You can use the second expression I wrote
#linear-algebra
pallid rampart
I don't think there's any other easier way
This is easy enough already
What are you trying to do?
wintry steppe
like write it in the form
$\begin{bmatrix} x \ y \ z \end{bmatrix} = $\begin{bmatrix} a \ b \ c \end{bmatrix} + t $\begin{bmatrix} e \ r \ u \end{bmatrix}$
stoic pythonBOT
polynomial:
Compile Error! Click the 
reaction for details. (You may edit your message)
wintry steppe
without a t infront
pallid rampart
then use the second expression I wrote
The second expression I wrote is obtained from the first expression
wintry steppe
why do we subtract though
pallid rampart
Hang on lemme draw a picture
So C=B-A, the line through the origin and C is obtained by tC (t is any real number), then the line through A and B is obtained by translating the line through origin and C by A, so the line through A and B is A + tC = A + t(B-A)
@wintry steppe
pallid rampart
Idk 
wintry steppe
@pallid rampart oh thanks i get your explanation 🙂
is this how would you write the equation for a line like that as well?
pallid rampart
So you get a line, ray, or line segment by restricting what t can be
But the equations are still the same
wintry steppe
i meant if you wanted to write an equation for a line
is that how you would do it
?
on an exam or w/e
pallid rampart
If I was to answer a question like that on an exam, I would answer the line is $$\begin{bmatrix}-1\3\2\end{bmatrix}+t\br{\begin{bmatrix}-3\5\-1\end{bmatrix}-\begin{bmatrix}-1\3\2\end{bmatrix}},\quad\text{where }t\in\bR.$$
stoic pythonBOT
Whoever:
hollow finch
is there a 'linear algebra way' to derive the determinant form of the cross product? i have a roundabout way using cramers rule but idk if thats the best way
odd kite
@hollow finch the reverse seems better to me as the determinant exists for all square matrices but the cross product requires 3d vectors. You can define the cross product as being the Hodge dual of the exterior product of 2 vectors in 3D.
the exterior product of 3 vectors in 3D will give you the determinant
So they're related sort of but the fact that the cross product appears like a determinant is sort of a mnemonic. It's not a real cross product since the top row has unit vectors instead of regular components
This also relates to the concept of parity of a permutation and Levi-Civita symbol
hollow finch
oof i dont know any of those things :C
but thank you for the answer i really do appreciate it @odd kite
odd kite
oh well, I thought that might be the case. Maybe something you can look into someday
but at least you get that the cross product isn't a real determinant right?
nimble raft
$$\begin{pmatrix}1&b+1&5&7\ 0&b&a+3&0\ 0&0&a^2-9&a+3\end{pmatrix}$$
stoic pythonBOT
Soup:
nimble raft
Im trying to find all 'a's and 'b's, so it has:
- no solutions, 1 solution, 1 free and 2 free vars.
So to make it have no solutions: Setting a = 3, will make the last row inconsistent.
I've also found (a = 4 or a = -3) and b = 0, will make it have no solution.
I've also noticed, there are a lot of ways it can have 1 solution.
Tho I'm not sure exactly how to find them all.
How and what other ones can I find?
nimble raft
Yea
opal osprey
so if you can guarantee you will have det ≠ 0 for certain values for a and b, then you should be good
you know a in { 3, -3, 4 } will yield an undesirable system
and iirc, you can do a laplace determinant (?) using the first column with the two zeroes
nimble raft
Right, so I thought determinant only tells u if its invertible or not?
opal osprey
hmm. let me check something
maybe I'm thinking of the rank
yes - so you need the matrix to have rank 3
(are you familiar with that concept?)
opal osprey
I think so
if the rank of the matrix equals the number of lines, then it should be a possible, determined system
(I'm not quite sure what the terminology is in english)
nimble raft
Right, I'll look into it and give it a go. Thanks!
opal osprey
hopefully someone a little more knowledgeable can chime in
gray dust
Ax=b has a solution iff rank(A|b)=rank(A), and that solution is unique iff rank(A)=number of variables
opal osprey
cool!
torn silo
if you want to be thorough you'll have to do cases
opal osprey
so soup should check rank(A) considering the restrictions on the value of a?
torn silo
like case a != 3 then divide the bottom by a^2 - 9
then use that to clear the third column and so on
opal osprey
well, since doing laplace determinant on that column yields b(a^2-9) = 0
imposing the previous restrictions on a gives b = 0
i'd say it's pretty clean
sour jetty
Have to find for which p this matrix is invertible. When I calculate the determinant and set it equal to zero, there is no p part of real numbers that works. So I suppose it isn't invertible for any p part of the real numbers? But the solution says that, because of this, all values of p part of real numbers will cause the matrix to be invertible. Am I in the wrong here or is the solution wrong?
The calculations give (p+2)² + 2 = 0
quasi vale
If you're setting the determinant = 0, then you're finding the values of p which give you a matrix that is non-invertible.
sour jetty
Wow really
quasi vale
Yes if det is 0 then the matrix isn't invertible.
sour jetty
That's huge, okay thanks a lot!!
true tiger
A^2+AB+B^2 => requires 3x multiplications and 2x additions. I need to convert this into only 2 multiplications.
All variables are matrices
true tiger
- The calculation of the expression A^2 + AB + B^2 requires 3 multiplications and 2 additions. Transform this expression so that 2 multiplications are sufficient for the calculation, and calculate it for the matrices A = {{1, 2}, {3, 4}}, B = {{3, 4}, {5, 6}}.
Oh, would that mean
dusky epoch
how many additions are we allowed
if at least two are allowed it is possible
(A+B)^2 - BA
true tiger
Yeah, it's probably that
dusky epoch
but it doesn't say how many additions we are allowed
two multiplications and no other operations will not suffice
true tiger
Yeah, it's probably 2 multiplications and x amount of additions/substractions.
rustic stump
Im guessing that they don't care about the number of additions since it can be done much faster than multiplication?
dusky epoch
if there's no limit it should explicitly say so
dire thunder
Im guessing that they don't care about the number of additions since it can be done much faster than multiplication?
still O(n^2)
nimble raft
Is this correct?:
(a - 3b) × (a + 3b) =
(a - 3b) × a + (a - 3b) × 3b =
(a × a) - (3b × a) + (a × 3b) - (3b × 3b) =
- (3b × a) + (a × 3b) =
(a × 3b) - (3b × a)
dusky epoch
yes
nimble raft
Thanks, do u think it could be simplified more?
dusky epoch
well it could be simplified to a × 6b
nimble raft
oh right
rustic stump
The fastest Matrix multiplication algorithm is O(n^2.3....), which has horrendous constants. Naive multiplication is O(n^3).
dire thunder
The fastest Matrix multiplication algorithm is O(n^2.3....), which has horrendous constants. Naive multiplication is O(n^3).
