also angle between m,n is pi/6
#linear-algebra
errant wyvern
is it equal to 1/2* |c|*|d| *sin(pi/6) or should i multiply before doing this
oh wait no i think i got it
is this correct?
topaz geode
Can someone help me understand this better?
I have been looking at the course materials, and I can understand it, but I don't know how to show it or make anything out of f...
I have been looking a lot about Affine Transformations.
e.g.,
[x'] [a b c][x]
[y'] = [d e f][y]
[w'] [0 0 1][w]
has a 6 digree of freedom
And I understood it as combining linear transformations with translations.
so will look like this for an image for instance:
┌────┐ /─────/
│ │ -> / /
└────┘ /─────/
I'm not sure if I got that 100% correct...
flat sedge
In a vector space, division isn’t an operation
wintry steppe
you can't divide vectors
oblique rune
Omg what's happening 
wintry steppe
is v a vector in R^n?
wintry steppe
if v is non-zero, try writing out the components of both sides
oblique rune
So $\alpha$ is a covector that transforms $\vec{v}$ to the real number line
stoic pythonBOT
Exynouz:
wintry steppe
oh
oblique rune
So is it possible now
pallid rampart
What is it then
wintry steppe
covector
wintry steppe
what's even happening in the picture; what's the context? (i mean, what is the goal)
oblique rune
I guess taking an inverse is kind of like dividing a vector but that's not what's happening
flat sedge
I don’t think mult inverse is apart of VS axioms
wintry steppe
it is not
wintry steppe
alpha is a covector
and presumably the epsilon^i's are too
with alpha_1, alpha_2 scalars
oblique rune
Ok so
$\alpha$ is a covector, a row vector (or a function) that takes a vector v as an input and outputs a scalar
$\alpha=\alpha_1\epsilon^1+\alpha_2\epsilon^2$
Epsilons are the dual basis vectors and $\alpha_i$ are the coefficients for them
stoic pythonBOT
Exynouz:
wintry steppe
ok so what is the goal in the first picture?
oblique rune
Wait
It is proving that property that $\alpha=\alpha_1\epsilon^1+\alpha_2\epsilon^2$
stoic pythonBOT
Exynouz:
oblique rune
wintry steppe
it is saying that because $\alpha(v) = (\alpha_1 \epsilon^1 + \alpha_2 \epsilon^2)(v)$ for all vectors $v$, $\alpha = \alpha_1 \epsilon^1 + \alpha_2 \epsilon^2$; two functions are equal if they agree at every point in the domain. note that the definitions of $\alpha_1, \alpha_2$ only depend on the covector $\alpha$ and on the basis vectors of $\mathbb{R}^2$
stoic pythonBOT
TTerra:
flat sedge
Agreed. Nice mini proof tho btw !
oblique rune
Oh so the functions act on the same vector and give the same output so the functions must be same
wintry steppe
for every vector, yes
vague cedar
stoic pythonBOT
jacob:
bold ivy
anyone willing to explain some stuff about linear algebra, If you have the time for it.
My question is about planes in linea algebra and cordinate systems.
My first question is, is planes cordinate systems?
void relic
I'll refuse to answer your question and say you can use any two linearly independent vectors to define a plane, as well as a coordinate system given by (x,y) for when v=x*e1+y*e2
bold ivy
so no, cordinate system is not a plane
winter oyster
In most contexts I would say no but since the question is being asked in linear algebra its difficult to determine what you mean by plane and what you mean by coordinate system
A plane can be interpreted as a surface embedded in some space or you may mean the cartesian plane which would be the "coordinate system" or vector space would be a more common word.
pure tangle
Could someone please help me 𝐅𝐢𝐧𝐝 if the points (2,3,-1) (5,-1,2) (-1,7,3) are collinear? I tried doing ab plus bc equal to ac and found that true but when I compare the slopes of ab and ac there’s no common multiple
pure tangle
So I got ab(3,-4,3) bc(-6,8,1) ac(-3,4,4)
median forum
ok so like
foreget my brain is melting for a second
that sum is always gonna happen
in euclidean space
regardless of colinearity
pure tangle
Fuck me
median forum
you need to find multiplicity indeed
pure tangle
I based it off a solution where they only used the sum
Should’ve kept my original answer:(
pure tangle
It’s not your fault
Thanks so much for the help. Based on seeing both methods of those I thought either was valid
At least I’ll know for the test lol
pure tangle
Yeah that makes sense
Thanks again
And I went back to the practice problem and saw they proved it with a coefficient
gritty python
Is the channel open for a question?
My question: is [1,1] , [2,1] , [-1,1] spam of R^2. My work: c1[1,1] + c2[2,1] + c3[-1,1] = [a,b]. into system of eq: (1) c1+2c2-c3=a, (2) c1+c2+c3=b. into a matrix:
into rref into system of eq: (1) c1+c3=-2b+3a, (2) c2-c3=-b+a
I can't get any of the c (c1 or c2 or c3) into terms of a and b, where I can subsitute said c into the other one to find the solution.
Should have I done Ref instead of RREF to get one of the c's into terms of a and b?
modest meadow
gray dust
quiet crane
guys quick q, does this mean to use specific numbers like 1, 2, 3, 4 or does it mean a variable like a, b, c?
wintry steppe
you can find specific matrices
it's asking you to find such matrices, so you don't need to worry about doing anything general
quiet crane
[-3 3] * [3 0]
[0 0] * [3 0]``` Ok I got thisthat's a nice example
quiet crane
Hmm I also realized that as long as all the non zero numbers are the same, (all 3's in this case) the solution will also be 0
If i changed all the 3's to something like 15 would that be too overboard?
wintry steppe
you could if you wanted to
quiet crane
Cool thanks
wintry steppe
since $$O = \begin{pmatrix} -3 & 3 \ 0 & 0 \end{pmatrix}\begin{pmatrix} 3 & 0 \ 3 & 0 \end{pmatrix} = 3 \cdot 3 \begin{pmatrix}-1 & 1 \ 0 & 0\end{pmatrix}\begin{pmatrix} 1 & 0 \ 1 & 0\end{pmatrix}$$
stoic pythonBOT
TTerra:
wintry steppe
so any scalar multiple of the rightmost part is also zero
devout void
hey can someone help me out what exactly are cofactors of a determinant
wintry steppe
it's 4 am for me so i don't want to agree to anything
but you're free to post your thoughts here
devout void
about why the traspozed matrices and their original have the same determinant
its fine
another time
wintry steppe
to be honest i don't have a good intuition for this other than "if you work out the sum for det(A^T) you get the sum for det(A)"
i think you can say something nice about it involving orthogonal complements (because ker(A) = im(A^T) perp)
lament briar
Hi can someone help me to understand spherical tensors?
