no
#linear-algebra
stoic pythonBOT
Ann:
dusky epoch
so it has a dimension
viscid apex
ok
dusky epoch
and from the fact that $\sum_{i=1}^{n-1} x_i v_i = u$ has a unique solution (this is taken straight from the problem statement), you can deduce right away that ${v_1, ..., v_{n-1}}$ is linearly independent, and so $\dim(\mathrm{span}{v_1, ..., v_{n-1}}) = n-1$
stoic pythonBOT
Ann:
viscid apex
ok
following you
ok its not a proof we have done yet but i can deduct it from a n+1xn matrix with a unique solution has a unique solution to every free variable.
uh i get what youre digging at
wait lemme type out what im thinking
$x_{1}v_{1}+ ...+x_{n-1}v_{n-1} = u$ has a unique solution, therefore $x_{1}v_{1}+ ...+x_{n-1}v_{n-1} + tw= u$ also has one unique solution t = 0, because it if had more it would imply a solution exists to $x_{1}v_{1}+ ...+x_{n-1}v_{n-1} = w$
lemme fix up the tex
stoic pythonBOT
kakamaika:
viscid apex
therefore they span the field
▮?
$x_{1}v_{1}+ ...+x_{n-1}v_{n-1} = u$ has a unique solution, therefore $x_{1}v_{1}+ ...+x_{n-1}v_{n-1} + tw= u$ also has one unique solution t = 0, because it if had a solution $t\neq 0$ we can do $\frac{1}{t}(x_{1}v_{1}+ ...+x_{n-1}v_{n-1} + u)= w$ and insert our unique solution to the first equation giving us a unique $x_{1}v_{1}+ ...+x_{n-1}v_{n-1} = w$
stoic pythonBOT
kakamaika:
viscid apex
I... think this is tight?
and then i just need to show a unique solution to that carries spanning the field since we havent proven that yet
man im dumb it cancels out when u moves sides
still works tho cus then it would imply t = 0 which we can easily show is not possible i guess
dusky epoch
$\mathrm{span} { v_1, ..., v_{n-1} } \subsetneq \mathrm{span} { v_1, ..., v_{n-1}, w } \subseteq \bbR^n$
stoic pythonBOT
Ann:
dusky epoch
$n-1 = \dim \mathrm{span} { v_1, ..., v_{n-1} } < \dim \mathrm{span} { v_1, ..., v_{n-1}, w } \leq \dim \bbR^n = n$
stoic pythonBOT
Ann:
dusky epoch
@viscid apex $\dim \mathrm{span} { v_1, ..., v_{n-1}, w }$ is forced to be $n$, and that can be argued as above
stoic pythonBOT
Ann:
dusky epoch
you really don't need any of those fussy arguments there lmao
viscid apex
kidna need it cus ithink your relying on proofs that we havent done yet. weve just touched spanning
dusky epoch
huh?
i mean
$V \subseteq W$ implies $\dim V \leq \dim W$ for any vector spaces $V, W$
stoic pythonBOT
Ann:
dusky epoch
that's really the most complicated thing i'm relying on
$\mathrm{span} { v_1, ..., v_{n-1} } \subsetneq \mathrm{span} { v_1, ..., v_{n-1}, w }$ follows immediately from the problem statement since $w \notin \mathrm{span} { v_1, ..., v_{n-1} }$
stoic pythonBOT
Ann:
viscid apex
i know and I understand the logic but if I put this on the exercise its gonna raise eyebrows a bit
dusky epoch
how come
viscid apex
because its obviously not something I could have come up on my own
$n-1 = \dim \mathrm{span} { v1, ..., v{n-1} } < \dim \mathrm{span} { v1, ..., v{n-1}, w } \leq \dim \bbR^n = n$
stoic pythonBOT
kakamaika:
viscid apex
this is the part thats giving me troubles
we havent really defined dimensions properly yet and neither span (of vectors)
anyhow thanks
wintry steppe
@dusky epoch
viscid apex
what would you define dimensions as
might have one and im not understanding the term
i know a span is all possible linear combinations
wintry steppe
the dimension of a vector space is the number of vectors in its basis (over the base field)
viscid apex
nah we dont have that
...its a bad program dont tell me about it its not out of choice that im learning here
well probably go over all this in a month or ill go over it next week if i have time
thanks anyhow ill insert my fiddly answer for now
lean aspen
@dusky epoch what 's the diff between $A^T,A^\dagger,A^\ast$
stoic pythonBOT
flimflam:
lean aspen
idk if there's actually any T, might just be a truncated dagger
@jagged pendant
@tranquil ermine are these all adjoint notations
jagged pendant
dusky epoch
@evеryone
viscid apex
yeah i took a peek in the book this is all talked about in a month or so
broken girder
applying the Gram Schmidt process is easy, but how do I check if the obtained set is orthonormal in a single matrix multiplication step, especially since the inner-product is not the standard dot product in R^n (and instead is the weighted inner-product) ?
(simply checking if the columns form an orthogonal matrix doesnt work in this case)
hang on, i think i've got it. we can check if E^T W E = Identity or not
wanton spoke
Yep, the diagonal elements of EtWE are (ei,ei) and the other elements are (ei,ej)
You could also just calculate (ei,ei) and (ei,ej) separately instead
wintry steppe
hey guys! how do you guys determine if vectors lie on a plane?
and how do guys determine if those vectors lie on a line?
ex.
slow scroll
v3 would lie on the plane formed by v1 and v2 if v3 is a linear combination of v1 and v2.
The vectors lie on a line if they are a linear combination of just one other vector.
as for actually finding out if they are linearly independent or not, I would just write v1,v2, v3 as column vectors in the matrix [v1 v2 v3] and put that matrix in row echelon form. That way, if there is any linear dependence, it will be nice and easy to see
wintry steppe
@slow scroll so basically, if vectors are linearly dependant, it lies on a plane
slow scroll
or a line
wintry steppe
and if v1 = cv2, it lies on a line?
slow scroll
yes
wintry steppe
so u can have vectors that lie on planes but not on lines?
slow scroll
sure. cv1 + dv2 where c!=0 and d!=0 are all vectors that lie on planes but not lines
wintry steppe
what is a way that i can say that those vectors do not lie on a line?
as in, since there are no scalar that can make v1 = v2, these vectors do not lie on a line
slow scroll
yea u could say that
v1 is not a linear combination of v2 or v3, and v2 is not a linear combination of v3, so none of the vectors are linear combinations of each other
slow scroll
if you put the the vectors in a matrix, and put the matrix in row echelon, one row will zero out if one of the vectors lie in the plane spanned by the other two vectors, and two rows will zero out if two of the vectors are linear combinations of each other (lie on a line) @wintry steppe
np
dull kettle
Hello guys, is II and III the correct statements?
dusky epoch
if (II) is correct then (I) must be as well 😛
dull kettle
Since A is a square matrix of order 4, can I say that its row space has dimension 4?