i spoke about addition
nimble raft
@rustic stump Isn't it generally done in hardware, making it not really "count" in real application?
dire thunder
most of matrix operations are bounded by O(n^2)
rustic stump
Yeah true.
dire thunder
$\Omega(n^2)$ is more correct
stoic pythonBOT
Commander Vimes:
nimble raft
lower bounded?
dire thunder
that's why for example in graphs it is more efficient to implement algorithms which work with adjacency lists
rustic stump
The individual adds/multiplies are done in hardware sure, and you might have circuits specifically optimized for matrix operations, but they still have a cost.
dire thunder
lower bounded?
yes
like for matrix-based algorithms we usually are satisfied with algorithms with n^2 complexity
rustic stump
I've mainly dealt with matrix operations in the context of algebraic number theory and geometry, but they very much have a cost.
nimble raft
Unless your doing some research with massive matrices for stat, or something, I don't imagine seeing anything costly for most calculations done at early uni level.
dire thunder
yes, but the good algorithm is not that works on small input
but which works on theoretically "infinite" input
rustic stump
Discriminants and resultants blow up quickly.
Taking hours for symbolic operations on higher degree polys if you aren't careful.
nimble raft
I feel even O(n) and O(1) amortized will slow at some point
dire thunder
no, O(1) is extemely good algorithm
nimble raft
Depends on the algorithm
rustic stump
O(1) is literally constant time.
dusky epoch
O(1) is the best an algo can perform
dire thunder
but f(n)O(1) is O(f(n))
nimble raft
e.g hash tables, which use linear probing, or bucketing
dire thunder
and if i've got O(1) it means i cannot do better now
torn silo
armotized is fine
nimble raft
If you got O(1) amortized, it means u can still do better.
Compare hash tables to arrays?
torn silo
no if you did armotiziation thats prolly the best you could manage under the given constraints
rustic stump
Besides, we want it to feel slow at very large numbers, not smaller ones. I don't want to wait 3 hours for 2 degree 5 polynomials.
torn silo
otherwise you don't go that route
nimble raft
Sparse array generally does the job a hashtable can do, and still does O(1)?
Just uses more memory.
torn silo
"just"
nimble raft
Memory is generally more expendable than speed nowadays.
torn silo
depends on the situation
rustic stump
Not necessarily at scale
dire thunder
btw, i just designed an algorithm for producing k-partition of graph and it seems to me that it has polynomial complexity (with big exponent but still) but i've read that it is NP-complete problem and now i am trying to find a flaw
nimble raft
congratulations!! u did it!!
rustic stump
Or you just solved the PNP problem
nimble raft
u win some million dollers or whatever
dire thunder
But for my general usage, imo memory is more expendable.
well, this part depends on the concrete task
nimble raft
yea depends on the situation and task
dire thunder
bruh i think there should be flaw in my reasoning, like i do not think nobody thought about my soltuion before
wintry steppe
[[0,1], [-1,0]] does this matrix have a special name?
quartz compass
I call it "rotation by 90 degrees matrix"
torn silo
R or something like Rot90 🤔
dire thunder
it is rotation by -90 degs no?
quartz compass
depends on if your convention is clockwise or not
austere sage
This paragraph from M Artin’s algebra is confusing me !
What does he mean by “consistent way” here. And in the next line, v1 is in R^1 ? Implying a 1 dimensional vector space ?? So confused
half ice
Convergence
For example
1 + 2 + 3 + 4...
What do we assign to that?
A lot of people would say -1/12 but we don't listen to stupid people
It's even worse if your vectors aren't from real numbers
So something like Span(all positive integers) is not obvious
Well, actually, bad example because they're all going to be integers. A much better example is Span(all rationals)
karmic oracle
I would say it as, infinite sums are fundamentally analytic in nature. They require limits, which in turn require an underlying metric or topology.
oak crater
all fin dim real/complex v spaces have a canonical topology
limber sierra
but we don't listen to stupid people
a lot of people in this server listen to me though
oak crater
@austere sage If you're still confused about that remark just don't worry about it and move on. Simply remember that in linear algebra all you care about are finite sums.
karmic oracle
Yeah, I was just saying, in the realm of algebra, you never get to take an infinite sum. You need to dip into analysis for that.
[I was not claiming you can't or shouldn't do that.]
@half ice I did finally see an argument from Tao that one can really meaningfully associate the value -1/12 to 1+2+3+...
half ice
You can but you have to be careful yeah
Not the kind of careful this book is looking for
And lol people shouldn't listen to me either
austere sage
@oak crater yeah, i think i'm gonna move on for now. thanks
I would say it as, infinite sums are fundamentally analytic in nature. They require limits, which in turn require an underlying metric or topology.
@karmic oracle I get what you're saying. but i connect it to what's written in the book.
ocean sequoia
Can anyone phrase this differently im a bit confused
Also if two vector spaces are isomorphic does that mean they share the same identity basis?
broken hawk
not sure what you mean with identity basis
what it means is that you can find a linear function that takes a basis of one space to a basis of the other
muted holly
T( S(T(v)) ) = T(S( Tv ) = I ( Tv ) supposing To S = I , then I ( Tv ) = Tv, so T(S o T)v = Tv so S o T = I
broken hawk
or if you prefer, that there’s a bijection between bases
broken hawk
the standard basis is only a thing in the vector space ℝⁿ (or 𝕂ⁿ for any field 𝕂) of column vectors
there’s no such notion in a general vector space
slow scroll
the only thing you can really say is that they have the same dimension
ocean sequoia
Im only dealing with definite vector spaces atm
slow scroll
i.e. same number of basis vectors
broken hawk
while every finite dimensional 𝕂-Vector space is isomorphic to some 𝕂ⁿ, they aren’t 𝕂ⁿ
like, “polynomials of degree ≤n” is a vector space (of dimension n+1)
but it’s not the same as ℝⁿ⁺¹
broken hawk
there are different ways you could compare it to ℝⁿ⁺¹
for example you could take an isomorphism that takes 1 to (1,0,0,0…,0), x to (0,1,0…) and so on
but you could also do many other ways and there’s really no natural way to go about it
different situations call for different ways to look at it
so the space of those polynomials doesn’t have a standard basis
and you don’t think of it as “an arrow in n+1-dimensional space”
ocean sequoia
this seems like i might have fallen into a rabbit hole i might want to avoid for now :/
broken hawk
it’s a good rabbit hole
it also doesn’t go too deep
unless you start looking at infinite-dimensional stuff
then all hell breaks loose :P
neat peak
knowing this is $P_n(t)$ and $P_0(0)=1, P_i(0)=0 \quad if i>0$ how to find $C_n$ in terms of $\lambda_n, \mu_n$?
stoic pythonBOT
ProphetX:
neat peak
it is a linear system of equations, especially a lower triangle one,which should work by solving eq 1 plugging into 2, then plugging into 3 etc.
