D is just a letter
#linear-algebra
dusky epoch
dusky epoch
what do you mean, "never left a matrix in that form"
i'm asking you to take the inverse of a matrix
are you not able to do that
wintry steppe
i am
dusky epoch
then what's the fucking issue
are you still somehow under the impression that since D is called D it has to somehow be diagonal even though in this case it very clearly ISN'T
wintry steppe
no
wintry steppe
i was thinking D had to be inversed to get this matrix D^-1 that i gave there in the pic, so why are we now inversing that matrix to get D ?
dusky epoch
D^-1 is known and we want D.
uneven bloom
Inverse of an inverse is the original matrix
dusky epoch
$D = (D^{-1})^{-1}$
stoic pythonBOT
Ann:
uneven bloom
Think about what an inverse is
ruby niche
u and v are unit vector and the angle between them are pi/3
what is (3u-4v) ∙ (u+5v)?
anyone mind to show me how solve this problem? the correct answer is suppose to be -23/2
gray dust
use dot product properties to rewrite that expression
ruby niche
do you by any chance mean this one (au + bv) · w = (au) · w + (bv) · w?
in this case is w = (3u-4v)?
gray dust
that looks useful
gray dust
this looks poorly written. what even is u^2
ruby niche
(3u-4v) ∙ (u+5) = 3uu - 4vu + 3u5v -4v5v
dusky epoch
·, not *
gray dust
$3u\cdot u-4v\cdot u+3u\cdot 5v-4v\cdot 5v$
stoic pythonBOT
RokettoJanpu:
gray dust
no prob
proven magnet
is it OK to create a vector space from the span of a line?
proven magnet
or reformulate to "vector space spanned by the direction vector of a line"
i know the conditions
half ice
A span is always a vector space, yes
proven magnet
so if I know a line goes through origin, "let a vector space be spanned by a line L that goes through the origin"
cuz idk if "direction vector of line" is proper
thoughts?
vast torrent
You span vectors, not spaces
proven magnet
so you can't span a line? ok i'll say "vector space spanned by the direction vector of the line L"
half ice
A line that goes through the origin is a vector space
vast torrent
Sure
half ice
In that all of the vectors going through that line form the space
vast torrent
a direction vector of the line
proven magnet
thx bois for clearing it up
half ice
Not by definition, you'll need to test some axioms on it, or find another clever way. I like the span idea
Like take (1,2) in R². The span of that is a vector space, which is also a line.
(-2,-4) is in the space
proven magnet
cloud cairn
if i have a line y = 3x
and i want a basis vector, that spans the line
solve for each variable in terms of just one variable gives me
(x, 3x ) ---> basis is (1,3)
so span of (1,3) represents the line
but what if the line is y= 3x + 2
does that 2 have an effect on the basis?
half ice
That's not a vector space at all
So you can't describe it with the same tools that you could with other vector spaces
half ice
One might call that an affine space. You could simply take out the 2, then add it back in after vector space stuff
cloud cairn
the reason i asked this question was cuz
i was given a plane, something like 3(x-2) + (y+1) + (z-5) = 2
so its not set equal to 0, but 2.
and i wanted to find a basis for the plane
am i going to run into the same problem as above?
cloud cairn
ok so on that first problem with the line equation. if i get the vector (1,3) as the basis.
and you say add the 2 now.
does that mean( span of (1,3) ) + (0 , 2) ?
half ice
Yeah something like that
woven geyser
I want to prove -v + v = 0 (where 0 represents the 0 vector). Is this proof correct: -v + v = v + -v (comm law of add) = 0 (add inverse law).
dusky epoch
sure
woven geyser
thanks!
my next proof differs from the book's, is this one valid too? prove 0 + 0 = 0 (where 0 is the 0 vector). my proof: 0 + 0 = (v+-v) + (v+-v) [add inverse law] = v + v + -v + -v [comm law of add] = 2v + -2v = 0 (add inverse law)
dusky epoch
you overthought that one a bit
you didn't need to introduce any new vector at all
0 + 0 = 0 follows directly from the definition of additive identity
woven geyser
oh snap!
dusky epoch
v + 0 = v for all v, and in particular for v=0
woven geyser
you're right, thanks for that insight. is my proof still valid even if overcomplicated?
half ice
I suppose lol. You drop the brackets, so it may be worth stating associativity
dreamy depot
Dumb question i've been practicing the row reduction algorithm and i'm not sure if this is right
$$
A = \left[\begin{array}{ccc|c}{0} & {1} & {-4} & {2} \ {0} & {0} & {1} & {-5} \ {1} & {-3} & {5} & {4}\end{array}\right]
$$
$$rref(A) = \left[\begin{array}{ccc|c}{1} & {0} & {7} & {8} \ {0} & {1} & {-4} & {2} \ {0} & {0} & {0} & {1}\end{array}\right]$$
stoic pythonBOT
Zophike1:
dreamy depot
For the row operations I took $-5R_{2} + R_{1} \rightarrow R_{3}$ then did $3R_{2} + R_{1} \rightarrow R_{1}$
stoic pythonBOT
Zophike1:
dreamy depot
then finally did $R_{3}/16 \rightarrow R_{3}$
stoic pythonBOT
Zophike1:
dreamy depot
Is what I did right because the calculator i'm using is showing me something different
cloud cairn
it doesnt look right
i think u shud have a pivot/leading 1 in each column
from what im seeing doing the work
dreamy depot
Then what should it be then for the rref(a) ?
