Yeah
#linear-algebra
half ice
half ice
That's possible. There's infinitely many solutions
Any basis for that plane works as a v1, v2
cold topaz
my v1=(5, 1, 4) and v2= (10, 1, 8).
but then the final answer, the parametric form, turns out different.
half ice
All points on the plane follow:
4x - 5z = 0
cold topaz
So..
cold topaz
...
half ice
You maybe meant (5,1,4) which does work
half ice
And v1, v2 form a basis on the plane, which should give you a working equation
cold topaz
so we have infinitely many parametric forms?
half ice
Yes
As long as you have no arithmetic errors, these two vectors will give you a right answer
cold topaz
I get my parametric through (0, 0, 0) + t1 <5, 1, 4> + t2 <10, 1, 8>
half ice
You k?
frosty vapor
uh
slow scroll
i wonder what = means here
slow scroll
o
half ice
Just pretend E1 is
a b
c d
Solve for what they have to be
digital garnet
thats literally all i got haha
can u show me how to do the first one
and ill do the rest accordingly
cold topaz
can <r, s, 3r+2s> be a solution to this system?
gray dust
if you let r,s be free you'll actually get ALL solns
gray dust
the system has infinitely many solutions, and if you let the general form of the solution be <r,s,3r+2s> with r and s being free variables, that will capture ALL solutions
austere cedar
Why are people deleting comments now?
So when people are trolling in other chats where are you deleting other peoples comments?
honest notch
Hey, I have a really basic linear algebra problem but i'm lost. Any help is greatly appreciated ty♡
half ice
Solve the system. Can be done quickly by adding both lines together
honest notch
im not sure how to find b or c from that
lusty carbon
Infinite is when two equations are exacrly the same
half ice
What are b and c?
lusty carbon
No solution is when thwy are parralel but diff y iny intercept
honest notch
i mean
b) Unique solution
c) Infinitely many solutions sry for confusion
also no i havent
lusty carbon
And unique is when they are not parallel
lusty carbon
cursive narwhal
The question is just asking you to consider the linear system consisting of 2 lines. So, each part has a geometric interpretation. When there are no solutions, the lines are parallel. When there is a unique solution, there is a point of intersection. When there are infinity many solutions, they are the same line.
@honest notch So, work from there. What are the values of s and t such that the two lines intersect and there are no problems?
spiral sonnet
how do you know whether or not a vector is in a span of another vector?
nevermind figured it out
Given 3 vectors in R3, v1, v2, v3. Find a vector W which exists in the span(v1, v2, v3) but are not equal to v1, v2, v3. How would I go about finding w? I'd like to think I have to setup an augmented matrix
dusky epoch
you don't
spiral sonnet
Really? How would I go about doing this then?
dusky epoch
are you given specific vectors or are you told to do this in the general case
dusky epoch
unless they're particularly badly chosen, v1+v2+v3 should do the trick
otherwise just go through other linear combinations of your choosing
spiral sonnet
okay will give that a try. So if had to find something not in the span of v1, v2, v3 would it be anything except whatever I get for W?
spiral sonnet
I'm also asked to find a matrix not in the span
dusky epoch
can you give the problem exactly as stated
spiral sonnet
dusky epoch
what are v1, v2 and v3? i'm assuming {v1, v2, v3} is LD
spiral sonnet
v1, v2, v3 are vectors and I believe v1 and v3 are linearly dependent
dusky epoch
no
what
are
v1, v2 and v3
you haven't given me them
i want to look at them
obv w must be a linear combination of your v_i, perhaps an integer linear combination if you want w to have integer components
spiral sonnet
v1 = [1, 2, 4] v2 = [4, 5, 1] v3 = [-1. -2, -4]
spiral sonnet
Any reason why 2? Could it be anything?
spiral sonnet
Also any reason for using v2 instead of v1 and v3?
dusky epoch
some linear combinations will not work but there's comparatively few of them and you aren't interested in exactly which ones work and which don't
you could do 2*v1 instead
or 2*v3
so no
it's all mostly arbitrary
spiral sonnet
So I'm not quite understanding how to find a k where there is only one solution, a k for a unique solution and a k for infinitely many solutions
broken hawk
a k where there is only one solution, a k for a unique solution
those two are the same thing
spiral sonnet
I set 4 - k != 0. This should find k where there is no solution because if 4 - k isnt 0 there would be no solution
Sorry I meant no solution
And there is no unique solution because X3 is a free variable
Nevermind..... had an error in my calculation
gloomy arrow
I know how to find the original transformation matrix and then invert is but is there an easier way to find T^-1?
broken hawk
for 2×2 matrices there’s an easy formula for the inverse of a matrix that you can just memorize
(I can never remember it correctly tho)
gloomy arrow
Yeah that was what I was going to do
I was just wondering if there is another way that is easier that I forgot
broken hawk
none I could think of outside of guessing the solution
gloomy arrow
So then just invert the matrix
[7 4]
[3 5]?
gray dust
[5 3]
gloomy arrow
whoops right
And the transpose of the inverse is the same as the inverse of the transpose?
vast torrent
How would you check if that's true?