another way is to use cramer's rule
but in both cases I get lost in the notations, and can't find the generalized c_n, i'll show what i've got so far
wintry steppe
how can I find a, b, c and d?
dusky epoch
there is a pretty simple formula for the inverse of a 2 by 2 matrix.
wintry steppe
where can I find it
dusky epoch
google lmao
wintry steppe
inverse of 2x2 or what? xd
dusky epoch
inverse of a 2 by 2 matrix
but you could also just write out the system of four equations in four variables manually
and arrive at the same formula yourself
wintry steppe
I am too stupid to find the formula lol
dusky epoch
too stupid to click the very first link?
i mean ok like
w/e
$\mat{x & y \ z & w}^{-1} = \frac{1}{xw - yz} \mat{w & -y \ -z & x}$
stoic pythonBOT
Ann:
dusky epoch
in your notation
dusky epoch
well
it's a two by two matrix
two by two matrices contain four entries
shocking, right
kindred ingot
A good exercise might be arriving at the inverse formula via row reduction. Then promptly cast row reduction into the sea and ask for whatever life it stole from you back.
I used to tell my students that I wouldn't row reduce in public, and they shouldn't either.
dusky epoch
what am i looking at
neat peak
the seemingly solution to the system posted above
but I'm not sure,would need verification
it should be the solution for c_n with condition t=0 if p_0=1,p_i=0, i>1 of this. even tho it is a lower triangular system, the computations are terrible
idk how to check tho if its ok
spare rune
i need help with rq=(1-s)(1-p)
dusky epoch
parts of the paper are cut off from the left and right
dusky epoch
point of intersection of what
dusky epoch
lemme see
storm python
@cursive narwhal i don't need help but i want to ping you
cursive narwhal
Oooh lala, i've never felt this wanted before
storm python
glad i make you feel special
zinc tapir
If i want to show a transformation is invertible do i have to explain injectivity/surjectivity
or can i just say the dimensions are equal and it must be a square matrix
rotund jetty
the condition for invertability is injectivity and surjectivity. If you have a theorem that says it follows from equal dimensions, then you can use it. If not, you must show the injectivity/surjectivity
helpful flowchart
so you must have some justification for claiming that the dimensions are equal and its a square matrix implies invertability
cursive narwhal
If i want to show a transformation is invertible do i have to explain injectivity/surjectivity
You have to show that the transformation is injective and surjective. In that case, it would be bijective and would have an inverse.
cursive narwhal
Yeap. A transformation also isn't the same thing as its matrix. A square matrix isn't necessarily invertible too.
It is true that vector spaces of the same dimension are isomorphic. You can establish a bijection between them. It doesn't mean that all maps between them are bijective.
zinc tapir
wow that clears up alot thank you
rotund jetty
^^^
and that's the wall you would've run into trying to claim that the transformation is invertible because it has a square matrix
also note you can have an invertible transformation between non-isomorphic vectors spaces
cursive narwhal
Okay, so, the first thing I would do is to draw a diagram of the situation. Something approximate will do but it might very well be helpful in visualizing what's happening.
Now, do you agree that $ax+by+cz = d$ is the general equation of the plane that contains all 4 of those points? Do you, then, agree that those points satisfy the given equation of the plane?
stoic pythonBOT
Abhijeet Vats:
cursive narwhal
Okay. So, from that, you can construct 4 equations by plugging each point into the equation of the plane.
spare rune
oh damn i overthought it
cursive narwhal
We have:
$$ap = d$$
$$a+bq = d$$
$$ar+b+c = d$$
$$bs+c = d$$
stoic pythonBOT
Abhijeet Vats:
cursive narwhal
Those are our four equations. From there, it's easy to see that:
$$ap-a-bq = 0 \implies bq = a(p-1)$$
$$ar+b-bs = 0 \implies ar = b(s-1)$$
stoic pythonBOT
Abhijeet Vats:
cursive narwhal
Can you take it from there?
spare rune
yeee
i just was in tunnel vision
i really dont know how to get out of that state of mind
cursive narwhal
Well, the good thing about that is that you can always choose to walk towards the light at the end of the tunnel.
Honestly, don't worry about it. Solve harder problems. You'll get better at doing it.
slow scroll
wrong channel @wintry steppe
ask in #prealg-and-algebra or something
wintry steppe
Oh, thank you.
zinc tapir
Ok makes sense ty @slow scroll
late sentinel
yeah its the 0 matrix
late sentinel
I'd use the determinant for a
zinc tapir
havent covered det yet
late sentinel
Ohh okay
late sentinel
yeah and the last line of your proof is a contradiction so it holds
since 0 isn't invertible
similar logic for b
zinc tapir
yeah but i have the identity times A = 0 which makes no sense
late sentinel
leading to A = 0 which is a contradiction to what you assumed in the beginning
that's your proof
since 0 isn't invertible
zinc tapir
how do i go from IA=0 to A=0 tho
late sentinel
it's a property of the identity matrix
elfin ingot
AI = A
zinc tapir
I'm gonna have to read up on full rank haven't got to that yet
We just got to change of basis
zinc tapir
soo uhh
half ice
It's not true. There's many A² = 0 that are not A = 0
Oh wait you're assuming A is invertible nvm I have the dumb
zinc tapir
Since B=! O we have that A = O and there exists an inverse of A such that A^(-1)A=AA^(-1)=I therefore A^(-1)A=O implies I=O
did I mess up here
half ice
Minor nitpick, you can only multiply on the right, or on the left. So writing
A'A² = 0A'
Is a strange move. It's correct since A and A' commute, but A'A² = A'0 is generally safer.
Yes your proof is correct
Using ' for inverse
Wait I don't see your next part
By assuming A has an inverse, we are forced to conclude that A = 0. But that doesn't have an inverse, a contradiction
rotund jetty
@wintry steppe sorry, I meant SUBSPACES of non-isomorphic spaces
zinc tapir
are you talking about b or a kaynex
im working on b atm
Since B=! O we have that A = O and there exists an inverse of A such that A^(-1)A=AA^(-1)=I therefore A^(-1)A=O implies I=O
did I mess up here
half ice
Oh lol I misunderstood let me see
You don't have that A = 0
We have a word for this. Matrix multiplication has "zero divisors". That is, matricies that are not zero that can multiply to make 0
half ice
Sadly no haha. This would be pretty easy if you could
normal quiver
If you have a ⋅ (b×c) u dot product with v cross product with w
can you rewrite that as b × ( a ⋅ c)
half ice
a•c is a scalar, so b×(a•c) makes no sense
zinc tapir
so i can just take the inverse of AB and the zero matrix
I get the identity times B = O
which implies that B=O
but we stated that B cannot be the zero matrix so contradiction
half ice
0 matrix has no inverse
And it's pointless to make a contradiction based on 0 having an inverse. It simply doesn't haha
zinc tapir
oh
half ice
And that seems like a problem!