cloud cairn
the first thing i did was:
swap row 1 and 3. (this puts a lead 1 in column 1)
then swap row 2 and 3 (puts a lead 1 in columns 2 and 3)
cloud cairn
ok so its in echelon form. that there tells you its nonsingular. and it can be row reduced with leading 1's and then 0;s in the other spots of each column (sorry idk how to say that last part)
im not a tutor btw
dreamy depot
ahhh okay I put in Echelon form then ?
cloud cairn
so next step id do is
4*R3 + R2
this means add to row 2, 4 times row 3
shud look like this after u do it
1 - 3 5 | 4
0 1 0 | -25
0 0 1 | -5
we did that step to get a zero in row 2, column 3
does this step make sense?
dreamy depot
Should I do that step after A is in echoloan form ?
cloud cairn
you do not have to solve A to echelon form, but sure its a very good guideline to try to follow
you basically end up solving echelon form in the process
and then yes do it after
dreamy depot
Because now looking at it I think I put in echelon form
cloud cairn
my echelon form matrix is :
1 - 3 5 | 4
0 1 -4 | -5
0 0 1 |-5
just check u got that when for yours
cloud cairn
this is to r2
so take 4times r3 and add it to r2
the notation i wrote for my step there
is how my book and professor do it
4R3 + R2
dreamy depot
I got it thanks 🙂
I've just been having trouble with this algorithm and applying iteffectively
clever cedar
is there a functional difference between rank and the dimension of a column space?
nvm
'''rank(A) = dim(row(A))'''
and since dim(row(A)) == dim(col(A))
then by transitivity it must be true
just saw it in my book
if a man can swim across a river in 10 minutes or (1/6 hours) and if the current flows at 4km per hour in the opposite direction, is it true that by multiplying 4(1/6) will get the resultant vector
gloomy arrow
Do you learn jordan canonical form in a first semester linear algebra class or an upper level class?
surreal shuttle
@wintry steppe iii is true right off the bat
wait hol up nvm
it is fals
as the only way that would be true is if it were v that was there instead of u as the thing which u was "orthagonal" to in the statement
as v and w form a basis for a plane which is itself orthagonal to u
the sum of two orthagonal vectors will be orthagonal to neither of them
false qed
the way to test these is to write out the vectors as something like w = <w1, w2, w3> etc. and then see the results
that would be an easy way to prove or disprove (i) for instance
surreal shuttle
rewrite all vectors as components sub 1,2,3 as outlined above, then compute the dot and cross products. compare.
for (i) at least
then you will be able to discern whether or not it is correct
wintry steppe
got it tysmm
gray dust
let S be the set of vectors given by (i)
for any real scalar c and any element X picked out from S, determine if cX is also an element of S using the condition that all elements in S must satisfy
if yes, S is closed under scalar multiplication. if no, S isn't
wintry steppe
oo okay
so the only one which would be under the scalar multiplication would just be (i) right @gray dust ?
gray dust
didn't check. just here to post tips
feral grove
why wouldn't ii be?
feral grove
well before guessing why should i be or not be closed under scalar multiplication
dusky epoch
if one of these sets fails closure under scalar multiplication then you should be able to present a counterexample
dusky epoch
(0,0) is not in (i)
that alone already disqualifies it
bc 0 times literally anything is (0,0)
wintry steppe
ohhim stupiddd
bitter root
Can anyone advise me as to how begin solving this?
https://i.imgur.com/l5DvIfN.png
dusky epoch
what am i looking at
bitter root
the operation AP - PB = 0
dusky epoch
where P is an unknown matrix, and A and B are known?
bitter root
prove similarity
dusky epoch
prove similarity ok
solving that ungodly complicated system is definitely not the way to go
dusky epoch
consider instead finding the eigenvalues of A and B
if A and B have the same eigenvalues, with the same multiplicity, and (if necessary) the same JNFs, then they're similar
bitter root
Oh, umm I missed part of the question actually. It wasn't just proving but proving by finding P and P inverse
stoic pythonBOT
Ann:
dusky epoch
so then you're gonna have $R^{-1}QAQ^{-1}R = B$
stoic pythonBOT
Ann:
dusky epoch
$P = R^{-1}Q$
stoic pythonBOT
Ann:
bitter root
Ohh I see,
https://i.imgur.com/DKxlX2N.png
bitter root
I am stuck, I cannot bridge the gap from what I figured out - about connecting similarity to eigenvalues and determinants - to calculating P. I am not sure what is going on in your first equation either, I haven't seen that form before with the whole A D B thing.
gray dust
do there exist scalars $c_1$ and $c_2$ such that $c_1v_1+c_2v_2=(-2,0,-1)?$
stoic pythonBOT
RokettoJanpu:
clever cedar
row reduce v_1 and v_2
then you can easily try scalars
to see which satisfy
@wintry steppe
@wintry steppe
u have two accounts
wintry steppe
specifically b and c
urban bough
So, say I have a matrix norm for a matrix Ax where A is PSD
because you can spectrally décompose A into QWQT, where W is the diagonal of all of A’s eigenvalues, the norm would just become Wx, right?