gloomy arrow
Take a random matrix and try it out
broken hawk
that can give you confidence that it might be true but it’s not a proof
you could just randomly have chosen a matrix where it just so happens to be true
gloomy arrow
I mean in not too keen on the proofs. I just remember whats on the invertible matrix theorem
broken hawk
if you know some basic properties of the transpose it’s a one-line proof
consider the expression $A^T (A^{-1})^T$, if you can prove that this is the identity matrix then you know that the second matrix is the inverse of the first, which is what you wanted to show
stoic pythonBOT
Sascha Baer:
gloomy arrow
Oh wait yeah I think that's what my professor did
spiral sonnet
What exactly is Nul(A)? A being a matrix
slow scroll
the set of x such that Ax = 0
spiral sonnet
oh. If someone asks for Nul(A) do I just find x?
slow scroll
the set of all x that satisfy that. For example, Nul(A) always contains 0 since A0 = 0
slow scroll
you know how sometimes a system Ax = b can have infinite solutions? You have to describe all of them.
spiral sonnet
oh okay
The answer would just be in terms of the free variable no?
or convert it to parametric form
slow scroll
to compute the null space, you just have to take A, put it in rref, and then solve for the pivot variables. Then the free variables are degrees of freedom in the solution
yeah so you get pivot variables in terms of free variables
So lets say you have $\begin{pmatrix}5&0&-2&0&1 \ 0&1&5&0&4 \ 0&0&0&0&0 \end{pmatrix}$ for your rref
stoic pythonBOT
kxrider:
broken hawk
The answer would just be in terms of the free variable no?
or convert it to parametric form
any description will do
so long as it’s complete
they’re all equivalent
slow scroll
then
5x1 = +2x3 - x5
x2 = -5x3 - 4x5
x3 free
x4 free
x5 free
and any vectors in the null space will have that form
spiral sonnet
Select all matrices A with the property that the matrix equation Ax = b has a solution for an arbitrary vector b exists R3. Does that mean just check whether or not the system is consistent or inconsistent for any b?
gray dust
yes consistent
south wadi
Can someone go over
subspaces and vector spaces with me
like how do I know something is a subspace
and when something isnt a subspace
normal canyon
yo peak can u help me with multi
gray dust
for subspaces of vector spaces, closure under addition and scalar multiplication are main criteria to check
normal canyon
yeah tru
gray dust
south wadi
lol
wispy perch
when we say "S spans V", does it mean span(S) subset of V or span(S) = V
gray dust
span(S)=V
storm carbon
How do you prove that all the points on the line project to the same 2d point on a given plane??
gray dust
gollum spaces exist too
south wadi
im only in lin alg 1
gray dust
Gollum rings too
south wadi
u have mastered lin alg
south wadi
what does that mean
gray dust
You should be ashamed to not know what Gollum ring refers to
gray dust
from your previous responses it seemed you were under the impression I was talking about a real linalg thing
I was also taking a jab at your spelling
collum, gollum
column space
gray dust
collogme spase
south wadi
????
gray dust
the vector space spanned by the column vectors of a matrix
gray dust
nah
south wadi
is abstract algebra easy?
gray dust
depends on you
south wadi
is galois and number theory easy?
gray dust
ask zoph
storm carbon
Sorry for posting again
How do you prove that all the points on a 3d line project to the same 2d point on a given plane??
south wadi
is it ez pz almond squeezy?
gray dust
lemon squeezy
sonic osprey
Yeah it's easy
wispy perch
how do i prove this? given that A is just any square matrix of order n
stoic pythonBOT
kxrider:
slow scroll
A^1 != 0 but A^2 = 0
void relic
does order n mean nxn?
slow scroll
square matrix cheeze said
warm flicker
Please ping!
I know how to solve this
But, I'm having trouble with the fundamental understanding of what this means
So is b a combination of these vectors?
I.e. adding/subtracting scaled or not scaled versions of a1 and a2 should bring us b?
So then, how does that geometrically correspond to the three linear systems of equations that the 3 rows are supposed to represent?
void relic
there's a plane
and (3 -5 0) + h*(0 0 1) is a line
you're just finding when the line hits the plane
warm flicker
Okay! That makes more sense
So then what about the 3 systems of equations way of looking at it? Where the first column is x1 and the second column is x2?
How does that relate to the vectors a1 and a2?
void relic
sorry what's x1 and x2?
warm flicker
So like a1, a2 and b form a 3x3 augmented matrix
And the coefficient matrix is 3x2
We write linear systems of equations as augmented matrices
So in this case, 3 equations "x1-5x2=3", "3x1-8x2=-5", and "-x1+2x2=h"
How do they fit into the plane because these 3 equations are in the xy plane but the plane formed by the vectors also has a z-part to it?
void relic
oof you can try to make sense of it I guess
but here you're looking at a 2d world where x1 and x2 are your coordinates
so it's kinda separate from the plane before
and the first 2 equations are lines that intersect, and you use that intersection (x1',x2') to plug into the 3rd equation for h
void relic
I think it's completely separate from the x,y, or z the original vectors are in
warm flicker
Okay..
Thank you for helping me out!