4 different vectors
dusky epoch
no, just because A is a square matrix of order 4 does not by itself mean its row space has dimension 4
dull kettle
because the row space is determined by the subspace?
dusky epoch
what you just said is nonsensical
dull kettle
sorry im quite lost, not sure why if A is a square matrix of order 4, it doesn't by itself mean that row space has dimension 4
dusky epoch
because the dimension of the row space of A is equal to the rank of A
dull kettle
what does rank of A means?
dusky epoch
i mean ok that's kinda the definition really
the rank of A is defined as either the dim of its row space or the dim of its col space (those are always the same)
but like for example the row space of the zero matrix (of any size) is {0} and so has dimension 0
or for a less trivial example, the dimension of the row space of $\begin{bmatrix} -1 & 5 & 1 & 3 \ -1 & 5 & 1 & 3 \ -1 & 5 & 1 & 3 \ -1 & 5 & 1 & 3 \end{bmatrix}$ is 1
stoic pythonBOT
Ann:
dull kettle
is that why order 4 doesn't necessary mean dimension 4
dull kettle
unless we know the values of the entries, we can't tell the dimension base on the order correct?
dusky epoch
beyond knowing that the dimension can never be greater than the order, yes, we cannot say anything
dull kettle
Thanks! For this question, what does orthogonality has got to do with subspaces?
timid ivy
Think R^3 first
dusky epoch
check whether V, as defined in the problem, is a subspace of R^4
timid ivy
Maybe it will help you understand the problem from a geometric perspective
dull kettle
how do I check without knowing the values
timid ivy
Two vectors $u,v$ are orthogonal if their innerproduct $\langle u,v \rangle = 0$, if I remember correctly
stoic pythonBOT
markus:
dull kettle
yes
timid ivy
So you need to check if the set ${\mathbf{v} \in \mathbb{R}^4 : \langle \mathbf{u}, \mathbf{v} \rangle = 0}$ is a vector space
stoic pythonBOT
markus:
timid ivy
subspace*
dusky epoch
to check whether something is a subspace you show that it satisfies the definition of a subspace
timid ivy
of R^4
dusky epoch
if you don't know what a subspace is, go back and read your notes
timid ivy
or just check the easiest: If v and w is in V, is then v+w also in V? Why/why not?
See here for an example of subspace check, you are basicly just checking certain axioms. Many of them are inherited by V just being a subspace of some bigger vector space
dusky epoch
subset
there are only two things to check for subspaces
closure under addition and closure under scaling
dull kettle
Show it is closed under addition.
Show it is closed under scalar multiplication.
Show that the vector 0 is in the subset.
dusky epoch
you don't need the third thing
it follows from the other two
(and nonemptiness)
it's a good exercise to prove that
timid ivy
That the 0 vector is in V is very easy to verify here tho
dusky epoch
it is, but it's extraneous to do so
timid ivy
Yeah ofc
Nice problem btw Ann ^
Or exercise, dunno if those words are synonyms or not in mathematics
dull kettle
how am I gonna check the conditions without knowing the exact values?
timid ivy
Take $u,v \in V$. What properties do they have?
stoic pythonBOT
markus:
dull kettle
dot product of u and v equals 0
timid ivy
Oh bad notation, because u is the vector everything in V is supposedly orthogonal to
so lets take v,w instead
dusky epoch
yeah u is fixed
timid ivy
Yeah, so if $u+w$ is also supposed to be in V, what should you verify then?
stoic pythonBOT
markus:
dusky epoch
v+w
dull kettle
dot product of v + w and u equals 0
timid ivy
Exactly
Now, how can you manipulate the expression $\langle v+w,u \rangle$ algebraicly?
stoic pythonBOT
markus:
dusky epoch
(this comes down to one of the axioms of inner products)
dull kettle
how the dot product is defined?
timid ivy
yeah, like how do you for example calculate $u \cdot v$ if u and v is in R^2
stoic pythonBOT
markus:
Compile Error! Click the 
reaction for details. (You may edit your message)
dusky epoch
$ \left< x+y, z \right> = \left< x, z \right> + \left< y, z \right>$
stoic pythonBOT
Ann:
timid ivy
Now you’re giving it away 😦
dull kettle
u1 v1 + u2v2 = 0
timid ivy
Yes, correct
timid ivy
So why should the formula Ann wrote over be true
It says that $(v+w)\cdot u = v \cdot u + w \cdot u$
stoic pythonBOT
markus:
timid ivy
for vectors $v,w,u$ in some euclidean vector space
stoic pythonBOT
markus:
timid ivy
And what can you then say about the dot product of v+w and u using that formula?
You are very close!!
dull kettle
that the condition for it to be subspace is satisfied
timid ivy
Yes, or it is closed under addition
Now, the only thing left to prove is that $(av) \cdot u$ is in V
stoic pythonBOT
markus:
dull kettle
it doesn't satisfy the condition
stoic pythonBOT
markus:
dull kettle
dot product of v and u equals 0 as defined earlier
but if its av and u, it might not be 0
?
timid ivy
No, remember what $(a \overrightarrow{v}) \cdot \overrightarrow{u}$ actually means!
stoic pythonBOT
markus:
dull kettle
oh, v is not fixed
timid ivy
It is $av_1u_1 + av_2u_2 + \dots$. Now, factorize out $a$, what happens
stoic pythonBOT
markus:
timid ivy
?
dull kettle
yeah but how do we determine its dimension
timid ivy
Okay, R^4 is hard to think of geometrically, so to get a intuitition for what the dimension should be, lets first go down to R^2
If you’re given a vector in R^2, what does any orthogonal vector to it have as angle to that vector?
what is the angle between two orthogonal vectors in R^2?
dull kettle
90
timid ivy
Yes! So what do that vector span out?
dull kettle
R^2?
timid ivy
No, thats not quite right.