zinc tapir
But it was stated that B cannot be the zero matrix
i can't just conclude the proof there?
half ice
That's the contradiction
If A has an inverse, then B must be 0
But B is not zero, therefore A doesn't have an inverse
zinc tapir
i mean thats what i said
but we stated that B cannot be the zero matrix so contradiction
normal quiver
is A ⋅ (B×C) = (A×B) ⋅ C a cross product property?
half ice
You can't take the inverse of AB
I'm thinking that's a typo now that I re-read it. You have the correct proof otherwise haha
zinc tapir
ok ty, sry if im asking many questions im just trying to get the ideas down
for this one should i prove phi is an isomorphism by proving that phi is bijective
elfin ingot
and a homomorphism
slow scroll
yes, or you can just find an inverse
zinc tapir
and a homomorphism
@elfin ingot i don't know what that is
slow scroll
linear transformation = homomorphism of vector spaces
elfin ingot
yea u have to show that
and its a bijection
an isomoprhism is a bijective homomorphism
and as kx
said
context of vec spaces homo means linear transofmration
wintry steppe
Homo
half ice
Morphism
elfin ingot
i dont know tho
if there is any difference in meaning
between linear transformation and iso
( bijective lienar trans that is )
like for me iso just means both act the same
while linear transformations means u can get one by just stretching and doing some linear stuff to the other
are these the same mehj
?
if you can get from something to another using transformations
are they the same
half ice
There's no difference in linear algebra no
elfin ingot
cool
half ice
Iso = invertible matrix always
half ice
But as you know, you can define this on other structures and it can get more complicated
elfin ingot
yea
yea
i wonder if ,like in vec spaces,
we can usue any pictures
to demonstrate 'being the same'ness
with other
thigns
is linear algebra 'nice' just cuz u can make sense with it with pictures
and hence cool shit happen?
or is that like in baby lin algebra where you deal with just R^3
half ice
Linear algebra is nice because you can essentially forget all the definitions of a vector space once you define the basis of it haha
oak obsidian
I mean that's one benefit, but if you start generalizing it you can't really picture it all that well anymore if you take arbitrary fields. Drawing a picture won't constitute a proof really, but it can give you nice intuition
half ice
Construct a very complicated structure that gives a lot of info but is difficult to work with. But then, find a vector space on it and a basis of that space, suddenly it's very easy to work with
limber sierra
not sure what you mean, just visualize them as shears/stretches of the underlying coordinate grid
fixing the vectors
elfin ingot
tes
yes
thats what i sometimes think of too
but its just too cool my brain just shuts down
half ice
It's nice to think of 2D arrows but it's rare that this picture is exactly right
limber sierra
¯_(ツ)_/¯
wintry steppe
Draw means doodle nicely
slow scroll
try to find an inverse for it. thats the easiest way imo
elfin ingot
injective homo <--> ker = 1
zinc tapir
uh
slow scroll
nahhh i wouldn't do that.
slow scroll
no thats nullity 0, but I think trying to compute rank and nullity is a waste of time. Bijective iff invertible is your friend
slow scroll
i mean, sure, computing the kernel is not that much work here, and since the domain = codomain, it follows that you have an isomorphism by rank-nullity
zinc tapir
yeah
im not sure how to go about the inverse way
i end up with AB=0
you said take the inverse of this right B^-1AB=0
slow scroll
you need to find some linear transformation $f : M_{n \times n}(F) \to M_{n \times n}(F)$ such that $f \circ \phi = \phi \circ f = \operatorname{id}$ where id is the identity transformation. In other words, $\operatorname{id}(A) = A$ for all $A \in M_{n \times n}(F)$.
stoic pythonBOT
kxrider:
slow scroll
as a hint, the inverse looks a lot like phi
slow scroll
right, that part looks fine.
cursive narwhal
ya i proved the linear trans part
@zinc tapir How about bijectivity? Have you been able to show that?
cursive narwhal
Well, look at the definitions of injectivity and surjectivity.
What do you need to do to show surjectivity?
zinc tapir
I have to look at the image and determine if rank = dim(im)
cursive narwhal
Let $C \in M(n \times n, \mathbb{F})$. You need to prove there exists an $A \in M(n \times n,\mathbb{F})$ such that $\phi(A) = C$
stoic pythonBOT
Abhijeet Vats:
zinc tapir
why do you need to set that equal to C
cursive narwhal
Well, you need to show that every element in the codomain has a preimage in the domain under the function.
Think back to the definition of surjectivity for a general function
zinc tapir
idk in my other examples i didn't have to set anything i just factored out and showed the vectors were lin indep
then i took the set of lin indep vectors and called that R(T)
then I take the dim of that and compare it to the the dim of the image
thats how im used to doing it
cursive narwhal
Well, then try to do it that way if you’re comfortable with that
Well, you need to show that every element in the codomain has a preimage in the domain under the function.
^I’m pretty sure this is easier though
zinc tapir
yeah but theres like nothing to show for lin indep for these matrices
like no mechanical way
so C in this case is just a matrix in the domain?
cursive narwhal
No. It’s a matrix in the codomain. The domain and codomain happen to be the same in this case
zinc tapir
okay
going to attempt to find that phi is one to one
BAB^−1= 0
i have this
since B is invertible i can take its inverse right
cursive narwhal
??? What about surjectivity?
zinc tapir
AB^-1=0
cursive narwhal
And no, that’s not how you show that it’s injective
zinc tapir
well im trying to show that it maps to the zero matrix
cursive narwhal
Urhmm why not work straight from the definition of injectivity?
Idk, it seems like you’re just randomly doing things without really thinking through what you’re doing or why
zinc tapir
naw im just an algorithm kind of guy, im trying to work through it using what i did in a previous example or something
cursive narwhal
Well, see how that goes for you. My suggestion would be to work straight from the definitions of surjectivity and injectivity.
dusky epoch
can i see the original problem again
dusky epoch
lmao ||phi has an inverse lul||
also that is a horrible way to handwrite lowercase f
zinc tapir
im using the text notation
dusky epoch
this was a jab at your handwriting
anyway
as i said
the "bijective" part of this thing takes no more than one line to prove
because phi HAS AN INVERSE
one that's EASY TO WRITE DOWN at that
@cursive narwhal did you suggest that chaf turn a blind eye to this
cursive narwhal
Turn a blind eye to what?