I think I'm going to go on geogebra and play with it for a while till I see it
wispy perch
@slow scroll huh true
void relic
oh sorry I guess you could draw the lines x1-5x2=3 and 3x1-8x2=-5 on the spanned plane
and they intersect where the line intersects
slow scroll
@wispy perch if it’s saying nilpotent order n then it is true
sonic osprey
n is the dimension of the matrix
wispy perch
i'll look into it again later
slow scroll
The statement would make more sense if n was referring to the index of nilpotency, yea
ionic dust
So part a and part c are areas I'm familiar with, but B is giving me some trouble, what I know is that the image of T is the span of column vectors of A denoted by im(T) or im(A)
but I do not really understand this theorem intuitively..
cursive narwhal
You're really just being asked to consider if there is a vector $x \in \bR^3$ such that $v_1 \cross x = v_1+v_2+v_3$
stoic pythonBOT
Abhijeet Vats:
ionic dust
okay so
how would I check for something like that?
I want to say there exists some vector that is crossed with v1 that does equal v1+v2+v3
but that's just a guess
I don't know how to prove it to myself mathematically
cursive narwhal
Well, we know that:
$x = \alpha_1 v_1 + \alpha_2 v_2 + \alpha_3 v_3$
$T(x) = \alpha_1 T(v_1)+ \alpha_2 T(v_2) + \alpha_3 T(v_3) = \alpha_2(v_1 \cross v_2)+\alpha_2 (v_1 \cross v_3) = v_1+v_2+v_3$
stoic pythonBOT
Abhijeet Vats:
cursive narwhal
So if you can find $\alpha_1, \alpha_2,\alpha_3$, then you can find $x$.
stoic pythonBOT
Abhijeet Vats:
cursive narwhal
Understand?
ionic dust
i've seen this somewhere before
not entirely but I think I can work it out from this point
cursive narwhal
If you want, I can give a more detailed explanation.
ionic dust
the scalars are called B-coordinates of x right?
because I found those scalars in part A
can I show you my work?
cursive narwhal
Lol I don't know what a B coordinate is.
Sure, I'll have a look once I get some breakfast
ionic dust
okay
Yeah I still don't get this, I'll post my work:
I just get back the question I had initially
cursive narwhal
Wot
$c_1 T(v_1)+c_2 T(v_2)+c_3 T(v_3) = v_1+v_2+v_3$
Since $v_1 \cross v_1 = 0$, we have:
$c_2T(v_2)+c_3T(v_3) = v_1+v_2+v_3$
stoic pythonBOT
Abhijeet Vats:
cursive narwhal
The question a matrix doesn't even come into play over here? You know how to get $T(v_2)$ and $T(v_3)$. You just have to find $c_1$ and $c_2$
stoic pythonBOT
Abhijeet Vats:
cursive narwhal
I mean, $v_1 \cross v_2$ is just another vector. $v_1 \cross v_3$ is also another vector. So, you have a linear combination of two vectors that is equal to another known vector. Finding $c_1, c_2, c_3$ from there shouldn't be too hard.
stoic pythonBOT
Abhijeet Vats:
ionic dust
There is a solution I can look at
This is a practice exam
would it be worth looking at it?
cursive narwhal
No.
cursive narwhal
You can't look at it during the exam. Instead, focus on trying to get this on your own.
ionic dust
yes.
cursive narwhal
Okay, do you want a detailed explanation?
ionic dust
yeah
cursive narwhal
Okay. So, remember, you're trying to determine if $v_1+v_2+v_3 \in Im(T)$
stoic pythonBOT
Abhijeet Vats:
cursive narwhal
Forget about matrices for one second, okay?
ionic dust
okay
cursive narwhal
Okay, so we assume that there is a vector $x \in \bR^3$ such that $T(x) = v_1+v_2+v_3$. We don't know if such a vector does actually exist but we can try to find it and hope that it leads us somewhere.
stoic pythonBOT
Abhijeet Vats:
cursive narwhal
Now, every vector in $\bR^3$ can be written as a linear combination of the basis vectors. So:
$x = a_1 v_1 + a_2 v_2 + a_3 v_3$
So, we actually have:
$T(x) = T(a_1 v_1 + a_2 v_2 + a_3 v_3) = a_1 T(v_1) + a_2 T(v_2) + a_3 T(v_3)$
Now, we have a clear formula that gives us the images of the basis vectors. That's the definition of the linear transformation.
stoic pythonBOT
Abhijeet Vats:
cursive narwhal
Okay?
ionic dust
yeah
cursive narwhal
Let me know if anything is unclear.
Okay nice
So, we know that $T(v_1) = v_1 \cross v_1 = 0$. Then, $T(v_2) = v_1 \cross v_2$ and we also have $T(v_3) = v_1 \cross v_3$. This gives us:
$a_2 (v_1 \cross v_2) + a_3 (v_1 \cross v_3) = v_1+v_2+v_3$
You can find each of those vectors easily. Then, it's just a matter of equating their components and obtaining $a_2$ and $a_3$.
stoic pythonBOT
Abhijeet Vats:
ionic dust
okay
so these vectors that I use the cross product on
v1xv2 and v1xv3
those are just the normals of those two vectors
I should be able to divide them and find c2 and c3
but wouldn't this be false? v1xv1 = 0 so c1 is also 0 then?
cursive narwhal
No. c1 doesn't have to be 0. It means that if c2 and c3 exist, then you actually have a whole bunch of solutions to the problem.
ionic dust
okay
It took sometime but I think I got it
these are just the normal vectors when I do the cross product
so I can simplify them
cursive narwhal
Mmmh
ionic dust
I got c2v3 - c3v2
cursive narwhal
Aight, then? What do you do with that?
