Alright thank you
#linear-algebra
stable urchin
slow scroll
npnp
stoic pythonBOT
robo™:
wintry steppe
By definition of the inner product, <v, w> = <w, v> and <w, v+u> = <w, v> + <w, u>
= <v+u, w>
Which in the case of
<v, e1> = <c1e1 +.., e1> = <c1e1, e1> + <c2e2, e1> +.. <cm * em, e1>
= c1<e1,e1> + c2<e2, e1> +.. cm<em, e1>
By definition, e vectors are orthonormal so the inner products all yield 0 except for <e1, e1>
Now apply this to all the other sums
@zinc tapir
dusky epoch
so long as you restrict to a convex set, it will not.
well except for singletons i suppose
delicate zealot
If I have a question about linear algebra should I ask here or in one of the questions chats?
dusky epoch
whichever one's free
if both are free it doesn't matter
but you might get more specific feedback in this channel as it's dedicated to LA
delicate zealot
Ok I just want someone to check my row reductions because I was asked to find two different echelon forms for a given matrix and then find the reduced echelon matrix of each but I don't get the same answer which is a big big issue...
And I typed it up in LaTeX so I was wondering if I could share my PDF and have someone else spot my error because I've been on it for much longer than I need to be
dusky epoch
yeah sure you can send your pdf here
ivory basin
can someone help me understand how to start problems like these
i read all my notes and lecture about the 10 axiom rules but im not sure how to use them in this situation
vague cedar
the axioms that usually fail are v,w in V => v+w in V and v in V => kv in V
0 in V is a common one that fails too
in the context of intro linear algebra you can usually just check those three and be fine
cursive narwhal
Well, in part a), they're really just asking you to show that $(0,0)$ isn't the identity element of $V$. So, what you can do is to see that:
$$(u,v)+(0,0) = (u+0+1,v+0+1) = (u+1,v+1) \neq (u,v)$$
because an identity element is supposed to do nothing to the vector after it is applied.
stoic pythonBOT
Abhijeet Vats:
cursive narwhal
@ivory basin
vague cedar
i'd be careful and state that you're trying to find the identity element of (V,+) specifically
cursive narwhal
What you should try to do is to read through the axioms again and to do simpler problems involving them so that you get some practise with using them whenever you wish. Then, these problems won't be so tough.
ivory basin
alright ill try that thanks for the help guys
cursive narwhal
You're welcome
Just to also give you give you a hint about part b), it's asking you to, then, show that $(-1,-1)$ is an identity element of $V$ with respect to addition. All you need to do, then, is to add it to another vector in the speciifc way defined above and prove that the result just spits that vector right back at you.
stoic pythonBOT
Abhijeet Vats:
wintry steppe
your name and your profs name and your course code is on it
delicate zealot
Ohhhhhhhhhhhhhhhh
wintry steppe
you might want to remove such personal info
delicate zealot
Good call thanks
delicate zealot
Yeah but my prof name
delicate zealot
Which I guess is also public info but ya know
pallid rampart
Just quick edit
delicate zealot
Boom fixed
Now I'm going to bed but I'll check this channel when I wake up and take a fresh stab at the problem
Probably some dumb arithmetic mistake =
pallid rampart
What’s your question tho
pallid rampart
Yes
stable urchin
Alright ty
delicate zealot
The question should be in the file... you're given a matrix and you have to find two different matrices in echelon form and then create a reduced echelon form from those echelon matrices that you chose
I got two different answers for reduced echelon form which is a big no no
wintry steppe
@delicate zealot
i checked what the final answer should be with wolframalpha
neither of your answers are correct
im not going to try to find any more errors, since if one of the first steps already has an error then you'll need to do things over
prime knoll
In a 3d vector what value of x y and z is required to make it have a magnitude of 1?
dusky epoch
what do you mean
prime knoll
A vector with all 3 values being the same and positive
dusky epoch
there are many vectors in R^3 with magnitude 1
oh
so your vector is (x,x,x) with x > 0, and you want its magnitude to be 1?
prime knoll
Yes
dusky epoch
do you know how to calculate the magnitude of a vector in general?
prime knoll
3d euclid
dusky epoch
are you having trouble solving the equation $\sqrt{x^2 + x^2 + x^2} = 1$?
stoic pythonBOT
Ann:
prime knoll
No I just figured it would be easier to ask
sqrt 1/3?
Now I feel like an idiot
Thanks
green knoll
The answer should be True right? Is there another vector can be orthogonal to itself other than zero vector?
muted holly
dusky epoch
yes
green knoll
so the answer should be True right?
dusky epoch
i already answered that question
take it up w your teacher tho. i think the question isn't worded as intended.
green knoll
ok thank you
according to teacher "Of course this queston is false. A nonzero vector cannot be the zero vector."
what a nice question
green knoll
why do professors ask these kind of stupid questions 😦
shy atlas
cuz they cant come up with thoughtful questions and have to rely on deception
pallid rampart
shy atlas
riot against ur profs
cursive narwhal
cuz they cant come up with thoughtful questions and have to rely on deception
||They should just become IMF agents if they want to rely on deception.||
half ice
Difference between applied and pure imo. Prof probably doesn't know that vacuous truths are a thing
green knoll
sorry what do you mean by a "vacuous truths" 😦 i am not mathematics student
do you think this professor is working applied or pure?
shy atlas
see your assertion is of the form "if P then Q"
and P is always false
so anything can follow
green knoll
oh the logic
green knoll
i understand it
it seems the prof interested in this field "Geometric and combinatorial group theory"
green knoll
i hope i can pass the class with good grade. i don't want to repeat the lesson. I like learning mathematics from my theoretical physicist professors. You mathematicians are not practical in your solutions 😦
sorry for that 😄
there is still no another vector that be orthogonal to itself other than zero vector in complex numbers right?
shy atlas
in any goddamn inner product space be it over a real field or complex field its true
half ice
By the definition of an inner product, no element is orthogonal to itself
half ice
Other than the zero I mean, oops lol
"If v is non-zero and orthogonal to itself, then v is made of cheese" is a true statement
shy atlas
😋
fickle tree
using the vectorial cross and knowing that A(0,1,0) B(2,0,3) C(9,-4,5) D(D5,-2,-1) verify that this is a trapeze and find a vector with the length 1 orthogonal of the image
This is what I did so far I said that BA is parallel to CD
I don't know what to do after
stoic pythonBOT
soαρ:
delicate zealot
@wintry steppe sorry fell asleep with my laptop open lol. I totally understand you not wanting to find errors if you found something super dumb in the very first step. I'm going to completely start over right now and see what happens. Thanks for finding that dumb idiotic mistake tho
sullen pollen
What is a linear variety and what is so important about it
wintry steppe
I thought we could only multiply a 3x2 by a 2x3. By extension of that logic shouldn't we be unable to multiply this 2x2 by a 2x1 as presented in the screenshot above?
tidal kiln
no in general you can multiply every NxR Matrix with any RxM Matrix, which results in a NxM Matrix where N,M,R are all natural numbers
wintry steppe
Think of each row as x y and z of vectors
If your matrix transforms x and y coordinates and the input matrix entirely consits of x coordinates, then the transformation cant be applied
There's probably a better explanation but im trying to make this intuitive
tidal kiln
If you where to look closer at the multiplication of 2 matricies, you will see that you multiply each row of your first matrix, with every column of the second matrix
=> row length of matrix 1 = column length of matrix 2
wintry steppe
Interesting, so it's just numbers of rows in A = number of columns in B?