His bad handwriting or the fact that it has an inverse?
dusky epoch
the latter
zinc tapir
i've seen worse handwriting on this server tbh anyway i just wanted to learn how it has an inverse and I couldn't figure that out so i went with trying to find that phi is one to one and such
cursive narwhal
Well, I just suggested working straight from the definition of bijectivity.
cursive narwhal
It’s not that torturous
dusky epoch
i just wanted to learn how it has an inverse
phi^-1(C) = BCB^-1 lol
dusky epoch
it didn't come from anywhere lol
anyway, letting $C$ be an arbitrary $n \times n$ matrix and solving $B^{-1}AB = C$ for $A$ gives you $A = BCB^{-1}$
stoic pythonBOT
Ann:
zinc tapir
so i get to AB=B^-1C am i allowed to take the inverse of B somehow?
idk guys sorry if im being slow here but i just got to this and yeah
dusky epoch
why WOULDNT you be allowed to take the inverse of B
zinc tapir
A is infront of it
dusky epoch
however what you did there is illegal
stoic pythonBOT
Ann:
cursive narwhal
$$B^{-1}AB = C$$
$$\implies B(B^{-1}AB) = BC$$
$$\implies (BB^{-1})(AB) = BC$$
$$\implies AB = BC$$
$$\implies (AB)B^{-1} = BCB^{-1}$$
$$\implies A(BB^{-1}) = BCB^{-1}$$
$$\implies A = BCB^{-1}$$
dusky epoch
no i mean you seem to doubt that B^-1 exists even though it's RIGHT THERE in the original eq so like
stoic pythonBOT
Abhijeet Vats:
cursive narwhal
You’re missing a B^{-1} on the RHS
Thank you, forgot about that
cursive narwhal
There's stuff I left out but you can just put that back in.
zinc tapir
i just thought the inverse operator was like multiply on the outside like regular algebra
dusky epoch
??
half ice
Let's talk multiplication on real numbers.
The inverse of 2 is 1/2. This is because 2(1/2) = 1
An inverse is a number that "brings it back to 1"
But in the matrix world, 1 is I
cursive narwhal
And Neo is the gatekeeper of the Province of Invertible Matrices
dusky epoch
because matrix multiplication is not commutative
you need to be careful whether you premultiply or postmultiply
pearl mirage
Does the following always hold true?:
Suppose $V$ is a finite dimensional vector space and $ v_{1}, ..., v_{m} \in V$
If $span(v_{1}, ..., v_{m}) = V$ then a list $ v_{1}, ..., v_{m} \in V$ is linearly independent
stoic pythonBOT
nicelagrangian:
cursive narwhal
Nope. If dim(V) < m, then the list is linearly dependent.
If dim(V) = m, then the list is linearly independent. If it wasn't, then there is a vector in the list which can be written as a linear combination of the other vectors in the list. So, you can remove all such vectors and still have the span of the remaining vectors be equal to V. But that means that dim(V) < m and that is a contradiction.
@pearl mirage
pearl mirage
Oh neat! Thanks @cursive narwhal
cursive narwhal
you're welcome
wintry steppe
Let V and W be vector spaces over a field F. Prove that if a linear transformation
f : V −→ W is one-to-one then, {f(v1), f (v2), . . . , f (vn)} is a linearly independent
subset of W whenever {v1, v2, . . . , vn} is a linearly independent subset of V
cursive narwhal
Okay, what do you need help with specifically?
dusky epoch
that's not specific now is it
cursive narwhal
What have you tried?
Do you know what linear independence means?
Surely you should know what you actually have to show? In order to show linear independence, what must you demonstrate?
wintry steppe
Linear independence, basically when a certain vector cannot be defined as a linear combination to others
cursive narwhal
Linear independence, basically when a certain vector cannot be defined as a linear combination to others
This is imprecise.
And, therefore, unhelpful.
Do you understand the image above? Is it clear to you?
wintry steppe
Yes it is @cursive narwhal
cursive narwhal
Okay. So, given the information in the question, can you prove that the implication above holds?
Do you have any ideas for how to approach the problem?
wintry steppe
No i dont
cursive narwhal
Do you have no idea how to go about it because you're actually confused about what the problem is asking or do you not know because you don't actually understand the terms in the problem?
wintry steppe
I'm actually confused about the problem
Everything actually
The corona didn't really help, because i had to come home from school and we didn't get to this part on school
cursive narwhal
Ah you got memed by corona huhh I got pranked by my teacher today too lol
Anyways, what are you confused about? Be a little specific, yes?
wintry steppe
I dont know how to go about it... You get me and i need to submit soon. Really wish you could just answer for me
wintry steppe
Thanks
cursive narwhal
you're welcome
wintry steppe
ok ignore the huge play button symbol, but you guys can see what's written underneath. how did he get to [2]7 from [9]7??
very confused
dusky epoch
9 is congruent to 2 mod 7
bc 9 and 2 differ by a multiple of 7 (specifically, 7 itself)
wintry steppe
wait so what math do i need to do to get 2?
9-7
oh, hm ok
another one is what is the simplest name for: [3]7
and the answer is [-3]7 = [4]7
so how did they get 4?
oh wait i got it nvm
@wintry steppe it's not 9-7, it's 9 mod 7
i know. you asked how to get 2
9-7=2
outer citrus
Do matrices and their transposed have the same eigenvalues and eigenvectors?
limber sierra
they have the same eigenvalues, but not the same eigenvectors
you can test this with some examples
(and the proof isnt particularly hard either)
outer citrus
I was confused because I was dealing with an example which does
M:=matrix([[1,0,0,0],[0,0,0,1],[0,1,0,0],[0,0,1,0]]);
quartz compass
it's probably to do with the fact that M^T = M^-1
so if you're looking at it after you've diagonalized it, you're raising the eigenvalues on the diagonal to the -1 power, and you have the exact same eigenvectors from this perspective
@outer citrus
quartz compass
you're welcome
pearl mirage
Statement: If some vector in a list of vectors in $V$ (with a field $F$) is a linear combination of the other vectors, then the list is linearly dependent.
My attempt:
Suppose $v_{1} , ..., v_{m} \in V$
Let $v_{m} = a_{1}v_{1} + ... + a_{m-1}v_{m - 1}$
Subtract $v_{m}$ from both sides and multiply the equation by -1
$0 = a_{1}v_{1} + ... + a_{m-1}v_{m - 1} - v_{m}$
$0 = -a_{1}v_{1} - ... - a_{m-1}v_{m - 1} + v_{m}$
Make a substitution for $v_{m}$
$0 = -a_{1}v_{1} - ... - a_{m-1}v_{m - 1} + a_{1}v_{1} + ... + a_{m-1}v_{m - 1} $
And now distributive property:
$0 = (a_{1} - a_{1})v_{1} +... + (a_{m-1} - a_{m-1}) v_{m - 1} $
Thus we see that there exist $a_{1}, ..., a_{m} \in F$, not all 0, such that the equation above holds, implying that $v_{1} , ..., v_{m} \in V$ is not linearly independent. Q.E.D
Am I right in this one? Thanks for possible attention 😉
stoic pythonBOT
nicelagrangian:
wintry steppe
Help please
@pearl mirage you're done after the first line, since if $$0 = a_1v_1 + \cdots + a_{m-1}v_{m-1} - v_m,$$ then you have a nontrivial linear combination of $v_1, \dots, v_m$ equal to zero and so the list is linearly dependent
stoic pythonBOT
TTerra:
pearl mirage
Shit, now that is what I call an overkill when it comes to my proof-writing abilities ehh
Thanks a lot
sick dragon
i tried using dot product formulas and subtracting their sums by y but i dont think its correct
quartz compass
that sounds right to me
once you subtract out the components projected onto u and v, you're only left with the component of y perpendicular to W
sick dragon
it was incorrect 🤔 maybe im bad
prime knoll
Is matrix inversion (not pseudo-inversion) only defined for n*n matrices?