Btw, I'm just relying on what you've gotten cos i'm not gonna go and compute this stuff lol
cursive narwhal
Okay, so you computed c2 and c3. By the way, c1 is not necessarily 0. You've just found a family of vectors whose image is v1+v2+v3.
But if you've done the computation correctly, you've shown that v1+v2+v3 is in the image of T
ionic dust
there could be more than one vector that gives the transformation that equals v1+v2+v3
I agree with that
it just seems like I'm unconsciously memorizing the procedures which is not good.
If I retake this course I'm going to read the book over the summer break
ionic dust
I'm in the course with applications only
they use this same book in the proof based course as well
cursive narwhal
Hmm, might I recommend a book that has benefited me greatly? Klaus Janich's Linear Algebra is fantastic for learning this stuff, in my opinion.
Only thing is that he's a pretty memey author.
ionic dust
At this point, it's worth a shot
I've been applying a brute force method approach to learning this content which is god awful
can't do this like I did in calculus
cursive narwhal
Yea. It'll fail at some point.
Only thing is that Klaus Janich's book doesn't have many problems to solve. You can rectify that by doing problems from the book above or you can try handling Problems in Linear Algebra by Ikramov
hoary agate
@cursive narwhal what all does your book cover?
cursive narwhal
Eh let me grab it and just state everything in the table of contents
cursive narwhal
- Sets and Maps
- Vector Spaces
- Dimension
- Linear Maps
- Matrix Calculus
- Determinants
- Systems of Linear Equations
- Euclidean Vector Spaces
- Eigenvalues
- The Principal Axes Transformation
- Classification of Matrices
Yea, translated into English
gray dust
I got scared by matrix calculus for a sec, I dont know why
gray dust
Not unlike you
cursive narwhal
And I quite like that. Also, the book does have proofs but he doesn't prove everything. Like, there's enough left for you to discover on your own and I quite like that.
Each chapter has a main section, a section for physicists and a section for mathematicians. I like that as well
hoary agate
oh ok
cursive narwhal
@ionic dust It just means that there's something that went wrong in your calculations.
@ionic dust Yeap, you get two values for c2 if you take the calculation to the end. So, that's a problem. Hence, the pre-image of v1+v2+v3 doesn't exist.
ionic dust
should I just drop my course now
I have till this friday deadline
I wanted to continue until failure,
cursive narwhal
Which courses are you taking now?
ionic dust
I'm taking an object oriented programming course in C++
discrete mathematics
and linear algebra
cursive narwhal
If your course textbook isn't working for you, then you should grab another textbook and try to learn from it.
Cos the trouble you're having is that your text doesn't seem to be explaining the concepts well enough. So, if it's not, then pick something that will.
cursive narwhal
Which book?
Oh klaus janich? Lol that's the only version of the book
He didn't get a 2nd edition out, i believe
wet pelican
anyone know what "m" means?
wispy perch
how do i show that S is either linearly independent or linearly dependent?
gray dust
a lin indep set of n vectors picked out from R^n serves as a basis for R^n
that was a light hint. don't give too much away abhi
cursive narwhal
hmm actually, i'll wait for them to respond before saying anything else
@wispy perch
wispy perch
OOOH
so the standard basis like (1, 0, 0, 0), (0, 1, 0, 0), ..., each and every vector in that standard basis should be able to be expressed as a linear combination of u1, u2, u3, u4. problem is, (1, 0, 0, 0) is not in the span so they are not the basis of R^4 hence not linearly independent?
cursive narwhal
If it was linearly independent, it would be a basis for R^4. So, it would span R^4. That's clearly not the case. So, it has to be linearly dependent.
gray dust
nothing special about the standard basis vectors of R^4 for this q. the point is if S is a basis for R^4 then EVERY vector in R^4 must be expressible as a linear combo of the vectors in S. (1,0,0,0) cannot be expressed this way, that's all you need
gray dust
you were basically already there with this bit
problem is, (1, 0, 0, 0) is not in the span so they are not the basis of R^4 hence not linearly independent?
i just wanted to clear up that the focus of the answer shouldn't be on the standard basis vectors, just the fact that not all vectors in R^4 can be written as linear combos of vectors in S
wispy perch
true, i overcomplicated it when it's already obvious that it should span R^4
gray dust
shouldn't span R^4 but yep
wispy perch
are all subspaces vector spaces?
dusky epoch
yes, a subspace of a vector space can be considered as a vector space in its own right.
wispy perch
can i have some hint on how to start
need to check whether S is linearly independent or not
storm python
quartz compass
these are row vectors, read it and weep
wispy perch
how do you prove this?
dusky epoch
which direction is giving you trouble?
wispy perch
actually both, but when i tried the "only if" direction, i tried to use proof by contrapositive
but idk how to continue from assuming that S union T is not linearly independent
dusky epoch
use the defn of linear dependence
and the linear INdependence of S and T individually
dusky epoch
uh,,,,, no not necessarily
you have a linear combination $\sum_{i=1}^k c_i u_i + \sum_{j=1}^l d_j v_j = 0$, where the $c_i$ and $d_j$ aren't all zero
stoic pythonBOT
Ann:
dusky epoch
notice that at least one of the c's AND at least one of the d's must be nonzero
(do i need to explain why this is the case?)
wispy perch
let me think
the latex that you wrote there, that is the definiton of the S union T being linearly independent right?