Imagine $u$ being the vector $u=(1,0)$. Then any orthogonal vector to u needs to be a vector on the form $(0,a)$. Agree?
stoic pythonBOT
markus:
timid ivy
okay, so what do all vector on the form $(0,a)$ span out in this case?
stoic pythonBOT
markus:
timid ivy
Where can you go by adding those?
dull kettle
straight line through origin
timid ivy
exactly
And what is a straight line to the origin
What vector space can we think of as that?
dull kettle
R^2?
timid ivy
No, what is a straight line?
Remember R^2 are tuples of two points
Remember, to span out a straight line you only need one basis vector. How many basis vector do R have, and how many do R^2 have?
dull kettle
R^2 has two basis vectors?
dull kettle
one
dull kettle
straight line
dull kettle
1
timid ivy
Perfect!
Okay, so know we know that in R^2 the vector space V consisting of orthogonal vectors to u have dimension 1.
Lets move up to R^3.
If $u=(1,0,0)$, and which form is the two other (linearly independent) orthogonal vectors to $u$?
stoic pythonBOT
markus:
timid ivy
Think geometrically of the axis
dull kettle
(0,1, 0) , (0, 0, 1)
stoic pythonBOT
markus:
dull kettle
R^2
timid ivy
Yes great job
dull kettle
wow
timid ivy
So what vectors are orthogonal to $(1,0,0,0)$ in R^4?
stoic pythonBOT
markus:
Compile Error! Click the reaction for details. (You may edit your message)
timid ivy
Yes great job dude!
dull kettle
thank you!!! can I ask a few more questions? these are the ones where I'm not sure
dusky epoch
go ahead
timid ivy
Just happy I could help @dull kettle
dull kettle
For this question, my answer is I is false, II is true but I'm not sure about III
broken hawk
what does it mean for u not to be in span(S)?
dusky epoch
yes indeed I is false and II is true
broken hawk
in terms of linear dependence, I mean
dull kettle
what does it mean for "u does not belong to V"?
broken hawk
V is a subspace of a larger vector space
so you can have vectors which are in the larger space, ℝ⁵, but not in V
dull kettle
oh, is statement III true then? since you want span(S) which has dimension 3 and u which has dimension 1
dusky epoch
@dull kettle $u \notin V$
stoic pythonBOT
Ann:
dull kettle
so what would span (S U {u}) be? 3?
broken hawk
span is not a number
it’s a space
the span of S union {u} is the set of all vectors you can get to by linear combination of those vectors in S union {u}
dull kettle
sorry i meant dimension of span (S U {u}))
broken hawk
well, that’s what we have to find out, right?
so, u is not in V
so u is not in Span(S)
(becasue V = Span(S) by definition of a basis)
right?
dull kettle
yes
broken hawk
so can u be linearly dependent with the vectors in S?
dull kettle
no
broken hawk
why?
dull kettle
its one of the condition for basis
broken hawk
yea but you don’t know if S union {u} is a basis yet
here’s a quick argument
let’s say S = {v1, v2, v3}, right?
dull kettle
ok
broken hawk
now, if {v1, v2, v3, u} was linearly independent
then you could find coefficients a1…a4 with
a1*v1 + a2*v2 + a3*v3 + a4*u = 0
by definition of linear dependence
agreed?
dull kettle
yes
broken hawk
now you can solve that for u, you’ll get
u = -a1/a4 v1 - a2/a4 v2 - a3/a4 v3
so u is actually in the span of S
but we know it’s not
contradiction
therefore, {v1, v2, v3, u} is linearly independent
and as such the basis of a 4-dimensional subspace
dull kettle
you mean "if {v1, v2, v3, u} was linearly dependent" instead of independent right?
dusky epoch
yes
broken hawk
yes, sorry
dull kettle
how does u = -a1/a4 v1 - a2/a4 v2 - a3/a4 v3 shows that u is actually in the span of S?
what does u have to be for it not to be in the span of S?
0?
dusky epoch
what does it mean for u to be in the span of S?
dull kettle
it was written above
u = -a1/a4 v1 - a2/a4 v2 - a3/a4 v3
so u is actually in the span of S```no, forget about that. can you give me the definition of "u is in the span of S"?
dull kettle
it belongs to the span, so span(v1, v2, v3, u)
sage mauve
what is a span?
dull kettle
consists of all the linear combinations
dusky epoch
so what does it mean that u belongs to the span of {v1, v2, v3}?
dull kettle
u consists of linear combination of v1,v2 and v3
sage mauve
and have you found such a linear combination?
dull kettle
yea sorry bad phrasing
dusky epoch
u = (-a1/a4)v1 + (-a2/a4)v2 + (-a3/a4) v3
dull kettle
and yes there's such a linear combination
For this question, I and II is true but i'm unsure about III
dull kettle
Statements I and II are correct, am I correct?
dusky epoch
does your course consider real/complex vector spaces only, or over arbitrary fields as well?
(if you don't know what i'm talking about it's likely the former)
dull kettle
former it is then
dull kettle
thanks! can you look at the question above the latest one, i'm tryna make sense of III
can't see how does random non zero vectors would give me basis for solution space of Ax = 0
dusky epoch
yeah they might not even be LI
and if they aren't LI they aren't gonna make a basis for anything
dull kettle
okay, in the same question, for statement I, is it false? since say if the solution space is {v1, v2, v3} then v1 + v2 would be not in the solution space anymore
dusky epoch
for \textbf{any} matrix $A \in \bbR^{m \times n}$, the set of solutions to $Ax = 0$ will \textbf{always} be a subspace of $\bbR^m$
stoic pythonBOT
Ann:
dusky epoch
and you should check that
dull kettle
sorry but how do I check it?
dusky epoch
use the definition lmao
let $x$ and $y$ be such that $Ax = 0$ and $Ay = 0$, and let $c$ be a constant.
stoic pythonBOT
Ann:
dusky epoch
is it true that $A(x+y) = 0$? is it true that $A(cx) = 0$?
