lodge
if you want a row vector, then you can just transpose the column vector
native lodge
you can think of it as an x y coordinate
dusky epoch
hh
native lodge
entry one would be the x coordinate and entry two is the y coordinate
dusky epoch
yeah ok so like.
with the "from the origin" thing
oh god. i'm not sure if i can explain it w/o getting too technical
dusky epoch
there's this concept in geometry of an affine space. which is essentially somewhat like a vector space except you forgot where the origin is
kinda
so like basically each affine space comes with a vector space acting on it
in the sense of like
basically the action of each vector is that of translation
and so you can draw a vector between any two points
and that'll be the vector whose action takes the start point to the end point
warm stirrup
Not sure if this helps, but as far as getting vector ZY, I'd just let X be the origin point. Then vector XY can be thought of as a position vector Y and similarly with a position vector Z. then it's just end-start to get vector ZY, or ZY=Y-Z.
so, Y=<4, -3> Z=<-2, 3>, then ZY=<4, -3>-<-2, 3>=<4+2, -3-3>=<6, -6>
quaint heart
@dusky epoch I haven't heard that explanation before for why we sometimes draw vectors as arrows. That's pretty nice
warm stirrup
I'll be honest, some of it goes over my head, but isn't it just basically saying that translations can be represented by the addition of vectors?
quaint heart
It's saying that addition of vectors can be viewed as composition of translation functions (and vice versa)
warm stirrup
I mean.... Yeah I'm sorry I'm just not getting why that's impressive? Not trying to be a dick just, vectors are kinda intimately linked with motion, and I kinda viewed them as translations by default
dusky epoch
vectors are generally like
an abstract thing that can manifest in many different ways
warm stirrup
Yeah, I'm vaguely familiar with the concept of vector spaces, but they all have to have the same basic properties of addition and scaling, though they might work differently
Wouldn't that inherently link it to a sort of translation, albeit potentially in a different way than what we might normally think?
dusky epoch
mmmm\
i mean like...
a vector space is just any collection of things that obeys the vector space axioms
warm stirrup
right
dusky epoch
and part of that is addition and scaling operations
idk what link to translation you're thinking of tho.
warm stirrup
i.e. that adding a vector to some other point(Represented as a vector) should give you that point translated along that vector
quaint heart
What do you mean "some other point"?
A vector isnt a particular way we represent an element
It literally just means an element of our vector space
warm stirrup
Perhaps I'm missing something, but to me that just seems like pedantry
quaint heart
It is lol, I just didn't like the fact you said represented as a vector
warm stirrup
You could just as easily remove the "some other point" from my argument and say "some other vector"
quaint heart
I'm a bit of a pedant
warm stirrup
It's just easier to say point instead of talking about two nameless vectors. That'd make it kinda hard to follow
slow scroll
i don't vectors inherently have anymore visual meaning than the fields they are over. You can think of just regular old numbers the way you are describing.
warm stirrup
Well of course not
quaint heart
You can have vector spaces over finite fields for example
And I think a lot of the intuition fails
In those cases
warm stirrup
They still have directionality and magnitude though, so does that not capture the very essence of a translation?
quaint heart
That's an extra stucture you have to impose on your space
warm stirrup
Isn't one of the axioms that you must be capable of multiplying vectors by a scaler? which literally...scales it?
slow scroll
a scalar is just an element of the field
quaint heart
A norm is a special function which acts on your space
warm stirrup
I.E., if you have a vector v and a scaler c, then cv is precisely c times larger than v
dusky epoch
no
like
vector spaces need not have a norm
and if your space isn't over R then it doesn't even make any sense to speak of vectors being "large" or "small"
fringe cave
Scalar multiplication is needed, but "magnitude" isn't
dusky epoch
"scalar" is just a name
warm stirrup
What is it it you mean by R? The real plane?
dusky epoch
the field of real numbers
warm stirrup
Complex numbers have magnitude as well, though
fringe cave
yeah but not all vector spaces do
warm stirrup
Right, but how does it not being over R inherently mean that it doesn't make sense to think of "size"?
fringe cave
I've been working recently in a vector space over the space of musical pitches. There is no "magnitude" but application of scalars (through inversion) is still well behaved
dusky epoch
in order to even be able to talk about size at all
you need an order relation
and in particular you need your field to be an ordered field
warm stirrup
If it isn't ordered then how can you even represent a vector....?
slow scroll
ordered means stuff like < > exists
fringe cave
You can represent a vector as (a, b) without there being an ordering for a < b or anything
Just because there isn't an ordering on the underlying field doesn't mean you can't have vectors
half ice
Take the vector space of polynomials. That is, the space with basis {1, x, x²,...} Something like x² - x + 1 doesn't have a "size" or "direction"
warm stirrup
Actually, that helps
quaint heart
You can impose a norm on it though if you want
warm stirrup
I see what you mean now. The only way I could think of trying to find a magnitude for it would be to transpose that onto some sort of structure which would define each basis as a certain length. But that's not necessarily included
quaint heart
Exactly
warm stirrup
And to give it a directionality would also require the same structure, but with the added specification of how the basis vectors relate to each other, direction-wise
half ice
But I see where the confusion lies, I remember learning vectors have "magnitude" and "direction" but they don't necessarily have to
quaint heart
They do if you're a physics student who only works in dimensions 1-3 lol
warm stirrup
I mean, I'd argue you could still define a directionality and magnitude in higher dimensions. I'd hate to be the poor bastard to do so, but yeah
quaint heart
It's pretty easy to define a norm on any R-vector space tbh
Assuming you have a basis
warm stirrup
oh yeah, the norm is easy
But the directionality I wouldn't even want to try to think about
There's probably plenty of people out there that could but yeesh
quaint heart
I don't even know what directionality means
warm stirrup
Is this a joke, more pedantry, or a serious statement?
quaint heart
A serious statement
dusky epoch
honestly yes "magnitude + directionality" is the most baby idea of vectors possible
quaint heart
Does directionality just mean you pick a point on your unit sphere and a magnitude?