prime knoll
Thanks
Also is there a name for non-pseudo-inversion
I'm inclined to say real inversion but I feel like that implies real numbers
wintry steppe
I guess it would be just inverse
prime knoll
alright thanks
If I were to have a n*m matrix and I wanted to be able invert it, would there be any side effect to just adding the appropriate number of identity columns/rows to make it an m*m matrix?
wintry steppe
well if u have a n*m matrix then i don't think u can just add a row/column to turn it into a square matrix.
sick dragon
this is wrong (somehow), reading the wiki it seems right but im blind, so who knows:
@quartz compass you think this will work?
wintry steppe
can someone tell me if this is true? if det(I+A) = 1 + det(A) ?
karmic oracle
It is not true
wintry steppe
and i just proved this is true but to make sure, det(AB^t) is equal to det(BA^t)
karmic oracle
yes, I agree with taht
sterile basin
question
evaluate the value of p if
50/2 - 65/13 + 40/2 = p% of 80
a. 30
b. 40
c. 50
d. 52
e. 36
<@&286206848099549185>
real stirrupBOT
sterile basin
oh my fault
wintry steppe
Lmao
wise inlet
are all transformation matrices square matrices?
slow scroll
nope
wise inlet
hmm, how would that work?
sorry, i just started last week and i have very surface level knowledge about this
sonic osprey
What exactly do you mean by transformation matrix?
wise inlet
a matrix you use to transform a vector?
sonic osprey
Then I mean
$\begin{pmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 \ 1 \ 1 \end{pmatrix}$
stoic pythonBOT
Zopherus:
sonic osprey
is a valid matrix to vector multiplication
wise inlet
wouldn't that result in a 2x1 matrix
sonic osprey
Sure, but that's still a vector
wise inlet
can i visualise this geometrically?
sonic osprey
Yeah
wise inlet
like, what does it even look like to visualize a 2x1 vector in 3d space
sonic osprey
Linear transformations can flatten things
Think about the map that takes (x,y) in R^2 to (x) in R
You can geometrically think of this as smashing everything vertically onto the x-axis
wise inlet
ah, that helps
dreamy iron
Could I borrow someone's kind attention please, to look over my proof:
https://docs.google.com/document/d/1ahZrhfDurtI2Vx9uCXk7kuJi_4Q2Xjw3Fpuyh-VJWew/edit
I start with a small toy example, and the proof follow directly after.
A. $$\vec{r}= \sum^{m}{j=1} \bigg( \sum^{k}{i=1} \alpha_i \lambda_{i , j}\vec{\mu}j \bigg) = \sum^{m}{j=1} \vec{\mu}j \underbrace{\bigg(\sum^{k}{i=1} \alpha_i \lambda_{i , j} \bigg)}_{\makebox[0pt]{\tiny these are only scalars and have no vector component}} $$.
stoic pythonBOT
ninnymonger:
dreamy iron
I pull a stunt like the above and I was wondering if that's intuitive. (It's not intuitive to me so I had to make the toy example to render it comprehensible.)
prime knoll
Sorry to ask a stupid question but what operation would take Vector A to the object space of Vector B
odd kite
what's object space
prime knoll
Relative space
Like the coordinates of Vector A using Vector B as the origin
Would it just be A - B
odd kite
yeah
prime knoll
Sorry about that, obvious questions are tripping me up lately
I'm learning about rotation matrices and nothing is definite in my head anymore
odd kite
but those aren't real terms like if you say relative space nobody is going to know what you mean
prime knoll
Object space is a game development term
grave pendant
Is it possible to prove $|x| < \sum_{i=i}^{n}|x_{i}|$ without comparing $<x,x>$ and the summation squared, i.e. ideally using cauchy/triangle inequality?
stoic pythonBOT
1a2b3c:
grave pendant
nvm I think i got it
wintry steppe
dot product, perpendicular, etc, etc
gray dust
i think it's more fun to derive the "formula" presented in that theorem rather than wield it in computations
gray dust
show work?
dry pulsar
If V is a vector space over the field K with V different from empty, then always exists a subspace W of V such that W is different from V
Is true o false?
sonic osprey
What do you think?
dry pulsar
I think is true because the addition of its components and multiplication with scalar are always in K
open pivot
Can anyone help me with this question?
karmic oracle
Prove the columns of AB are a linear combination of the columns of B
(nice problem)
open pivot
How can I show that without knowing the matrices?
storm python
just give them arbitrary column vectors
slow scroll
worth noting that the columns of AB would be linear combinations of the columns of A.
open pivot
Can I just use any random vectors?
slow scroll
u could say B = [v1 v2 v3 v4 v5] for some arbitrary vectors in R3 for example
open pivot
wouldnt B be in R5?
slow scroll
B is a matrix, but since it has 3 rows, the columns are vectors in R3
open pivot
Ah right
Im having trouble showing what AB would look like using the arbitrary vectors
slow scroll
By the definition of matrix multiplication, if B = [v1 v2 v3 v4 v5] then AB = [A(v1) A(v2) A(v3) A(v4) A(v5)]
i.e. AB is the matrix you get by applying A to each of the columns of B
open pivot
Right so I have to show A(v1) ... A(v5) is a linear combination of v1 ... v5?
rotund jetty
@open pivot you can't pick a specific vector, because the statement asks you to prove for all
slow scroll
well, so you want to show that the set {A(v1), A(v2), A(v3), A(v4), A(v5)} does not form a basis for R5, i.e. the set is not spanning or not linearly independent
these are arbitrary vectors nicholas?
rotund jetty
nono replying to >can I just use any random vectors
slow scroll
ah ok
rotund jetty
there are statements that are true about some things that are not true for all things
for example, I claim all numbers are even
I will pick a random number
I pick 123456
this is even
QED
open pivot
how can I show they form the basis if they're arbitrary?
slow scroll
well, for your next hint. Look at the columns of B. Are they linearly independent? The columns of B are made up of 5 vectors in R3
open pivot
so they're linearly dependent right?