stoic pythonBOT
Ann:
dull kettle
yes and yes
dusky epoch
voilà
dull kettle
thank you! @dusky epoch
broken hawk
oh btw @dusky epoch the reason you have to check 0 in a subspace is to ensure nonemptyness
random wasp
A(0) = 0 o:
broken hawk
yea I know, but I listed three properties to check that sth is a subspace and ann complained that 0∈W is redundant
cause you can prove it from x-x
but what if there is no x
the empty set isn’t a subspace
random wasp
fair fair o:
broken hawk
and I just remembered why you need it
so yea technically you don’^ have to check 0∈W, but you have to check something ∈ W
and 0 is usually the easiest anyway, so you might as well
random wasp
yeah in linear algebra 1, our definition of a vector subspace was for it to be non-empty + other two
whereas in linear algebra 2, our definition of a vector subspace was for it to contain 0 + other two
which as you mentioned, are obviously equivalent
I wonder if there's any reason to why someone people prefer one over the other
broken hawk
our definition was “a subset which is a vector space with the same operations”
the other characterization was a theorem
random wasp
I would have prefered that tbh, makes more sense chronologically o:
.> makes me wonder how many other 'definitions' I know are just conveniences
broken hawk
our definition for a matrix was “a rectangle with numbers”
wintry steppe
lol
dusky epoch
our definition for a matrix was "a map {1, ..., m} x {1, ..., n} -> R"
where R is the base ring
broken hawk
wait what
proper crescent
That's a way to formalize the same idea
The image of (i,j) is the i,j coordinate of the matrix
broken hawk
oh
yea, fair
did you do linalg over general rings?
ie did your linalg course just straight up do modules?
dusky epoch
we started with modules yeah
broken hawk
well that’s certainly a curious approach
dusky epoch
and then used that module theory to build our way up to JNF and all that jazz
proper crescent
Jordan form = trivial corollary of STFGMPID
broken hawk
at my school you only really study modules if you take commutative algebra
dusky epoch
STFGMPID?
dusky epoch
what an abbreviation
broken hawk
wehre we proved the classification theorem for finitely generated modules over PIDs
which may be STFGMPID come to think of it
proper crescent
Yup
broken hawk
what’s the S? structure?
proper crescent
Yeah
broken hawk
I’ve seen different versions of that interestingly
proper crescent
Lmao I love how I can just say that and people can guess what it means
broken hawk
the one we proved was that $$M \cong R^n \oplus R/(a_1) \oplus \dots \oplus R/(a_m)$$ where $a_i$ are neither units nor zero, and $a_1 \mid a_2 \mid \dots \mid a_m$ and $m, n, (a_i)$ are all unique
stoic pythonBOT
Sascha Baer:
broken hawk
the proof was really weird tho
and I couldn’t really follow it
it suddenly involved a bunch of linalg over PIDs, which we’d of course never done
proper crescent
Smith normal form?
broken hawk
he did a thing where he found some matrix representing a module homomorphism, and then did something akin to gaussian elimination to it
proper crescent
Yeah that's Smith normal form
It's where you have a "diagonal" matrix and a_{ii} divides a_{i+1 i+1}
random wasp
@dusky epoch wait, you started linear algebra with modules? o.0
that sounds... really unconventional
dusky epoch
we also had rings before groups 😛
gray glen
in soviet russia 
random wasp
tbh, we had fields before groups
but it's not like we talked about any real field theory, just field theory about the reals
... I'm sure that makes sense lol
proper crescent
Spivak chapter 1? 😛
broken hawk
we also had rings before groups
in Algebra
but we’d already seen both in linalg and analysis anyway
we just did ring theory first
bleak cairn
Can anybody here explain what M-matrix and Stieljes matrix are? they confused me 😦
broken hawk
we had like…
in linalg:
-groups, rings and fields: definitions and most basic properties
-vector spaces
in analysis:
-ordered fields
in algebra:
-ring theory
-group theory
-field extensions and modules (only very brief intro each)
-galois theory (ongoing)
the linalg and analysis stuff was in parallel and took place within the first two weeks of the course
algebra was one semester, except galois theory which is algebra II
dense holly
Can three vectors be LD in r4?
frosty vapor
wintry steppe
of course; even a set of one vector can be linearly dependent
lean aspen
@wintry steppe only if it's zero lol
@dense holly one salient point is ever pairing can be lin independent but the set as a whole can still fail to be linearly independent
gray glen
matrices and vectors and all that shit
slow scroll
algebra
its algebra, but its linear
well a lot of LA is derived from the concept of creating linear functions between vector spaces. Linear functions are functions that satisfy f(ax_1 + bx_2) = af(x_1) + b(fx_2)
pastel aspen
and those functions are the main link between most subfields of linear algebra
like matrices for example
slow scroll
.... and vector spaces are just sets along with the operations of scalar multiplication and addition. An analogy would be f(x) = x^2 + x as a function which takes a real number input under the operations of addition and multiplication. (Sets under multiplication and addition are called fields.)
Think of mathematical vectors and vector spaces as like a super generalization of physics vectors
Yeah a function is just something that assigns an input to a unique output
hoary osprey
quiet ferry
can i ask question here? i saw the almost full moon today. and i was wondering if there is a way to calculate where on earth i have to stand to look up and be directly under the moon when it's completely full. i'm sorry if i ask it in the wrong place. just really curious
barren plank
sounds like #geometry-and-trigonometry
but als you would need the moon's ephemerides
i.e numerical data on where the moon is
rotund quail
lean aspen
the basic idea with spectral theorem is selfadjoint operators on a nice enough Banach space are unitarily similar to a diagonal operator?
for complex ips
@honest swift is this the gist of it?
or is it not that simple once you're not fdvs
honest swift
It is not that simple. Without an inner product what is unitary? Are your operators bounded?
Even defining self-adjointness in general requires some care when operators are unbounded, as they often will be.
(Note that symmetric and self-adjoint are not interchangeable terms in this setting.)
austere roost
does anyone know riemann notation for distance between 2 points in a non euclidean space?
honest swift
take the infimum of the lengths of all smooth curves joining your points, with curve length being defined using the Riemannian metric and integration.
austere roost
its some kind of ds=dx²+dy²+dz² but a liitle more complex with news terms
austere roost
mb
lean aspen
@honest swift was gonna just assume compact operators, and you still have an inner product
honest swift
You said Banach, not Hilbert.
honest swift
Compact self-adjoint is the nicest case basically.