If you are multiply B and A, yes
Also that is nice for my intution, thanks robo @wintry steppe
Ahh so I got it backwards @wintry steppe
tidal kiln
no i meant:
numbers of entries in every row of A = numbers of entries in every column of B
wintry steppe
In general the number of columns in your first matrix = the number of rows in your 2nd matrix
In order to multiply matrices
tidal kiln
@wintry steppe So i got it backwards? with the rows and columns, oh man
wintry steppe
Yah
tidal kiln
but phrased like the second time, should be right, right?
number of entries in row = number of columns
wintry steppe
Yes
tidal kiln
thanks 🙂
tidal kiln
You too
tiny grove
can someone check my answer to the first part of the q:
w is in column space, row space, and null space of A
this is rref of A
tiny grove
i used matlab
tiny grove
wdym
torn silo
well try to build the vector w with just rows of A
tiny grove
ok, ill pick the transpose of last row of A
wait, does w have to a row in the first place? @torn silo
to check whether it's in rowspace or not
torn silo
🤔
tiny grove
seems like it lol
torn silo
that would make sense no?
you could also transpose A and keep thinking of w as a column
tiny grove
yeah
ocean sequoia
so is an affine subspace simply : U is a subspace of V then for every U + v would be affine(?) to U?
to use non-technical terms so i maybe dont botch it so badly
affine subspaces are when a subspace is shifted by some constant
So im watching a video and the guy says the affine combinations of a vector must be equal to 1 so it forms a line in R^2 couldnt the affine combinations just be equal to some c
ocean sequoia
also would 2x + 6 be an affine subspace of 2x?
wintry steppe
"2x + 6 is an affine subspace of 2x" doesn't make sense
ocean sequoia
is 2x + 6 a affine subspace of R^n?
so i know that 2x would be a subspace it obviously satisfies all conditions then i thought an affine subspace is a subspace shifted by some constant
wintry steppe
by 2x do you mean the set {(x, y) : y = 2x}?
because if you are talking about subspaces then you should be more careful
wintry steppe
that's a vector subspace of R^2
ocean sequoia
yea
wintry steppe
the set {(x,y) : y = 2x + 6} is an affine subspace of R^2
because you can think of it as the set of all vectors in the first set with (0, 6) added on
ocean sequoia
ok so an affine subspace is defined on R^n? Not as a subspace of the vector subspace?
not sure if that makes sense
sorry trying my best to make my terminology as precise as it can be
wintry steppe
one moment let me get on my computer
ocean sequoia
ok
wintry steppe
affine subspace is defined on R^n? Not as a subspace of the vector subspace?
im not quite sure what you mean. if you have a vector space (here it is R^2) and some vector subspace (here it is the set of all (x,y) with y = 2x), then you define an affine subspace by adding a vector to all of the vectors in that given subspace
ocean sequoia
I guess an affine subspace is not a subset of the vector space?
i.e. its viewed as its own thing
wintry steppe
i use "vector subspace" to mean a subspace in the vector-space sense
wintry steppe
no problem
ocean sequoia
i appreciate the help
wintry steppe
geometrically you can think of an affine subspace as a vector subspace translated so that the origin is at a different point
an affine subspace will not generally be a vector space, but it is a vector space which has undergone a translation
that's how i see it at least
ocean sequoia
no that totally makes sense
wintry steppe
the notation $$U + v$$ for the affine subspace $${ u + v : u \in U }$$ is suggestive of this geometric interpretation
stoic pythonBOT
TTerra:
wintry steppe
might be something worth keeping in mind
wintry steppe
glad to help
delicate zealot
@wintry steppe thanks for the linear algebra help last night. I went back to do it again and got the answer :)
I unfortunately didn't have time to submit the correct answer tho so i submitted what I sent last night and still somehow got a 7/10 lol
dreamy iron
I'm getting bored of Axler, am I'm worried grinding through any more of it is going to cause burn out.....
I want to do a mini digression into Einstein Summation Notation and learning how use the Levi-Cevita symbol to prove identities. would that be profitable use of my time, for future Lin. Alg learning? do you guys encounter them in Lin. Alg, ever?
wintry steppe
i've not seen it so far in any linear algebraic stuff
i would assume that stuff mostly comes up in differential geometry / tensor-related-things
if you're on axler's ladr then you dont need it
unless you really really want a way to make some things involving sums shorter, there is no point
the indices don't get so bad in ladr-level linear algebra that it's worth using einstein summation
(although i must admit i like the proof that matrix multiplication is associative that uses einstein summation)
$$ ((AB)C)^i_j = (AB)^i_k C^k_j = A^i_l B^l_k C^k_j = A^i_l (BC)^l_j = (A(BC))^i_j$$
stoic pythonBOT
TTerra:
wintry steppe
but thats just like a fun application of it. there's no serious need to use it
quartz compass
I say if you think it's fun and interesting and you're trying to avoid burning out, go for it
I think it would be worth your time if you appreciate that kind of stuff, I find it to be much nicer to work with personally
dreamy iron
cool guys, great advice. Thanks! I'll go on my little digression then.
i'll have to.read that proof about the associativity of matx. mult. when i get to it, but it does look much neater with out all the Sigmas getting in the way.
quartz compass
just write in the sigmas and you get the regular proof lol
wintry steppe
Hi, I'm confused on to answer this properly
I'm trying to understand how im(A) and null(A) works
Is the green stuff correct and the red stuff incorrect?
Apparently it is
I don't get how the first part is wrong
This might help
slow scroll
the image part is right. the first vector you have written down for null(A) clearly doesn't work, but it is not too hard to see how you can combine the first two columns to get the zero vector without putting it in rref.
wintry steppe
Like this?
slow scroll
its like you mixed up some stuff between the two wrong vectors in your null space. They are both almost right, but its like you mixed the answers somehow. weird
wintry steppe
col 1 and col 4
slow scroll
if you have already put A into rref then dont bother doing it again. The (2, 1, 0, 0, 0) you have written down is saying: "if I take 2 times column one plus 1 times column two then I get the zero vector." This is wrong, BUT there is a way you can combine columns one and two to get the zero vector
same thing for (-5, 0, 1, 0, 0). i.e. there is a way to combine columns one and three to get the zero vector, but NOT by taking -5*column one + column 3
slow scroll
once you figure it out, you'll notice its as if somehow you switched the numbers around for your answers. Interesting mistake :p
wintry steppe
Can you walk me through it?
slow scroll
c(1, 2, 1) + d(5, 10, 5) = (0,0,0)
what should c and d be?
slow scroll
yes. since (1,2,1) is the first column, c = 5 goes in the first entry. Since (5,10,5) is the second column, d = -1 goes in the second entry. i.e. (5, -1, 0, 0, 0) is in the null space of A.