It seems pretty vague to me
warm stirrup
I mean, directionality as an English word just means "having the property of being directional or maintaining a direction"
dusky epoch
ggdfhjdghkdghd
slow scroll
i hear eigenvalues explained as vectors that have the same direction after being acted on by a linear operator. Maybe you can just say that all scalar multiples of a particular vector belong to an equivalence class of vectors with the same direction lol
quaint heart
That would fit with my definition
dusky epoch
now that just sounds like projective space to me
quaint heart
Yep
warm stirrup
But the actual definition is just Tv=lamda v, though, so I don't see how that would necessarily require a direction either
dusky epoch
it's more like
vectors being scalar multiples of one another still makes sense no matter what your space si
is*
warm stirrup
Right, so eigenvectors still make sense and can exist in any vector space
dusky epoch
yes eigenvectors are a thing everywhere
warm stirrup
And since vector spaces don't necessarily require a definition for direction, it follows thinking of eigenvectors as keeping their direction doesn't work for more abstract vector spaces
half ice
Going back to the polynomial thingy again, differentiation is a linear operator. What's the eigenvectors?
quaint heart
The constant functions I guess
slow scroll
never thought about that. I guess it wouldn't have any. Could it even have complex eigenvectors?
warm stirrup
The constant functions should be eigenvectors, though. They'd have an eigenvalue of 0
quaint heart
Yep
slow scroll
o yea
half ice
Am I doing a dumb? My original thought was the Taylor series of eˣ
slow scroll
wait do we even count that?
warm stirrup
Well, given that one of the basis vectors is 1, I'd say we count it
slow scroll
i don't think we count the null space of a matrix as an eigenspace
idk ;alskdkl;asdjfkl;ajsdf
quaint heart
We do not
slow scroll
i thought about e^x, but it would take an infinite matrix to differentiate that
quaint heart
That's not an element of the polynomial space
slow scroll
right.
Just to be clear, eigenvalues are non zero always, right?
quaint heart
But yeah, those are the solutions in the vector space of differentiable functions
They can be zero
re^cx
dusky epoch
vectors in the kernel are eigenvectors with eigenvalue 0
warm stirrup
I mean, the derivative matrix should already be infinite, so I guess we could use the e^x family of functions
And by that logic... anything else?
quaint heart
It all depends where it acts on lol
dusky epoch
matrices only make sense in findim spaces
quaint heart
I think they make some sense in general
warm stirrup
Why? I've seen infinite matrices used before
half ice
Sorry, what's the largest power that belongs to the polynomial space?
warm stirrup
Unironically this. XD
God I always feel like an idiot trying to keep up with these conversations
dusky epoch
fringe cave
tbh
half ice
Every space has a basis. Except the polynomial space, not that one.
fringe cave
quaint heart
half ice
Inb4 one of jaco's alts can apply choice to a basis
royal timber
$A=\begin{pmatrix}
1 & 2\
0& 1
\end{pmatrix}
\
Solve \ the \ equation \ X^{n}=A $
stoic pythonBOT
boilhats:
royal timber
yes, find all roots
quasi urchin
Sorry, new here---is this something you'd like help solving (without us giving away an answer), or is it a puzzle for us to think about together? ^^
royal timber
I'm just curious about the method used to solve it I don't necessarily need an answer.
quasi urchin
Well, looking at products of matrices of the form
$\begin{pmatrix}
1 & x\
0 & 1
\end{pmatrix}$
suggests that the roots should also be of this form.
stoic pythonBOT
jesyspa:
dusky epoch
i'd be surprised if all the roots looked like this
royal timber
If I raise the matrix to the nth power I get $X^{n}=\begin{pmatrix}
1 & nx\
0& 1
\end{pmatrix} $ And I find the root x=2/n, but how do i find the other complex roots
stoic pythonBOT
boilhats:
dusky epoch
p sure if you scale that matrix by exp(2πki/n) you'll get n-1 more roots
quasi urchin
Good point, should not have said they'll all be like that.
Are there any theorems on how many roots we should expect there to be?
royal timber
Find a formula for calculating $A^{n} (n\geq 2)\ if \ A=\begin{pmatrix}
a & b\
c& d
\end{pmatrix}
\
A\in M_{2}(\mathbb{C}), det(A)=0 $
stoic pythonBOT
boilhats:
wintry steppe
rare barn
so ik that for it to be a basis it needs to be a spanning set that is linearly independent. is the only way to solve this by checking through each one?
maiden lintel
Well first of all, what is the dimension of the vector space?
dusky epoch
once you know the dimension of the space, you can rule out some of the answer options as not having the right size
maiden lintel
What about a
maiden lintel
Another easy thing to check for is to see if a vector is a multiple of another
half ice
There's a fairly clear problem with b
rare barn
c is LI
dusky epoch
well there you have it then don't you
rare barn
got it, thanks
half ice
Note there's a 0 in the bottom right of every vector in b.