Banach with its specified norm induced by ip = Hilbert more precisely.
lean aspen
is the norm ever not ip induced
honest swift
yes, not all banach spaces are hilbert spaces
austere roost
thanks charles catto
lean aspen
oh ya 
broken hawk
can you induce the supremum norm with an inner product?
honest swift
nope, try to prove this.
honest swift
🐱
proper crescent
You could use Riesz rep and say try to show it isn't reflexive
Wait actually actually LMAO
I have the best proof for this
Everyone get some popcorn
So, subspaces of reflexive spaces are reflexive, right?
Well, it turns out every separable Banach space embeds as a closed subspace of C[0,1]
So to show C[0,1] is not reflexive, we just have to show that some separable Banach space isn't. L^1
@honest swift what do you think?
honest swift
Lol that's a good one.
proper crescent
It's definitely gonna be easier to use just the parallelogram law if you wanna show C[0,1] isn't Hilbert but is it easier to show that it's not reflexive?
I guess you can use Eberlein-Smulyan and ask whether the unit ball is weakly sequentially compact
wintry steppe
i'm dumb someone please check #prealg-and-algebra
proper crescent
Please don't message channels other than the one you posted your question in
wintry steppe
:(
proper crescent
Anyway I think the dual of C[0,1] is the space of signed Borel measures by Riesz rep? So we'd wanna find a sequence of functions f_n, let's say of sup norm 1, so that no subsequence converges weakly
Wait hmm so to converge weakly you need that every measure works, so let's be ballsy and say x^n with Lebesgue measure
stoic pythonBOT
Darisal:
proper crescent
Tbh this idea now that I think about it isn't too far off from embedding L^1 in it
honest swift
Yeah will think about this another day @proper crescent , have just gotten home after 6 hours of grading hooman noobery.
proper crescent
Lmao, isn't this a break from that? 😛
But yeah I'm just thinking out loud more or less
Also wait I love how this is in linear algebra, some poor first year is gonna wonder what it's all about and then a barrage of words flying around
honest swift
Haha kind of, but I don't want to look at any math for a little while :p.
Lmao yeah.
proper crescent
I don't like thinking of the dual of C[0,1] so I'm gonna try something else
Wait actually if C[0,1] were reflexive, then it would be separable + reflexive, so the dual would be separable as well
That would imply the unit ball of C[0,1] were weakly metrizable which feels very wrong, also I just don't think the space of signed Borel measures on [0,1] is separable
Oh wait a sec
LMAO
Just take dirac masses
Yeah gg
honest swift
🐱 👍
warped sinew
is this where we deal with vectors?
i just wanna check if i'm doing some things right
round wharf
Yeah there are vectors in linear algebra haha @warped sinew
A lot of the basis of linear algebra is based on vectors
slow scroll
yes the basis hAha
round wharf
xddddddd
flat pasture
hey guys i have a question regarding eigenvalues
so my eignvalues are 2 and 3
and i got x1=1 and x2=1
but x2 is supposed to be negative one
<@&286206848099549185>
this is for my eigenvalue of 3
upper token
i got
,, \begin{pmatrix}-1&1\ 0&0\end{pmatrix}\cdot \begin{pmatrix}x_1\ x_2\end{pmatrix}=\begin{pmatrix}0\ 0\end{pmatrix}
stoic pythonBOT
0031:
upper token
idk
upper token
same
upper token
not me
upper token
i watched like 2 vids on this so idk
round wharf
Doesn’t that mean the solutions are on a plane or something?
the set of solutions is a plane
i think
winter reef
X1=x2 So V(3)=span of 1,1
lean aspen
do you ever care about the complex conjugate untransposed or the transpose not complex conjugated with complex matrices?
also is there any standard notation if so
bitter root
I have two vectors D and N (direction & normal_of_collision)
I want to calculate a correction vector that is perpendicular to N and is based on the direction D points.
Does someone know how to do this?
wintry steppe
ummm
you want a vector in direction orthoganal to N
but you also want it to be based on the way D points?
this seems contradictory my dooooood
bitter root
Figured it out
project D onto N and subtract it from D
broken hawk
@lean aspen just like, $\overline{A}$ for conjugation
stoic pythonBOT
Sascha Baer:
Compile Error! Click the reaction for details. (You may edit your message)
broken hawk
I don’t think you would care about it a lot, but I mean it can still be useful to have notation occasionally
conjugate-transpose doens’t have stnadard notation, I’ve seen at least three different notations
$A^*, A^H, A^\dagger$
stoic pythonBOT
Sascha Baer:
dusky epoch
$\overline{A}^T$
stoic pythonBOT
Ann:
dusky epoch
broken hawk
$\overline{A^T}$
stoic pythonBOT
Sascha Baer:
broken hawk
dusky epoch
graceful summit
slow scroll
this is not linear algebra xd
multiply both sides by 7
limber sierra
More proof that "linear algebra" is an incredibly unfortunate name
Though I don't really have a better proposal
"matrix algebra" is too specific and obscures the motivations, while also not really being correct
slow scroll
they should just change the name of this channel to "vector space math" or something
limber sierra
Maybe "algebraic representations"? Nah nvm that sounds clunky af and too broad
Uhhh
wicked trellis
Anyone have an idea how to do 1?
half ice
@wicked trellis
The left side is A¯¹b
So, you're given that
[1]
A¯¹b = [2]
[3]
Multiply on the left by A:
[1]
b = A[2]
[3]
And thus, that's your solution to Ax = b
wicked trellis
Thanks 😃
half ice
Np. Feel free to ask if you have anything else!
wicked trellis
So multiplying a matrix by its inverse will equal 1?
wicked trellis
Oh ok
solid basalt
Hey guys I need some help on my hw, here is a picture of one of my many questions I’m confused about
wicked trellis
Ok
solid basalt
half ice
Have the characteristic polynomial?
solid basalt
umm lol should that be given?
half ice
If you don't that's k we can find it
solid basalt
I don't know it
half ice
A - Ix =
[2-x 0 -3]
[-1 -1-x -3]
[1 3 6-x]
The characteristic polynomial is the determinant of that.
stoic pythonBOT
Requested by Kaynex#7520
half ice
There we go. That's the polynomial
solid basalt
okay so from there where do we go in my problem?
stoic pythonBOT
Requested by Kaynex#7520
half ice
Also going to need that
Now, I don't know what lemma 1 is. Your book must make a reference to it
half ice
Well, it's easy to check if these matrices satisfy the lemma, but I'm not sure where they come from lel
Let me look it up
Oh thank god Ann is here
dusky epoch
@wicked trellis the identity matrix plays the role of 1 in the world of matrices
half ice
Know how to get E1 and E2 from the info above?