Now try the same thing for columns one and three: c(1,2,1) + d(-2,-4,-2) = (0,0,0)
wintry steppe
c = 2
d = 1
slow scroll
yep, so this means what vector is in the null space of A?
slow scroll
hmm, idk what you mean by that. We are looking for some vector (u,v,x,y,z) such that A*(u,v,x,y,z) = (0,0,0,0,0). Check what I did here:
since (1,2,1) is the first column, c = 5 goes in the first entry. Since (5,10,5) is the second column, d = -1 goes in the second entry. i.e. (5, -1, 0, 0, 0) is in the null space of A.
wintry steppe
im(A) should be (1,2,1) and (-2,-3,0)... right?
slow scroll
well, those are vectors in the image, yea, but that is not what I am talking about.
We are looking for some vector (u,v,x,y,z) such that A*(u,v,x,y,z) = (0,0,0).
Okay, so I made a mistake, but I fixed it in the above quote (sry about that). Anyway, when u compute A*(u,v,x,y,z) , you are taking u times column one + v times column two + x times column three + ....
We want to find the u,v,x,y,z such that when doing that operation gives (0,0,0).
We know that 2(1,2,1) + 1(-2,-4,-2) = (0,0,0). So this is the same as A(u,v,x,y,z) = (0,0,0) for which values of u,v,x,y,z?
wintry steppe
(2,1,0,0,0) , (5,-1,0,0,0) and (-5,0,0,-2,1)
slow scroll
not quite for the first one: which column is (-2,-4,-2) in?
wintry steppe
3rd
slow scroll
therefore the 1 goes in the third entry (2, 0, 1, 0, 0).
wintry steppe
ah my apologises
slow scroll
so like, when you see this on a test or something, and the null space vectors are nice (aka have lots of zeros in their entries), its not too hard to fact check yourself
e.g. you get a wrong answer like (2,1,0,0,0): "does 2(1, 2, 1) + 1(5, 10, 5) = (0,0,0). Oh no it doesn't, there is a mistake. Oh it looks like it should have been -5(1, 2, 1) + 1(5, 10, 5) = (0,0,0). Therefore (-5, 1, 0, 0, 0) is the correct vector"
slow scroll
the image is fine. the columns of A form a perfectly valid spanning set for the image
half ice
The image, much like the image for a function, is the set of all outputs your matrix can make. Think about it for a bit, it's natural that the columns will span this
You can't make anything that the columns can't make
delicate zealot
Is there a way to tell whether a linear system is consistent just by looking at the matrix or do you have to simplify it?
half ice
You can't tell just by looking at it
delicate zealot
That's why I ask cause it says don't completely solve but the only way I can figure it out is by completely row reducing.
So I was wondering if I was missing something.
delicate zealot
Thank you sir
Why multi color pens for $10 when you can buy a stylus for $70
And that's why I'm a math major folks 😂
the joke was the flawed logic
wintry steppe
it means don't do back substitution I think
just rref and check if there is a contradiction
oh well..
delicate zealot
At the end of the day I'm happy if I understand the math
wintry steppe
that's probably not satisfactory nvm
delicate zealot
I know how to identify a consistent/inconsistent system so if I can do that IDC about losing a point on one HW question for not answering the problem
Especially when my professor didn't dock me massive for way less work
fluid linden
Hi there, could someone point me in the right direction please? I'd like to prove that if A^2 = I, eigenvalues can only be +- 1
brittle juniper
the polynomial X²-1=(X-1)(X+1) annihilates A so that should tell you some things
quartz compass
I'd write Av=kv then multiply by A again to get v = k^2 v
shy atlas
i would say A^-1 = A. now if k is an eigenvalue of A then 1/k is an eigenvalue of A^-1 = A. so k = 1/k
quartz compass
heh
shy atlas
welp that doesnt work 
or does it
k and 1/k are the eigenvalue corresponding to the same eigenvector. they should be equal to each other
shy atlas
nice 😌
torn silo
well what is e raised by infty?
gray dust
this is #precalculus or #calculus
stray coral
thanks! i'll move there
uncut hull
Hey there, I have a question about linear dependency/independency I'm hoping someone can help explain
Within a matrix: If a column is a multiple of another column, it is extremely clear that the first column is dependent on the other. However, what about when it is not obvious that the columns may be multiples of each other? Is there an algebraic way to determine dependency without having to just visually inspect each element?
gray dust
say you're seeing whether {v1,...,vn} is linearly dependent (LD). you want to see if there exist scalars c1,...,cn, not all 0, such that c1v1+...+cnvn=0. the c's can be found in any way you want, which probably includes rewriting the eqn as a matrix eqn and row reducing. if there exists a solution where not all the c's are 0, then for sure {v1,...,vn} is LD
uncut hull
Okay, I see. Is there a name for that theorem or rule? That if there are scalars c1, c2…,cn (not all 0) then c1v1+...cnvn=0?
gray dust
definition of linear dependence
uncut hull
Huh. I suppose that makes sense lol
I have been seeing references to that rule in my book but at literally no point is it defined or explained as 'this is the rule for the thing'
Anyways that's super helpful, thank you
gray dust
very suspicious if that defn wasn't introduced explicitly before but sure no prob
uncut hull
Yeah this book isn't great. They explain LD in terms of individual columns as values themselves can be dependent or not in the first chapter or so. But when you fast forward to matrix multiplication there is no definitions provided for the rule as it applies to matrices
gray dust
i don't think it's great introducing the 1st few weeks of linalg material in the context of matrices. like when i say vector, it's most of the time best to just depend on its general idea as some object contained in a set (you'll learn later as a vector space) where there exist notions of adding and scaling stuff. then on a related note, from this idea one eventually can view certain sets of matrices as being vector spaces with matrices themselves being the vectors in question
uncut hull
Yeah, I suppose that would make more sense. I am in an introductory LA class, but we did do Vectors in Calculus 2 and Calculus 3, so the first week of my current class was more of a review and introducing new defs.
torn silo
on the other hand starting with the matrix view gives the student something more accessible
uncut hull
Now that we are on to matrices, there are a couple of missing points, so I imagine your way would be better
Okay I actually have one more question @gray dust . So given that definition, if there are 3 vectors in 3d space, with the first two u and v being dependent on each other, but the third vector w is independent, does that mean that there cannot exist a plane with these three vectors?
I have this example where u = [1, -1, 0] and v = [0, 1, -1] and w = [0, 0, 1]. Since b cannot equal zero in this case, my book is laying it out like no plane can exist between them. However, in the second example where they replace w with vector w* = [-1, 0, 1], now there can exist a plane.
gray dust
when you say lin dep or lin indep, you're always referring to sets of vectors, so "u,v dependent" reads as "{u,v} is lin dep", but idk what "w is indep" means
from the example i'm guessing you mean {u,w} & {v,w} are LI
uncut hull
I suppose I shouldn't say it's independent. I am only given that there are three vectors, with w not being able to be in a plane with the other two until the values are altered so that all three vectors result in b = 0
gray dust
& idk what's b
uncut hull
If u v and w make up matrix A, then their result is b1 + b2 + b3 = 0
gray dust
idk what this says either
uncut hull
All good. I'll figure it out. Thanks for helping me
gray dust
i mean the problem is the b's came from nowhere, they were introduced with no context, so idk what's going on
uncut hull
It says: All linear combinations x1u + x2v+x3w* lie on the plane given by b1+b2+b3=0
That is all that is given to me. I assume they are referring to the b = 0 as in c1v1+...cnvn=0 or the result in Ax=b
gray dust
ok looking at the example, just ignoring whatever the b's mean, {u,v,w} is LI, equivalently meaning there doesn't exist a plane containing em
however by inspection, i see {u,v} is LI, so u,v "span" a plane, and -u-v=w* so w* lies in that plane
uncut hull
Okay… just so I’m understanding, that means a plane can only exist between 3 vectors only when all 3 vectors are LD?
gray dust
remember linear dependence talks of sets of vectors. an equivalent rephrasing of the defn of linear dependence is that the set is LD if any of the vectors in the set is a linear combo of the other vectors in the set
uncut hull
Okay, I think I get that. So then if {u,v} are LI, why wouldn’t they span all of 3d space? Why only a plane?
gray dust
{u,v} is LI and contains 2 vectors and so they span a 2 dimensional subspace of R^3, what you visually call a plane
uncut hull
OH
Okay, I get it now.