Ergo, b doesn't span the set you want
rare barn
oh yep
rare barn
If H is a subspace of V and B is a basis for H. Then is there a basis for V containing B?
dusky epoch
why wouldn't there be
broken hawk
there is, it does't seem entirely obvious to me
but there is, it's a theorem
I forgot the name of the theorem though
given a set of linearly independent vectors, you can always extend it to a basis
rare barn
spanning set theorem?
broken hawk
(note: the theorem only concerns finite dimensional vector spaces, though I'm fairly sure it also applies in the infinite dimensional case. but the infinite dimensional case is ugly when it comes to bases)
rare barn
havent gotten to that yet
broken hawk
(you need quite strong tools to prove a basis even exists and may not be able to ever write it down)
yea just covering my bases :P
pun intended
rare barn
haha
broken hawk
here's an algorithms for making a basis btw (only works in the finite case):
-let (v1, ..., vm) be a linearly independent set, m<n = dim(V)
-let W = span(v1, ..., vm)
-pick v in V\W
-then v is linearly independent from (v1, ..., vm). Therefore, (v1, ..., vm, v) is linearly independent
-repeat until you've got n vectors
this proof relies on the well-definedness of the dimension, and on the fact that a vector not in W cannot be written as a linear combination of the vectors in W (which is fairly obvious)
I don't actually remember how one proves that the dimension is well-defined though (that is, all bases have the same number of elements)
dusky epoch
something something transfinite recursion
wintry steppe
In the derivation of the method of least squares one step is to go from
stoic pythonBOT
Darnock:
wintry steppe
to
stoic pythonBOT
Darnock:
wintry steppe
What allows us to perform this step?
stiff surge
1 1 | 0
1+√3i 1-√3i | 1
is the matrix i had
and i needed to find x1 and x2
1 1 | 0
0 2√3i | 1
I reduced the matrix to this
then I moved the 2√3i to the other side to get
1 1 | 0
0 1 | 1/2√3i
so I thought that
x1 = -1/2√3i
x2 = 1/2√3i
but apparently the i is in the numerator
how is that so...? 🤔
half ice
@stiff surge
Are you using row reduction? "Moving 2 √3i to the other side" is a no go
humble rapids
Hello, I'm having trouble understanding how intersections of subspaces of different dimensions work.
native lodge
like the null space and row space stuff, or more general cases?
wait, null and row are in the same space lol
humble rapids
More general cases
So like, if we have U, V in R3, and U = span(e1,e2), V = span(e1,e2,e3), would the intersection be span(e1,e2)?
native lodge
it has to be, because R3 has an extra entry that doesn't have a correspondence in R2
the only common vectors are 2D vectors
or planes when considering the span of the basis vectors
humble rapids
Ok
So basically, the dimension of the intersection would be the same dimension as the smallest one?
quaint heart
It has max the dimension of the smallest one
native lodge
from our small thought experiment that appears to be the case, and I would say that is most likely true in most cases we will come across. It makes sense for sure and for our typical notions of what a vector space is this should work. But I'm not very sure with spaces like polynomial ones, as I haven't gotten that far myself into the abstract aspects of linear algebra
stiff surge
@half ice apologies wrong wording - I divided both sides by 2sqrt(3)i
austere owl
fringe cave
smh
2 times as many, so 2^n more (depending on how they want it answered)
since 2^(n+1)=2(2^n)
fringe cave
How much more is B(n+1) than B(n)
austere owl
I would be able to answer that but I was caught up in what they were describing
how it has length n and digit n
which doesn't make sense to me
fringe cave
it has n digits
so it is of length n
those are exactly the same
it is n digits long
austere owl
but 2^4 doesn't have 4 digits
warm stirrup
2^4 is the number of possible binary numbers with 4 digits
dusky epoch
bitstrings.
fringe cave
no but there are 2^4 numbers in binary with 4 digits of length
warm stirrup
For a simple example, with 2 digits there's 2^2=4 possibilities, specifically 00, 01, 10, and 11
austere owl
did you have prior knowledge to this, because it didn't state that
fringe cave
it did state that
dusky epoch
state what
warm stirrup
^I knew binary before but it does state that B(n) tells you the number of "binary code words" with length n
dusky epoch
binary code word = bitstring
warm stirrup
I would just call it a "number" but whatever
dusky epoch
i'd be wary of identifying 001010 with 1010
fringe cave
yeah but a number would mean 0001=1 which isn't the same here
austere owl
okay I was confused with the terminology
dusky epoch
pseudo oak
what's an easy way to remember how to multiply matrices
native lodge
take dot products with rows by columns
or think of matrices on the right as acting on the rows of the matrix to its left
or matrices on the left as acting on columns of the matrix to its right
the way I personally do it is draw horizontal lines to separate the rows of the left matrix and vertical lines to separate the columns of the right matrix, and then do dot products with each column using the first row, then the second row, and so forth
half ice
AB
pseudo oak
aight
half ice
You put the second one "up"
It quickly gives the size of AB, and if the dot products aren't equal, then you can't multiply the matricies
wintry steppe
can anyone explain whats going on here.. im abit confused at the bit with the P
I don't get the line from let P = .... to the next line detP=...
muted orchid
@pseudo oak it wouldn't be BA since the multiplication does not have matching row and column dimensions. B_2x3 * A_4x2 => any row vector in B is not the same number of elements as any column vector in A
pseudo oak
Yeah ur right. Thanks
jagged pendant
@wintry steppe it's just verifying that P is infact an orthogonal (rotation) matrix
when u multiply a row by c, it multiplies determinant by c
if u multiply the whole matrix by c
here it's 2×2 so there's 2 rows u multiplied
so it multiplies determinant by c²
c=1/√2, c²=1/2
then det of the other thing is 1×1 - (-1)×1 = 1+1 =2
multiply together to get 1
the other condition is that the columns are unit vectors and are orthogonal (dot product = 0)
(1,1)•(-1,1) = -1+1=0
rare barn
Let A be an m×n matrix, and suppose we know that rank A is either 4 or 5 and dim Nul A is either 3 or 4. Which of the following dimensions for A is possible?