Let's imagine we want to calculate some property $a$ of operator $T$ $a = \textbf{v} T \textbf{v}^T$, where $\textbf{v}$ is a vector and $T$ is 2nd rank tensor. In cartesian, I do a traditional matrix multiplication and obtain a final scalar.
However I am not sure how to proceed in spherical representation. I know that I can reduce $T$ to 3 irred tensors of rank 0,1,2 and vector itself is rank 1. I found a rule how to do a product of two tensors through their components (aka $2k+1$ components for tensor of rank $k$) but I have no idea how to put those components together to obtain a final scalar as in cartesian formalism
stoic pythonBOT
Kubik:
neon spear
One question: Why do eigenvectors change, if I swap the columns of a matrix. Aren't the matrices similar, i.e. represent the same linear transformation under different bases (swapped basis vectors), and consequently have the same characteristic polynomial?
native rampart
They don't represent the same transformation
If you want to get a matrix of linear transformation in another basis,just swapping columns won't work
neon spear
Then, how does the matrix of a linear transformation look, if I permute the basis vectors, compared to before?
neon spear
Yes
native rampart
Use the change of basis rule
It will be more complicated than just swapping 2 columns
neon spear
Ah ok thanks
half storm
Man, something as some subtle of that does change the transformation. They aren't the same transformation but rather they have the same span.
I guess it should be obvious that they aren't the same but in my head i was associating as same span as same transformation which is not true.
oblique rune
In Einstein notation, is $(a_{ij}+a_{ji})x_ix_j = a_{ij}x_ix_j+a_{ji}x_ix_j$?
stoic pythonBOT
Exynouz:
obsidian bluff
can someone explain to me BV(Omega), it is the space of functions with bounded variation but i dont know what this means; context is image processing
pls tag me
void relic
variation is just how much the function goes up and down @obsidian bluff
so something with infinite variation would be something that wiggles a lot and gets fuzzy, like
BV(omega) just rules out those weird cases
I guess in image processing when you reduce noise some methods can't get rid of infinite variation, so you focus on the finite ones
gray dust
@oblique rune yes bc distributivity, nothing to do with einstein
stoic pythonBOT
Exynouz:
oblique rune
In that notation
blissful pewter
can anyone explain this question to me? does it mean the 1's are not all 1's but now like 3, 5, 2, 1?
ember wedge
hey guys
can someone explain this solution to me
at the end when they spilt 10a and a2-9 but they only used a2-9 to get the values of a
what happened to the 10a
half ice
@blissful pewter
A is the matrix
A² is a different matrix, given by A×A
0 is the zero matrix.
For the A that they gave you, it happens to be true that A² = 0
Is there a non-zero symmetric matrix B such that B² = 0?
blissful pewter
it could be [ 1 1]^2 [-1 -1]
half ice
That one does square to give zero, but it isn't symmetric
blissful pewter
oh so it has to be positive on the left and negative on the right?
half ice
Don't be afraid to Google definitions you don't know haha.
A symmetric 2×2 matrix would look like this:
a b
b c
If you flip it over the diagonal, it stays the same
blissful pewter
hmm it doesnt look like it has anything that equals to 0
subtle walrus
for a nilpotent matrix all eigenvalues are 0
real symmetric matrices are diagonalizable
therefore every nilpotent symmetric matrix is similar to the 0 matrix and thus itself the 0 matrix
blissful pewter
oh so if and only if the matrix is [a b] [b c]then only and only if they are 0, their squares will also be 0?
half ice
"their" is an odd choice of wording. Such a matrix, when squared, will never be the zero matrix. Unless of course, it was the zero matrix to begin with
blissful pewter
oh one more quick q, to show that it is idempotent, i would just do A^2 to the matrices here right?
Or is there more to the proof?
wintry steppe
you do calculate A^2, yes
and, if it all went right, you should get A^2 = A
you're just checking those matrices against the definition of an idempotent matrix, given to you there
blissful pewter
gotcha
marble lance
@ember wedge grouping
karmic ginkgo
hey i've been doing this question for a few hours now and its really got me frustrated. I did the operation algebraically and it revealed that it is impossible to have a negative diagonal. If someone can guide me in the right direction I will greatly appreciate it
marble lance
Show your calculations
karmic ginkgo
ember wedge
@marble lance so i factored it a different way and i got (z^6+1)(z^3-2)
i dont know where to take it from here
ember wedge
so do i find the all the third roots of 2+0j?
marble lance
2+0i, yes
marble lance
@karmic ginkgo Are you disallowing complex matrices?
karmic ginkgo
no complex matrices, although i understand that it could possibly be solved with complex ones
marble lance
Well, you are right that it's impossible for real matrices
karmic ginkgo
ok well its good to know haha, i kept thinking there is because it never states to prove if it isnt
thanks
winged belfry
Hello all, is there a reason why I can't do
$R_2 - 3R_1$ ?
stoic pythonBOT
Apollo:
marble lance
Who says you can't?
winged belfry
because when I tried that I got $x_2 = 14/5$
stoic pythonBOT
Apollo:
marble lance
Is the correct answer -2?
winged belfry
yea
marble lance
Your arithmetic is not good
marble lance
Do it again
marble lance
No 1 - (-6) = 7
lusty flame
Can I ask a question in here?
wintry steppe
yes, ask
so long as they are related to the channel name and the channel is not actively being used, you may always ask questions
obsidian bluff
@void relic thank you :3
proper mauve
Can anyone show me good youtube videos on how to do this. Its been a while since I done this stuff
https://gyazo.com/49923eee6ddf1af6eb57a1d81345f1c4
spice storm
@proper mauve are you doing the det?
proper mauve
whats dat
spice storm
determinant
spice storm
Ahh, what is the section from?
problem
so I can make sure I tell you what method to use
void relic
yea
n columns with m coordinates each
The row and column spaces are subspaces of the real spaces Rn and Rm respectively.
gaunt field
ah okay
so basically its just because the way they defined the column spaces is that they are n columns with m coordinates?
void relic
yea, the columns are vectors of size m, so they span a subspace of R^m
void relic
the null space is just the vectors that map to zero
so find out what size vectors are allowed to be multiplied like Av=0
gaunt field
its not clear to me how that is an answer to my question
so it's in R^n because the null space describes solutions and n correspond to the variables in the system of equations?