Thank you for explaining this to me.
Like 3 things just connected in my brain
gray dust
You’re welcome 
delicate zealot
I have a question about the row reduction algorithm (I'm getting a pic of what I'm talking about rn)
wintry steppe
@uncut hull you need 3 vectors that arent coplanar to span all of R^3
And two(or three) vectors that are coplanar are necessarily linearly dependent
uncut hull
Yes, I get that now. Thanks!
delicate zealot
So in this case does that mean you get the matrix into echelon form and THEN reduce to rref?
gray dust
And two(or three) vectors that are coplanar are necessarily linearly dependent
No, not all coplanar pairs are LD
gritty frigate
Lest say I have a vector from (3, 4) to (8, 6)
Then I get another vector from (0, 0) to (5, 2) Which is the same as te one above
The magnitude of both of them. Has to be the same right ?
Forget what I was saying.... I did not copy the right exercise..
torn silo
🤔
gritty frigate
I was not arriving to the answer because I was doing the wrong exercise hahaha
ivory trout
Hello, quick question: is it possible to have other values than 0 or 1 in a reduced row echelon form?
ivory trout
thank you 🙂
pallid rampart
For an entry that is not on a pivot column and is to the right of a pivot, the entry can be whatever number you want it to be
zinc tapir
yo sup my people
if im given a span W in C^3 and im told to find an orthonormal basis of W and its orthogonal complement
i find the norm, normalize the vector and i got its basis right?
half ice
Ya gotta be far more clear, haha.
What is a "span W" here? Are you given vectors that form a basis for a subspace W?
If you do have such a basis, gram schmidt is your boi
pallid rampart
You can use Gram Schmidt's process
zinc tapir
W = span {(i, 0, 1)}
half ice
Cool, okay so you've got a basis for W, which happens to be the span of one vector
zinc tapir
ive got that the orthonormal basis is {(i/sqrt(2),0,i/sqrt(2))}
half ice
Since there's only one vector, it's already orthogonal! You just need to make it normal.
half ice
Yaya. Gram is the common procedure for that
wintry steppe
No u
delicate zealot
So I just found general solutions but I used my calculator to check my work and it got all 0s in the bottom row, anyone know where I went wrong?
half ice
R3+R1 you said 5 + 7 = 7
delicate zealot
Oh yeah that'll do it
God I feel like I go sooooo slow and still mess this up
I feel dumb
half ice
It's very computational
delicate zealot
Yeah
Just need to talk out loud or in my brain when I do that I do fine
Is what I put in the box a good conclusion or am I making too many assumptions by saying that?
half ice
I'm thinking you want to describe the solutions
sonic osprey
We call it equivalence because they generate the same topology
wintry steppe
hope im not interrupting anything
but how did they get the image span of [1 -2]
this is what i got but i feel like im finding the kernel
half ice
This is a transformation R⁴ → R²
You can get a sense of what it does by multiplying in the basis vectors
wintry steppe
well yes that's what i did
gritty frigate
If I say that a vector exists on the xy axis
And the vector is R3
Then z must be zero. Is that right ?
wintry steppe
you can either check that T doesn't satisfy some property of normal operators
or you can compute the adjoint and see if T is normal
in this case you can probably just compute, since the matrix of T is easy to find
zinc tapir
i mean i know how to check if its self adjoint but im having trouble computing if its normal
wintry steppe
what is your definition of a normal linear operator
wintry steppe
"T commutes with its adjoint"
wintry steppe
can you compute the adjoint of T
i feel like you can do so very easily using the matrix of T
delicate zealot
Sorry for so many linear algebra questions but I promise this will be the last one for the evening. In this problem, don't you need to know what the nonzero values are in order to figure out if a system is consistent or inconsistent?
Wait. I'm dumb. Already figured it out.
zinc tapir
uhh i dont think i learned up to that @wintry steppe im trying to read up on it
wintry steppe
up to what?
representing linear maps as matrices?
if you're using the book i think you're using (friedberg insel spence) then that's chapter 2 stuff
wintry steppe
anyways
there is probably a theorem in the book you can use, let me take a look
theorem 6.10 in my edition, the one that has to do with matrices of adjoints
you can use that here
zinc tapir
so basically the adjoint of the matrix is equal to the matrix of the adjoint
yeah thats the one i had trouble understanding during lecture
wintry steppe
yes that's it
in that theorem it's after a choice of basis, but you have a canonical basis on R^3 so who cares
anyways
some computation and you should get that T is not normal
wintry steppe
how did you go from A to the thing on the right?
zinc tapir
thats the cofactor of matrix A
wintry steppe
uh
the leftmost matrix is the matrix of T
not whatever the thing on the right is
zinc tapir
i thought you said find the adjoint of the matrix
zinc tapir
basically i took the 2x2 det of each cofactor
wintry steppe
my bad
i have literally never used that adjoint so i interpreted it as having the same meaning as that for linear operators
but you read the theorem right? the one that says you can calculate the matrix of the adjoint of an operator using the conjugate transpose?
use that
spice storm
@delicate zealot A is inconsistent because if you try to reducing it you’ll get back the same value back and forth
zinc tapir
ok since im inthe reals its just the transpose
wintry steppe
so @zinc tapir, all you have to check is if A commutes with its transpose
zinc tapir
alright ty, so whats the difference between the adjoints we were talking about
wintry steppe
and you know that's equivalent to T commuting with its adjoint
https://en.wikipedia.org/wiki/Adjugate_matrix you might have been thinking of that
which is also called "classical adjoint"
and then there is the adjoint of a linear operator as your book defines it
"S is an adjoint of T if <Tu, v> = <u, Sv> for all u, v" or something like that
its unique and it necessarily exists if V is finite dimensional, etc. etc.
ivory basin
Can someone please help me, my teachers lectures arent clear and I have no clue how to do these types of problems
do i make matricies or do I use axioms or something like that?
wintry steppe
if you know about dimension then you can get rid of one of them easily
@ivory basin , what do you think linear independence is?