5x8, 4x9, 3x7, 6x6
cant n be either 8, 7, or 9?
how do you find m?
rare barn
<@&286206848099549185>
quaint heart
So n can be either 7, 8, or 9, but there are some restrictions on m as well
rare barn
not sure how to find these restrictions
quaint heart
if rank A is 5 then how can the m be 4 for example?
we know the rank <= both n and m
so that eliminates the 4x9 option
@rare barn does that make sense
quaint heart
with the 3x7 case we know rank + nullity = n so we have rank=4 and nullity = 3
cause thats the only combination that works
but then the rank is too damn high!
rare barn
rank is 4 which means that nullity is 3
rare barn
and for the 5x8 its either rank 4 and nullity 4 or rank 5 and nullity 3
and it works for 4 and 4
quaint heart
yep
rare barn
got it! 👌
quaint heart
either works really
slow scroll
i can't find where my book talks about this... do all operators rank = n have n eigenvalues counting multiplicity?
proper crescent
No, that'd imply diagonalizability
quaint heart
There are a lot of obvious counterexamples if you think about it for a bit
slow scroll
i haven't messed around too much with this. To be clear, I'm talking about complex eigenvalues.
quaint heart
The identity matrix
quaint heart
Yeah I just realized
proper crescent
@slow scroll so, square matrices A and B are similar if there's some invertible matrix P such that A = PBP^{-1}. A matrix is diagonalizable if it's similar to a diagonal matrix
What you had in mind with respect to multiplicity basically boils down to the existence of a basis of eigenvectors, right? Well, that's equivalent to the matrix being diagonalizable
If you like working with minimal polynomials, it's equivalent to the minimal polynomial factoring into distinct linear factors
(For reference, if the minimal polynomial just factors completely over your field, that's equivalent to triangularizability)
(So matrices over C, even those with low rank, are triangularizable)
slow scroll
hmm i might ask another question about this later, but thanks!
proper crescent
That I think corresponds more to "generalized eigenbusiness"
wintry steppe
Woog I sort of understand now, thanks
lean stag
I am so confused, I don't know how to do this, could anyone help? I'm Swedish so I am translating it, some mathematical terms might therefore not be completely accurate, and I apologize beforehand.
The question is:
"The planes 3x - 2y + z -6 = 0 and x + y - 2z - 8 = 0 as well as the line (x, y, z) = (1, 1, -1) + t(5, 1, -1) are given."
"a) Show that there is a point in common (mutual, unified, in common?? Dont know the accurate translation here) for the planes and the line. Decide the coordinates for this point."
"b) Decide the equation for the plane that contains the intersecting line between the given planes and the given line. Present the planes equation on affine form ( ax + by +cz + d = 0 )"
No Cross Products allowed...
Thank you kind soul whomever might help 
broken hawk
just use wedge products instead of cross products
native lodge
Part one, solve the system of three variables for the single point, if it exists
Wait, there’s a line involved
Ew
lean stag
Yeah that's where I am confused; because of that line
lean stag
Found the solution c:
upper jewel
Just a question to fix my visualization.
What I'm assuming is that having 3, 3 component vectors. Will fill out A piece of 3d space, and not the whole 3d coordinate system?
dusky epoch
upper jewel
Let's say we have 3 columns with 3 components. And let's say these columns are a(1,2,3) + b(2,2,5) + c(1,1,1). These three columns will combine too form a 3d space, but will they fill the whole 3d coordinate system, as in every possible coordinate in 3d space?
Also assuming the scalars a,b,c can be any real number of course.
slow scroll
if they are linearly independent, then yes. In general, to span a space of dimension n, you need n vectors. However, if one or more of the vectors are linearly dependent on each other, then the space could span a subspace of lower dimension.
The intuition here is that if you have 3 vectors in R^3, and if one of the vectors lies on the same line as another vector, or in the same plane as the other two vectors, then those three vectors would have to span a plane instead of 3 space.
upper jewel
ah I see. Also sorry for the late response had too do something.
ashen scroll
so why would the orthogonal vector to 3x+y=4 on r3 be (3,1,0)? is it because the equation is technically 3x+y+0z=4? where does the 4 fit in?
dusky epoch
it doesn't. if that 4 were anything else you'd still have the same normal vector
and yes it's bc the z coefficient is 0 that your normal is (3,1,0)
native lodge
One way of defining a plane in R3 is by specifying the vector normal to it and a point in the plane
As Ann is saying, you will always have the same normal vector but the choice of point is arbitrary
broken hawk
(which is bad\™ as it only works in exactly 3 dimensions, but it can be handy)
native lodge
But you should still end up with 4 every time
ashen scroll
ah
the question was "Find a vector that is orthogonal to the plane 3x+y=4 in R3" and the answer sheet said (3,1,0) was the answer. So the normal is an orthogonal vector?
wait whoops
the normal is one number
dire bluff
uhh how do you define the trace of an operator?
potentially infinite dimensional operator
proper crescent
In infinite dimensions a priori there's no good notion of trace
broken hawk
https://en.wikipedia.org/wiki/Trace_class seems like not every operator has one, but if it does it's exactly what you'd think
proper crescent
Maybe a compact operator on a Banach space with a countable Schauder basis
broken hawk
see also the linked article nuclear operator, which seems relevant too for non-hilbert spaces
(note, I don't know this stuff, this is simply what I found within 10 seconds of googling)
(and yes, this is me calling you out for not doing so)
dire bluff
operators of mass destruction
I thought there was some elegant concept for traces in infinite dimension 😔
proper crescent
A lot of stuff in finite dimensional linear algebra dies in a fire once you get to infinite dimensions
Functional analysis is a desperate attempt to recover some of it
broken hawk
another fun example: the derivative on the space of analytic functions (with monomial basis) can be represented by the matrix $$\begin{bmatrix} 0 & 1 & 0 & 0 & \dots \
0 & 0 & 2 & 0 & \dots \
0 & 0 & 0 & 3 & \dots \
\vdots & \vdots & \vdots & \vdots & \ddots
\end{bmatrix}$$
stoic pythonBOT
Sascha Baer:
broken hawk
this is an upper triangular matrix with 0s on the diagonal, but it has eigenvectors of eigenvalue not equal to 0
such as (1/0!, 1/1!, 1/2!, 1/3!, ...)