wicked trellis
Makes sense :))))))
solid basalt
I have no idea
this class has put me in the fetal positions and kicked me to the point right before death
half ice
And as quickly as Ann arrives, she leaves. Mysterious
solid basalt
lol
half ice
Lel I'm jk of course
dusky epoch
what are the E_i in Lemma 1?
half ice
We already have the partial fraction decomp, but I'm not sure how to use that to get E1 and E2
I've been trying to look that up but can't find it?
dusky epoch
half ice
solid basalt
Im sorry guys it looks like I've stumped ya but thats what fun about math, right?
half ice
Of course! It's an interesting question and I've never seen the method before
solid basalt
I was able to understand this class until wk 9 but now i'm in week ten and completely lost
half ice
Yeah, if you have any name attached to this, it would make it ez to look up
dusky epoch
strange
solid basalt
Does this help?
this is the best I got
I tried to look this up but I think my professor makes his own material
solid basalt
so is that what you needed?
dusky epoch
oof sorry i don't have the energy to do this rn
solid basalt
thats okay can I come back later?
dusky epoch
sure
solid basalt
okay thank you
sand hedge
slow scroll
For b) I can see that the matrix would have to have two columns, but its not really obvious to me whether such a matrix with these properties exists, and if so, how i would find it
dusky epoch
rank-nullity theorem
slow scroll
Okay then what?
rank = 1
nullity = 1
rank + nullity = 2
it seems like all rank-nullity can do here is tell me how many columns the matrix would have to have, not whether it exists or how to find it :/
dusky epoch
yeah, and yet your matrix is supposed to be 3 by 3
slow scroll
why cant it be 3x2?
slow scroll
in other words, if my matrix is 3x2, the span of its columns is a subspace of R2? Im confused.
dusky epoch
sorry, i mixed it up
i meant its nullspace
a 3x2 matrix represents a linear transformation from R^2 to R^3
static patio
Can gauss elimination have different solutions at the end or is there only one solution?
slow scroll
The rref form of a matrix is unique if that’s what ur asking
half ice
No, since rref is another ref
static patio
dam that's sound logic lol thanks
broken hawk
similarly, the inverse which you can compute via gaussian elimination is also unique
and LU-decomposition is unique if you force one of the matrices to have only ones on the diagonal, iirc
dusky epoch
LDU decomp is unique if L and U are forced unitriangular then
winter reef
big help pls
I subtracted lambda at diagonal
but I get scuffed polynomial
-λ^3 + 6λ^2 - 12λ + 8
when checked in calculator the eigenvalue is 3.26 which is weird
it doesnt distribute on linear components 
maybe I screwed up matrix B: I typed it as: first row(2,2,0), 2nd row(1,2,2), 3rd(0,-1,2)
winter reef
yeah
barren plank
and a matrix M
winter reef
I made matrix psi in standard basis
winter reef
first row(2,2,0), 2nd row(1,2,2), 3rd(0,-1,2)
and like to find Jordan matrix of B I would need like an eigenvalue
barren plank
what's B
winter reef
M(psi)st
barren plank
never seen this
barren plank
which basis, the only basis here is the canonical basis of R^3
winter reef
yeah
winter reef
M is just notation for matrix of transforamtion psi in these basis
barren plank
what
winter reef
its not a multiplication
barren plank
then what does "and let M = ..." mean
winter reef
yeah the notation here is scuffed, the other matrix should be named differently, its needed for another question
barren plank
yes that seems correct
winter reef
yeah, so it has only 1 eigen value 3.26
barren plank
what are its roots?
barren plank
that's not correct
wintry steppe
Let V be a subspace of Rn. Suppose that B = (b1; ..., br) is an orthonormal basis of V . Prove that for all v, w ∈ V , [v]B * [ w]B = v * w.
how do? 
barren plank
@wintry steppe split v and w into a linear combination of elements of B
then use orthonormality
and linearity of *
winter reef
So I have one exercise where I need to determine whether there exists basis A, such that matrix of transofmation in that basis equals M(a given matrix). We also are given matrix of the transforamtion in standard basis, lets call it B. I checked that B and M have the same Jordan matrix, so there exists such basis A, but how do I find vectors of this basis? R3
slow scroll
Okay so on 7.14, it seems like it is obviously possible for complex matrices, since 7.13 serves as an example (assuming i did the question right), but I am having trouble with real matrices. If I can just find a symmetric matrix such that A^2 = 0, then this would be easy, but I can't come up with one nor prove that there can't be one. Any hints?
full palm
the set that contains all Mn that are invertible, has a null element? (n = nxn, n being a integer)
wintry steppe
in (b) my immediate intuition is to claim that Qv = - reflV*reflV so Q(v)=-v
is that correct?
dusky epoch
when 1 is not an eigenvalue of A
broken hawk
spec being the eigenvalues?
dusky epoch
yes
broken hawk
hm, actualy, yea wait, that seems reasonable
so let’s assume A is diagonalizable first
then $A = Q^{-1}DQ$
stoic pythonBOT
Sascha Baer:
dusky epoch
why not jordan it lol
broken hawk
fair actually. repalce D with J
dusky epoch
$I - A = Q^{-1}(I - J)Q$
stoic pythonBOT
Ann:
broken hawk
and so $I - A = I - Q^{-1}JQ = Q^{-1}Q - Q^{-1}JQ = Q^{-1}(I - J)Q$, which is invertible iff J has no 1 on the diagonal
stoic pythonBOT
Sascha Baer:
dusky epoch
indeed.
i mean there's a less nukey argument tho
A - I invertible iff det(A - I) != 0 iff 1 not in spec(A)
broken hawk
determinant doens’t play well with addition tho so how does that follow?