void relic
yea that's right
(we're talking about being subspaces of R^n and R^m btw, the null space doesn't have to be exactly all of R^n)
gaunt field
yeah
i really have a hard time understanidng row picture and column picture
is row picture just like a * x = b
and column picture is x(something) + y(something) = b?
robust pond
it was mostly just a stab in the dark but im having trouble seeing why my answer is wrong, i got
just doing basic algebra, not that thats the correct way
stoic pythonBOT
jan Niku:
robust pond
not sure what other method youd use on such a general problem
gray dust
how do you define dividing matrices
void relic
you reversed the order of matrices somewhere
dusky epoch
what the fuck is $\frac{CD}{A-BC}$ supposed to mean \thonk
stoic pythonBOT
Ann:
gray dust
which is why i asked how you define such a thing
void relic
what was the first step you did?
robust pond
multiply both sides by (D+BX)
robust pond
post
dusky epoch
ok so that gives you what
robust pond
AX=C(D+BX)
void relic
ur answer is just off because it should be CB
robust pond
how CB?
dusky epoch
why wouldn't it be CB
dusky epoch
surely you don't insist that C(D+BX) = CD + BCX for whatever reason
robust pond
not anymore i dont 
so actually paying attention to order gets the correct answer easily 😄
🙇♂️ thanks
errant wyvern
Soo if i have a set of vectors and need to add vectors to that set to create a basis how do i approach it? I know that the set to be a basis in a space of dim of N has to have N vectors, and those vectors need to be linnearly independant
native rampart
Choose an arbitary vector
Say v1
Now choose vector v2 such that v1 and v2 are linearly independent
Now choose v3 , such v1,v2,v3 are all linearly independent
And extend till you have a basis
errant wyvern
yeah but lets say the exercise gives me 2 vectors and my task is to expand said set to a basis, how do i go about finding a linnearly independant vector?
lets say (1, -1, 4) and (2,2,0)
half storm
There are a variety of different ways to go about doing this.
Say that you have a linearly independent set of vectors and you want to have a basis that contains that specific set of vectors, there is a method that you can use to get such a set by taking the columns of a preexisting basis that you are already aware of, adjoining the vectors of that basis and your set together into a matrix and then finding the reduced row echelon form of the matrix.
So like one of the preexisting basis that you are already aware of should be the standard ordered basis right.
errant wyvern
yeah the standard basis for R^3 its span{(1,0,0), (0,1,0), (0,0,1)}
half storm
Take the the set $ \beta = {e_1 , e_2 , e_3 } $ and then adjoin them into a matrix with the vectors of your linearly independent set so you get a matrix $$\begin{pmatrix} 2 & 1 & 1 & 0 & 0 \ 2 & -1 & 0 & 1 & 0 \ 0 & -4 & 0 & 0 & 1 \ \end{pmatrix}$$
btw as a side note $\mathbb{R}^3 = span{ (1,0,0), (0,1,0), (0,0,1) }$. The basis itself is the set $\beta = { (1,0,0), (0,1,0), (0,0,1) }$.
stoic pythonBOT
TheDon:
errant wyvern
yeah i see i phrased that horribly
half storm
Not really, but it's just a matter of the terminology.
You're fine.
Actually I kind of messed this matrix up.
errant wyvern
yeah i see
half storm
It needs to be that the last two columns are the first two columns.
errant wyvern
so now we just row reduce it? and then take whatever vectors are left?
half storm
Kind of, let me fix my matrix really quickly.
ahh the rows of the last three were the opposite of what I thought they would be.
One more sec.
stoic pythonBOT
TheDon:
half storm
Cool.
Alright so then you wnat to row reduce this matrix using Gaussin elimination.
So you're only allowed to use elementary row operations.
knotty raven
I have a true or false question
all linear algebra problems are reducible to row reduction.
half storm
After doing that the first column and the second column will turn into $e_1, e_2$ and then another one of the vectors will turn into $e_3$ that tells you that those three vectors are linearly independent and so they form a basis for $\mathbb{R}^3$
hopes that helped.
stoic pythonBOT
TheDon:
knotty raven
row redjuice
half storm
This question is a bit more winded than it might bseem. What do you mean a LA problem? There are certain LA problems that have nothing to do with matrices row reduction.
So I guess that answer is a no.
half storm
Ok lol.
knotty raven
he said it in his rank nullity video
half storm
I'm sure he didn't mean that in the literal sense.
wintry steppe
if only
it would make my functional analysis course so easy; everything is on a vector space so i can call it linear algebra and be done
just row reduce lmao
knotty raven
yo ez
errant wyvern
$$\begin{pmatrix} 1 & 2 & 1 & 0 & 0 \ -1 & 2 & 0 & 1 & 0 \ 4 & 0 & 0 & 0 & 1 \ \end{pmatrix}$$ to $$\begin{pmatrix} 1 & 0 & 0 & 0 & \frac{1}{4} \ 0 & 1 & 0 & \frac{1}{2} & \frac{1}{8} \ 0 & 0 & 1 & -1 & \frac{-1}{2} \ \end{pmatrix}$$
knotty raven
functional row reduction
stoic pythonBOT
Lonac:
errant wyvern
ok so just help me interpret this, this would mean adding (1,0,0) would make it a basis right?
half storm
So the first 2nd and 3rd column of the original matrix are linearly independent.
Yup
you're done.
errant wyvern
ight thanks for helping out
proper mauve
For this question do I have to find the deteminent first?
https://gyazo.com/0a5d3df0018f8c94885610db01466cad
half storm
No problem.
No.
Make an augmented matrix and make sure that it's consistent.
I.e. there is no nonzero entry in the last column.
Then you're done.
Or you can just solve it like a regular linear system like you learned in probably high school.
proper mauve
Gaussian Elimination?
half storm
@errant wyvern There is another way to do this right. You could think of an arbitrary vector existing in $\mathbb{R} ^ 3 = \begin{pmatrix} a_1 \ a_2 \ a_3 \end{pmatrix}$. Make a matrix from the columns again and then compute the determinant of the matrix. You want to find the values of $a_1,a_2$ and $a_3$ such that the determinant is zero.
stoic pythonBOT
TheDon:
half storm
so det $\begin{pmatrix} 2 & 1 & a_1 \ 2 & -1 & a_2 \ 0 & 4 & a_3 \ \end{pmatrix}$
stoic pythonBOT
TheDon:
half storm
if the determinant is nonzero, then the matrix is invertible and so it's a bijection. Meaning that it's spans all of $\mathbb{R}^3$. And you know it's a basis beacuse it follows as a corrollary from the exchangability theorem that any spanning set containing exactly the same number of vectors as the dimension of the space forms a basis.