dusky epoch
being linearly dependent
wispy perch
oh right, aren't all zero
@dusky epoch at least one of the c's and d's must be nonzero because we still have to follow that S and T are both linearly independent?
dusky epoch
well,,, yes. if all the d's were zero it'd contradict the linear independence of S, and vice versa
so you have $u + v = 0$, where $u = \sum_i c_iu_i \in U \setminus {0}$ and $v = \sum_j d_jv_j \in V \setminus {0}$
stoic pythonBOT
Ann:
dusky epoch
so you get u in U cap V
wispy perch
yeah which means it's not only {0}
sorry but
well,,, yes. if all the d's were zero it'd contradict the linear independence of S, and vice versa
this one still confused me
do you mind explaining more
dusky epoch
if all the d's were zero you would have a linear combination of S summing to 0
dusky epoch
it's actually easy to reverse this argument
for a nonzero vector w in U cap V, set S = {w} and T = {2w}
wispy perch
sorry idk if i'm complicating it, but if all the d's were zero, it contradicts the linear independence of S. so i can say there exists a c_x and a d_y such that these two are not zero, right?
but wouldn't that mean S and T are both linearly dependent?
dusky epoch
no, S and T individually are LI, but we assumed their union wasn't
wispy perch
oooh, suppose if there is cx and dy, cx ux + dy vy = 0
wispy perch
true
i'll just say there exists
and then you're summing all the ci ui together and they cannot be zero
since they both can't be zero, why does U cap V contain u?
wintry steppe
Just ask your question
limber sierra
matrix multiplication is not generally commutative
so you'll have to do it the other way around
since:
digital garnet
ohhhh
limber sierra
$MN = A \
M^{-1} MN = M^{-1}A \
N = M^{-1}A$
wintry steppe
For two matrices A,B
digital garnet
@limber sierra ohhhh makes sense tysm!
since i got you here
for this one
i got the answer and it says interpret the results
so is it just saying that that the total value of shares in ur portfolio for EACH of the scenarios
like the answer is [15500,18350,12250]
is that the total value in ur portfolio in scenario 1 2 and 3
stoic pythonBOT
Namington:
limber sierra
oh wait
misread what youre aasking
uh, i havent seen your lab but from the looks of it
yes, thats exactly what it means
digital garnet
beautiful
srry last question quick one!
for this question i did the way u told me to
and i got N = ....
do i row reduce now?
cuz im given this
limber sierra
i'd assume you just decode using that key
oh, it does ask you to row reduce
so yeah, i guess you do that
limber sierra
are you uh, sure they want you to row reduce fully
limber sierra
here?
like if you decode that
its sensible
and fits the context of the question
digital garnet
but like what would i write for Z-26?
limber sierra
i dont see 26
digital garnet
limber sierra
i see 19, and then 5, and then 12, and then 12
yes, thats the key
so use that
to translate each entry of the matrix
into letters
digital garnet
yea im super confused lol
digital garnet
x1?
digital garnet
ohh
yeah but there isnt a 26
OHHH
i think i get it
OHHH
lmfoao
that was so easy idk what i was doing
tysm lol
digital garnet
@limber sierra
hey boss a quick question
do u know how to do these LDU factorization problems
dusky epoch
you can compute $\mat{1 & 0 \ a & 1} \mat{b & 0 \ 0 & c} \mat {1 & d \ 0 & 1}$ explicitly...
stoic pythonBOT
Ann:
half ice
There's no problem with the size
X is a 2×2
But it's not obvious that there's a 2×2 that solves this
half ice
Hmm, that is a very big contradiction, yeah
I don't see any mistakes
Question seems to imply that such an X exists
slow scroll
you could try to compute a left inverse for A
vocal breach
@slow scroll hmm, let me try
slow scroll
oh nah im dumb that wouldn't work
yea there is no solution. both columns of AX are linearly independent from the columns of A.
slow scroll
yes. more specifically, the system you got at the end is "inconsistent," which only ever happens in the situation i described: when b is linearly independent from the columns of A in Ax = b
hot flame
can someone explain why this is wrong
pallid rampart
$\begin{bmatrix}2&0\0&1\end{bmatrix}=\begin{bmatrix}\sqrt{2}&0\0&1\end{bmatrix}\begin{bmatrix}\sqrt{2}&0\0&1\end{bmatrix}$
Rip bot
Here is a counterexample to that statement
hot flame
But thats not LU decomposition tho
hot flame
pallid rampart
Diagonal matrices are upper triangular and lower triangular
Therefore it is LU decomposition
So the only thing you can conclude is that the matrix will be diagonal
wooden forum
Ok so I'm a little lost when it comes to the definition of cross product of two vectors
it seems to be only defined for 2 vectors in 3-dimensions
and the result is a vector value
but in other places I've seen it defined for two dimensions as the determinant of the matrix where 2 2D vectors are slotted as rows (hopefully I've used the right terminology)
I understand how the geometric explanation works for the 2 vectors in 3D definition (perpendicular to the plane that the two vectors span)
but I'm not sure how those two definitions align
and this kinda makes things just all the more confusing
Never mind all this, I just had to finish the 3b1b video I was watching, sorry for clutter.
ruby glacier
lime marlin
any of you guys ready for the linear algebra test tmr?
dusky epoch
what linear algebra test
lime marlin
math 214
dusky epoch
that says precisely nothing to me
not everyone here is from your particular class at your particular institution, shockingly enough
austere cedar
You seem surprised
lime marlin
I didn't realize that the server was a cummulation of all unis across america
austere cedar
🙏 Lord have mercy
lime marlin
was hoping to get some help before my exam tomorrow
sonic osprey
you can
hoary agate
lmao not just america, it has people from all around the world
slow scroll
im a bit confused by the indexing: does the vector you choose have to have c(i) as its i'th coordinate or something?
im trying to say: let c(i) be an arbitrary scalar.
there exists a = c1a1 + c2a2 + ... + ciai + ... cnan
where c(i) = ci and <a, ai> = c(i)? is that the question?