sharp lodge
If you know that adding I increases the eigen value by 1, yot can say it nore elegantly without det
dusky epoch
@broken hawk how does what follow
broken hawk
I don’t know how to show that without invoking JNF
the fact that det(A-I) != iff 1 not in spec(A)
aka adding identity changes all eigenvalues by 1
broken hawk
oh. right. and we just added another I in, so λ+1
clever relic
Hi I'm doing linear independence, but I'm stuck on doing gauss jordan elimination when there's a variable involved... my vectors are (1, 1, 2, -4), (1, 2, k, -5), (1, 5, 3, -8), I made a matrix
[ 1 1 1 0]
[ 1 2 5 0]
[ 2 k 3 0]
[-4 -5 -8 0]```Doing elemination until
[1 1 1 0]
[0 1 4 0]
[0 k-2 1 0]
[0 0 0 0]
``` but now I'm stuck<@&286206848099549185>
dusky epoch
try to make your matrix diagonal
you can for example subtract the first col from the second and from the third, thereby zeroing out the first row
and then zero out the second row by subtracting 4 * the second col from the third
you'll end up with $\begin{bmatrix} 1 \ & 1 \ & & 9-4k \ & & & 0 \end{bmatrix}$
stoic pythonBOT
Ann:
clever relic
how does $a_{23}$ become 0
stoic pythonBOT
GyroW:
clever relic
the one left of 9-4k that is
dusky epoch
err
you're right sorry it doesn't
i messed up
well basically it all now hinges on whether 9 - 4k is zero or not
if it is zero then your vectors are LD bc the rank of this matrix is 2
if it is not zero then you can zero out a_23
dire bluff
moin
wintry steppe
plus
dire bluff
I have an exam in which I'll have to compute some least-squares polynomial fits
that gives a linear system of equations
but I can only do that with a simple calculator that doesn't uhhh have much stuff tbh
is there an approximate method to solve such a linear problem?
something that I could do more or less easily with a simple calculator
slow scroll
HYPE, but uhmm wrong channel lmao
rare barn
for any given nxn matrix, is the detA^TA always >0 ?
brittle juniper
$\geq 0$
stoic pythonBOT
Tuong:
brittle juniper
but not always >0
rare barn
when would it be 0
brittle juniper
When A isn't inversible
rare barn
oh well yea mb, i was asking for whenever its invertible
can some1 help me find an example of this: a 3×3 invertible matrix A such that A^3=0
brittle juniper
I don't think there exists any
rare barn
me neither i cannot think of anything where this would work
brittle juniper
if A is inversible and A³=0, then
A¯¹×A³=A¯¹×0
A²=0
A¯¹×A²=A¯¹×0
A=0
but 0 isn't inversible!
rare barn
yes
ok, i just one more, im a little confused by the wording of the question: cant there be no determinant for an mxn matrix, what do they mean by working through m=1>?
brittle juniper
There's no question in this
Do you have to prove the statement?
brittle juniper
I've found a counter example
A = [0, 1]
with $A=[0\ 1],\ \ \det(A^TA)=\begin{vmatrix}0&0\ 0&1\end{vmatrix}=0\ \
\det(AA^T)=|1|=1$
stoic pythonBOT
Tuong:
rare barn
but isn't there no determinant for mxn matrices anyways?
rare barn
yeah yoiur counterexample makes sense
i just thought we could immediately say its false because its an mxn matrix, but i get it
half ice
AA' and A'A are always square
rare barn
oh i see
ruby crow
Hey y'all, anyone able to help me out with this?
I managed to show that they commute but nothing more
Any help would be really appreciated :)
half ice
B = xI + yP
A = aI + bP
AB = (aI + bP)(xI + yP)
= (ax + by)I + (ay + bx)P
= x'I + y'P
So if two matricies are of the form aI + bP, then their product will be. Multiplication by A can be seen as a linear transformation:
[a b] [x] = [x']
[b a] [y] [y']
The inverse of that matrix can be seen as a multiplication by A¯¹
@ruby crow
ruby crow
OH. I don't know why I didn't factor by I and P lol. That's perfect thank you
half ice
Oh, and you need to recognize that I is of that form, that helps
Np. Feel free to ask if you have anything else
ruby crow
dull kettle
Hi guys, none of the statement is true, am I right?
dull kettle
how does i affect its orthonormality
dull kettle
but what if u1's coordinates consist of negative number
wintry steppe
in this case yes, because c1 is positive
in general it has the same sign as c1
dull kettle
thanks!
dusky epoch
@dull kettle $w \cdot u_1$ is literally equal to $c_1$
stoic pythonBOT
Ann:
dull kettle
wow
dull kettle
thanks @dusky epoch overlooked that part in my notes
winter reef
how do I tell how many blocks will be in a jordan matrix? and also their size
dusky epoch
oh
there's actually a really beautiful method to do that
so what i assume is that you've already fully factored your char poly
(and if you haven't you should bc the method makes use of that)
@winter reef are you here? this is gonna take a little while to fully explain so i want to make sure you're here
winter reef
yep!
dusky epoch
alright good
winter reef
I remember it being about the rank
winter reef
ohh so like rank of subspace of eigenvalue
dusky epoch
so just to fix some notation, let $A$ be the $n \times n$ matrix whose jordan form you're interested in, let $\lambda$ be an eigenvalue of $A$, $s$ be $\lambda$'s algebraic multiplicity (i.e. its multiplicity as a root of the characteristic polynomial of $A$), and $B = A - \lambda I$
stoic pythonBOT
Ann:
dusky epoch
we're going to be interested in the ranks of powers of B
so what you want to do now is make this table:
$\begin{array}{c|c} k & \operatorname{rk}(B^k) \ \hline 0 & n \ 1 & \ 2 & \ \vdots & \vdots \end{array}$
stoic pythonBOT
Ann:
dusky epoch
and fill in its right column
you should get a descending sequence of numbers
and you should continue that until it stabilizes at $n - s$
stoic pythonBOT
Ann:
winter reef
umm I remember doing similiar but with a kernel
dusky epoch
this is basically using images instead of kernels but it's essentially the same just dualized
broken hawk
well you can brute-force it by just finding elements in the kernel of B^k until you can’t find any more
but it’s tedious and to get the change of basis matrices you have to find cycles
dusky epoch
anyway, continuing with what i wrote, $\operatorname{rk}(B^{k-1}) - \operatorname{rk}(B^k)$ is the number of Jordan blocks of size at least $k$
stoic pythonBOT
Ann:
winter reef
hmm so I kinda see, but using kernels we were multiplying the matrix by itself until it was a zero matrix, then take any vector and find a vector that is in kernel of the one before minus the kernel of the next one kinda
so I see the similiarity of your way
and why do we multiply the matrix until n-s?
ohh is it because then it will be a matrix diagonal on n-s x n-s and rest are zeroes?
dusky epoch
not quite.
so here's the thing
if two matrices are similar (i.e. are related via a change of basis matrix), then their rank sequences are the same
and if you write out the jordan form of A, then the blocks corresponding to λ will take up s positions
and accordingly the jordan form of B will have s zeroes where A has λs
so that block of the jordan form of B will be nilpotent
but the other blocks will stay full-rank no matter what power you raise them to
and so won't give you any rank reduction
winter reef
wait what does jordan form of B will be nilpotent mean?
dusky epoch
you parsed my sentence incorrectly
i said that that block of the jordan form of B is what's going to be nilpotent
winter reef
ohh ok I found it, I just didnt know the term 'nilpotent' but I see now
yeah kinda makes sense, I'm gonna have to think about it more to get it, but thanks for your time!
half ice
What do you mean n×n (0,1)?