So then you're done.
errant wyvern
but wait as someone said, shouldnt i then be looking for such a1 a2 a3 that det isint equal to zero?
half storm
Oh snap. sorry you're right.
stoic pythonBOT
TheDon:
errant wyvern
let me do an example first then i will form the question if i dont figure it out on my own
half storm
No problem, the first method is useful if you have two sets of vectors that have certain properties that you want because they might be useful to you for some reason and getting a basis out of them.
The 2nd method is useful if you just need to find a basis for the space.
errant wyvern
Im trying to see what would i do in case where the third column isnt the solution
errant wyvern
i think i got, i just needed to freshen up with gauss jordan method
sorry to keep u waiting, thanks for helping out
ember wedge
guys quick question, does jRe(5-5j) mean that you get the real value within the bracket which is 5 then multiply it by j
so would u get 5j for that then?
gray dust
yes @ember wedge
wintry steppe
the line y = x?
subtle edge
You are reflecting across the y=x line
So the matrix transforms the r2 vector that way
subtle edge
so what you do with y=x is that you take the transformation A(x1 x2)^T=(x2 x1)^T
where T is transpose
the job is to find a 2x2 matrix A such that we get this result
ie we switch the x1 and x2 coordinates
think of what you can put in there to get you that result
a hint is to try and make sure there are no x1s in the top and no x2s in the bottom ie some of the entries of A are 0
the standard bases work yeah
cool!
looks good to me!
try it out with a vector of (x1 x2)^T
youll see you get x2 x1
t
wintry steppe
less helpful answer: there is a formula for the matrix that gives reflection in a line in the standard basis out there that you can use
so if you're ever asked to compute such a matrix, you can check your answer against it
subtle edge
^^
bleak ginkgo
Can someone explain me what happened from step 2 to step 3?
subtle walrus
its basically the formula for a finite geometric sum
bleak ginkgo
ty
ocean sequoia
lusty flame
can i ask a question in here
ocean sequoia
How do you know what R matrix to choose?
Also what the steps to get it to that rref form I honestly cant figure it out lol
R3 is obviously R1 - R2
half storm
You could do it as a system of equations right?
Like, if you already have C like in here, then it reduces down to that.
ocean sequoia
ah
thats clever
sorry my factorization of matricies is really weak its why I want to go through this book tbh
half storm
$\begin{bmatrix} 1 & 3 & 8 \ 1 & 2 & 6 \ 0 & 1 & 2 \ \end{bmatrix}$ = $\begin{bmatrix} 1 & 3 \ 1 & 2 \ 0 & 1 \ \end{bmatrix}$ $\begin{bmatrix} a & b & c \ d & e & f \ \end{bmatrix}$
ocean sequoia
yea and just solve through?
half storm
Yea multiply out the right hand side and two matrices are equal iff their components are equal.
stoic pythonBOT
TheDon:
ocean sequoia
gracias
lusty flame
anyone can help me with this https://gyazo.com/d5651b0f27b4f1114d6804931c6bf742
ocean sequoia
do you know what a unit vector is?
lusty flame
yes
ocean sequoia
sometimes numbers can help
ocean sequoia
we dont know if they are orthogonal so be careful
lusty flame
ah
ocean sequoia
cause then the dot product is zero
so the first one i believe will just be negative cos($\theta$)
stoic pythonBOT
brzig:
ocean sequoia
do the next one and foil it out see what it comes too
lusty flame
i dont get how you got negative cos(0)
ocean sequoia
do you know what the formula for dot product is
lusty flame
yeah
ocean sequoia
what is it
lusty flame
a dot b = a1b1 + a2b2
ocean sequoia
yep or |a| *|b| * cos(theta)
ocean sequoia
they are the same formula
oh wait im dumb its just - cos(theta) = - 1
magnitude is always positive
ocean sequoia
im pretty sure the next one should just be 0
lusty flame
the absolute value of a * b
ocean sequoia
no its the length
lusty flame
oh the length
ocean sequoia
yea
lusty flame
wouldnt that be ll and not just l
ocean sequoia
im pretty sure the next one is zero
because its going to be u * u - v * v
= 1 * cos(0) - 1 * cos(0)
do you see why?
lusty flame
im trying
ocean sequoia
and im if wrong hopefully someone will let me know
thats the magitude of u?
its just one so, u*u = 1
ocean sequoia
the first one is negative one you see why right
i realized i never gave you the full answer just cos(theta)
lusty flame
reffering to 2a?
ocean sequoia
im trying to work on my own stuff and help you let me focus on you here
thoughts are super scattered
ok so its going to be u * -u = | | u | | * - | | u | | cos(theta)
but that becomes 1 * -1 * cos(0) = -1
right because two vectors have a cos(0) between them
lusty flame
yes
ocean sequoia
the second one is because its going to be u * u - v * v
= 1 * cos(0) - 1 * cos(0)
and we saw that right?
ocean sequoia
do you think you can solve the last one oon your own?
post it here and I can check
this way you get some practice too 🙂
ocean sequoia
i just foiled em
ocean sequoia
proper mauve
anyone know how to do this
https://gyazo.com/7463642a32ac1a3ddfc6ecb7ebee34df
subtle edge
is this microeconomics?
gray dust
@wintry steppe yes & add the dots as you said
strange crystal
the hint is wrong right?
(1,0) rotated -3pi/4 clockwise should be (-1/sqrt(2), -1/sqrt(2))
or am i reading this wrong
or is it possible saying that it's negative of the clockwise direction not that the negative is saying go clockwise
torpid portal
it seems to be saying turn 3pi/4 rad clockwise
then reflect across the x axis
are you forgetting to read the
reflect across the x axis part
perchance
i.e. (1,0) rotated clockwise is what you said
but thats before the reflection across the axis
tulip basalt
If a matrix is positive definite, does it follow that its transpose is positive definite, too?
quiet adder
If positive definite means symmetric then yes because of the definition of a symmetric matrix, but I'm not sure if positive definite = symmetric
strange crystal
@torpid portal True yea okay I was processing that the rotation of e1 = -1/sqrt(2), 1/sqrt(2)
strange crystal
yes i am thank you
torpid portal
alright awesome
tulip basalt
@quiet adder awww positive definite does not mean symmetric
quiet adder
ahh sorry then
tulip basalt
so this does not generally hold?