@pastel saffron hint: choose a so that most of the terms go away
hint: recall that <a,0> = 0
i thought you said it didn't have to be orthonormal
right, but we can still sort of manipulate the outcomes if we choose the vector carefully. You can make a lot of the terms go away by choosing a vector such <a,0> = 0 will apply to a lot of the terms
yep yep thats close. Just think about the constant you need so that you get c(i) in the end
yea so lets say you choose a = ci ai
Then <a, ai> = <ci ai, ai> = ci <ai, ai> = ci | ai|
so the ci you choose needs to cancel with the | ai | that comes out in the end
yea, different variable ofc
any #❓how-to-get-help channel or this channel is just fine. As long as you don't interrupt anyone else c:
npnp
slow scroll
@pastel saffron hey um i just realized i wrote |ai| instead of |ai|^2 earlier. hopefully you correct for that lol
fossil oar
multiplying transformations was the same as multiplying their matrixes?
cursive narwhal
The composition of linear transformations corresponds to the multiplication of the matrices that represent them.
cursive narwhal
?
fossil oar
i have to prove x^n is eigenvalue of T^n
quartz compass
are you given that x is an eigenvalue of T?
dusky epoch
are you saying you don't know what T^n refers to when T is a linear operator
quartz compass
pick some eigenvector v with eigenvalue x, then Tv=xv
now apply T to both sides, show me what you get
fossil oar
ok thanks
quartz compass
wot
does that mean you're working on it or you have no idea what's going on and think I just gave you the answer
dusky epoch
inb4 they reply with "yes"
quartz compass
lol gone like a ghost
dusky epoch
👻
fossil oar
working on it
dusky epoch
and you haven't even answered my question
which i... don't appreciate, to say the least.
wintry steppe
Can someone explain what transpose matrices have to do with dual spaces?
broken hawk
well
let’s say we’re in ℝ³. a vector of ℝ³ is a column with three numbers
a vector f in its dual can be written as a row-vector, in such a way that applying f to a vector v is just matrix multiplication
dusky epoch
take a linear map $A: V \to W$. take a basis $\mathcal{B} = {v_1, v_2, ..., v_n}$ for $V$ and another basis $\mathcal{C} = {w_1, w_2, ..., w_m}$ for $W$. construct the dual bases $\mathcal{B}^$ and $\mathcal{C}^$ for $V^$ and $W^$ as follows: $\mathcal{B}^* = { \hat{v}^1, \hat{v}^2, ..., \hat{v}^n}$ such that $\hat{v}^i(v_j) = 1$ when $i=j$ and $0$ otherwise (then extend by linearity), and ditto for $\mathcal{C}^$. finally, consider the dual map $A^: W^* \to V^$ defined by $A^(\psi) = \psi \circ A$ for all $\psi \in W^*$.
with all that in mind, you have that the matrix of $A^$ with respect to $\mathcal{C}^$ and $\mathcal{B}^*$ is the transpose of the matrix of $A$ with respect to $\mathcal{B}$ and $\mathcal{C}$.
stoic pythonBOT
Ann:
dusky epoch
this is probably verbose and hard to comprehend
but hopefully it sheds some light on the matter
@wintry steppe
wintry steppe
@dusky epoch this is pretty verbose and hard to comprehend
hehe
ill try my best to understand what this is saying
thank you for helping
wait but why is that true
coral ferry
hey guys
do you have any insightful/comprehensive/clear group theory textbooks that might be supplemental to first year linear algebra
slow scroll
not quite sure how group theory is going to help with first year linear algebra
coral ferry
ive heard concepts from group theory are very similar to linear algebra
slow scroll
I mean, thats true in the sense that group theory is algebra,
but I'd say its more like knowing concepts from linear algebra will help you in group theory. Anyhow, there are many resources to learn group theory.
Algebra by Artin
Algebra by Dummit and Foote
Algebra by Judson
Harvard lectures on abstract algebra (that follows Artin).