(also known as e^x)
dire bluff
lol
can you really represent derivatives like that?
I mean, I get it's the derivative of the Taylor poly
broken hawk
yes, which is why I said on the space of analytic functions
dire bluff
but does it always work for analytic functions?
broken hawk
only makes sense if you have the taylor poly
analytic means has a taylor series with infinite radius of convergence
in which case in particular derivative and summation may be interchanged
dire bluff
that's wicked
I knew analytic functions had all these well behaved properties
but not that one
broken hawk
seing as you're a german learner, grab yourself einsiedler's notes on analysis :P
you'll find that theorem eventually
quaint heart
@broken hawk
analytic functions are locally represented by power series, and can be represented by different power series in different open sets
So I don't buy your claim
Where did you get that an analytic function has the same Taylor expansion everywhere?
proper crescent
I think Sascha was referring to the real analogue of "entire" functions
And Taylor expanding at 0
But I'll let him clarify since I'm not sure
quaint heart
I guess I would buy that
wintry steppe
I am trying to show that for three vectors u, v and w, if u is orthogonal to both v and w, then u is not in the span of v and w
some help would be appreciated
slow scroll
go for a contradiction. what if u could be represented as some av + bw where a, b are scalars?
wintry steppe
would you then take the dot product of u.v and u.w, but substitute av + bw in for u?
that didn't get me far
slow scroll
you're gonna have to use the linearity of the dot product
or well... inner products in general
wintry steppe
ahhhhhh I think I see
(av + bw) . v
goes to (av).v + (bw).v
which is a(v.v) + b(w.v)
and any vector dotted with itself is 0
so that is just b(w.v) = 0
dusky epoch
and any vector dotted with itself is 0
wrong
only the zero vector dotted with itself gives 0
dusky epoch
no
the dot product of a vector with itself actually gives its length squared
not 1
dusky epoch
what website
wintry steppe
nevermind I just read it wrong - it was talking about the dot product of unit vectors with themselves
yeah my bad again
hahaha
long lily
In what scenario would you use a jordan cycle to find solutions?
dusky epoch
what's a jordan cycle 
warped bluff
@wintry steppe fam i have the stuff for you
wintry steppe
😮
warped bluff
2 secs
wintry steppe
I'm trying to show that if u is orthogonal to v and w (i.e. u.v = 0 and u.w = 0) then u isn't in the span of v and w
so I supposed that it is in the span, so u = av + bw
wintry steppe
gimmie gimmie 
warped bluff
i'll just give you the link
wintry steppe
ok
warped bluff
my teacher's username and password is
JESSICAW2301
jessi246
click on the precalc book
and go to lessons 8.3 / 8.5
you might find something usefyul
wintry steppe
username and password not working
slow scroll
,rotate
stoic pythonBOT
Couldn't find an attached image in the last 10 messages
warped bluff
password is jessi246
stoic pythonBOT
warped bluff
np
wintry steppe
such a joke that loads of high schools do linear algebra but in my highschool we didn't TOUCH it and so at uni lots of people already know stuff
rare scarab
can I ask a question here or does it strictly have to be in questions?
warped bluff
you can ask it her if it pertains to the topic
rare scarab
Ah thanks OwO
so
basically I have to prove that the determinant of a matrix, in which the sum of the elements in each row is zero, equals zero, but I can't use eigenvalues
warped bluff
I am sorry; I am not an expert with matrices, I thought I could just provide some help with the previous question
rare scarab
ooof
the thing's that the teacher said that we should use cofactors
but my classmates and I have been trying to solve it and we have tried everything and yet no clue what we should deliver
if anyone comes up with anything at all that can help it'd be so amazing honestly, literally almost anything
wintry steppe
did you learn lin algebra and matrices etc. at highschool or uni?
wintry steppe
ok good, I feel the same because lots of my uni classmates learnt linear algebra at highschool
slow scroll
why not do row operatations on the transpose?
rare scarab
do what what with the what what?
sorry
English isn't my first language
:c
I'm not sure how some operations are called in English
slow scroll
hmmm ill write it out one sec
wintry steppe
ye
rare scarab
ah danke schön my dude, muchas gracias
slow scroll
im not sure how u would so this with cofactors tho
nahh d00d np
i don't know what to do next, im useless
half ice
@wintry steppe
With your orthogonal question?
wintry steppe
yes 😦
half ice
I am trying to show that for three vectors u, v and w, if u is orthogonal to both v and w, then u is not in the span of v and w
wintry steppe
that's it
rare scarab
@slow scroll Neither do we https://drive.google.com/file/d/1uSk1SMGU8fE4OEe9Z7cmkbwRcojim89w/view
half ice
If u.v = 0 and u.w = 0, then u.(av + bw) = 0. That is, the vector space spanned by v and w is orthogonal to u. Obviously u is not orthogonal to itself, so u is not in the span. @wintry steppe
wintry steppe
@half ice ahhhh. How did you derive the equation u.(av+bw) = 0 ?
wintry steppe
yes but I am not sure where that came from in the sense of the question you see , maybe I am missing something
half ice
Because it's always zero!