Like a matrix over mod 2 arithmetic?
The determinant gives invertability of a system. So stuff breaks when matricies have zero determinants
As well, it's easy to find the determinant of a product of matricies, without even needing to compute the product itself, so you can reduce computation time with clever implementations
Np. I also have other examples lol. Matricies control a lot of computation
rare barn
if A is an nxn invertible matrix with entries that are all integers, and its determinant is 1, does that mean all entries of the inverse matrix of A are also all integers? how do i show tht
proper crescent
There's a cofactor formula for the inverse
The idea is that you divide by the determinant, so if it's just +/- 1 you're fine
rare barn
@proper crescent ik it works if its +-1, but here its just saying that its 1
its not including -
proper crescent
That's a special case
proper crescent
@formal knoll just ask your question and if someone can help they'll answer
formal knoll
let's say i have A dot B =C which are all in r3. And we have b and c, how would you solve for vector A?
proper crescent
A dot B is in R, not R^3
formal knoll
a and b are in r^3
proper crescent
Okay so, the next issue is that there's more than one vector A which works
proper crescent
No problem! 👍
cloud veldt
Is there a extra problems floating around regarding college functions and models?
cloud veldt
Yeah
proper crescent
So that question should go under #prealg-and-algebra instead
quaint harbor
wintry steppe
can anyone help me solve q13? not sure what i'm suppose to do here. ty
<@&286206848099549185> ^ sorry it's a pretty easy question
proper crescent
@wintry steppe still here?
wintry steppe
ya hi n.n
proper crescent
So for 13a, do you notice something about these vectors?
Or actually better idea
Look at 13b as a hint
What does 2:00 have to do with 8:00?
wintry steppe
aren't they opposite
wintry steppe
8 is -2?
proper crescent
Yeah
(To avoid ambiguity we should write x:00 since 8 and -2 could be interpreted as scalars)
But alright, so now when we add all the vectors on the clock, if v appears, -v appears later, right?
wintry steppe
i'm really not sure how to get them into vectors in the first place, really new to this, is 12 = (0, 1) and 3 = (1, 0), then 1 = (1/3, 2/3)
proper crescent
Yeah technically you'd get them in coordinates but for the sake of this problem the better thing is to just say that we know that if x:00 corresponds to some vector v, then (x+6):00 will correspond to -v, and that v+-v = 0
So this way we don't have to do 12 different computations, you know?
wintry steppe
by x:00 what do you mean
proper crescent
x o'clock, what you call x
wintry steppe
oh lol sorry
proper crescent
I don't like referring to 2:00 as 2 because the issue is that 2 is also technically a scalar and there's cause for confusion
tall bridge
Anyone able to help?
proper crescent
Chances are that won't become an issue for us here, but in any event yeah that's why I'm using the notation I'm using
tall bridge
proper crescent
@tall bridge we're in the middle of something here, and that's not linear algebra, ask in #prealg-and-algebra or a questions channel
wintry steppe
oh i think i get it
proper crescent
Also for fuck's sake post in one channel and then stop
wintry steppe
oh wait so the answer would just be the zero vector
proper crescent
Yup! 😄
glass sluiceBOT
🆙 | ronnie has given @proper crescent a reputation point!
proper crescent
😃
Somehow 😃 on discord looks more like :D than 😄
Anyway tackle 3b using the same geometric ideas
wintry steppe
@proper crescent wait so with b), the reason it sums to 8:00 is there's no -8:00 (2:00) to negate it, and the rest of them sum to 0?
proper crescent
Exactly
wintry steppe
👌 ty n.n
proper crescent
Now part c is basically asking the angle between the 12:00 vector and the 2:00 vector, right?
proper crescent
Close, remember the full circle is 2pi, not pi
wintry steppe
but isn't 2:00 just 30 degrees?
proper crescent
It's a third of the way to 6:00, right?
wintry steppe
i'm not sure I understand that part
proper crescent
My point is that it's gonna end up being 60 degrees instead of 30. Do you buy that the angle between 12:00 and 2:00 should be one third of the angle between 12:00 and 6:00?
proper crescent
What's the angle between 12:00 and 6:00?
proper crescent
180 is pi
wintry steppe
omg sorry i'm being stupid tonight lol
proper crescent
It's fine, so now you're with me that 2:00 is (cos(pi/3), sin(pi/3))?
wintry steppe
isn't that the angle between the horizontal and 1
proper crescent
Right right I was measuring off of 12:00 for some reason
You were right the first time
Sorry
proper crescent
Yup you're good
wintry steppe
👌 great ty for all the help
surreal loom
hello, how can I prove that given a permutation matrix P, there exists a number k, so that P^k = Identity matrix?
dusky epoch
how familiar are you with group theory
surreal loom
somewhat familiar
dusky epoch
and have you encountered what are called permutation groups?
surreal loom
you mean cyclic permutations for example ?
stoic pythonBOT
Ann:
dusky epoch
sound familiar?
surreal loom
yes that rings a bell, in matrix (3x3) permutation the group of all permutations has 3! matrices I believe
and I understand that if you apply one permutation over and over again, eventually you will end up at the starting position
or is this not always the case?
dusky epoch
well i mean... it's a consequence of Lagrange's theorem in group theory that in a finite group G of order n, raising any element to the n'th power produces the identity
in your case, your group is the group of all permutation matrices, of order n!, which is of course still finite
surreal loom
I'm still not sure how I can prove this mathematically. I am attending an introductory linear-algebra course and I guess it should be quite simple