sonic osprey
Yeah it really wouldn't help
knowing linear algebra is helpful for learning group theory
but the other way around isn't super helpful
vector spaces have a lot more structure than groups, and so the theory of groups doesn't really help in studying vector spaces
broken hawk
though some concepts are relatively similar and having seen them in one place makes them easier to understand the second time round
e.g. quotient spaces
solar osprey
Any help for part 3-4
vast torrent
It you prove the domain and codomain are the same dimensionality, then linearity proved in 1. has that condition 2 and 4 imply each other @solar osprey
So 1,2 and 5 together imply 4
vast torrent
And 3 implies 5,I'll let you think about why
solar osprey
(aA+B)X = aAX + BX
For part 2
I simply said
The only matrix that can be associated with a zero transformation
Is the zero matrix
Hence Kernel(L) = {0}
Hence L is one to one
And 3 implies 5,I'll let you think about why
@vast torrent by definition of Isomorphism I think
vast torrent
It's just going through the matrix multiplication definition
crystal pagoda
A is already in "row vector" form
vast torrent
Not too interesting
crystal pagoda
just multiply and show that two representations are the same
solar osprey
U talking about part 3? @crystal pagoda
crystal pagoda
yeah
broken hawk
for part 3, take any vector v, write it as a sum of basis vectors, then apply the matrix multiplication and see what happens
crystal pagoda
write any vector as a linear combination of basis vectors, then multiply
it will come out exactly like the other representation
solar osprey
I see
To make sure I understood the exercise
We’re associating for an m x n matrix
A linear transformation from Rn to Rm
vast torrent
And 5 and 1 together imply that surjectivity and injectivity imply each other
crystal pagoda
your notation is kinda confusing me
solar osprey
Let me write it
crystal pagoda
i get what you're trying to say though
matrix multiplication is a representation of linear transformation
solar osprey
crystal pagoda
yeah
solar osprey
Is there any point
In doing that
I don’t really understand this whole idea behind associating a matrix to a linear transformation
crystal pagoda
hm
let me think of a good explanation
i personally haven't thunk of the matrix representation of a linear transformation for a while
lol
crystal pagoda
it's a more concrete and concise way to represent it i guess
solar osprey
Btw can you see my part 2?
Is it enough to say that for L(X) to be equal to the zero transformation
X has to be the zero matrix
Okay I think I have found a more mathematical argument for that
Assume there exist a matrix M that is non -zero and such that it is associated with the zero transformation
This means that there must be at least one ij-entry in M that is non-zero
crystal pagoda
you're down the right path
crystal pagoda
i would say just write L(x-y)=0 implies x=y
solar osprey
But arent we assuming that L is one to one by saying that?
crystal pagoda
no, injective says f(x)=f(y) implies x=y
im only using the fact that L is a linear map
rearrange we have L(x)-L(y)=0, and use linearity
man i can't spell
LOL
crystal pagoda
hm?
solar osprey
L(x-y) = L(x) - L(y)
crystal pagoda
by linearity yeah
solar osprey
AHHH
my man
I thought u were
Proving 1-1 from scratch
😂😂
Btw its a given theorem that if Kernel is trivial then its 1-1
crystal pagoda
uhhh
the following fact
if Ax=Bx for all x, then A=B
implies if Ax=0 for all x, A=0
then null space is trivial
i think that should be good
@solar osprey
crystal pagoda
np
solar osprey
crystal pagoda
man
i can't imagine all the students at my university taking linear algebra online next quarter
that's gonna be hella difficult
barren panther
thats me rn
I have question
for a set of all eigenvectors of a matrix, is the set closed under vector addition?
barren panther
what do you mean by that?
vast torrent
I mean "the space of all eigenvectors" needs 0 to be a vector subspace
but once you shove in 0 then yes, it's closed
fallow jolt
How would you prove that? I am intrigued
crystal pagoda
that eigenspace is a vector space?
fallow jolt
No, that its closed under addition
vast torrent
exercise for the reader: for any given eigenvalue, the set of all eigenvectors associated with that eigenvector, with the 0 vector added in, form a subspace
fallow jolt
I know that
crystal pagoda
if you prove it is a subspace then it has to be closed
crystal pagoda
that's the "field" part implies
fallow jolt
What field??
crystal pagoda
uh?
fallow jolt
I've never heard that term before
crystal pagoda
oohh
vast torrent
because a vector space needs closure under addition and scalar multiplication, that's part of the definition
crystal pagoda
sorry, the definition of vector space is a set over a base field
solar osprey
Wassup
crystal pagoda
a field is closed under addition and multiplication
vast torrent
a vector space needs to be closed under addition and scalar mult.
vast torrent
that's flawed
barren panther
wouldnt the eigen values in front of v1 and v2 be different?
crystal pagoda
he specified lambda to be AN eigenvalue
fallow jolt
eig is the eigenspace
barren panther
crystal pagoda
yeah that should be fine
fallow jolt
Would the eigenvalues in front of v1 and v2 differ @vast torrent ? Because if so, you cannot factor it out on the last step.
vast torrent
yeah he doesnt need to say that though
because he's not saying v is an eigenvector
I take back my objection
solar osprey
i can't imagine all the students at my university taking linear algebra online next quarter
@crystal pagoda lmfao
vast torrent
eig(lambda) is the eigenspace corresponding to the eigenvalue lambda
solar osprey
I dont even follow what the prof says
crystal pagoda
yeah, plenty of them don't have any background in abstract algebra either
so they're kinda screwed
solar osprey
Thats my second semester in college and all ive been doing is self learning
vast torrent
so the eigenvalues cant change, it's only lambda
barren panther
I thought those were the same?