Once I got that, the question was done
Any vector in the span of v, w is orthogonal to u
wintry steppe
So if v and w are both orthogonal to u, then any linear combination of v and w is also orthogonal to u
is that right?
wintry steppe
Ok, I think I understand. Thanks @half ice and @dusky epoch
slow scroll
@rare scarab lmaoo those whiteboards and the a's hahahaha
rare scarab
Yeah, we were trying to use cofactors, it just doesn't work
EDIT: I mean, it may work, but if you fill 3 whiteboards trying to solve it yet you're still far from the answer, it probably isn't worth it
If he doesn't admit he answer the way you told me
I'm gonna report him to the mathematics department of my college, I think all of my class has had enough
quaint heart
@wintry steppe the set of all orthogonal vectors to some subspace is in fact a vector subspace called the orthogonal complement
The set of all vectors orthogonal to some vector can be viewed as the orthogonal complement of the one dimensional subspace generated by v
slow scroll
im pretty sure the answer is yes but im not sure i completely understand why.
dusky epoch
(1,2) and (1,1) form a basis for R^2
and you've just specified what your transformation does to them
slow scroll
wait i changed my mind. The matrix representation of a linear transformation isn't unique. So why would this?
dusky epoch
The matrix representation of a linear transformation isn't unique
what do you mean by this
that a linear transformation will have different matrices in different bases?\
slow scroll
yes
dusky epoch
yes, but if you fix a basis, then each linear transformation admits one and only one matrix
slow scroll
the act of choosing eigenvalues (1,2)^T and (1,1)^T fixes a basis for the transformation?
dusky epoch
okay no look
a linear transformation is uniquely determined by what it does to a basis
this is a fact that doesn't even involve matrices in any direct way
do you accept this
slow scroll
yes
dusky epoch
ok
so now our linear transformation from R^2 to R^2
what we're told about it is that it sends (1, 2) to (1, 2) and (1, 1) to (3, 3)
so we know what it does to (1, 2) and (1, 1)
{(1, 2), (1, 1)} is a basis for R^2
therefore there exists one and only one such transformation
does any of this need explanation?
slow scroll
I can't represent the same transformation using a different basis?
dusky epoch
what do you mean
you're essentially asked to construct the matrix of this transformation in the standard basis
slow scroll
suppose i wanted to represent this transformation (call it T) in the standard basis. Could I not define a change of basis G((1, 2)^T) = (1, 0)^T and G((1,1)^T) = (0,1)^T where
H = GTG^-1.
Is H a different transformation? I must be missing something really simple 🤦
dusky epoch
okay so like...
if you want that
then T is the matrix of your transformation in the basis {(1,2), (1,1)}
and T = diag(1,3)
quaint heart
@slow scroll so you know how to do diagonalization right? Well suppose that you have two distinct matrices with this property. Then you can diagonalize both of them and end up with the matrix of the eigenvalues.
long lily
whoever came up with matrix inversion needs to be gunned down in the street alongside their ugly family
quaint heart
But the matrix you used to diagonalize them is the matrix with the eigenvectors in them
So that's the same
So assuming you know about diagonalization this should make sense
slow scroll
ok i understand why transformations are unique now. Had a brain fart moment.
@quaint heart Its not obvious to me that you can't use some scalar multiple of the eigenvectors (since those are also eigenvectors w/ the same eigenvalue) to create a different matrix.
^ like that
quaint heart
Im not sure what your doing
dusky epoch
whoever came up with matrix inversion needs to be gunned down in the street alongside their ugly family
lolwat
slow scroll
instead of using (1,2), i use 3(1,2) = (3, 6) and instead of (-2, -2), i use -2(1, 1) = (-2, -2) for the eigenvectors in my change of basis matrix. these vectors are in the eigenspaces of the same eigenvalues but I thought may produce a different matrix since, well, they create different matrices.
When you multiply everything out though, you get the exact same answer, even though there is seemingly no relationship between the matrices (1 & 1 \\ 2 & 1) and (3 & -2 \\ 6 & -2)
quaint heart
If A is the matrix of eigenvalues for two matrices of eigenvectors P and Q then P^-1 A P = Q^-1 A Q
Is that what your saying?
This is pretty well known I think
slow scroll
taht is?
quaint heart
It's an easy corrolary of diagonalization
slow scroll
wait wait i think i see why
quaint heart
I guess I mean PAP^-1 actually
And QAQ^-1
@slow scroll do you see it?
This is assuming that the eigenvectors form a basis btw. Otherwise I couldn't take the inverse
slow scroll
i thought somehow you might make the argument that there exists an invertible matrix B such that P = QB, but im not really sure tbh
quaint heart
Wait actually I just realized the statement is wrong
quaint heart
Yeah what I said is that if two matrices have the same eigenvalues and they also both have a basis of eigenvectors then they are the same
Which is not true
I think I meant to say that something about if you scale an eigenvector and got carried away
Basically my statement is that you can scale the vectors in the change of basis matrix P by any nonzero constant without changing P A P^-1 if A is a diagonal matrix
@slow scroll is that better?
slow scroll
i mean, i still don't know why you can do that. P with scaled vectors and P without scaled vectors have different entries and are therefore different matrices.
quaint heart
This theorem is actually pretty easy
I think
Multipyling a column of a matrix corresponds to multiplying on the right by a diagonal matrix
All diagonal matrices commute
So the diagonal multiplication matrix and it's inverse cancel out
slow scroll
Diagonal matrices commute with other diagonal matrices, but not arbitrary matrices. Just tested it
slow scroll
oh wait i see
slow scroll
i had this written out eariler:
QBAB^-1 Q^-1 = QAQ^-1
and wasn't sure how to proceed xd
quaint heart
Did that resolve your question?