slow scroll
why
open pivot
you cant pivot in every column
slow scroll
sure okay. Using what we know about the columns of B, try to show that ${A(v_1), A(v_2), A(v_3), A(v_4), A(v_5)}$ is also linearly dependent. In other words, there exists scalars $c_i$, at least one of which is nonzero, such that $$\sum_{i} c_i A(v_i) = 0 .$$
After unpacking the definitions, this is what you have to prove.
stoic pythonBOT
kxrider:
slow scroll
np
note that this is a very direct way of proving this. It is actually equivalent to the fact that ker(B) \subset ker(AB)
dreamy iron
is what I'm doing here "kosher"? (i'm worried i've made a grave error somewhere)
The end result should be this, which I hope is also okay: $$\vec{r}= \sum^{m}{j=1} \bigg( \sum^{k}{i=1} \alpha_i \beta_{i , j}\vec{\mu}j \bigg) = \sum^{m}{j=1} \vec{\mu}j \underbrace{\bigg(\sum^{k}{i=1} \alpha_i \beta_{i , j} \bigg)}_{\makebox[0pt]{\tiny these are only scalars and have no vector component}} $$.
stoic pythonBOT
ninnymonger:
odd kite
me eyes
dreamy iron
(you'd hate to see the source), the underbraces are the only thing keeping my head oriented to what's going on.
odd kite
and yeah, it looks fine to me, all you are doing is switching the order of the sum
dreamy iron
i can't find a source on when switching the order is legal, and when it's not legal.
this is my attempt at math ad lib
odd kite
well, it's not legal when k depends on j, but you probably knew that
dreamy iron
can you explain that please?
odd kite
this post here is an example of how things are a little more involved when m depends on i, you can sometimes still swap the order but you have to change things to make it work: https://math.stackexchange.com/questions/1876828/how-to-change-the-order-of-summation/1876838#1876838 This paper deals with how it is legal to swap indices when the limits don't depend on the index. http://www.math.ubc.ca/~feldman/m321/twosum.pdf
I think there's a typo on page 2 though
and yeah I meant "m depends on i" or "k depends on j" depending on which way you are going. The limits of the innermost sum can't depend on the summation variable of the outer sum, if you want to swap with no other changes.
dreamy iron
my beta variable depends on both
odd kite
so?
dreamy iron
so i had to keep it in the inner most nested sum
odd kite
yeah
dreamy iron
tyvm for these resources!
oh, lol. thats the same MSE page im going trough trying to comprehend all the answers.
shy atlas
~~hi ninny 
~~
wintry steppe
Says the ninny
dreamy iron
So in my proof, I don’t ever mention linear independence! Is that a mistake?? Do I need to appeal to linear independence some way or another?
cursive narwhal
A list is linearly dependent if it is not linearly independent. Any mention of linear dependence would be linked automatically with linear independence.
@dreamy iron
stoic pythonBOT
Abhijeet Vats:
dreamy iron
@cursive narwhal thank you
cursive narwhal
you're welcome
pallid rampart
Linear dependence doesn’t imply the coefficients sum to 0, or $\sum_{i=1}^m\lambda_i=0$, because you can have $-1\cdot(0,1)+1\cdot(0,1)=(0,0)$ but $\lambda_1+\lambda_2=-1+1=0$. What you need to mention is that if $\sum_{i=1}^m\lambda_i=0$ then $\sum_{i=1}^m\lambda_i\vec{v}i=\sum{i=1}^m\lambda_i=0$ then by linear independence of $\vec{v}$’s we have $\lambda_1=\cdots=\lambda_m$ contradicting the fact that not all coefficients are 0. Therefore $\sum_{i=1}^m\lambda_i$ is not zero, which allows you to divide by that on both sides
stoic pythonBOT
Whoever:
pallid rampart
@dreamy iron
dry pulsar
If A_nxn is a invertible matrix then A is product of elementary matrix of order n
Is this statement false o true?
dry pulsar
Thank you 😄 yes i just do it
spare rune
oh
cursive narwhal
and if you have multiple planes, you have PI_1, PI_2 and so on
spare rune
weird, google didn't give me a result
sick trail
what kind of results exist for norms of products of matrices?
like i know |AB|_2 <= |A|_2 |B|_2
but can we say anything about 1-norm?
where 1 norm is sum |a_ij|
wintry steppe
lazy answer: show that your norms are equivalent, and then work with your inequality |AB|_2 <= |A|_2 |B|_2 a little bit
upbeat blaze
Quick question, how would I read this out loud?
wintry steppe
context?
upbeat blaze
Just a true or false question, I think I might've read over how to voice this in the book and forgot it
I got the answer right, but I blanked on how to read that out loud lol
wintry steppe
ive never seen that kind of notation before so i have no clue 😐
upbeat blaze
Dang, it has something to do with the change-of-coordinates matrix stuff
I think it is matrix P B to C, given that the arrow is pointing from B to C lol
But I could be wrong
on how to read it out loud I mean
wintry steppe
if that's the case then you could just say exactly that
"change of basis matrix from B to C" or some equivalent sentence
upbeat blaze
Alrighty, thanks :D
wintry steppe
as long as the person you're talking to understands what you mean
upbeat blaze
Fair enough, I'm sure my teacher will be able to understand what I am getting at, especially if I have the notation in front of us lol
wintry steppe
i don't think that notation is very standard so you might want to be careful using it outside of your class
upbeat blaze
What is the standard for this?
wintry steppe
but if it's for your class then you might as well use it
uh
i think it would just be "let P be the change of basis matrix..."
upbeat blaze
Ah okay
wintry steppe
my first year LA book used something like $$P = [I]_B^C$$
stoic pythonBOT
TTerra:
wintry steppe
then again, i could be wrong, im only saying it might not be standard just cause ive never seen it before
upbeat blaze
Interesting, I haven't seen that kind of notation, what was the book? I'm using my math lab for my class
Some Pearson book
wintry steppe
"linear algebra" by friedberg, insel, spence
upbeat blaze
Oh okay
wintry steppe
it stands for the matrix of the identity transformation $I : V \to V$ relative to the bases $B$ and $C$, which is exactly what a change of basis matrix is
stoic pythonBOT
TTerra:
upbeat blaze
Interesting, I need to study up more on this stuff haha
untold citrus
can someone help me (b)
gray dust
i see it often enough, though C<-B is written as a subscript, not entirely under P
wintry steppe
just curious, where have you seen it?
gray dust
occasionally on this server when helping out with LA hw
wintry steppe
yeah that makes sense, thanks
guess i shouldnt jump to "that might not be standard notation" so fast
gray dust
guess not but it was good of you to ask duke to clarify what they read
upbeat blaze
Sorry for not giving enough context at first
wintry steppe
it's fine, this sort of thing is more common than you think
@untold citrus what does the hint for part b say?