wild fern
Hello my friends what is THE BOOK for linear algebra? (Im a CS student)
void relic
Linear Algebra Done Right
wild fern
Linear Algebra Done Right
@void relic this book is a little bit basic I think do you recommend any other book which is more heavy ?
limber sierra
halmos' finite dimensional vector spaces
if you want something really heavy, at least
linear algebra isnt like real analysis; there's no "standard" that every other text is compared to
koffman & kunze's linear algebra is a fairly typical rigorous introduction
that you see recommended a lot
wild fern
halmos' finite dimensional vector spaces
@limber sierra Oww, thanks I'll give it a try
jolly rivet
@limber sierra whats the standard for real analysis
limber sierra
rudin's principles of mathematical analysis
colloquially "baby rudin"
but yea as mentioned, halmos is very heavy
and its from back in the 50s so it like
predates most modern pedagogy standards
the content is good but i definitely wouldnt recommend it to students who don't have both:
- experience in linear algebra
- experience in pure math (proofs, etc)
wintry steppe
How do you visualize the determinant? I’ve read the determinant of a 2x2 matrix is a parallelogram, but why?
gray dust
@wintry steppe det(A)=unique oriented volume of the hypervolume spanned by A's cols, and it's not so much an interpretation as it is a definition if you wish to define det as such before further formulating it
from scratch you can define the oriented volume W of a set of vectors x_1,...,x_n in R^n as having some properties we think are nice to have. W is linear in each slot, W=0 iff the x_i's are linearly dependent, and W=1 for at least one orthonormal basis of R^n. then through a somewhat lengthy journey you can prove some other nice properties W has as a result of the prior properties you postulate W has, then finally write an explicit formula for W(x_1,...,x_n) which turns out to be the leibniz formula for det
gaunt field
How do I express the null space as a column space of another matrix?
gray dust
write the vectors in a spanning set for the nullspace as the cols
errant wyvern
So on my exam i had a question to prove that a Linear operator of an odd dimension has a real eigenvalue. Quick thinking on my end, i just wrote that because the charactheristic polynomial is of an odd degree, it always has a real root, meaning a real eigenvalue. I got 3/4 of the points on the exercise, because the answer wasnt formal enough but correct. How would formal phrasing sound like?
dusky epoch
maybe they wanted you to pick a basis for the space, consider the matrix of your operator in that space, and take the charpoly of that
errant wyvern
allright, idk im in a shitty situation since he doesnt want to hold a review of our exams so we really cant ask him questions about the scoring :/
So in the space $$P_\le2 $$ of polynomials with degree less or equal to 2, we are given $$S={g\in P_\le2: 3g(0) + g(1) = 0} $$ We need to check if it is a subspace of said space, we need to find its dimension and a basis.
stoic pythonBOT
Lonac:
errant wyvern
From here so far found that dimS=2, and that the basis could be $$ a_0, a_1, -4a_0-a_1$$
stoic pythonBOT
Lonac:
errant wyvern
is this correct? i got it by using the equation given in the definition of S, and putting $$a_0+a_1*t+a_2*t^2$$ and putting 0 instead of t in the first part and 1 instead of 2 in the second mention of g.
stoic pythonBOT
Lonac:
dusky epoch
$P_{\leq 2}$
stoic pythonBOT
Ann:
dusky epoch
also your description of the basis makes no sense
errant wyvern
yeah the basis would be for $$ a_0 =1, a_1=1$$ $${1, t, -5t^2}$$ right?
stoic pythonBOT
Lonac:
errant wyvern
obv i think i was wrong, i think dimS=3, just need conformation i didnt mess something up so far
errant wyvern
allright im listnening
dusky epoch
well you messed up royally, what other confirmation are you looking for
i can't see your work
errant wyvern
$$S={g\in P_{\leq 2} : 3g(0) + g(1) = 0} $$ we know that g(t) has the form $$ a_0+a_1t+a_2t^2$$ then we use that in the form given us in the definiton of S. So $$3*(a_0 + a_10+ a_20) + a_0+a_11 + a_2 1= 0$$ From here we get $$ 4a_0 + a_1 +a_2=0$$ we can exrpess $$a_2=-4a_0 - a_1$$ I picked $$a_0=1; a_1=1$$ Then i got $$a_2= (-41 -11)=-5$$ so g(t) has the form $$ g(t)= 1+1t-5t^2$$ and from there we get {1, t, -5t^2}
stoic pythonBOT
Lonac:
Compile Error! Click the 
reaction for details. (You may edit your message)
dusky epoch
this doesn't make any sense
you got that 1+t-5t^2 is an element of S, at best
doesn't mean that any of its constituent terms are also in S
in fact they are not
errant wyvern
i see, so how do i get a basis then?
dusky epoch
well you have expressed a2 in terms of a1 and a0
so you can set (a0, a1) = (1, 0) and get one element
and then (a0, a1) = (0, 1) and get another
errant wyvern
ok so for $$ (a_0, a_1) = (1,0)$$ we get $$g(t)= 1-4t^2$$ and for $$ (a_0,a_1) =(0,1)$$ we get $$g(t) = -t-t^2$$
stoic pythonBOT
Lonac:
errant wyvern
ohhh so the basis would be $${ 1-4t^2, -t-t^2}$$
stoic pythonBOT
Lonac:
tulip basalt
how do I prove $A\in\mathbb{R}^{n\times n},$ then $A$ is positive definite if and only if its symmetric part is positive definite?
stoic pythonBOT
emphatic_wax:
subtle edge
have u tried using the xAx>0, A=Atranspose and using the norm to show it is positive?
subtle edge
sorry that equals should be +
tulip basalt
This is what I have so far:
$A_S=((\frac{1}{2}(A+A^T))x,x)=\frac{1}{2}(Ax,x)+\frac{1}{2}(A^Tx,x)$
stoic pythonBOT
emphatic_wax:
subtle edge
you have to show xAx=x(0.5(A+A^T))x>0 unless you have another defn of symmetric part
oh ok
uhh
tulip basalt
I know that 1/2(Ax,x) is positive but Idk if 1/2(A^Tx,x) is positive
or what guarantees that its positive
subtle edge
gotcha
so are we doing hte direction of if A in pos def then symm part is pos def?
is*
subtle edge
if A is pd then AT is also pd right
tulip basalt
yes that's one direction of the problem
subtle edge
yeah so u want the other way?
tulip basalt
both directions since the problem is biconditional
subtle edge
yeah
but the defn is that the matrix A is symm too right
so that cdn is already met
A^T=A and since A is pd then that direction works
the other direction is if 1/2(A+AT) is pd then A is pd right?