Of why you can't just multiply the eigenvectors by a constant
slow scroll
i need sleep as well. I have more questions coming tomorrow morning haha
slow scroll
im central time. my sleep schedule has deteriorated as well
thin bloom
Can someone explain to me what it means for a vector to be orthogonal to a subspace. Is a 3d vector that is orthogonal to a plane (which is a subspace) considered orthogonal to that subspace which is the plane?
half ice
Take any vector in the subspace. It is orthogonal to v. Therefore, v is orthogonal to the subspace
thin bloom
So all vectors in the subspace must be orthogonal to the vector v
half ice
Yusyus
thin bloom
Thanks I had the idea but its difficult to think about it in higher dim
half ice
You can think of this as a normal vector to a plane, but the intuition breaks down in higher dimensions
thin bloom
yup, and the calculation is just dot it with the parametric form of the subspace span right>
?
and check if its zero for all vectors
half ice
Exactly
thin bloom
cool beans
half ice
Actually, just dot it with each member of the basis
thin bloom
wat
half ice
Since the dot product is linear, it's enough to check the basis of the subspace
Either way is fine and pretty much the same thing, but you've got options
thin bloom
oh since any linear combination of the basis spans the subspace and with the property if v*(e1 + e2 +... e3)=ve1 +ve2 + ... = 0
Well something like that
half ice
Yes yes
thin bloom
tanks alot
half ice
Np. Feel free to ask if you have anything else
thin bloom
Sorry I need some help again. If a subspace W is orthogonal to another subspace W' then every vector in subspace W is orthogonal to every vector in W'
would the basis vector checking thing work here to ?
But I have to check for each basis with each other
so for example let {e1,e2} be the set of basis vector for the subspace W.
Let {b1,b2} be the set of basis vector for the subspace W'
To check if W is orthogonal to W' do I do b1e1 = 0 b1e2 = 0 and b2e1 = 0 b2e2=0?
wintry steppe
that works yes
thin bloom
ok
Thanks
Still trying to digest it
Is this then a legitimate proof?
this is still a draft but would this proof be valid?
dusky epoch
this would get you a "multiple minor wording flaws" from me
and also the inappropriate use of the zero vector where there should've been a zero scalar, in the first line
thin bloom
Oh yea
So wording wise, should I have used more mathematical notation? or is this preference?
dusky epoch
Let $\vec{x} \in W \cap W^\perp$. Since every vector in $W^\perp$ is orthogonal to every vector in $W$ by the definition of $W^\perp$, we have that $|\vec{x}|^2 = \vec{x} \cdot \vec{x} = 0$. Hence, $|\vec{x}| = 0$, and since the only vector with zero norm is the zero vector, $\vec{x} = \vec{0}$.
stoic pythonBOT
Ann:
slow scroll
Hint on the second part?
second part = proof that $A\bar{\mathbf v} = \bar\lambda \bar{\mathbf v}$
stoic pythonBOT
kxrider:
quaint heart
Can't you just break v up into it's real and imaginary parts?
slow scroll
i tried that one sec...
quaint heart
And then it should immediately follow, I think
slow scroll
slow scroll
i don't even know what to do after that.
My goal was to prove that c = u and d = -w
it didn't look like it was going anywhere, so i stopped
maybe ill just try to do A(v+m) real quick
ehh i don't like that idea either
quaint heart
Suppose lambda=a+ib
Then A(u+iv) = Au + iAv = (a+ib)(u+iv)
So you can find Au and Av by just looking at the real and complex parts of that
Then once you have Au and Av you can find Au -i Av very easily
slow scroll
the complex terms correspond to Av and the real terms correspond to Au, right?
slow scroll
hmhmmm
quaint heart
I did it out on paper, it's not too hard
Just algebra
I guess the big insight is that real matrices are linear operators on C^n as well so you can treat any complex number as a scalar
slow scroll
got it. thank you!
ashen scroll
So the matrix that reflects any vector about the xy axis in r3 is
1 0 0
0 1 0
0 0 -1
but how do I get that?
slow scroll
is it possible for the zero vector to be the eigenvector of a nonzero eigenvalue?
dusky epoch
the zero vector is never an eigenvector of anything
slow scroll
i understand that by some accounts the zero vector doesn't count as an eigenvector (i don't think my book specifically mentions it), but is there any reason that the kernel of A-\lambda I couldn't be {0} for nonzero lambda?
dusky epoch
that just means lambda isn't an eigenvalue of A
slow scroll
oh wait i remember a theorem like that....
dusky epoch
if ker(A - λI) = {0} then det(A - λI) isn't 0
slow scroll
oh yea oh yea thats right i forgot, thanks
native lodge
This is calc question though
dusky epoch
no, it is not
wanton glade
Could someone confirm the answer is 4/5 √5 ?
hallow trail
guys
for vector
if i have 2 vecotors
on the same line
but one of them is x2 that of the other
do they still have the the same vector?
or is one going to be like
2(x,y,z) = (x,y,z)
<@&286206848099549185>
broken hawk
so you have two vectors pointing in the same direction from the origin, but one is twice as long as the other?
then they define the same line, but the vectors are different
one of them would be (x,y,z) and the other (2x, 2y, 2z)
(I am making the extremely reasonable assumption here that you are not working in a projective space with homogenous coordinates, I think if you were you'd know)
(if you don't know what those words mean then everything is fine)
@hallow trail
astral junco
To prove this I need to show that the top two matrices multiply to produce the bottom one, correct?
half ice
Yes. You'll need the sum identities
astral junco
ok i think i tried but couldn't produce the identities. one sec i'll post what i did
i must have made a mistake along the way? possible to make row one col 1 of the last matrix into cos(theta_1+theata_2)?
half ice
And you're done, lol. Just need to apply the sum identities
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
sin preserves the sign
cos does not
astral junco
ah darn it. my trig cheat sheet didn't work too well. thanks for the help. much appreciated 😃
forgot about the sign
wintry steppe
How do I find the minimal polynomial? I know how to find the characteristic, but what to do after that?