@ivory basin easiest way is to see if the determinant of the matrix whose columns are the vectors is nonzero
However, you can definitely eliminate some options
quartz compass
don't give it away @wintry steppe
ivory basin
im really not too sure my prof uploaded a lecture but it didnt have audio
wintry steppe
lol
ivory basin
so i only saw what he was writting which didnt help
wintry steppe
Sorry my bad
quartz compass
I'd do a,b,c by different tricks each personally
ivory basin
he typed this but didnt delay the hw so i have to do it without the lecture
wintry steppe
@ivory basin still, you should tell us what you know
we shouldn't tell you what the answer is but it's ideal that you show us where you are in your understanding
ivory basin
i know it has to do with what the vectors span
and linear dependant is nonzero scalars that equal a zero vector
i think
wintry steppe
so that's the issue; it seems you don't know what linear dependence is or independece
try seeking a formal definition from a book or online resources
ivory basin
do you guys know any youtube lecture i could watch then cuz i really need to learn in under 2 hours
alright ill look around i guess
hollow finch
really odd question, sorry:
is there a concise/symbolic way to denote the matrix created by replacing a column of the identity matrix with some column vector v
alternatively, is there a concise/intuitive way to prove that the determinant of that matrix will be the k-th entry of v if v is replacing the k-th column of the identity
wintry steppe
do (k-1) column operations to stick v in the first column, then (k-1) row operations to get v_k in the (1,1) entry of the matrix. you've done an even number of row and column operations so the determinant is the same. now the matrix is lower-triangular, so the determinant is the product of the diagonal entries, which are v_k and then (k-1) 1's, so the determinant is v_k
that might be wrong, i am pretty tired rn so lemme know if theres an error
actually it can be simplified: you can just switch the kth col and the 1st col, then switch the kth row and the 1st row. the determinant switches sign twice, so it is unaffected
hollow finch
Aha that's smart
wintry steppe
lemme do an example to make sure im not making a silly mistake here
hollow finch
Getting it into the form of a triangular matrix seems the best way to go about it so regardless, thanks for that idea. It sounds like it should work though
wintry steppe
once you swap the kth column and the 1st column, the 1 that was in the kth column is now in the (1,k) entry
then you swap the kth row and the 1st row, that stray 1 goes to (k,k)
so its fine, the matrix is lower triangular
i tried a low dimensional example and it worked fine
wintry steppe
delicate zealot
Did I set up this system of equations correctly and if so can I have a hint on how to better set up these systems of equations?
And then in order to figure out if it is possible to get an equal plate I would solve the system and see if the resulting variables are equal
dusky epoch
this is not linear algebra
wintry steppe
and what
dusky epoch
you're in the wrong channel
wintry steppe
sorry
storm python
||then what are those linear equations doing there ann? checkmate||
dusky epoch
this is in reference to a now deleted question, pub
stiff frost
@delicate zealot In your first equation, what does 1 stand for? What does F stand for? etc. I would recommend you write out units for equations like this. By units I mean "grams of fat" or "ounce of pulled pork". If you write all that stuff out, I think you'll be able to set up equations with confidence
vale arrow
Hey
What bilinear and multilinear algebra books do you recommend?
keen patrol
Can't you set up the vectors as [5 5 1], [0.5 0 1], [1 2 6] in 3D nutrition space, right?
hollow finch
is this something that needs to be proven as a lemma or is it just understood as a definition/result of basic matrix multiplication
$$B\begin{bmatrix}\vec{c}_1&\cdots&\vec{c}_n\end{bmatrix}=
\begin{bmatrix}B\vec{c}_1&\cdots&B\vec{c}_n\end{bmatrix}$$
stoic pythonBOT
nix:
gray dust
following rules of matrix multiplication
hollow finch
alright cool
is there a name for a system Ax=b where A is invertible since consistent doesnt necessarily imply that?
dry pulsar
If kerT= y=-x/2, is the dimkerT=2?
limber sierra
what does "kerT= y=-x/2" mean
the kernel of T should be a set
do you mean that kerT is the set of vectors (x y) such that y = -x/2?
if so, what do you think a basis for that set would be?
dry pulsar
This is the linear transformation
$T\begin{pmatrix}x\ y\end{pmatrix}=\begin{pmatrix}5x+10y\ 2x+4y\end{pmatrix}$
stoic pythonBOT
alef0:
limber sierra
so you DO mean that kerT is the set of vectors (x y) such that y = -x/2
can you find a basis for that set?
dry pulsar
I think but in confused because the options i can write are (0,0), the line y=-x/2 or cartesian plane
limber sierra
the line y = -x/2 is the set of vectors (x y) such that y = -x/2
those mean the same thing
a line is just a set of vectors (formally the span of a single vector)
anyway, whats the dimension of a line?
dry pulsar
1 because we only need "x" togenerate all line right?
limber sierra
uh, i think i know what you mean
anyway yes
the dimension of a line is 1
indeed the vector (-1, 1/2) spans ker(T)
dry pulsar
Thank you!
plush raven
I am not sure where to begin. Any help would be appreciated. Thanks.
wintry steppe
there is a simple formula for the inverse of a 2x2 matrix that you can use here
plush raven
would it be (2,7,1,-3) *(0,1,1,0)? the identity matrix?
wintry steppe
you can use this to find A
how one would get this formula is a different matter. for now, just try applying it in the most straightforward way
(hint: whats the inverse of the inverse of a matrix?)
plush raven
ok, i will mess with it.
stable urchin
Inverses are so much more annoying when it’s 3x3
vague cedar
the move then is usually just reducing (A | I) ~ (I | A^-1)
same for 2x2's for me usually just bc i dont like memorizing formulas
gritty frigate
Lets say you have two vectors on R2, A and B. You know that |A| = 5 and that A direction is the opposite of B direction. Also B = (1/3, -2/3)
I found A direcion by doing tg-1((1/3)/(-2/3)) and then I added 270°
So the opposite angle is 116°33°54,18°
For finding the components of A I did:
x = sen(b)*5
y = cos(b)*5
b = Angle of the triangle made of the opposite angle of A
But I do not arrive to the correct answer
Where am I failing ?
plush raven
Not sure how we get from step 1 to step 2. Any help would be great.
wintry steppe
Do I have to learn about transformation matrix to do this ?
@plush raven can you come to one of the question channels?
hollow finch
@wintry steppe Technically every matrix is a transformation. But here you just need to know about elementary matrices
stiff frost
@gritty frigate There's a much easier way to understand "one vector's direction is opposite that of another" than to calculate angles with trig.
Like can you draw a vector that goes in the opposite direction of (1,-2)?
gritty frigate
With the angle between two vectors ?
hollow finch
Even simpler than that
hollow finch
Yea
gritty frigate
Keeps its magnite change its direction
hollow finch
Absolutely correct
gritty frigate
I was trying to find a hard solution hahahaa, it was pretty simple
I m really interested to know if a vector is (x, -y);
Then it makes a triangle like this
So the angle a is tg-1(x/y) + 270 ??
stiff frost
@gritty frigate The triangle you drew makes me think of (-x,y), not (x,-y). If x and y are positive, then tg-1(x/y) is the a in the diagram, but usually when we think about "the angle of a vector", it's measured counterclockwise from the positive x-axis. If you have questions about this sort of trigonometry stuff, maybe #geometry-and-trigonometry or perhaps #precalculus (just because you're talking about vectors) would be good if it's free.
gritty frigate
I know that it is measured counterclockwise, I m just making a graph of the 4rth quadrant
So that angle I m making needs to be added 270°
In order to to get the real angle of a
hollow finch
tg-1(x/(-y))=-tg-1(x/y)
so i believe you would subtract a from 360
gray dust
don’t crosspost
half forge
viscid kernel
What does a complex matrix represent ?