subtle edge
i thought that was part of the defn of pd
hmm
that is i thought it was hermitian
for pd defn to occur
i might be wrong a bit of time since i did this
tulip basalt
i don't think they're equivalent. I will look. because in the bok we use it's not necessarily symmetric
tulip basalt
stoic pythonBOT
emphatic_wax:
wintry steppe
when A and x are real yes. you can prove this quickly by writing
(Ax, x) = (Ax)^T conj(x) = x^T A^T conj(x) = x^T conj(A^Tx) = (x, A^Tx)
wintry steppe
conj is conjugate
$$(Ax, x) = (Ax)^T \overline{x} = x^T A^T \overline{x} = x^T \overline{A^Tx} = (x, A^Tx)$$
stoic pythonBOT
TTerra:
wintry steppe
@tulip basalt
something something we use the fact that A and x are real in this example
tulip basalt
Thank you so much! This is great.
wintry steppe
(well just A real works for this specific calculation, but in the thing you posted i think the second equality requires x be real)
i am assuming this is the standard inner product on C^n btw, that's what the context made it seem like
more generally $$(Ax, y) = (x, A^{}y)$$ for all $x, y \in \bC^n$, where $A^{}$ is the hermitian adjoint (complex conjugate) of $A$. when $A$ is real, that's just the transpose
stoic pythonBOT
TTerra:
wintry steppe
what is wrong with me
wintry steppe
Can someone help me with this question?
dusky epoch
what have you tried @wintry steppe
wintry steppe
I tried to sub in a + bj for z, but I got stuck for the absolute value part
@dusky epoch
wintry steppe
Mb lo
dusky epoch
|x+jy| = sqrt(x^2 + y^2)
dusky epoch
omg no
z - 3 = (a-3) + bj
|z-3| = sqrt( (a-3)^2 + b^2)
the -3 goes with the real part since yknow
it's real
dusky epoch
i mean you could but i'd probably hold off on that until absolutely necessary
'sides, you can look at the imaginary part of the left-hand side and notice that it must be zero
wintry steppe
0 right
so if thats the case then does it become 3a =sqrt((a-3)^2)
or is it 3a =sqrt((a-3)^2+b^2)
dusky epoch
well b = 0 so what do you think
bold ivy
what I don't get is that if you us dot product you will get a parrarellogram but here they just add vectors they will get pararellogram it doesn't make any sense
gray dust
if you us dot product you will get a parrarellogram
this is nonsense
bold ivy
@gray dust sorry I mean cross product
subtle walrus
there is no cross product in R^2
gray dust
the parallelogram can be seen if you add u,v tip to tail, draw u+v & v+u
winged belfry
for matrices, is the 1st number in the row called the pivot?
Curious if this is correct to get into RREF
this was the prob
i ended with (x1, x2) = (2, 2) and (x1, x2) = (7,4)
icy ruin
umm what chapter is that from?
i think I had that same textbook so it would be helpful to just copy paste the formula im talking about
icy ruin
So the problem you posted is basically referring to this picture here. It's asking how do I reach point A via using a combination of u and v? So for example A is equal to u-2v. Now it is asking for the rest
errant mist
I was wondering for the reason why the dimension of the vector space of polynomials of degree <= n is given as dim P_n(t) = n +1 and not simply n? Is it because we count n+1 terms from i.e the polynomials (1,t,t^2,...,t^n)?. These polynomials form a basis for P_n(t) and therefore we can take their dimension.
icy ruin
@bold ivy basically this problem shows you that you can reach all of these points and an infinite amount of more points via linear combinations
limber sierra
precisely yes @errant mist
if youd like you can think of it as
the constants are 0-degree polynomials
and a basis must contain one vector for each degree (since a maximal linearly independent set is a basis)
so a vector for degree 0, degree 1, ... degree n
this is n+1 such vectors
errant mist
@limber sierra Thanks! Im used to seeing n+1 being linearly dependent (i.e given a vector space V of finite dimension n), so was at first uncertain.)
limber sierra
yeah, admittedly it is a bit weird at first
bold ivy
@icy ruin I see, but how? they've done linear transformation. idk how to do that in that condition.
modern palm
how to prove V^A is closed under addition?
let g,h be in V^A, so g(x) and h(x) is in V
modern palm
g(x) + h(x) is in V?
but Isn't that using the fact that vector spaces are closed under addition in the process of proving that vector spaces are closed under addition?
limber sierra
you're given that V is a vector space
so you already know V is clsoed under addition
youre not proving anything about V
modern palm
ohhhh
limber sierra
youre proving something about V^A
modern palm
I get it
thanks
another question, what is pointwise addition?
is it just saying that for addition, g(x) + h(x) = g+h(x)?
modern palm
got it
robust pond
how bad is this
from this prompt
do i really have to invoke the whole identity matrix linear combination thing
native rampart
You could show if $Cx=0 \forall x$ C=0
stoic pythonBOT
DrunkenDrake:
robust pond
i get that like
native rampart
Then use if Ax=Bx
(A-B)x=0
robust pond
oh hrm
i saw a proof that tried to do that i think but i think they said there was an error
do you think what i showed was not sufficient?
my teacher isnt gonna like it
native rampart
Well,You haven't proved a linear transformation can't have 2 matrix representations (Ax=Bx need not imply A=B)
robust pond
oh
because there are vectors for x that would make this trivially true?
i might have to think about that a little more
native rampart
Show that
||Take basis vectors and apply T on them, and contrast with multiplying with matrix form||
dusky epoch
what does it mean for a function to be in $\mathbb{U}_e \cap \mathbb{U}_o$?
stoic pythonBOT
Ann:
dusky epoch
well...
dreamy iron
I have s in the intersection...
dusky epoch
you started with the assumption $s \in \mathbb{U}_e \cap \mathbb{U}_o$
dreamy iron
It’s both an even and odd function
stoic pythonBOT
Ann:
dusky epoch
and you ended up with $s(x) = 0$ for all $x$
stoic pythonBOT
Ann:
dusky epoch
therefore s is in fact the zero function
therefore you've proved that IF $s \in \mathbb{U}_e \cap \mathbb{U}_o$ THEN $s = 0$ and hence you've proved that $ \mathbb{U}_e \cap \mathbb{U}_o = {0}$
stoic pythonBOT
Ann:
dreamy iron
Can we make the proof a touch more gory please.