native lodge
is this like eigenvalue stuff? reading wiki on this as fast as I can lol
humble rapids
Weird question, but can 3rd dimensional volumes in 4th dimensional space have intersections which are planes?
native lodge
what was your result for 9a?
native lodge
so graph that and you're good to go 
jaunty fern
oh really, why does it say 0≤n≤10?
native lodge
because those are the values of n they want you to use
to let the x axis be n and the y axis be C
lusty robin
People are getting this channel mixed up for linear functions 
slow scroll
linear algebra deals with linear functions
lusty robin
This is more matrices and stuff
quaint heart
Matrices are linear functions
inner isle
native lodge
god tier screenshot
use window + shift + S to copy sections of screen onto clipboard for high quality screenshot you can paste
timid ivy
If V is a finite dimensional vector space, then finding the adjoint, T*, of a linear operator T is easy because $[T^]\beta = ([T]\beta)^$. However, what if V is infinite-dimensional? Is there then some way to prove that T* exists other than actually trying to find it explicitly?
stoic pythonBOT
markus:
empty copper
Depends on the type of vector space. You can't even define adjunction without having some sort of inner product, unless you delve into the realm of dual operators
timid ivy
Sorry, but I meant an inner product space
timid ivy
Interesting! However, this goes a bit over my head tbh. What is a Hilbert space?
lyric sphinx
A Space with an inner product that is complete for the norm induced
broken hawk
that is, it's a vector space with an inner product, and the additional condition that every cauchy-sequence converges, i.e. if you have a sequence $(a_n)$ with the property that for every $\varepsilon > 0 \exists N \in \mathbb{N}$ such that $|a_n - a_m| < \varepsilon$ for any two $n,m > N$ , then this sequence has a limit $a$ in your space and $|a_n - a| \to 0$
examples include $\mathbb{R}^n$ with the usual inner product, $\ell^2$ the space of sequences in $\mathbb{R}$ which are square-summable, and $L^2$, the space of functions over $\mathbb{R}$ whose square is integrable, up to almost everywhere equivalence
stoic pythonBOT
Sascha Baer:
broken hawk
thanks texit but what about the other one
stoic pythonBOT
Sascha Baer:
broken hawk
aight, read those in the opposite order and ask if you have questions
timid ivy
That is clear. Thank you! 😊
native lodge
for small $\theta$, $\sin(\theta) \approx \theta$
stoic pythonBOT
⚡Amphy⚡:
gray glen
$\sin(ax^n) \approx ax^n$
stoic pythonBOT
CaptainLightning:
gray glen
also get out of this channel 
ember pilot
slow scroll
i understand this is just a matter of constructing matrix with an eigenvalue of algebraic multiplicity > geometric multiplicity but im completely lost on how to do this 😦
quaint heart
Do you know about Jordan normal form?
slow scroll
nope
slow scroll
you wouldnt happen to know of a way to solve this with only knowledge of diagonalization?
quaint heart
Just play around a bunch
It seems like that's probably what the question wants you to do anyway
Or you can Google Jordan normal form
slow scroll
how do you construct a nilpotent matrix? I know from looking ahead to the next problem that those aren't diagonizable lol
quaint heart
It's pretty easy actually
Pick a basis
Map one of the basis elements to 0 and the others to the first basis element
@slow scroll
So for example the matrix
[0 1 1]
[0 0 0]
[0 0 0]
Is nilpotent
That's an easy way to make one
slow scroll
i see. thanks!
quaint heart
You can make higher order one's as well
slow scroll
yea something like A=
[0 1 1]
[0 0 1]
[0 0 0]
would be A^3 = 0 i think
quaint heart
Yep
Is the highest you can make 3?
For the 3d ones?
Think about it
Wlog you can assume the first column is all 0's btw
There's actually an easy classification using jordan normal form
wanton zenith
can someone discuss eigen-decomposition please? is there a particular method to do it or is it all analyzing the system case by case?
slow scroll
is that just diagonalization?
wanton zenith
yea, diagonalization with eigenvalues across the diagonal
slow scroll
you find the eigenvalues, then you use the eigenvalues to find the eigenvectors
wanton zenith
is there a special case when eigenvalues are not all distinct?
generalized eigenvectors?
slow scroll
they don't have to be distinct, but the dimension of the eigenspace has to equal the algebraic multiplicity of the eigenvector
wanton zenith
ok, thanks
slow scroll
np
for part a, i got that all of the eigenvalues are 1. This means the diagonal matrix is the just the identity which would imply that the diagonalization of this transformation is just itself.
Does the case where A = P^-1 I P count, or does that fall under non-diagnoalizability?
oh shoot i think i messed something up....
forgot about antisymmetric matrices
okay so now the eigenvalues are 1 (dim(eigenspace) = 3) and -1 (dim=1). So that would be diagonizable as far as i can tell. My question still stands though. What do we say about diagonalizability when a matrix is similar to the identity?
quaint heart
What do you mean?
slow scroll
err yea oops thats what i meant xd
quaint heart
Why is the dimension of the subspace of antisymetric matrices 1?
dusky epoch
antisymmetric matrices of what size?
slow scroll
2x2
quaint heart
I wouldn't think that would be true
dusky epoch
here's a one-element basis: { [ 0 1 ; -1 0 ] }
slow scroll
P^-1 I P = I
it seems strange to me that that happens tho. its kinda like we divided by zero or something
yea thats what i did
quaint heart
Oh right, I forgot antisymmetric matrices are 0 on the diagonal
oof
@slow scroll If you think about what the identity map does, it shouldn't seem so strange
slow scroll
idk it does for me. It doesn't quite make sense to me that a transformation with eigenvalues all equal to 1 disables us from talking about the transformation in terms of a basis of eigenvectors