Lets say we have a complex matrix in wich the columns represent matrixes. If you look at your first column then lets say you have a vector with coordinates ( 3 , 2i , 6 ) ?
dusky epoch
a complex matrix is a complex matrix ¯_(ツ)_/¯
it could be viewed as representing a linear transformation from C^n to C^m
same as a real matrix just with the base field now being C rather than R
viscid kernel
Doesnt seem intuitive tho. Is there a way to visualise it ?
dusky epoch
¯_(ツ)_/¯
if you want to make a geometric visual for the space C^2 it's gonna require four real dimensions so probably not
viscid kernel
Oooh ok I think I get is it cuz the “ i “ represent a 90 degree rotation. It kind of like creates another dimension ? Am I thinking right ?
normal quiver
help on how to prove this?
elfin ingot
whats (v)_s
wintry steppe
what is v_s?
normal quiver
i think its the basis vector of v
wintry steppe
the easy way to prove this would be to notice that the linear isomorphism taking vectors to their coordinate/basis vectors takes {v,w} to {v_S, w_S}, and then...
eh, not even. write out what it means for {v,w} to be linearly dependent, and then try translating that to coordinate vectors wrt the basis
(it'll end up being the same as the first way i recommended, just less general)
cursive narwhal
@normal quiver
normal quiver
ok so: @wintry steppe @cursive narwhal
- For {v,w} to be linearly dependent, at least one vector in the set can be defined as a linear combination of the other vector
IE: ifa_1 * v + a_2 * w = 0 - You said that i need to "translate that to coordinate vectors wrt the basis" im confused as to what this means
stoic pythonBOT
TTerra:
wintry steppe
(your i.e. about linear dependence (originally said independence) is incomplete but im going to assume you know how to finish the sentence)
normal quiver
uhh does it imply that the v_s and w_s are linearly independent?
gray dust
sanity check, reread the q
gray dust
damn i might need to thonk you too
wintry steppe
😔
gray dust
it's ok i missed it so it's on me
wintry steppe
i mean a statement about linear dependence is the negation of a statement about linear independence
:^)
thonk aside, @normal quiver can you answer the question?
fixed my typo
normal quiver
yee its that v_s and ws_ are also linearly dependent
wintry steppe
yes
im not gonna ask you to tell me why exactly that follows
but you can convince yourself
should be a straightforward computation using the definition of a coordinate/basis vector
normal quiver
yeee okey im gonna work on that :o ty
wintry steppe
ocean sequoia
how bad would it be to skip the exercises regarding products of vector spaces? It seems kinda boring in LADR?
i also havent seen products of vector spaces that i can remember in a more computational setting
wintry steppe
its rule #3 in #❓how-to-get-help
i personally dont think it's rude, it's just against the rules and one of the mods might not like it
it can waste people's time answering a question when it was already answered
but this one was open ended anyway so I don't really think that applies
some people would get spammy if they could post their question in every channel, and even though an occasional post of a subjective question over 2 channels doesn't harm anything I can see, it is a server of thousands of people and it is easier to make a blanket rule rather than deal with every individual case
So I feel like I understand this conceptually but how can I solve without knowing the transformation or the dimensions of the vectors involved?
So since i'm projecting one vector onto another i'm going from R^2 to R^1
wintry steppe
And I just realized that rereading it, big dumb for that one lol
limber sierra
do you know what it means to "project a vector onto another"
wintry steppe
hmm
did it a lot in calc 3
but maybe I'm missing the linear algebra nuance of the operation?
something magnitude times consine(theta)
wintry steppe
I don't think you need to touch any of those things
You are projecting a 2 dimensional space onto a 1 dimensional space
fickle citrus
I'd almost read it as this:
Let f be a function from R^2 to another member of R^2.
What is the domain of f? What is its range?
Only under kernel do you need anything from LinAlg
limber sierra
uh
you're missing an important word @fickle citrus
"projects"
its not just an arbitrary map between vectors
its a projection of x onto v
fickle citrus
oh it's a projection not a transformation
wintry steppe
Its essentially takes x in R^2 -> y in span{v} @wintry steppe
fickle citrus
Wait nvm, that still doesn't do much change
The projection only makes it idempotent right (true but useless here)
Oh yeah you're right, you can narrow the range to the span
That's what make the Kernel useful oops
I was thinking of a 0-dim Kernel
wintry steppe
Exactly
fickle citrus
TBH this is all fancy speak 😦
I prefer the terms
Parallel-part vs Perpendicular/orthogonal part
wintry steppe
A projection essentially gets rid of the orthogonal part in ur terms
fickle citrus
Yeah
limber sierra
from an abstract perspective, if V is your vector space, then there is a direct sum V = range(P) + ker(P)
this holds for all projections P
fickle citrus
IDK if asker really should be introduced to arbitrary vector spaces
A geometric perspective on the question would solve it fully
limber sierra
ok yeah im just telling you this since it clarifies why we use the term "projection" and kernel/image
rather than the more geometric terms
fickle citrus
It stems from Rank-Nullity I guess, which I think for most people is a 'given' rather than 'you need to prove this!'
limber sierra
er
its just by definition really
any x in V can be written uniquely as P(x) + (x - P(x)) = P(x) + (I-P)(x)
fickle citrus
I see
Not sure if this sounds dumb but existence of I is guaranteed under the vector space axioms right
limber sierra
I is just the identity function here, it always exists
you dont need vector space axioms or whatever
it just maps things to themself
fickle citrus
I suppose what I'm asking is if the identity function is a linear transformation or if there is an identity transformation
limber sierra
the identity function is linear, yes
"function" and "transformation" are used interchangeably in mathematics fwiw
although "transformation" often specifically implies a function between two things (like between two vector spaces)
fickle citrus
Hmm for me function/mapping is a little more abstract
limber sierra
in any case, the identity function satisfies the criteria of linearity
since Id(x+y) = x+y = Id(x) + Id(y)
and Id(rx) = rx = r*Id(x)
sorry i changed "I" for "Id" here, habit
fickle citrus
Yeah it's ok I get what you mean
Essentially it only needs to fit + of the vector space and * of the underlying scalar field
limber sierra
yeah
fickle citrus
Good to check :X
frank robin
hey i dont remember how 2 do this, does anyone can tell me the process, or any example in order to do it by my own?
normal quiver
wouldnt there not be any c that can satisfy the set to be true?
quartz compass
why do you say that @normal quiver
normal quiver
you cant ever match a constant out to be equal to variable x right?
i thought at first that you could set x=0, and it would be c= -c, which means c and -c are multiples.
So saying c=0 would make the set independent. but im not sure
quartz compass
well P_2 is a 3 dimensional vector space
if you're trying to make a basis with only 2 vectors, you're doomed from the start
wintry steppe
maybe it meant "for which c is the set linearly independent?"?
dry pulsar
Suppose V is a vector space of finite dimension, 𝑆,𝑇,𝑈:𝑉→𝑉 linear transformations. Suppose further that 𝑆𝑇𝑈=𝑖𝑑𝑉 Show that T is invertible and determine the inverse of T
Im stuck
1=det(STU)= det s det T det u so all of S, T and U have nonzero determinants and are invertible but i dont know how to find now the inverse of T
gray dust
STU=id leads directly to a formula for T^-1
wintry steppe
you can also use the fact that in finite dimensions, the condition TS = id implies T and S are invertible (i.e. left inverse = right inverse)
which ends up giving you a formula for T^-1
so i might have just said the exact same thing rokabe did
dry pulsar
Thabk you! I think i have now an idea let me check
gray dust
the idea is starting with STU=id and doing easy manipulations to get T^-1
ocean sequoia
@wintry steppe row reduction should work
normal quiver
for this proof do i just need to prove that all 10 vector space axioms hold?
limber sierra
"abstractly" yes @normal quiver but there's a nice "shortcut" for subspaces
while you could spend your time verifying commutativity, associativity, whatever
the thing is: you "inherit" these properties from the space you're a subset of
ie M_23
so to verify something is a subspace, the only thing you actually need to verify is that
it's closed under vector (in this case matrix) addition, scalar multiplication, and taking additive inverses
but additive inverses follow from the first 2 statements
(since -1 is a scalar in every field)
so the ONLY things you need to check are:
- the subset is closed under addition +
- the subset is closed under multiplication by scalars *
well okay, technically you also need to make sure it's nonempty
but that's obvious in this case
normal quiver
i really just spent 20 minutes proving all 10 just to find i only need to prove 3?
oh bruh
shoot me
:(
limber sierra
note though that this is not actually a subspace since its not closed under addition
$\begin{bmatrix}0&0&1\0&0&1\end{bmatrix} + \begin{bmatrix}0&0&1\0&0&1\end{bmatrix} = \begin{bmatrix}0&0&2\0&0&2\end{bmatrix}$
stoic pythonBOT
Namington:
limber sierra
which is not in the set
in other words, addition is not well-defined in this set
so it cant be a vector space
(and therefore is not a subspace)
fickle citrus
Yeah usually the 'constant' part means non-closure under vector addition
If you're a more geometric person - the constant parts usually means the origin doesn't exist and you need an origin in vector spaces
wintry steppe
Can a 2 row by 4 col matrix have an upper triangular matrix ?
ocean sequoia
no
upper triangular must be squares
in other words and upper triangular matrix is a special kind of square matrix and a 2x4 is not a square matrix
Im a bit confused here does a linear functional have to make an vector equal to one? Or is that just how we are defining the linear functional here?
@normal quiver wait are you done? sorry didnt mean to interrupt if you werent
wintry steppe
just how you're defining it there
normal quiver
no im good! you can fire away :)
wintry steppe
brzig recall that a linear map is completely determined by what it does to a basis
so it's fine to define these functionals by what they do on the chosen basis
ocean sequoia
ok so following on that tho
why is this functional taking ej to 1 and not to ej
since it took xj to xj
wintry steppe
a functional maps into the underlying field of the vector space
e_j is an element of F^n, not of F
but F has an identity element to which e_j maps via phi_j
what do you mean by "took x_j to x_j"? the functional takes the tuple (x_1, ..., x_n) belonging to F^n to x_j
x_j is an element of the field F
ocean sequoia
I thought xj was an element of the tuple (x1...xn)
wintry steppe
it is
the tuple here is a point, defined by n coordinates
that's what the elements of F^n are
ocean sequoia
i might have to go back and reread the linear maps chapter
just to make sure i dont misremember that a linear map uniquely changes the basis right?
wintry steppe
i dont even know what that means
could you clarify "a linear map uniquely changes the basis"
ocean sequoia
yea so
if we have a linear map T: V->W
and v1...vj as a basis of V and w1...wj as a basis of W
Tvj -> wj
and vj is the only vector that gets map to wj
ocean sequoia
thanks i realized i didnt make that clear
gray dust
wait that's your explanation of "linear map uniquely changes basis"? still hazy
ocean sequoia
let me make one more edit
wintry steppe
i dont really see how this relates to your picture
ocean sequoia
the one on linear functionality? i just wanted to make sure my definition of a linear map was correct sorry
wintry steppe
a linear map is a function with T(ax) = aT(x) and T(x+y) = T(x) + T(y)
gray dust
just say homomorphism between vector spaces /s
wintry steppe
nothing more
you asked why, if $\phi_j$ is the $j$th projection map on $F^n$, does $\phi_j(e_j) = 1$
stoic pythonBOT
TTerra:
wintry steppe
that's just the definition
moreover, $\phi_j$ cannot send $e_j$ to itself because $\phi_j : F^n \to F$
that is my answer to the question following your most recent image
ocean sequoia
wait but didnt it send x_j to itself
stoic pythonBOT
TTerra:
ocean sequoia
yea sorry got sidetracked
wintry steppe
$\phi_j$ is not defined on the domain in which $x_j$ lies. $x_j$ is an element of the field $F$. $\phi_j$ is a function defined on the product $F^n$
stoic pythonBOT
TTerra:
wintry steppe
$(x_1, \dots, x_n)$ is a tuple containing $n$ elements from $F$
stoic pythonBOT
TTerra:
wintry steppe
also...
earlier you were asking about doing the exercises on products of vector spaces
this is kind of that
ocean sequoia
yea i did a few but i guess i clearly need to do more
wintry steppe
i think your confusion just comes from the definition of phi_j. just be a little more careful
ocean sequoia
will do thank you
wintry steppe
is this right? i tried to put the matrix in echelon form
now i need an extra step to make it into reduced echelon form correct ?
I need to get rid of the 4 above the 1
yes
gray dust
-2R2 doesn't match your 2nd row op
ocean sequoia
yea im a bit confused as to what your steps were
gray dust
oh that works lmao, my bad
wintry steppe
oh lol
gray dust
i need to copswing both of us
wintry steppe
you can also just multiply the second row by -1
but thats one way to do it
lol true
gray dust
so yea get rid of the 4 in R1 to get rref
normal quiver
uhhh quick question. For this problem is my way of looking at ax4 prove that the property doesnt hold for this custom addition?
gray dust
@normal quiver not really. the point isn't to test if (0,0) is the 0 vector wrt the weird addition defn. use the defn of the sum, find a v that satisfies u+v=u, ie v is an expression for the 0 vector. recall the ax says "there must exist a so called 0 vector such that u+0=u for all vectors u", so v mustn't depend on u if v is to be the 0 vector
normal quiver
ok, so is the zeroth vector in this case: (-v_1, -v_2)
gray dust
how'd you get that
normal quiver
i got it from the (u_1 + v_1) /2 , (u_2 + v_2) /2
by setting each term equal to zero?
gray dust
reread ax4 or defn of 0 vector
normal quiver
uh. 'theres an object 0 in V, called the zero vector for V, where 0+u = u + 0 = u for all u in v`
gray dust
so setting the components to 0 doesn't help at all, that means you're solving u+v=(0,0), not u+v=u as i already suggested before where v is an expression for the supposed 0 vector you seek
half ice
Note that, especially in this case, the 0 vector is not actually related to the number 0
I can see why this is confusing haha
gray dust
now you're ready for the next logic step
recall the ax says "there must exist a so called 0 vector such that u+0=u for all vectors u" (carefully note the logical flow of this) so v mustn't depend on u if v is to be the 0 vector
normal quiver
wait so if it depends on u, that mean v no good -> v cant be the 0 vector
which means ax4 doesnt hold?
gray dust
more like a 0 vector doesn't exist so R^2 under the weird addition defn violates ax4, so isn't a vector space
normal quiver
yooooooooooooooooo :o
i think i was also getting confused because i wrote u & v's are similar
lol
tribal lodge
can someone walk me through on how to do a change of basis for a function?
wintry steppe
Well under change of basis means you want to tranform each basis vector by P
To go from B' to B
So for the first vector in B'(say b'1)
Then [b'1]_B = c1 * b1 + c2 * b2
Where b1, b2 are the basis vectors of B
And [b'1]_B means b'1 after being transformed by B
tribal lodge
ok
like this?