Since The norm is unitarily invariant?
clever cedar
Q. AX = 0 where A is an n x n matrix and X != 0, prove that det(A) = 0.
Can i prove this with a contrapositive?
AX = 0
AA^-1X = 0
IX = 0
X = 0
QED
is my proof done there if its a contrapositive?
dusky epoch
i'd rather call this a proof by contradiction
clever cedar
oh alright ty
dusky epoch
"assume det(A) != 0 and hence that A is invertible, then we have <your work>, contradiction, therefore det(A) = 0"
wintry steppe
Hi,
How to compute the square of a matrix and vector product.
I have a matrix M (mxm) and a vector n (m x 1), and the product is defined as
$M ; n$
stoic pythonBOT
beepbopbot:
stoic pythonBOT
beepbopbot:
wintry steppe
The answer is $n^T , M^T , M , n$
stoic pythonBOT
beepbopbot:
wintry steppe
How to get here?
stoic pythonBOT
Ann:
dusky epoch
do you have any context for this
wintry steppe
Okay I will elobarate
I have $\theta(x + \delta x) = \theta(x) + \nabla \theta(x) \delta x + O(\delta x)$
stoic pythonBOT
beepbopbot:
stoic pythonBOT
beepbopbot:
wintry steppe
I am computing $|\theta(x + \delta x) - \theta(x)|^2$
stoic pythonBOT
beepbopbot:
wintry steppe
So in that expression, I get $(\nabla \theta \delta x)^2$
stoic pythonBOT
beepbopbot:
wintry steppe
??
hoary zinc
Proof of Proposition 2.2.1
Can somebody elaborate on how we get that system of equations
Not sure how those three equations come from the orthogonality of the rows
daring solstice
Are these notes good? http://joshua.smcvt.edu/linearalgebra/
They seem pretty solid to me so far
minor galleon
@daring solstice i use https://dec41.user.srcf.net/notes/ this set of notes myself
it really has everything for first 1-2 years of uni math
lunar sable
I have a question pertaining to diagonalizing a matrix. How do I know which one I set to lambda 1 and lambda 2?
dusky epoch
it doesn't matter
as long as you don't mess up in pairing up eigenvalues with eigenvectors
lunar sable
Okay, so the eigenvalue obviously has to pair up with their respective eigenvector
easy enough
wintry steppe
can anyone help me find a matrix for the reflextion across the line y=2x?
gray dust
the matrix corresponding to a reflection about the line $y=\tan(\theta)x$ is $\\\begin{pmatrix}\cos(2\theta)&\sin(2\theta)\\sin(2\theta)&-\cos(2\theta)\end{pmatrix}$
stoic pythonBOT
RokettoJanpu:
wintry steppe
thanks
i dont really get why the slope becomes tan(theta)
tan(theta) = y/x is that why?
$\tan(\theta)=2\implies \theta=\arctan(2)$. Is there a way to evaluate $\cos(2\arctan(2))$ without a calculator?
stoic pythonBOT
xyz:
minor galleon
@wintry steppe a more rudimentary method to do it would be, imagine any random point x,y
reflect it across the y=2x axis and express the new coordinates in term of x, y
find the matrix since you have M X= X'
wintry steppe
then solve for M?
gray dust
you could try a double angle identity but i doubt you'd get too far without resorting to calculator
wintry steppe
$M\begin{bmatrix}-2\1\end{bmatrix}=\begin{bmatrix}2\-1\end{bmatrix}$
stoic pythonBOT
xyz:
gray dust
for a test maybe, but it's always nice to know how to derive it
wintry steppe
hmm okay
solar terrace
wintry steppe
The graph shows how many kronor one day one had to pay for English pounds at a currency exchange office. Draw the function that shows how the number of kronor (y) depends on the number of pounds (x).
don't know how to do this, anyone willing to help me out?
solar terrace
@wintry steppe whats the whole question?
minor galleon
@wintry steppe yup solve for M
so basically you say, the perpendicular distance of (x,y) from y=2x is a certain distance d,
this lets you know the position of (x', y') after reflection
and you just do MX=X' and sort of intuitively guess at a solution to M
silent dune
How do I do this
half ice
@silent dune
Have you tried to expand the brackets?
silent dune
Not sure how to do that
Like multiply it out I'm assuming but like I know order matters so I'm still not really sure how to
half ice
X(A + B) = XA + XB
For example
Note it's not AX, it's XA
You can't flip order, like you said
silent dune
How do I deal with the ^-1
silent dune
Mhm, I've skipped it for now but yea
minor galleon
protip @silent dune @half ice (A-I)=A(I- A^{-1})
silent dune
How do we know that
silent dune
Hm
wintry steppe
if two linear transformations are invertible, then their composition is invertible right?
because $g^{-1}\circ f^{-1}=(f\circ g)^{-1}$ right?
stoic pythonBOT
xyz:
minor galleon
yh
silent dune
How would I prove that A(A-I)^-1 is A then?