So is this right:??
s(x) = 0 for all x IMPLIES s in {0}
?
(I know what I did, lol, I don’t believe that I did what I think i did....)
The step I wrote that’s in the red, I don’t think that’s the right logical step.
$\textbf{is this right}$
$$s(x) = 0,,\forall x~~ \implies s \in {0}$$
stoic pythonBOT
ninnymonger:
dreamy iron
¿¿And the next step is
$$s \in \mathbb{U}_e \cap \mathbb{U}_o \implies s\in \left{\vec{0}\right} $$??
stoic pythonBOT
ninnymonger:
wintry steppe
well you really did needlessly make this proof "gory"
yes, if s(x) is zero for all x, s is the zero function
and yes, sure, that's the next step
so U_e cap U_o is a subset of {0} (by way of what the most recent picture says)
this is entirely summarized in what ann wrote
dreamy iron
THANK YOU!!
I’m a noob and a physics main, also, this is all self study, I’m not getting feedback from TAs and graders....I have to hold myself to a higher standard if I want this to be passable in a university setting.
robust pond
@native rampart checked the spoiler, we did that but you're saying we need to do that in order for the proof to be sufficient?
sorry to ping
native rampart
Yes
robust pond
oh i guess you can multiply still with the generic form
just use indeces for everything
native rampart
Yes
sharp estuary
Ok, this is kind of out-of-the-blue, but does anyone here happen to know a good deal about idempotent matrices over general ring and not just a field? Is it true that if P is idempotent I can claim ker(P)= im(1-P) or is that nonsense?
dusky epoch
so like... you're considering P as an operator on A^n where A is your ring?
sharp estuary
Yep
dusky epoch
do you assume it to be a commutative ring with unity
dusky epoch
IBP?
sharp estuary
Invariant basis property
sharp estuary
I mean, it's true in the case of vector spaces right?
Do I at least have that?
To give context, I'm trying to determine K_0 of rings exclusively in terms of idempotents
dusky epoch
yes
would have to recheck the proof tho
to see how much of it transfers to the ring case
sharp estuary
Gotcha. Well thank you!
dusky epoch
x in A^n, P idempotent, 1-P will be idempotent too
(1-P)^2 = 1-2P+P = 1-P
1 stands for identity operator obv
sharp estuary
Right
sharp estuary
Right that makes sense
dusky epoch
ker(P) obv contains im(1-P)
sharp estuary
yep
dusky epoch
now for the other inclusion
Px = 0
then i think x = (1-P)x?
yeah
so the other inclusion still holds
independently of the ring being a field or not
so yeah looks like you still have that in your case
sharp estuary
I see.
Actually it's kind of obvious in retrospect when you just remember that module elements will still distribute over operators.
So yea
that makes sense
Cool. Thank you again!
autumn beacon
Is this correct?
viscid kernel
Is it possible to solve a system of nonlinear equations with a matrix ?
pallid rampart
Probably not
A matrix is a very linear algebra object
And nonlinear equations are not linear
dusky epoch
Okay, the center of the interval [x, y] means (0, 0, 0)^T.
are you sure
cause what you've written is like... at best, it's a very circular and notationally overloaded way to prove all linear maps map 0 to 0, and at worst it's just nonsense
errant wyvern
so if $$U= span{ \begin{pmatrix} 1 \ 1 \ 1 \ 1\end{pmatrix}, { \begin{pmatrix} -2 \ 1 \ -1 \ 4\end{pmatrix} , { \begin{pmatrix} 8 \ -1 \ 5 \ -10\end{pmatrix} } $$ and $$V= span{ { \begin{pmatrix} 2 \ -1 \ 1 \ 0\end{pmatrix} , { \begin{pmatrix} 1 \ -2 \ 0 \ 1\end{pmatrix} }$$ how do i find a basis for $$U\cap V$$
dusky epoch
@wintry steppe perhaps the center of [x,y] is actually 1/2 (x+y)
stoic pythonBOT
Lonac:
Compile Error! Click the reaction for details. (You may edit your message)
errant wyvern
anyone willing to at least point me into a direction of how to solve this?
surreal thistle
could someone explain why (1subG, b) is the kernel
i dont understand how that is the identity element of set G
native rampart
The function Takes in (g,h) and returns g
oblique rune
How do people represent rank 3 tensors?
It's represented as a cube.. in Google images..
Is that how people represent it usually?
wintry steppe
surreal thistle
yo i dont really understand why Z2 x Z2 isnt isomorphic
is it because (0,0) and (1,1) arent distinct?
but like z2 x z3 also have nondistinct elements
so idek
pallid rampart
Assuming we are talking about the additive structure, Z_4 has an element of order 4, but Z_2 x Z_2 does not have an element of order 4
spiral star
@errant wyvern i hope that helps. note that [A -B] is a block matrix, i didnt clarify that in my text.
surreal thistle
wait so @pallid rampart i get that you cant make an element of order 4 cause if u try to add them then it still is additive mod 2
but why does z_2 X z_3 work out
pallid rampart
Well it is because 2 and 3 are coprime
and the reason why z_2 x z_3 works out you'll have to work out the exercise yourself
👀
but just saying, in general Z_a x Z_b is isomorphic to Z_(ab) if a and b are coprime
pallid rampart
yes
brisk fractal
@lusty flame |U - V|^2 = (U - V) \cdot (U - V) = U \cdot U - 2 U \cdot V + V \cdot V = |U|^2 + |V|^2 -2(U \cdot V)
combine that with (a) and you have your answer
midnight hedge
can someone help. I was working on a problem and so far I think there is no answer
i've reduced it to
1 3 0
0 1 (h+16)/19
0 0 1
jagged gulch
one way to do it would be to put those vectors as columns in a matrix (which you probably did), take the determinant, and set it to zero
then solve for h
midnight hedge
ah didnt think of that
@jagged gulch when I set the determinant = 0, I get 1 = 0
does that mean the there are no values for h to be linearly dependent
jagged gulch
yeah I guess so
midnight hedge
jagged gulch
I checked it myself, and yes, if you set the determinant equal to 0 then you get no solution
midnight hedge
when i find the determinant with what is given. i get 304, with the h's canceling each other out
god, my professor really would put that on the quiz. had no idea how to do it; guess there was no answer
limber sierra
if we have a linearly dependent set of vectors, then one vector is a linear combination of the others
similarly, if one vector in a set is a linear combination of the others, then it must be linearly dependent
for example, $\left{\begin{pmatrix}1\0\0\end{pmatrix}, \begin{pmatrix}0\2\0\end{pmatrix}, \begin{pmatrix}-2\6\0\end{pmatrix}\right}$ is linearly dependent, since [-2\begin{pmatrix}1\0\0\end{pmatrix} + 3\begin{pmatrix}0\2\0\end{pmatrix} = \begin{pmatrix}-2\6\0\end{pmatrix}]
stoic pythonBOT
Namington:
limber sierra
i.e. that vector is a linear combination of the others
furthermore, if we "order" our set of vectors
then at some point we'll be able to "cut off" the list
and have that vector be linearly dependent on everything "to the left" of it
as long as at the first vector is nonzero, that is
this last fact is kind of obvious from the first sentence
but i'd assume it'll be used in some proofs
midnight hedge
its basically saying that if we have a vector that is a multiple of the other vectors, then it is linearly dependent. if you go back to the definition of linear dependence you see this
from the same textbook
limber sierra
franticcelery thats not exactly what its stating
" we have a vector that is a multiple of the other vectors, then it is linearly dependent" this is either the definition of linear dependence, or completely obvious
midnight hedge
so if
v1 = [1, 2, 3]
v2 = [2, 4, 6]
v3 = [3, 6, 9]
we can get v2 and v3 from multiplying v1 by 2 and v1 by 3
limber sierra
"multiple" and "linear combination" are not synonyms
bold ivy
@limber sierra but -2 6 0 is combination of the other two. BUt you are saying -2 6 0 is linear dependent
limber sierra
yes epic, correct.
that is what the theorem says.
if a vector is a lin. combation of other vectors, the set is linearly dependent
midnight hedge
whoops am i looking at somehting different
midnight hedge
yeah my bad i thought it was referring to this in the book
bold ivy
what is thereom 8 mean?
midnight hedge
its saying that if there are more vectors then entries then it linearly dependent.
v1 = [2,4]
v2 = [2,5]
v3 = [3,7]
as can be seen we have 3 vectors and 2 entries in each . so it is linearly depenedent
Or you could say in an augmented matrix; if there are more columns than rows, its linearly dependent
scenic vapor
How do i find out if a stochastic matrix has a unique steady state vector?
bold ivy
I dont understand why you would us dot product times cross product to find out the volume of a parrarelped when you can just find the determinant through 3x3
like what
wintry steppe
when u are calculating the determinant you are basically doing a dot product times the cross product
quiet crane
Can anyone check if my answer is correct? It asks me to find the elementary matrix's inverse:```
[0 1]
[1 0]
= [0 1 | 1 0]
[1 0 | 0 1] R2 <=> R1
= [1 0 | 0 1]
[0 1 | 1 0]
A^-1 = [0 1]
[1 0]
My logic: I need write down the identity matrix on the right of the vertical line and manipulate the left and right side until the left side is the identity matrix
modern palm
try multiplying your original matrix with the inverse
if you get the identity, then it's right
but yeah its right
half ice
Wolfram Alpha can do quick checks for you, if you only need to check answers
If you have questions on the process come to us ofc
quiet crane
I hope that its possible to check it on a physical calculator, I be writing down the wrong numbers sometimes
modern palm
how can I prove that the set is closed under scalar multiplication?
quiet crane
@modern palm Can I ask 1 more quick question?
modern palm
sure, i'll try my best to answer
quiet crane
So I have a elementary matrix [2 0] [0 3] is it illegal to do 2 row operations at once to get the elementary matrix's inverse? I think that R1/2 and R2/3 gives me the right answer
modern palm
do what you had in mind before. the left side of the matrix is your original matrix, the right side the identity. Try to change your left side into the identity and your right side should give you the inverse
doing two steps at once, my profs didn't really care for it
quiet crane
Ok I got it thanks
Ohhh so that means checking the answer by multiplying the identity matrix and the elementary's inverse does not work
modern palm
how to prove closure under scalar multiplication
wintry steppe
check that if c is a scalar and f : A -> V is a function, then the function cf is a function from A to V
modern palm
yeah, but that's what i'm having trouble proving
wintry steppe
the function cf is defined by (cf)(x) = c f(x); of course, you have to check that c f(x) is always in V
you know f(x) is in V, so...
modern palm
why would cf(x) be in V?
wintry steppe
look at the definition of a vector space
modern palm
i don't get it...
wintry steppe
a vector space is, among other things, closed under scalar multiplication
modern palm
this is how my book introduces a vector space
modern palm
wait, then this thing I'm trying to prove is literaly given in the definition
lol sorry let me think about this again,
ohh I get it
thank you
median forum
not exactly
you implicitly use two definitions
that one and the elementwise product
modern palm
elementwise product?
oh pointwise scalar multiplicaiton?
maybe i'll just ask my prof tomorrow, we havent even had our first class yet
and I swear this is the first time I've heard of those terms
quiet crane
quick question
to do EA = B
can i use the assumed matrix like this? [a c] [b d]
or can it be this too?
[c d]```
rocky wharf
Can anyone help me?
jagged gulch
sure
@rocky wharf I'm kinda late but I'm pretty good with linear, I'll probably be able to help you
also I'm not the first person to say this, but the answer to the question "can anyone help me" is always yes in this server
Someone will help you eventually, and if that takes a while you can always tag helpers
bold ivy
@wintry steppe but why do you use dot product why cant you just take cross product aka base times height aka c in this case ?
Right now you are taking dot product of c between cross product but why get the projection in normal vector when you could just the length of c
jagged gulch
if you want the volume of a parallelepiped, it's the area of the base times the height of the parallelepiped
the length of c is not the height of the parallelepiped
the height is the projection of c onto the vector that's perpendicular to the base, and that vector is the cross product of a and b
bold ivy
Matthew how in earth is c length not the height of the parapeellied pls explain , im a retard.
jagged gulch
nah it's not exactly obvious
same reason why the length of b isn't the height of the parallelogram at the base
bold ivy
But what makes it different when you do a projection of norm vector?
jagged gulch
I think it's most helpful to think about the 2d case then go up to 3d
bold ivy
Sure
jagged gulch
the area of a parallelogram spanned by the vectors a and b is $||a||*||b||*sin(\theta)$
stoic pythonBOT
mathew_soto78:
jagged gulch
where $\theta$ is the angle between the two vectors
stoic pythonBOT
mathew_soto78:
jagged gulch
the reason why is that b is not the height of that parallelogram
the height of that parallelogram is $||b||sin(\theta)$
stoic pythonBOT
mathew_soto78:
bold ivy
You are using dot product now right?
jagged gulch