2x + e^x = k_1(x) + k_2(e^x)
5x - 2e^x = k_3(x) + k_4(e^x)
wintry steppe
Yes but not the same k1 and k2
Then (k1, k2) and (k3,k4) will be ur vectors in P
Actually wait other way around
It says B'->B
@tribal lodge
tribal lodge
okay i think i fixed the k_3 and k_4
wintry steppe
Essentially u want B = PB'
Where P is a matrix with column vectors (k1, k2) and (k3, k4)
@tribal lodge
Then if u find k1 through k4 you will get P
wintry steppe
@tribal lodge we can find a matrix that takes B->B' and then find its inverse
Lets call it P^-1
So we have 2x + e^x = k1(x) + k2(e^x) => 2x = k1(x), e^x = k2(e^x)
So k1 = 2, k2 = 1
Now you can solve k3 and k4
And P^-1 has column vectors (k1, k2) and (k3, k4)
tribal lodge
ooh i see
so then:
5x = k_3 * x
-2e^x = k_4 * e^x
k_3 = 5
k_4 = -2
@deft wedge ?
normal quiver
i got this?
tribal lodge
do i just inverse that matrix?
wintry steppe
Yes
You can do that by row reducing [P^-1 | I]
^ thats the augmented matrix of P^-1 and the 2x2 identity matrix
This will row reduce to [I | P]
So in other words you want to apply row operations to turn P^-1 into I and in the process turn I into P
@tribal lodge
wintry steppe
@normal quiver so we have a matrix that takes B->B'
But we want a matrix that takes B'->B as the problem wants
So we need to invert the matrix that takes B->B'
tribal lodge
@wintry steppe im not quite sure what you mean by [P^-1 | I]
wintry steppe
Its an augmented matrix
Containing P^-1 (the matrix you found earlier) and I (2x2 identity matrix)
tribal lodge
so this
wintry steppe
Other way around
((2, 5) | (1,0))
((1, -2) | (0,1))
Now row reduce the left part of it so that you get
[I | P]
tribal lodge
sorry i havent seen this notation before. so if i reduce that matrix i get this
tribal lodge
oh you mean reduced row ech?
wintry steppe
Yes
wintry steppe
Yes
Notice how the denominator is det(P^-1)
Theres a general formula for inverting a 2x2 matrix but this process doesnt involve u memorizing anything
But yea that looks good
tribal lodge
@wintry steppe thank you for being patient ❤️ , ive never done one of these with equations before 😮
wintry steppe
Of course :D
gritty frigate
What happens with two vectors with the same direction and its proyection ??
wintry steppe
@gritty frigate do you mean projecting a vector onto a vector in the opposite direction
wintry steppe
Well projection always involves scaling and changing the direction of the vector
So if we project u on to v
We will point u in the direction of v and then scale it by 1/|v|
gritty frigate
But if you consider that u = va
And you want to find the proyection of
v and va
wintry steppe
Well the same rule applies then
gritty frigate
a being a constant
wintry steppe
Its already pointing in the direction of v
So now we just need to scale it down by 1/|v|
@gritty frigate
fossil gust
I need help explaining why it not a vector spaces
dusky epoch
(1,1) ∈ W_1 but 2 * (1,1) ∉ W_1
wise inlet
are all square matrices in reduced row echelon form identity matrices?
quartz compass
put the 0 matrix in RREF and tell me what you get
stoic pythonBOT
TTerra:
wintry steppe
also, if that were the case, then every square matrix would be invertible (why?) (this is true assuming that what you asked is true, if interpreted as "every square matrix's rref is identity")
and i'd like to think at least a few of them are not invertible
wintry steppe
@olive gazelle not every square matrix
olive gazelle
Wrong ping I assume
olive gazelle
Np
wintry steppe
Take $\begin{pmatrix}
5 & 5\
2 & 2\
\end$
gray dust
$\m{5&5\2&2}$
stoic pythonBOT
RokettoJanpu:
crystal oracle
Please help me prove this one thing. It's obvious, but I am struggling with technicalities. I want to use this lemma to solve a problem from Linear Algebra Done Right.
Suppose V is a real or complex inner product space. Suppose v_1, ..., v_M is a linearly independent list in V. Define linear functionals φ_1, ..., φ_m on V: φ_i(u)=⟨u, v_i⟩.
Prove that φ_1, ..., φ_m are linearly independent.
dusky epoch
suppose $\sum_{i=1}^m \alpha_i \phi_i = 0_{V^*}$. can you show that $(\forall i \in 1:m)(\alpha_i = 0)$?
crystal oracle
Or, alternatively, prove that Ф: V->𝔽^m defined as Ф(u)=(φ_1(u), ..., φ_m(u)), is surjective, whichever is easier.
@dusky epoch Is V^* the dual space?
stoic pythonBOT
Ann:
dusky epoch
yes, V* is the dual space.
stoic pythonBOT
Phi:
crystal oracle
I've got a followup question. In the setting given above, I want to show that if $\phi_1, ..., \phi_m$ are linearly independent, then $\Phi:V\to\mathbb{F}^m$ defined as $\Phi(u)=(\phi_1(u), ..., \phi_m(u))$ is surjective. I found a proof on math stackexchange which uses the standard inner product on $\mathbb{F}^m$ and goes like this:
Suppose $dim range \Phi < m$. Then there is a vector $w∈\mathbb{F}^m$, which is orthogonal to $range \Phi$. Then $\forall u \in V \ \langle \Phi(u), w \rangle=0=\sum_{i=1}^m \overline{w_i} \phi_i(u)$. Then $\phi_1, ..., \phi_m$ is linearly dependent. QED
But I have a feeling that it's bad style to use an inner product on a vector space, if the theorem itself doesn't say anything about this inner product. Is this feeling correct? Can we prove it without using an inner product on $\mathbb{F}^m$?
stoic pythonBOT
Phi:
crystal oracle
Instead of talking about inner product on 𝔽^m, I can just say that since span Ф ≠ 𝔽^m, there is a linear functional ψ which annihilates range Ф. And this linear functional can be represented in the standard basis, and then I get the same thing.
crystal oracle
@wintry steppe Yes, it talks about an inner product space on V. But after some point, I just use the fact that φ_1, ..., φ_m are linearly independent, and don't use inner product on V anymore. And also, noone said anything an inner product on 𝔽^m.
wintry steppe
did i say something copswing worthy
i might have
yeah lol i said something completely false it's true in context
no clue what i was thinking
my bad
cursive narwhal
?
wintry steppe
scroll up