fallow jolt
lambda is the eigenvalue tho
barren panther
sorry Im a bit lost
vast torrent
it's only considering one eigenvalue at a time
barren panther
so this has to be one eigen value that satisfies the condition
vast torrent
eig(lambda) depends on a particular lambda
vast torrent
you should add "for some basis {e1,e2,...,en}" explicitly
vast torrent
oh, I missed that
solar osprey
He did mention e1...en being the standarb basis
vast torrent
nevermind then
solar osprey
Yes
vast torrent
sorry
vast torrent
it doesn't actually have to be the standard basis of course, just any basis
barren panther
so my professor is asking me to prove that the set of all eigenvectors is closed under vector addition, but in that case how would that work
pallid rampart
Orthonormal basis is the best
solar osprey
Rixia Mao u didnt tell me how do I proceed from there
solar osprey
vast torrent
so
solar osprey
Okay I think i got it
vast torrent
fusionvictini
barren panther
dropped my notebook oops
crystal pagoda
@solar osprey then you consider Ax
solar osprey
@crystal pagoda yes yes
crystal pagoda
and you show that they are equal
vast torrent
let's temporarily consider $\mathbf 0$ an eigenvector for $\lambda$ because it's in $\text{eig}(\lambda)$ by definition even though it isn't an eigenvector
stoic pythonBOT
gfauxpas:
barren panther
ok
vast torrent
so the definition of eigenvector is that a matrix/lin. operator A acts on v by $\mathbf{Av} = \lambda \mathbf v$
stoic pythonBOT
gfauxpas:
vast torrent
so we're asking
solar osprey
Ever since that first Iron man movie, ive always wanted to know what Eigenvalue means
vast torrent
if $\mathbf {v,w}\in \text{eig}(\lambda)$
stoic pythonBOT
gfauxpas:
solar osprey
Dont spoil tho
vast torrent
is $\mathbf{v+w}$ an eigenvector?
stoic pythonBOT
gfauxpas:
crystal pagoda
the geometric intuition is the ratio that you stretch or compress a vector in one direction corresponding to the eigenvector
somewhat
barren panther
ok
barren panther
that A(v+w)=lambda(v+w) ?
vast torrent
yeah exactly
barren panther
ok nice
vast torrent
our question is, is that true if v and w are eigenvectors?
barren panther
no right?
vast torrent
well by definition A is linear
meaning it can be distributed
and of course "multiplication by a number" is also linear
so it can be distributed
barren panther
which means its closed under multiplication right?
barren panther
Av+Aw = lambda(v)+lambda(w)?
vast torrent
Av = lambda v, right?
barren panther
yea
vast torrent
Aw =?
barren panther
lambda w
solar osprey
@crystal pagoda Rixia rixia
crystal pagoda
sup
vast torrent
$\mathbf{Av+Aw}\overset{?}{=} \lambda \mathbf v + \lambda \mathbf w$
stoic pythonBOT
gfauxpas:
barren panther
yes
vast torrent
alright cool
barren panther
thats it?
vast torrent
that proves that if v and w are eigenvectors, v+w also is an eigenvector
is that enough to show a set is a vector subspace?
crystal pagoda
@solar osprey by part 3 you have a representation of A as T
think about the implication
solar osprey
Wait
barren panther
wait
crystal pagoda
yes, but what does part 3 say?
vast torrent
you need to show that if v is an eigenvector
cv is also an eigenvector, for any constant multiple c
barren panther
yea I did that
solar osprey
yes, but what does part 3 say?
@crystal pagoda Ahhhh yes right I thought about something else
vast torrent
ah okay then yes
barren panther
so thats to say that the set is closed under scalar multiplication right?
vast torrent
you've proven eig(lambda) is a vector subspace
solar osprey
Idk why I imagined part 3 to be another question LOL
vast torrent
yes and what we did with v+w is to show closure under vector addition
crystal pagoda
lol my professor's homework would just be part 5
solar osprey
@crystal pagoda are u a math major?
vast torrent
which you can just say as "addition", it's implied
vast torrent
"multiplication" doesn't imply "multiplication by a scalar" but "addition" applies "vector addition" in linear algebra
barren panther
ooh right ok that makes sense
solar osprey
@crystal pagoda anything academic has nothing to do with IQ tbh
barren panther
ok well thank you very much! Im new to linear algebra and this discord so this is good
crystal pagoda
i guess
but it is evident that some people are better at forming intuitions
im not very good at that
took me some time to understand linear algebra
vast torrent
discord is a great resource
solar osprey
How much the subject interests you matters as well Imo
vast torrent
especially now that physical colleges are centers for pestilence and death
solar osprey
Lol
crystal pagoda
i have great difficulty doing computational stuff
solar osprey
It feels like I havent talked for weeks tbh I forgot what my voice sounds like
vast torrent
being persistent and willing to work hard is more important than iq
solar osprey
@vast torrent yes
crystal pagoda
yeah
solar osprey
All about neuroplasticity
vast torrent
I got into NYU, one of the best schools in the world, not because I'm smart, but because I constantly went to professors outside of class to bother them with questions and to help get better
I really don't know why I got in tbh, I didn't get into my safeties lmao
barren panther
woow congrats
vast torrent
well not the better half of my safeties
I was like "hell, I'll apply to NYU,why not"
solar osprey
Congrats
solar osprey
Ive never been to office hours before lol
barren panther
yea this whole online things been wierd half my professors have been struggling with it
solar osprey
Anyone on codeforces here?
crystal pagoda
LOL
i shouldve gone to purdue or ohio state
uc irvine math department kinda lackluster in terms of resources
vast torrent
honestly I think it was like divine providence that I got in, I don't think I would lhave let me in lol
there are so many people here smarter than me
I know I just said smart isnt as important as work ethic
but damn there are so many people here smarter than me
crystal pagoda
being smart definitely helps