cause im assuming the (I-A^-1) needs the A infront to = I
stoic pythonBOT
mike0x81:
stoic pythonBOT
mike0x81:
clever cedar
@silent dune
half ice
@silent dune
Let X = (A - I)'
(A - I)X = I
AX - X = I
AX = I + X
So A(A - I)' = I + (A - I)'
Oop that's not a useful result. Sorry for the ping
silent dune
lol all good anything helps its the last question I gotta do now
thorn robin
they already showed you how to do it above
namely this:
A(A-I)^{-1}(I-A^{-1}) = A(A-I)^{-1}(A-I)A^{-1} = I
thorn robin
it is equal to I - A^{-1}
toxic pendant
If I'm given 3 vectors and told to calculate the volume of the parellelpiped, can I use gaussian elimination to simplify them or does that not work
toxic pendant
A question is asking true or false, if the det=-3 then the volume of the parrallelepiped is 3
clever cedar
use dot product
half ice
Yes you did lol
toxic pendant
Assuming my assumption works I am able to prove that it is true
half ice
Oh good point
clever cedar
Let A, B, C be any three vectors in $R^{3}$ space then the volume of parellepiped = <(A x B), C>
stoic pythonBOT
mike0x81:
toxic pendant
Yes that is the case
steady fiber
yes, determinant is the volume of a parallelepiped
clever cedar
@north sierra does it satisfy subspace test?
toxic pendant
Can I row reduce
steady fiber
you can derive that using the properties of the determinant
namely that it is bilinear
half ice
The question isn't asking if H is a subspace (though you'd have to assume this) don't go with the subspace test
Instead, what are the properties of a basis?
toxic pendant
Sounds interesting
toxic pendant
I'm going to go and try and prove the determinant thing then, ty
half ice
Linearly independent, good. It also has to span the set that it is a basis of.
Luckily, they show you that v1 and v2 DO span H
steady fiber
you just need the properties of the determinant:
- swap two rows -> determinant becomes negative of what it was before
- multiply a row by a scalar -> determinant multiplies by scalar
north sierra
yeah since those two are met, isn't a basis? for some reason my book says it's not a basis
north sierra
why not?
north sierra
how do i do that?
half ice
Oh lol fair
north sierra
no
half ice
2v1 + v2 = [2,1,0]
So [2,1,0] is in Span(v1, v2)
north sierra
oh
half ice
But obviously this isn't in H, so Span(v1, v2) has things that aren't in H
half ice
They are
half ice
H is the set of vectors sv1 + sv2
Span(v1,v2) is the set of vectors av1 + bv2
These are not the same set, and therein lies the problem
half ice
And Poros did notice this, in that H is a one-dimensional space
So it can't have two basis vectors
You may not know about dimension yet but it's useful
north sierra
yeah i dont know about dimensions yet
assume this is a matrix
could we say this is a basis of R^3 because there are three pivots in the rows (so it spans R^3) and it is linearly independent ?
half ice
It shouldn't be a matrix lol
half ice
If you do turn it into a matrix and row reduce, you'll find all three are pivots
So yes it's a basis for R³
north sierra
but i could still make it into one if i wanted to row reduce right? though i really dont have to in this case
i dont really have to row reduce tho
the pivots are shown
its in ref
north sierra
so could i say what i said earlier about the three pivots in the row?
i mean three row pivots
cause number of pivots in rows = span right
half ice
Yes it's a basis for R³
half ice
Number of pivots is the number of vectors that are linearly independent
So you've proven all are independent
half ice
Np. Feel free to ask if there's anything else
lunar sable
I have a question
north sierra
Sure
lunar sable
about diagonalization of a symmetric matrix using orthonormal matrix
It seems to matter what order I put in the eigenvectors before I calculate the orthonormal matrix
how do I know which vector is x1 and which is x3
*x2 not x3
The ordering only seems to matter in this case
lunar sable
hmm
I do get that as I asked earlier today and the person said it doesn't matter which of course is true
but I'm diagonalizing for a symmetric matrix using orthonormal matrix
and to check if they are orthogonal I multiply the orthonormal matrix by itself
and when i do the multiplication on either configuration, the answer is different
If you look at the first paper I multiplied the orthonormal matrix and got an answer different from the bottom orthonormal matrix, which was correct
Nevermind I didn't do the transpose when I multiplied (QQ^T)
north sierra
@half ice
hi
what if we have a question where we have a 4x4 matrix
and all the columns are lineraly independent
and it asks us to check for basis for R^3
would there still be basis for R^3?
half ice
Any basis for R³ will have 3 vectors in it
Since you have 4 lin ind vectors, there's a problem
north sierra
can we pick any three vectors and make a new set?
half ice
@north sierra
It also doesn't help that the vectors are each in R⁴
solar terrace
if A is a 2x4 matrix And E=A*C^T -B^T is a 2x5 matrix then what is the order of B and C
dusky epoch
what have you tried
dusky epoch
@solar terrace it's been 80 minutes and you've either neglected or refused to provide any feedback
solar terrace
@dusky epoch I just saw your comment, I'm not sure what your question entails, I think I solved it im just wondering if anyone else comes to the same answer
dusky epoch
my question entails sending me your attempt and pointing out exactly where you got stuck
if applicable
minor galleon
can someone help me verify how many distinct solutions
we get
i got 8
cuz u just get |det(P)|=|det(Q)|=|det(R)|
and you get 2 equations from each equality
so 2^3 =8 solutions ideally
i did all of that, but im not sure whether anything has gone wrong
mint sentinel
i got the following exercise: Find the polar form and sketch the following complex numbers in the complex plane
a) for example is 3-3i
what am i supposed to do? i cant remember that we've gone through this in class
minor galleon
@mint sentinel polar form is basically r cis(theta) or r e^{i theta}
r is the magnitude i.e. √(a^2 + b^2)
while theta is tan(theta)=b/a
mint sentinel
okay thanks
stoic pythonBOT
yami:
wintry steppe
ok how do i show this
sonic osprey
What are you confused about?
wintry steppe
im not even sure if i understand the problem honestly
sonic osprey
Well okay, what don't you understand?
wintry steppe
ok so V and W are just some random vector spaces with arbitrary many v and w elements respectively yes?
sonic osprey
yes
wintry steppe
what does v' and w' mean
i dont understand the operations tbh, the lambda thingy looks like a scalar
wintry steppe
okay i think i have somewhat of a grasp about it now
north sierra
wintry steppe
then how do i go about showing that it results in another R vector space?
north sierra
what does the M stand for in question 17 and 18?
sonic osprey
well, what does something need to satisfy for it to be a R vector space
wintry steppe
is that about the 8 axioms
sonic osprey
yes
wintry steppe
hmm okay so the associativity and commutativity seem trivial
the 0 vector i assume exists because theyre spaces over R?
sonic osprey
no
wintry steppe
oh
half ice
Does this need to be over R? You get a vector space regardless right?
stoic pythonBOT
yami:
half ice
Is (v + 0, w + 0) the zero vector in V×W?
wintry steppe
probably not
half ice
Without getting too into it, (0,0) is the zero vector in V×W
Oh wait, that's what you're getting at, isn't it?
half ice
(v,w) + (0,0) = (v,w)
Which shows that (0,0) is the zero vector
wintry steppe
that sounds like i just made it up on the spot and cant be right, why is it right
half ice
The issue is - is (0,0) a member of V×W?
half ice
You can construct (0,0) by using the zero vector in each set
V and W have a zero vector because they are vector spaces.
The tensor product has a zero as well, by combining their zeroes
wintry steppe
right
V and W have a zero vector because they are vector spaces.
this kind of opened my eyes
half ice
Oh? Well that's good I'm glad. Why?
wintry steppe
because instead of looking at them as vector spaces and thus having a zero vector by definition i kept trying to look at R
half ice
A hint, you will likely need "because U and V are vector spaces" for 99% of this
wintry steppe
and how somehow because theyre over R they would have a zero vector
sigh
i can make the same argument for the inverse element no?
half ice
Yup. U and V have an inverse for every element, U×V does as well
steady fiber
Just saying [Insert number] axiom or [Insert number] theorem is almost always useless, you can number the axioms however
wintry steppe
right
Compatibility of scalar multiplication with field multiplication
talking abt this one
so what im thinking is
,tex $ \lambda \cdot \mu(v,w) = (\lambda\mu v, \lambda\mu w) = \lambda (\mu v, \mu w) = (\lambda\mu) \cdot (u, v)$
steady fiber
are lambda and mu scalars?
wintry steppe
i think theyre supposed to be elements of R
since V, W are vector spaces over R
the concern that i have is that it doesnt show anything does it
stoic pythonBOT
yami:
cobalt tartan
I am having trouble understanding what these two questions are asking for
Could anyone explain to me what they want me to do?
sonic osprey
Do you know how to show something is a vector space?
cobalt tartan
Closed under scalar multiplication and addition and that the 0 vector is in it?
sonic osprey
No there's more
sonic osprey
Yes
cobalt tartan
So what part a is asking for is to show that for any v, z, w in V, where V is a subset of C, that it is also a vector space over R?
sonic osprey
V is a subset of C?
sonic osprey
no
cobalt tartan
Oh
zinc tapir
Anyone good with images and curves
wintry steppe
Verify that $U=\left{(x-3y,2x+4y,3x+y)\right}$ is a subspace of $R^3$
stoic pythonBOT
xyz:
wintry steppe
how can i do this using image?
how do i do i even do this using the definition?
because i dont have an equals sign so i cant see if its also in the set after addition
nvm i figured it out
rustic panther
True or false, the inverse of an invertible upper triangular matrix must be a lower triangular matrix.
rustic panther
I went with false
steady fiber
it's actually true
paper egret
what is geometric multiplicity
i don't get it
and algebraic multiplicity
what do those mean intuitively
gray dust
@paper egret matrix A has an eigenvalue L. algebraic multiplicity of L is how many times L shows up as a root in A's charpoly. geometric multiplicity of L is the dimension of the eigenspace corresponding to L
gray dust
idk what you consider special about this stuff
if alg mult of L > geo mult of L, L is called "defective" and there's some slightly interesting phenomena when that happens
paper egret
oh
half ice
The normal basis is {1,t,t²}
half ice
Nvm, let's go a different way. If that vector can be expressed in B, then it's in the span of B. That means:
a(1 + t) + b(1 + t²) + c(t + t²) = 6 + 3t - t²
For some a,b,c
vast torrent
So find a,b,c
north sierra
idk how to
half ice
Simplify. Collect like terms.
vast torrent
Two polynomials are equal iff the coefficients on each term 1,t,t²,t³,... are equal
So you should be able to get 3.equations in 3 unknowns that aren't expressed interms of 1,t,t²
north sierra
so i row reduced the top matrix into the bottom one
and for some reason when i plug in the values from the augemented column into the original equation, I don't get 1,1
why is that?
half ice
What are the values?
half ice
That's what I'm saying lol
north sierra
what do you mean
half ice
What are you plugging in?
north sierra
5 and -2
half ice
Into what?
half ice
Looks like you're plugging in (5, -2, 1)
north sierra
yeah
half ice
Why 1?
half ice
So you can set that to anything and it won't change the result?
half ice
So your augmented matrix is
1 0 -5 | 5
0 1 1 | -2
north sierra
yeah
half ice
Note this is actually a system of equations
north sierra
ye
half ice
If the vector we're plugging in is (x, y, z) then this reads:
x - 5z = 5
y + z = -2
It definitely does not say
x = 5
y = -2
north sierra
im still confused
so when i plug 5 into x
and -2 into y
into the original equation
i only work with the first two vectors?
and not the last one?
so since there's no z i dont include that?
vast torrent
What are you trying to do
vast torrent
Okay
vast torrent
What do you mean into x and y
north sierra
i dont get 1 on either of the equations
north sierra
doesn't 5 correspond x and -2 correspond to y?
vast torrent
No?
vast torrent
Not sure what you mean by correspond
half ice
If the vector we're plugging in is (x, y, z) then this reads:
x - 5z = 5
y + z = -2
It definitely does not say
x = 5
y = -2
north sierra
i dont know what that means ^
vast torrent
This has a free variable, the solution isnt going to be a single 3-tuple
north sierra
every time i checked my answer, i always did what i did and got the answer but for some reason its not working now
half ice
Why are you plugging in x = 5, y = -2?
north sierra
im so shocked lol
i always thought thats how you check your answer
ive been doing it since september
vast torrent
I don't think you're clear in what youre trying to do when you reduce that augmented matrix
vast torrent
Youre not going to find a triple of numbers that solves it
You know that before you start
half ice
If your matrix were
1 0 | 5
0 1 | -2
Then that reads x = 5, y = -2
north sierra
lol
half ice
But it isn't, and it reads very differently
vast torrent
Because it's 2 eqns 3 unknowns
vast torrent
Here you are probably going to find a whole line of solutions
Before row reduction that should be your guess
north sierra
so whats the way to go about this ? when you have 2 eqns and 3 unknown?
yeah my goal is to find 2 solutions to 1,1
half ice
Think solving for the nullspace
half ice
👍
vast torrent
Now to emphasize the solution is a line
Write x3 as a variable, like t or s
And then the line of solutions is parameterized by that variable
half ice
t is a good replacement
obsidian rapids
Hey everyone, I have a question that is probably really simple to answer.
vast torrent
You don't have to but you should
north sierra
is there a reason?
north sierra
true
obsidian rapids
Define the unit square as the collection of points in R^2 with vertices (0,0), (0,1), (1,0), and (1,1).
Would this just be, R^2={(0,0), (0,1), (1,0), (1,1)} ?
vast torrent
It traces out a line
half ice
Np, feel free to ask if you have anything else
@obsidian rapids
It's any point (a,b) where 0 ≤ a,b ≤ 1
obsidian rapids
So what I put would be fine then. Thanks.
half ice
Yeah the copy paste definitely got that wrong
This reads
"R² is the set that contains (0,0), (0,1), (1,0) and (1,1)."
What do you want to say?
obsidian rapids
This is the question I was answering
half ice
You only want to sketch it
obsidian rapids
Is there any easier way to do this than trial and error?
obsidian rapids
anyone help me out here? Haven't been able to find info on this topic.
craggy thistle
how have u started
obsidian rapids
really unsure how to even start with this one
craggy thistle
it says u want "transformation matrix", so we must start off with ..?
obsidian rapids
I get that 0,0 needs map to 0,0 and the corresponding unit square coordinates but unsure what to do with that.
craggy thistle
do you have matrices?
obsidian rapids
No I was just given the vertices
vast torrent
How many vectors do you need to define a parallelogram?
obsidian rapids
4
vast torrent
Really?
vast torrent
How many arrows
obsidian rapids
2? This is all still very new to me.
vast torrent
Yes. The sides will be u, v, u+v, u-v
So.you only need two vectors
See what i mean?
obsidian rapids
yeah I see. You get everything you need from 2
craggy thistle
4 worked fine for me
vast torrent
They want us to find the matrix based on the usual rectangular coordinates
obsidian rapids
right
vast torrent
You just need to find the one on the x axis and y axis
And those will be the first and second column of your matrix
Draw a picture
vast torrent
I think that's the easiest way. You have a better way nox?
obsidian rapids
[2 1]
4 3
pretend the brackets extend down
yeah that works. Is there some sort of equation in the background for this? I'd like to know how this works.
dusky epoch
not everything needs an "equation"
craggy thistle
stack all the points for unit square in one matrix, and stack the given points in another matrix, then figure out how to do elementary row ops to get from one to the other
obsidian rapids
The "show your work" mentality always has me thinking i'm skipping steps I need to show.
dusky epoch
i mean
vast torrent
The idea is that to define a linear transformation all you need to know is how it acts on the basis vectors
dusky epoch
"here's a construction i pulled out of my ass and here's why it works"
is a perfectly valid way of showing your work
like showing your work is good and i'm all for that. but sometimes you just gotta make that leap of faith and pull something out of thin air.
vast torrent
So you need to.think
How does this function act on (0,1) and (1,0)
Linear transformations always map 0 to 0 so considering that won't help
obsidian rapids
right
vast torrent
You could consider (1,1) but you dont need to because (1,1)=(1,0)+(0,1)
Notice also that (2,4)+(1,3)=(3,7)
dusky epoch
How does this function act on (0,1) and (1,0)
i mean that's the basic idea behind matrices in the first place
that you can specify a linear transformation completely only by saying what it does to the vectors in a basis
and everything else just follows by linearity
and you never run into ambiguity due to the definition of "basis"
obsidian rapids
so, knowing we are going to have 0,0 with a transformation. We also know that, (0,1) = A*[2/4] and that (1,0) = A*[1/3] which gives us our matrix. not one third, 1 over 3 in the matrix that I can't do on discord
formatting
why
dusky epoch
you can use matlab-style notation for writing matrices in plaintext
separating rows by ; and entries within a row by ,
like this:
\verb+[1, 2, 3; 4, 5, 6; 7, 8, 9]+ for $\begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{bmatrix}$
stoic pythonBOT
Ann:
obsidian rapids
Did my logic make any sense there?
vast torrent
What does slash verb do?
half ice
It's a very good strategy, usually, to find your basis, and find what it maps to. The matrix is immediate if you can do that
obsidian rapids
and to do that you just need the vertices that correspond to the unit square points of (0,1) and (1,0) right?
dusky epoch
the symbol immediately after the verb is the delimiter, and everything between it and the next occurrence of it is rendered in monospace font verbatim as typed
half ice
Ya ya
obsidian rapids
i'm just at a loss on how to explain this homework style haha
vast torrent
"It suffices to find how the operator acts on the basis { (1,0),(0,1)}"
Also called the standard basis
obsidian rapids
Well one of the students asked for help on it and this is what the instructor said, which makes it seem like he want's more.
vast torrent
And then explain how you found T(1,0) and T(0,1)
half ice
There's more than one way to skin a cat
What your prof mentioned will work, but it will merely find the same matrix that you've already got
obsidian rapids
gotcha
half ice
Our method works in general. If you know what a transformation does to a basis, then you know what it does to everything
sullen hinge
I actually came because I have a question about a transformation
Im supposed to describe this transformation geometrically but tbh Im not sure what it does. We are given the forms for rotations and reflections but this doesnt seem to follow either of them, so Im trying to figure out what it does
obsidian rapids
Acceptable?
half ice
Similar idea, what does it do to the basis?
obsidian rapids
op format mess ups haha, fixed those
half ice
T(1,0) = (1,1)
T(0,1) = (1,1)
sullen hinge
so does it just squish everything onto the line y = x? is there any better geometric description?
half ice
That's pretty much all it does
The basis gives you a pretty good idea of how the squish happens
You can see the matrix isn't full rank, the output loses dimensionality
sullen hinge
things like basis, rank, dimensionality are outside of the scope of the course Im taking, I think. I wonder if there is a different way to determine what it does geometrically. I mean, for matrices like
well maybe Im just thinking of it wrong
half ice
Try test points? Have a question about where a certain point goes? Throw it in
north sierra
if a set of vectors is linearly independent then the determinant should be non-zero right?
gray dust
det of what?
north sierra
determinant of the matrix of the set of the vectors
gray dust
sure
north sierra
okay thanks
north sierra
oh yeah it has to be nxn right
dusky epoch
i.e. as many as the dimension of your space
gray dust
square
dusky epoch
otherwise your matrix isn't square
sullen hinge
Im having some issues here
the matrix to represent a rotation counterclockwise by 90 degrees should be:
0 -1
1 0
if Im not mistaken, labeled A, and the matrix to represent a reflection across the y=x line should be:
0 1
1 0
labeled B. And AB then is:
-1 0
0 1
which represents a reflection across the y-axis, but shouldnt rotating by 90 degrees then reflecting across the y = x line result in a reflection across the x-axis?
it seems like maybe A is wrong, that seems like a clockwise rotation? but the book Im reading seems to have that notation
sullen hinge
Im just thinking because A[x, y]' = [-y, x], so if the x axis becomes the -y axis, and the y axis becomes the x axis, isnt that a clockwise rotation by 90 degrees?
dusky epoch
so if the x axis becomes the -y axis, and the y axis becomes the x axis,
that's the wrong way of thinking about it
sullen hinge
yeah, youre probably right
oh yeah I see now, youre right. man this stuff can be confusing sometimes lol. thank you and nice Homestuck profile pic
sullen hinge
sure yeah
dusky epoch
south sedge
need a little help with linear subspaces
How do I know that V is a subspace?
should i look at this as a vector
that is in r3?
sonic osprey
Sure that's a vector in R3
south sedge
but why equal to 0 ?
sonic osprey
What
south sedge
sonic osprey
x_1 is the first term in x
south sedge
i know
but why is the whole expression equal to 0?
what would be different if it was equal to 1, for example?
would it still be a subsapce of r3?
south sedge
so it's a zero-vector?
quaint heart
Also @south sedge do you know what a nullspace/kernel is?
south sedge
im afraid not :/
quaint heart
Oh okay
So one of the vector space axioms is that the 0 vector has to be in the space
Just check the axioms for a subspace lol
@south sedge btw there's a very easy theorem you can prove that every subspace of R^n is of this form (the solutions of a system of linear equations)
south sedge
would that help me with intuition? because that is what i feel im lacking
for example
i dont know what this is telling me
so x is in r3 such that the first or second element in the vector must be zero
what do i make of that in terms of subspaces
wintry steppe
quick sanity check
$$p_T\left(\lambda \right)=\lambda ^2$$ this is an example of a characteristic polynomial that is NOT split
right?
stoic pythonBOT
apples:
sonic osprey
What does it mean for a characteristic polynomial to split
wintry steppe
it can be written as
$$p_T\left(\lambda \right)=\left(-1\right)^n\left(\lambda -\alpha _1\right)\left(\lambda :-\alpha _2\right)...\left(\lambda ::-\alpha _n\right)$$
stoic pythonBOT
apples:
wintry steppe
for some alpha_i in the real numbers
@sonic osprey
or whatever your favorite field is
sonic osprey
Then no that splits just fine
clear spoke
Hello
Kronecker cappeli theorem tells us if rang(a) = n than there is a single solution
if rang(a) < n then there are infinite solutions
But what if rang(a) > n?
dusky epoch
that's impossible
the rank of a matrix literally cannot be larger than its size
you'd have to find more than n linearly independent columns
but by definition you don't even have that many columns in the first place
how can a 3 by 3 matrix have rank 10
feral grove
so A is clearly nilpotent, and it seems clear the way to approach this problem is to show that there exists a basis B s.t. [A]_BB is the nilpotent matrix on the right
sonic osprey
Yeah, think about eigenvectors
feral grove
0 and any vector in ker(A)
well not 9
0
by definition
so like i could construct a vector as Av !=0 which would be an eigenvector of A
clear spoke
@dusky epoch What if you have a 4 by 3 matrix
its rang can be 4 and n can be 3
but I guess thats not something that shoud happen
dusky epoch
no
the rank of a 4 by 3 matrix cannot be 4
the rank of a matrix is at most the smaller of its dimensions
clear spoke
hmm
vast torrent
what's the question
wintry steppe
"Prove the equality"
wintry steppe
Transposed Vandermonde
sonic osprey
no you can do this not too badly
Think about determinant rules
what things can you do to a matrix that doesn't change its determinant
wintry steppe
Add rows/columns?
sonic osprey
More than that
wintry steppe
switch two pairs of rows
sonic osprey
more than that
sonic osprey
think more
wintry steppe
i can add 0 (i.e. a matrix with two same rows or a null row)
fuck fam I don't know
sonic osprey
You can add any scalar multiple of a row to any other row
Furthermore, everything you said for rows, can also be done for columns
wintry steppe
oh, yeah, that's what I meant by adding rows
forgot the scalar part
Which is probably the most important part of that rule 😅
still can't figure out how I would make this work trying that
That was my first approach
Then I tried expanding along the first row and trying to make some factorizations work / make other determinants but couldn't get closer at all
tranquil trail
Can anyone help my for numbers 3-9
vast torrent
I hope you mean expanding along the first column
wintry steppe
yeah, yeah, column, fuck me, I've been at this problem for too long and I'm starting to lose my sanity
@sonic osprey please expand on what approach you meant
sinful heron
i have a question for dot product. I don't understand what this formula is for... what is the cos for if you just add the two vectors
tranquil trail
@sinful heron you use that formula to find the angle
wintry steppe
it's not just adding
sinful heron
oh
atomic ember
theta is the angle between the vectors
tranquil trail
@sinful heron so if you’re interested in finding the angle between u and v
wintry steppe
and the dot product is a scalar not a vector
sinful heron
okay
so this is used for determining if two vectors are perpendicular to each other right?
if the dot product equals zero
gray dust
sure
sinful heron
thanks guys
wintry steppe
@tranquil trail for 9. you can use the fact that a*a =|a|^2
and separate the parts of the left hand side into two dot products
and the equation immediately follows
wintry steppe
ye
sinful heron
kk
wintry steppe
You can also get to that without memorizing the formula(though it is easier)
(2i + j + 0z) * (i -2j + 3z)
sinful heron
i read in my textbook that if the dot product is <0 then its obtuse i dont get why that is though
wintry steppe
when is cos theta < 0
wintry steppe
since you're looking at the smaller angle you'll never get past 180 degrees
But yes
And quad 2 is 90 degrees to 180 degrees i.e. the angle is obtuse
sinful heron
ohhh wow i get it now
wintry steppe
or pi/2 to pi if you wish
sinful heron
i see yeah that makes sense
pliant berry
What are you supposed to do? Find a, b, c?
wintry steppe
prove the equation
pliant berry
Oh, okay
quartz compass
are you familiar with what rules you have for manipulating determinants?
for instance you can add a row to another row without changing it