wintry steppe
if you did the calculations i think you did for part (a) then you should be able to figure out the null and column spaces of B
untold citrus
null space is just making it = 0
in 3(a) is really long i did
the null space of B =
[1 1 -1] = 0
[1 -1 2] = 0
[1 0 1] = 0
``` to find the null space right?you are supposed to solve the equation $Bx = 0$ for $x$, which can be done by row reducing (which i am assuming is what you did in part (a))
stoic pythonBOT
TTerra:
untold citrus
yes to find b^-1
rref i got is
[ 1 0 0 1 1 -1]
[0 1 0 -1 -1 3]
[0 0 1 -1 -1 2]
of B
B^-1
[1 1 -1]
[-1 -1 3]
[-1 -1 2]
well, one way you can solve $Bx = 0$ is by multiplying both sides by $B^{-1}$
stoic pythonBOT
TTerra:
wintry steppe
iirc you can also do it by row reducing the augmented matrix (B | 0) (i'm not gonna try to format that)
but you already know what B row reduces to by part (a)
which should make it straightforward to find the null space
untold citrus
row reduction
goes to
[1 0 0]
[0 1 0]
[0 0 1]
okay, now to find the null space you can solve the equation $rref(B)x = 0$ for $x$
stoic pythonBOT
TTerra:
wintry steppe
Does anyone here know how to unlock a protected ms word file?
i dont think this is the right channel
untold citrus
squash99 this is not right channel for that go to programming channel mostly likely they can help you on that
wintry steppe
can someone help me visualise this
i cant really visualise it as it's 3 unit vectors
since they're 1 each
1 + 1 -1 = 1
limber sierra
how do you define 'unit vectors'?
components of an arbitrary basis? of the standard basis?
just a vector of length 1?
wintry steppe
i would define them as vectors with a length of 1
limber sierra
not sure what you mean
untold citrus
tterra i believe my professor want to use the NM4 in 3b how would that apply there?
wintry steppe
i dont know because i dont know what that is
untold citrus
oh
limber sierra
"NM4" is not a standardized term
wintry steppe
i also have to go very soon so ill let someone else help you
good luck!
hmm
dot product of any vector by itself is
just 1
limber sierra
well, these arent the same vector
wintry steppe
well specifically unit vectors
untold citrus
wintry steppe
there, 2 and 3 are equivalent
since B row reduces to the identity, its null space, by 3, is the zero vector
limber sierra
^
wintry steppe
im sure you can figure out the row space calculation and the linearly independent columns now
untold citrus
i do it and come back see if is correct by you guys is that ok?
wintry steppe
so this is what i see in my notes:
Notes on Permutations
For the equilateral triangle, the permutation a = (123) represents a
120-degree rotation about an axis perpendicular to the triangle, while
the permutation b = (23) represents a 180-degree rotation (or a reflection) that interchanges vertices 2 and 3. The product ba (a followed
by b) of these two permutations, (23)(123), is equal to (13). Why?
• (123) takes 1 to 2, then (23) takes 2 to 3.
• (123) takes 2 to 3, then (23) takes 3 back to 2.
• (123) takes 3 to 1, then (23) leaves 1 alone.
• So the net effect of ba is to interchange 1 and 3.
Similarly, the product (13)(12) is equal to (123). Why?
• (12) takes 1 to 2, then (13) leaves 2 alone.
• (12) takes 2 to 1, then (13) takes 1 to 3.
• (12) leaves 3 alone, then (13) takes 3 to 1, .
• So the net effect is 1 → 2 → 3 → 1.
i am very very confused
what the logic with the first four bullet points that is the proof to "(23)(123), is equal to (13)"
limber sierra
do you know how cycles work?
wintry steppe
no, what are cycles?
limber sierra
uh
cycles are the way you write permutations
so (23)(123) is a cycle
(13) is a cycle
etc
wintry steppe
oh ok
limber sierra
the idea is, they tell you where you send each number
in each indivudal bracket, each number is "sent to" the one to its right
so in (123), 1 is sent to 2, 2 is sent to 3, and 3 is sent to 1 (it loops back around)
if you have a product of cycles, then you continue this right-to-left through the entire product
so for example
in (23)(123)
we go right-to-left so start by applying (123) and then (23)
i.e.
1 gets sent to 2, then 2 gets sent to 3 [so 1->3]
2 gets sent to 3, then 3 gets sent to 2 [so 2->2]
3 gets sent to 1, then 1, since it's not written, gets sent to 1 [so 3->1]
hence this is the cycle that sends 1 to 3, 3 to 1, and leaves 2 fixed
this is the cycle (13)
wintry steppe
wait what do you mean [so 1->3} in the first line
limber sierra
look at (23)(123) and try and track where 1 goes
first we apply the right cycle (123)
so 1 gets sent to 2
then we apply the left cycle (23) to 2 [since 1 was sent to 2 already]
so 2 [previously 1] is sent to 3
so overall the thing that started as 1 ends up at 3
so this permutation puts the thing in the first position into the third position
wintry steppe
are you in (23) for the second line?
limber sierra
2 gets sent to 3 by (123) and then 3 gets sent to 2 by (23)
so 2 ends up... right back at 2
i.e. the thing in the 2nd position doesnt move
wintry steppe
but that was already written at the end of the first line
so that's why im confused
but ok let me read on
dreamy iron
@dreamy iron
@pallid rampart Regarding this:
dreamy iron
I tried to understand what you were saying. I'm not sure I understood it 100%, but that's what I took away from your comments.
(do you wannna see the lemma i referenced?)
pallid rampart
What I'm saying is, not all $\lambda_i$'s are zero does not mean $\sum_{i=1}^m\lambda_i\neq0$
stoic pythonBOT
Whoever:
pallid rampart
Well the rest was just an example and how the proof is supposed to go, but it seems like you also got the proof
So you're good
dreamy iron
thank you kindly sir!
fickle agate
when you say
A = A*
that means every entry of matrix A has to be equal to every entry of its adjoint matrix right?
pallid rampart
Yes that is what it means for two matrices to be equal
fickle agate
Hmm ok
elfin ingot
prob similar
limber sierra
"similar" here in the sense of row equivalence
i.e. you can get from A to C (and vice versa) via elementary row operations
(alternatively put: they have the same RREF matrix)
fickle agate
i see thanks
shy atlas
wtf is this
i have so many questions
can anyone explain what the point of even defining this is
such a weird and confusing definition
sonic osprey
I mean, it furthers the relationship between spaces and their dual maths
Given a map between two spaces, you get a map between their dual spaces (in the opposite direction
and this goes the other way too
shy atlas
sonic osprey
does this make sense or
tiny grove
shy atlas
@sonic osprey it did but i had to draw a hecc ton of diagrams to get why T' was in the opposite direction
$\mathcal{M}(T \circ S) = \mathcal{M}(T)\mathcal{M}(S)$
stoic pythonBOT
soαρ: