Orthonormal basis and orthogonal basis are completely different things right. If I'm asked for an orthogonal basis of a matrix, I don't have to normalize after using the gram Schmidt process right? That'll give me an orthogonal basis
#linear-algebra
restive hound
subtle walrus
not completely different, but yeah
in particular every orthonormal basis is also orthogonal
restive hound
Ok ty
restive hound
How am I supposed to identify a eigenvalues geometric multiplicity?
I know the matrix is diagonalizable
The eigenvalues are 1,1,4,9
Geometric multiplicity is the dimension of null space right?
So then can I find the dimension of
A-λi for each lamda by plugging in each lamda?
So for 1 I'd end up with 2 free variables, for 4,9 I'd have 3 free variables, thus is the geometric multiplicity for 1,4,9
2,3,3 respectively?
But at most the geometric multiplicity is less than or Equal to the algebraic multiplicity right? So I'm kinda confused
(I don't have the original matrix)
bitter glade
yes
coral ferry
i need some help with a proof
if a real symmetric matrix is positive definite
ive read that if v is an eigenvector, then v^t(A)(v) = v^t(lambda) v
why is this true?
lol dumb question oops
i mean tot ask why v^t(A)(v) is greater than 0
gray dust
But at most the geometric multiplicity is less than or Equal to the algebraic multiplicity right?
yes
Geometric multiplicity is the dimension of null space right?
more concisely, geo mult of an eigenvalue lambda is dim(ker(A-lambda*I))
for that example above, idk how you got those numbers and it's gonna be problematic if you don't even have the original matrix to show what you're doing @restive hound
slow scroll
@coral ferry I think thats just the definition of being positive definite
slow scroll
np]
coral ferry
what are the conditions i should work towards
to show that A is diagonalizable?
haha I have no idea
slow scroll
oops wait nevermind
coral ferry
i was thinking about just computing A^2
but then I can't split A from A^2
uhh i think im gonna ned another hint
wait is that a question for me
im dumb, i don't know anything
slow scroll
@wintry steppe for your question, yes. Coincidentally, the trick I was trying to apply for snoopy's problem works for yours. It diagonalizes by the identity P = PIP
coral ferry
why would A(Av) = v
oh im so dumb lol
ok hmm idk if i can mention eigenspace
i haven't learnt that yet
hmm so the eigenspace is the range
because A^2 v always equals v
but i still don't know the conditions A has to meet for it to be diagonalizable
in my case
coral ferry
@wintry steppe i just wanna know like the second last step that im working towards
thats what i mean by what condition implies that A is diagonalizable
cold topaz
Linear Combination is k1v1 + k2v2 = -k3v3 or k1v1 + k2v2 = v3?
cold topaz
if we have two vectors <a b c> and <x y z>, to figure if they are independent, can we do cross product and if it comes out 0i+0j+0k, call it dependent?
coral ferry
@wintry steppe crap i still don't get the proof
like i can show that A = A^-1 from A^2 = I
but idk if thats useful at all
wanton zenith
Not sure if this is the section I should ask, but could someone help me understand the discrete nodal domain theorem and how to determine one from a graph? I can find the eigenvalues and eigenvectors of a graph. I'm just unsure how to assign signs to vertices of a graph.
$\begin{bmatrix} -1\ 0 \ 1 \end{bmatrix}$
For example, if this is an eigenvector for my graph's Laplacian, do the components' signs give the sign to the vertices? Vertex1 would be negative, Vertex2 would be 0, Vertex3 would be positive?
stoic pythonBOT
TheRad1:
warm dove
May somebody please help me with this question ?
gray dust
gray dust
recall the criteria for a subset of a vector space to be a subspace
wintry steppe
yes
gray dust
extra pings is discouraged
wraith owl
sorry
gray dust
you listed the criteria for subspace but you didn't actually check that p+g or sp satisfy the condition for polynomials to be in S
wraith owl
a+a', bx+bx', cx^2+cx^2', dx^3+dx^3'
this shows that im still in the dimension of p3 which shows that if i add two vectors, it will still stay within my dimension. Because the dimension of my answer is 4 which is P3
which still stays in P3
thats show closure during my addition aswell
gray dust
no it's not enough to show the sum is in P3, you need to show the sum is in S
wraith owl
how would i do that
gray dust
start with explicitly saying "let p,g be arbitrary vectors in S", so right off the bat we know p(-1)=0, g(-1)=0
wraith owl
everything else is right?
gray dust
no
wraith owl
so what do i do next
gray dust
show that p+g is in S
wraith owl
what do you mean?
gray dust
first do you know what "p is in S" means?
wraith owl
no
for any polynomian in the set S with parameter -1 will it always give me 0?
I don'tunderstand when professors get g(x) when proving closure under addition
where do they get it
gray dust
no i'm making sure you fully get the beginning before doing anything else
ie knowing how to read $S=\brc{p\in\mathbb P_3:p(-1)=0}$
stoic pythonBOT
RokettoJanpu:
wraith owl
1 sec my friend will join in, we are doing the same assignment
how would i read this for matrix, vector, polynomial etc?
also what does left and right sides mean
gray dust
first off you need to know how to read setbuilder notation
gray dust
i'm not a prof
wraith owl
oh
wintry steppe
hes a mad scientist
wraith owl
lmao
gray dust
that, maybe
wraith owl
so how do i read set builder notation
gray dust
if i have a polynomial p, p(-1) means the evaluation of p at -1, in other words the value of p when i plug in -1
gray dust
not talking about setbuilder yet
gray dust
going back to middle school algebra for a sec, if i give you a polynomial p defined as p(x)=3x^2+4x, what's p(-1)?
wraith owl
-1
gray dust
say i define a set $A=\brc{x\in\bZ:x>4}$, how do you read this?
stoic pythonBOT
RokettoJanpu:
dense spindle
there exists x such that x is an element of all rational numbers where x is defined to be greater than 4
gray dust
wrong
dense spindle
what is it?
gray dust
and don't answer for viper
dense spindle
I'm his friend I wasn't in the discord so he was typing for me
because i had to wait 10 min
wintry steppe
nice pfp
dense spindle
so bold Z is the set of integers so is the rest correct?
gray dust
nope
dense spindle
it's because that's how my professor says it and he's really bad so I'm determined to learn so if it's wrong I accept that
so which way is correct?
gray dust
A is the set of all integers x where x>4
dense spindle
can you give me another example?
gray dust
$B=\brc{z\in\bC:0<|z|\le1}$
stoic pythonBOT
RokettoJanpu:
dense spindle
B is the set of all complex numbers where z is greater than 0 and the absolute value of z less than or equal to 1
gray dust
"z is greater than 0" is wrong
dense spindle
0 is less than z?
stoic pythonBOT
RokettoJanpu:
dense spindle
how would we read 0 < |z|
gray dust
in a similar fashion to reading $|z|\le1$
stoic pythonBOT
RokettoJanpu:
dense spindle
absolute value of z is greater than 0
gray dust
better
dense spindle
thank you
gray dust
did you read everything i said to viper about polynomials?
dense spindle
yes, because I didn't understand the concepts of closure under addition and multiplication. I know to prove a subspace we need to prove those things and that our set is non-empty. But, eventhough it may seem straightforward, I still have a difficult time proving closure under addition, multiplication, and that our set is non-empty
gray dust
how do you read $S=\brc{p\in\mathbb P_3:p(-1)=0}$?
stoic pythonBOT
RokettoJanpu:
dense spindle
S is the set of all polynomials p of third degree polynomails where p of -1 is equal to 0
gray dust
nope that's not what $\mathbb P_3$ means
stoic pythonBOT
RokettoJanpu:
dense spindle
dim(P3) = 4?
gray dust
that's true but not what i asked about
dense spindle
it's third degree polynomials?
gray dust
P_3 is the vector space of polynomials of degree less than or equal to 3
dense spindle
oh yeah <=3 thats what I was missing
gray dust
so what's S actually defined as?
dense spindle
S is a set defined as all polynomials p in the vector space of polynomials degree less than or equal to 3 where p of -1 is equal to 0
gray dust
k yeah, evaluation of p at -1 is 0
dense spindle
okay awesome!!
gray dust
now to actually showing S is a subspace of P_3
dense spindle
We need to satisfy those 3 things to show
and my professor said to use the 0 polynomial because it's easiest to show our set isn't empty
dusky epoch
yeah and showing that 0 ∈ S is downright obvious
if you know what the zero polynomial actually is
dense spindle
yes, but in the case of the problem it may seem outright obvious so is there some type of systematic way to show that 0 is an element of S for any dimension including matrices, vectors, polynomials, etc.?
And what would that way be?
wintry steppe
its in the definition of a subspace
dusky epoch
check the zero vector against the definition of whatever set it is that you're considering
dense spindle
In this case, how would we show it's nonempty
gray dust
it's as you said, you can show S contains the 0 polynomial
dense spindle
so the 0 polynomial exists so p0 of (-1) = 0, so our set is nonempty
dusky epoch
there were better ways of wording that but yes
dense spindle
what is the better way?
dusky epoch
"evaluating the zero polynomial at -1 gives 0, therefore the zero polynomial is in S"
"therefore S is nonempty"
dense spindle
and for closure under addition what is the systematic way of proving that. Sometimes my professor just assumes g(x) is in our set then adds it to f(x). So, I'm not too sure how closure under addition works for any set
dusky epoch
closure under addition means that if you take two arbitrary elements of your set and add them, the sum is also an element of your set.
gray dust
let two arbitrary polynomials p_1, p_2 be in S. show p_1+p_2 fits the condition to be in S
dense spindle
what would that condition be? p(-1) = 0 is the condition? And would taht work for any set that defines their right side differently?
dusky epoch
it's the same as proving any "for all" statement
gray dust
the condition for a polynomial p to be in S is if p(-1)=0
dusky epoch
$p_1 + p_2 \in S$ means that $(p_1+p_2)(-1) = 0$
stoic pythonBOT
Ann:
dusky epoch
is that what you were asking
dense spindle
thank you Ann and RokabeJintarou I understand much better
gray dust
last one. can you tell how to show closure of S under scalar multiplication?
dense spindle
assume s is an element of all real numbers. sp(-1) = 0, s(0) = 0 closed under mult
gray dust
let s be a scalar, let p be in S. sp(-1)=s(0)=0. yes done
dense spindle
Also, sorry for going back, but how would we show p1(x) + p2(x) satisfies our condition
dusky epoch
let me spell that out
p_1 in S therefore p_1(-1) = 0
p_2 in S therefore p_2(-1) = 0
(p_1+p_2)(-1) = p_1(-1)+p_2(-1)
dense spindle
Also, if you don't mind me asking how did you guys learn this so well
dusky epoch
practice, practice, and practice
dense spindle
I can't seem to understand some of these concepts even after reading the textbook partly because my professor isn't the greatest teacher
I do find an interest in this and I want to learn
pallid rampart
by doing copious amount of exercises
dusky epoch
and a little bit of good luck wrt having a good environment for mathematical learning
pallid rampart
You don't do it until you get it right, you do it until you can't get it wrong
gray dust
btw make sure your classmate viper reads this convo over if they haven't already
dense spindle
ok also he's a great friend but he was just typing out what I was saying. He wanted to help me out by finding great tutors like you guys
Also, is it okay if I ask a few more questions?
gray dust
sure
half ice
Honestly in many cases, this stuff retains so well because we use it in useful things. Vector spaces are really, REALLY nice structures. It's handy to quickly know when you're dealing with one
dense spindle
Yes, I want to learn how would we use these for all types of problems
half ice
It might be more useful to findwhy something like
{S ∈ P3, p(1) = 1}
Is not a vector space
dense spindle
I'm not sure where to start on this one
it looks like it deals with coordinate vectors
dusky epoch
B is a basis for R^2, therefore it contains 2 vectors. call them v_1 and v_2. now write the two equations you're given in terms of those.
it looks like it deals with coordinate vectors
you'd be right to say that btw
dense spindle
It feels like I'm going backwards on this one
and my professor never did any examples like this in class
dusky epoch
did your professor never do any exercises involving coordinate vectors at all?
dense spindle
never
dense spindle
and he assigned us this
I did he told me to do examples in the textbook and I did
none of them looked like this
dusky epoch
did he at least tell you what coordinate vectors are
dense spindle
you solve RREF
dusky epoch
dense spindle
right side is the r1, r2, ...rn
dense spindle
he didn't tell us what coordinate vectors were
dusky epoch
well ok then that is a problem
then you should skip this exercise and email your prof
saying that the exercise uses concepts not yet covered
dense spindle
he just showed us how to find coordinate vectors
everyones teaching theirselves in this class
dusky epoch
k well alright
dense spindle
so I have to do the same because he doesn't answer any questions through email
this is my last hope
dusky epoch
so if you want me to add in my two cents
dense spindle
yes please
dusky epoch
let's step back for a moment
and suppose we have a vector space V with a basis
an ordered basis B = {b_1, b_2, ..., b_n}
since B is a basis, by definition every vector in V can be expressed as a linear combination of the vectors in B
uniquely, at that
does that make sense
dense spindle
yes, that makes sense
dusky epoch
yeah so alright
dense spindle
wait
dusky epoch
they're elements of B yes
dense spindle
or are they "everything"
gray dust
are you aware of what basis means?
dense spindle
is our basis a set?
gray dust
rip
dusky epoch
dense spindle
I know LOL
dusky epoch
what do you think a basis is if not a set
half ice
Ahh, reminds me of my linear algebra class. This is how me and the other engineers learned it
gray dust
a basis B for a vector space V is a set of vectors picked out from V such that B is linearly independent and span(B)=V
dusky epoch
yeah so every vector v ∈ V can be expressed as a linear combination of b_1, b_2, ..., b_n
its coordinate vector with respect to B is just a column listing the coefficients of said linear combination
in order of course
half ice
You know how we sometimes write 3D vectors like
3i - 2j + 5k?
In this case, the basis for this vector space is {i,j,k} because all vectors can be written as ai + bj + ck.
half ice
Aight have a good sleep
dense spindle
gn
half ice
A basis is like "the smallest way to express every vector"
half ice
Oh, mb. I'm not very good at time zones
wintry steppe
did i do this right?
dense spindle
okay, if I understand correctly a basis would be the smallest way to represent anything. So, if I have a polynomial (x-1)^2+(x-1)+1 the basis is x^2+x+1. If something is a basis, it is linearly independent and spans a vector space
half ice
A basis for all polynomials in P2 is:
{1, x, x²}
Because all polynomials in P2 can be expressed as a(1) + b(x) + c(x²)
wintry steppe
dense spindle
wintry steppe
did i get this matrix right?
dense spindle
B is a basis of R2
half ice
The very important thing to take from this, is that there's more than one basis on a vector space. We normally use {1,x,x²}. But, you don't have to
dense spindle
so that would mean B is lin ind and spans R2
half ice
I actually don't know the notation. @gray dust
Help! What does [v]_β mean?
dense spindle
that's the basis is equal to the coordinate vectors i think
gray dust
$\m[b]{3\2}_B$ refers to the coordinate vector of $\m[b]{3\2}$ wrt $B$
stoic pythonBOT
RokettoJanpu:
half ice
Oh I see, so that's [3,2] in B's basis
gray dust
yar 
gray dust
half ice
ALRIGHT So then for some two vectors in the basis, we have:
3B1 + 2B2 = 1i + 1j
If I'm understanding the process here
Where i and j are the standard basis on R²
That is, i = [1,0] j = [0,1]
gray dust
let $B$ be an ordered basis of $\bR^2$ where, say,
$$B=\brc{b_1,b_2}$$
then the following equation
$$\m[b]{3\2}_B=\m[b]{c_1\c_2}$$
can be alternatively read as
$$c_1b_1+c_2b_2=\m[b]{3\2}$$
stoic pythonBOT
RokettoJanpu:
let $B$ be an ordered basis of $\bR^2$ where, say,
$$B=\brc{b_1,b_2}$$
then the following equation
$$\m[b]{3\\2}_B=\m[b]{c_1\\c_2}$$
can be alternatively read as
$$c_1b_1+c_2b_2=\m[b]{3\\2}$$
Oh fuq. I've never been amazing at this change of basis stuff
dense spindle
its kind of weird because it asks to find the vectors in the basis
so would this be the first steps?
half ice
Weird is an interesting choice of word
dense spindle
lol
half ice
Basically
b1 + b2 = (3,2)
-b1 + b2 = (1,-4)
This can be rearranged into the matrix equation
(1 1)(b1) = (a)
(-1 1)(b2) (b)
Where
a = (3) b = (1)
(2) (-4)
dense spindle
why is a = 3?
half ice
a = (3,2)
I just tried a lazy way to do vertical vectors, it didn't work out great
dense spindle
b1 and b2 just represent elements of the basis?
half ice
Easy way to solve that is to take the inverse of the matrix
Yeah, they're the two basis vectors
dense spindle
and those basis vectors are equal to their respective coordinate vectors
half ice
Every basis on R² has exactly two basis vectors. So, we say the dimension of R² is two
dense spindle
thank you
gray dust
you might not need to invert a matrix 
half ice
That's very true. You can just add both equations together and quickly get
2b2 = (4,-2)
Blowing your basis wide open
b1 = (1,3)
b2 = (2,-1)
gray dust
very good. painless
half ice
Now, for my own sanity.
3(1,3) + 2(2,-1)
Isn't anything important?
gray dust
wdym
gray dust
where'd 3(1,3) + 2(2,-1) come from?
i mean that'd be (3,2) but in the "point of view" of B, so that'd give 3(1,3) + 2(2,-1) in the POV of R^2's standard basis
long blade
this question doesnt make sense to me
it says the line
but why do they give 2 equations
and why does it look like the equation to a plane
rather than a line
a bit confused
can someone please explain what this represents
gray dust
you got the eqns for 2 planes. their intersection is a line
long blade
hmm, im struggling with this question.
x= 1- z
2(1-z) -y + z = 0
2- 2z -y + z = 0
2-z=y
let z=t
r = (1,2,0) + t(-1,-1,1)
this is what i did
but i dont know what to do after this
this is the equation of the line i got
r = (1,2,0) + t(-1,-1,1)
quasi vale
@long blade Since the plane passes through the point M and the line l, it contains the point (1,2,0) which is a point on line l.
With these two points you can find a vector contained in the plane, cross it with the direction vector of line l, you get a vector that's perpendicular to both, or the normal vector of the plane.
long blade
Or so
Ur saying
M - (1,2,0)
(1,3,-2) - (1,2,0)
(0,1,-2)
Then cross product with (-1,-1,1)
(-1,-2,-1)
So this is the normal vector
-x-2y-z +d=0
Sub (1,3,-2)
-1-6+2 +d=0
D=-5
Hmm
I think d was meant to be -3
In the awnser
Did I do it wrong
@quasi vale ?
quasi vale
I think it is
long blade
Okay
quasi vale
by your working it looks like d=5
quasi vale
so what's the answer and what are u getting
long blade
awnser is
x - 2y -z + 3 = 0
what im getting is a bit off
i'll try working it out again
from the begining
quasi vale
check now if it's okay
quasi vale
cool
quasi vale
Yes
quasi vale
and
quasi vale
the nromal vector is <-1,2,1>
long blade
yeah
quasi vale
won't matter if you make it <1,-2,-1>
long blade
true
true
cool
um do u mind
helping me with b)
what would be find the vector that is orthogoal to L?
quasi vale
post it again
quasi vale
No
long blade
by chance
quasi vale
This is more easier
long blade
wait
a minute
since we know that um
the direction vector of L is
(-1,-1,1)
it would just be a vector that is perpindicular to this?
quasi vale
No..
long blade
lol there goes that theory
quasi vale
What do we need at the very least to find an equation of a plane
long blade
yes
quasi vale
And the normal vector?
quasi vale
what is it
quasi vale
Yes!
quasi vale
👍
hardy stone
could someone possible help me with number 4
gray dust
where are you stuck
novel elbow
what prereq knowledge do i need for linear algebra?
subtle walrus
pretty much nothing
subtle walrus
does that teach you how to add and multiply numbers?
you don't need calc or trig
maybe some basic trig for a few examples
novel elbow
great thanks. wonder why all my state colleges have prereq of calc 2 for linear alg?
subtle walrus
probably because it's a more advanced linalg class
and they want students to be more (mathematically) mature
novel elbow
understood. thanks for the info.
quartz compass
it's good to have already some experience working with linear operators like derivatives and integrals so that you can work with polynomials as an example of a vector space where the vectors aren't the typical sort of pointy arrow in 3D space
gray dust
i'll stab YOU with a pointy arrow
storm python
pliant fox
<@&286206848099549185> just want some clarifications
are all orthogonal transformations that starts in $\mathbb{R}^n$ of form $$T: \mathbb{R}^n \to \mathbb{R}^n$$
stoic pythonBOT
TheJohn:
dusky epoch
,calc 41^2 + 29^2 + 23^2
stoic pythonBOT
Result:
3051
there you have it
dry spear
oh i did 14 instead of 41 lol
i dont know where i messed up on this q
r is just the unit vector of v
a and b are orthonormal vectors to r
[C]S<-B is the column vectors a, b, r
[R]B<-B is the rotation matrix for pi/3
[C]B<-S is the transpose of [C]S<-B
[R] = [C]S<-B [R]B<-B[C]B<-S
dry spear
<@&286206848099549185>
gray dust
let $B=\brc{b_1,b_2}$ be an ordered basis of $\bR^2$, then the following equation
$$\m[b]{3\2}_B=\m[b]{c_1\c_2}$$
can be alternatively read as
$$c_1b_1+c_2b_2=\m[b]{3\2}$$
stoic pythonBOT
RokettoJanpu:
cursive prawn
I want to check my math, how can I calculate the area of a parallelpiped in R^4, given 3 vectors?
I've done it using projections, but I want to double check
We're trying to find the "hypervolume" created by 4 vectors in R^4 (pretending that the determinant doesn't exist)
dense spindle
aren't there infinitely many solutions rokabe
because c1 and c2 represent any constant?
cursive prawn
you are solving for the b1 and b2 in the above example
you have a vector, and its representation in another basis, but you need to find out what that basis is
gray dust
use what i said to rewrite the eqns you were given as a sort of system in b1,b2
dense spindle
gray dust
err no
maybe it wasn't clear but i said let B={b1,b2} be the basis for R^2. B is a set of vectors. b1,b2 are not scalars
gray dust
the B subscript indicates a desire to rewrite (3,2) in the "standard" basis which is {i,j} into a basis from the point of view of B
once you write out b1+b2=(3,2) you're saying the (3,2) is in the standard basis and putting the B subscript is wrong
gray dust
ye the math is pretty painless. you just need to really get to know what coordinate vector means
dense spindle
Thank you man I’m really understanding this after two days
gray dust
@cursive prawn ik you said you're ignoring dets but my thoughts on using that: kinda like how the cross product of a,b in R^3 involves a 3x3 det where the top row is the R^3 standard basis and the 2 rows below are the entries of a,b, and the cross prod produces some c where ||c||=volume of parallelogram spanned by a,b, a 4x4 det where the top row is the R^4 standard basis and the 3 rows below are the entries of your vectors in R^4 a,b,c, and it produces some d where ||d||=volume of parallelepiped spanned by a,b,c
cursive prawn
@gray dust what would the top row be precisely?
I'm unsure what the standard basis in R^4 means
gray dust
do you know what a basis even is? @cursive prawn
dense spindle
B = {b1, b2, b3, b4} where leading 1s are in each row ?
That’s the standard basis I think
cursive prawn
yeah
gray dust
basis means a bit more than that
cursive prawn
but how can the first row of the determinant be a set of vectors
a basis is a spanning set of vectors that is linearly independent in R^n
it is essentially a mapping of vectors
gray dust
k yeah, then you know what the standard basis of an R^n space is?
cursive prawn
the identity, right?
gray dust
er nope, for R^2 it'd be {i,j} and R^3 {i,j,k}
gray dust
for bigger R^n spaces you'll typically use e_1,...,e_n so you don't run out of letters
all i mean is the 1st row is e_1,e_2,e_3,e_4
cursive prawn
yeah
but that gives you an expression, not a scalar
or a vector
depending on how you look at it
gray dust
the det gives a vector d, and note i also said ||d|| is the volume of the parallepiped spanned by a,b,c
cursive prawn
ah yeah
how would you solve this? It seems I've done it incorrectly until this point
We know that the determinant is the volume of an n-dimensional parallelotope
but we haven't actually proven that, so we're trying to find the volume of a 4 dimensional parallelpiped without using the determinant
this is difficult, because the cross product isn't useable
steady fiber
cross product wouldn't exist in 4 dimensions
big sad
one of the biggest sads in math tbh
cursive prawn
the area of the parallelogram can be found using $||A - Proj_B A|| * ||B||$
stoic pythonBOT
Heasummn:
cursive prawn
the area of the parallelpiped is a bit more difficult
we have a vector C, we want to find a vector orthogonal to the parallelogram and project C onto that
or, we can skip that step and know that $Proj_{plane} C = Proj_{B} C + Proj_{orthogonal to B} C$ (I believe)
stoic pythonBOT
Heasummn:
cursive prawn
this isn't a formal proof, we just can't use the determinant
that gives us an answer, finding that projection and multiplying by the area of the parallelogram
but it's not the same as taking the determinant with the standard basis
cursive prawn
yeah if I use the determinant and then reapply this logic, I get the right answer
but I can't get the volume of the parallelpiped
oh
it's not the projection time's the base
it's C - projection
pallid swallow
@raven cargo Just see where your basis vectors go
pallid swallow
yeah the other 4 are clearly false
wait but doesn't $A^2B=A(AB)=A(BA)=(AB)A=(BA)A=BA^2$?
stoic pythonBOT
Element118:
raven cargo
yeah, i missed that. thanks man
bronze gale
This might be really dumb question...
Let A be a nonsquare matrix of size n by m. Let T be a square matrix of size m by m, such at A * T = 0.
Then (A * T) * T^{-1} = 0, but A * (T * T^{-1}) = A. Can someone please tell me where I'm being stupid?
unkempt robin
The first lemma is self explanatory. Substitude A * T = 0.
Second one relies on inverse matrix definition (here, I'm assuming T^{-1} is the inverse of matrix T). If X is an inverse of A, then XA = I and AX = I, for I is the identity matrix. If you do some substitution, you can conclude that TT^{-1} = I. From there, you can probably work out how the second lemma is true.
I assume T is invertible, so what I wrote is wrong if that isn't the case.
slim bridge
Hey <@&286206848099549185>
for this question i need to convert the plane to cartesian and then sub in the x,y,z values (in terms of t) from the line and solve for t right?
then sub t into the line equation and get an answer right?
quasi vale
Try it and see what happens.
slim bridge
idk how to convert to cartesian...
@quasi vale
im trying letting line = plane
and then using gausian elim
to solve for t, lambda and mu
and then i guess i sub t into the line equation and a vector pops out?
quasi vale
uh hopefully, for the cartesian equation of the plane, do the cross product of the two direction vectors you have
that'll be the normal vector
quasi vale
?
slim bridge
why do i wanna findt he normal vector?
quasi vale
so you can find the cartesian equation of the plane
quasi vale
If a plane has a normal vector <a,b,c>, then the cartesian equation of the plane is ax+by+cz = d
where d is some constant
slim bridge
oH
i see i see
righty lettme try that
wait for the cross product
is there a formula without theta
quasi vale
there is a way to compute the cross product yes w/o that theta formula you're talking about
quasi vale
Yeah
The first row will have elements i,j,k
the other two rows will be your given vectors
bronze gale
@unkempt robin I know that... I'm saying that (A * T) * T^{-1} should be equal to A * (T * T^{-1})
quasi vale
Did that come from the cross product?
slim bridge
yea i did Det(IJK,V1,V2)
slim bridge
right the "a"
quasi vale
Idk what u mean by that
unkempt robin
@bronze gale They should. So wouldn't that mean A is the zero matrix.
bronze gale
I don't have a specific example but I don't think A has to be 0
quasi vale
Yes that
dusky epoch
no
Find a standard basis vector for R^3 ...
there are only 3 such vectors, which is quite a long shot from infinity
@cold topaz @wintry steppe
do you not understand what "standard basis vector" refers to
the standard basis.
yes it does
and the problem asks to find one among these that together with the given v1 and v2 makes a basis for R^3
dense spindle
Hey guys you have time to check over some problems I have? Sorry to butt into the conversation
Just 2 problems of proof work
dense spindle
dusky epoch
uhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
hol up
tf is up with 1b
{1, x+1, (x+1)^2} is not a basis of S by any means
1 isn't even in S!!!!
dense spindle
Lollll wait how would I disprove it’s not a basis
dusky epoch
S = {p ∈ P_3 | p(-1) = 0}
dense spindle
Because it’s in the form ax^2+bx +c
dusky epoch
imponed that doesn't change the fact that 1 is not in S
S is {p in P_3 | p(-1) = 0}
the polynomial 1
just
ISN'T
IN THIS SET
AT ALL
the exercise is just fucked up
you're asked to prove a false statement
dense spindle
I think it’s talking about some arbitrary set S not specifically the one from #1
Because I thought the same thing at first
dense spindle
It wouldn’t make sense because the dimensions aren’t even the same
dusky epoch
this problem set is fucked up, period
dense spindle
Because I’ve read that proving basis some set S = {v1...vn } is a common notation I think
Yeah man so what do you think I should do?
Also what do you think of #2?
dusky epoch
you should shove that shit back in your prof's face and tell him to check the wording in his problems
okay, #2. lemme take a look at it.
dense spindle
Bro trust me I wish I can do that he’s so bad
I like linear but it’s a bitch learning it with this prof
dusky epoch
i'd appreciate not being called bro, thank you
If S is linearly independent, that means its solution is trivial
uh oh stinky
dense spindle
Sorry
dusky epoch
ok so
you wrote a bunch of words there
a good bunch of words indeed
the only downside is that you've said precisely nothing
dense spindle
If it’s linearly independent it’s solution is trivial
dusky epoch
you kinda just
sets don't have solutions
you misquoted the definition of linear independence
dense spindle
What is it then?
dusky epoch
a set S = {v_1, v_2, ..., v_n} is said to be linearly independent when the EQUATION r_1v_1 + r_2v_2 + ... + r_nv_n = 0, where the r_i are scalars, has ONLY the trivial solution r_1 = r_2 = ... = r_n = 0.
what i'm saying is
the thing you wrote? it's all complete bullshit and needs to be redone from scratch.
i can't put it any other way.
dense spindle
Okay sorry but I wasn’t taught the correct way
dusky epoch
...
dense spindle
That wasn’t necessary to be outright completely rude
dusky epoch
what
how was that rude
there is no putting it gently when someone produces work that makes no sense
"not correct" is too vague.
dense spindle
Okay there’s better ways of phrasing that wasn’t nice in just saying
Not correct and tell me why I’m not correct
I’m here to learn and not argue
dense spindle
I’m not going to say that your a good person because your nice enough to help
I just would respect if you didn’t curse at my work and call it bs
dusky epoch
yeah like... i mean, i already told you exactly where you went wrong
you attempted to do something with a fucked-up version of the definition of linear independence
nothing good can ever come of that
you did say, correctly, that you need to show S is linearly independent and spans V
the first bit you're already given
the second is the "interesting" part so to speak
dense spindle
Ok so the definition of linear independence would be that the eq r1v1...rnvn = 0 is trivial?
dusky epoch
no
you're trying to shorten the things i say but you keep omitting too many words for the remainder to make any sense
dense spindle
Yes so showing it spans would mean you can take a linear combination of the vectors in the set?
dusky epoch
no
you want to show that any vector in V can be written as a linear combination of S
dense spindle
What would be the first step to showing that?
dusky epoch
ok give me some time to actually write out the proof in detail
dense spindle
Jinkies are you in college?
I’m having difficulty grasping these conecpts my professor is way past these concepts but for some reason I have trouble grasping the information
slim bridge
can anyone explain why for a cartesian form of a plane, why is a,b,c also coincidentally the normal vector?
dense spindle
Linear algebra the gateway to mathematics
Very
And I’m an undergrad second year coming from calculus classes
pallid swallow
@slim bridge entire question with context, in #❓how-to-get-help
dusky epoch
i can't think of a proof that wouldn't be completely nuking it
pallid swallow
oh wait, I think I see what you are asking about @slim bridge
dusky epoch
if you're allowed to use the fact that any non-spanning LI set can be extended to a basis, then it becomes trivial. but i can't seem to recall how to prove that from scratch without nuking it.
dense spindle
I’m totally new to proving and the professor confuses me
pallid swallow
For a plane equation (a, b, c) dot n = k...
dusky epoch
okay @pallid swallow @slim bridge can y'all move to a diff channel please
dense spindle
I find it difficult to prove this I’m not sure where to start at all
slim bridge
@dusky epoch kk
dense spindle
Other than knowing a basis is Lin ind and needs to span a vector space
slim bridge
i pmed u @pallid swallow
dusky epoch
there's got to be some theorems you're allowed to use
proving this from scratch can be done but it's nasty
like, if you can do this:
any non-spanning LI set can be extended to a basis
then you can say something like
"suppose for the sake of contradiction that S doesn't span V. extend S to a basis S', which will thus necessarily have more than n vectors in it. this contradicts dim(V) = n."
dense spindle
Did you all see 1a? What do you think?
I said since the 0 polynomials exists it’s non empty
dusky epoch
"since the 0 polynomial is in S, we know that S is nonempty"
dense spindle
So if there is anything you recommend I’m happy to hear you out
dusky epoch
try to avoid using the word "it" at all early on
dense spindle
Ok
dusky epoch
i've seen way, way, WAY too many people use the word "it" way too often
and as a result confusing themselves as to what they're even saying
dense spindle
Thank you for the help Ann and Jinkies I learned something new today
hollow finch
Hi can someone give me an example of an application for weighted euclidean inner products?
sonic osprey
Can you clarify what you mean
hollow finch
So an inner product like
$$\langle (u_1,u_2),(v_1,v_2)\rangle = 3u_1v_1+7u_2v_2$$
stoic pythonBOT
nix:
hollow finch
Is there any application where that is useful?
Ive never found any uses of it and idk what to tell the linear algebra students I tutor when they ask me about them
sonic osprey
yeah I mean if you take your basis vectors to be like 3i and 7j, this is what your inner product translates to
thick gull
iirc, there's some machine learning algorithms that take advantage of weighted euclidean inner products (but it's been a year since I studied those, so I might be wrong)
hollow finch
@thick gull It would make sense if certain parts of a task are more important than others. Maybe for measuring error.
It definitely sounds like a field which could make good use of it.
thick gull
that sounds right.
viscid kernel
How solve a system of equations with matrices if the determinant is 0 and/or if you are dealing with a non square matrix
dusky epoch
well you'd still do gaussian elimination as best you could. and then either you end up at your system being inconsistent or you can make the non-pivot variables free and express the pivot ones in terms of those to get your general solution
viscid kernel
Wdym with non pivot and pivot ?
viscid kernel
Nope
dusky epoch
uh
viscid kernel
I know what it is tho
dusky epoch
do you know what RREF (reduced row-echelon form) is
viscid kernel
Yup
dusky epoch
yeah so
the pivot columns are the ones where there's a 1 in one position and 0s elsewhere
everything else is non-pivot
viscid kernel
Ah ok
slim bridge
can anyone explain what this question is asking for?
is it saying im looking for v so that vB = [-4, -1]?
but like thats not possible right? cause B is a 4x2 matrix
dusky epoch
is it saying im looking for v so that vB = [-4, -1]?
no
slim bridge
row 1 of the matrix product AB
@wintry steppe but u can multiply these matrices no?
they are different rows/columns?
hm any idea?
gray dust
the product AB is very much defined
slim bridge
is the product a 5x4 matrix?
gray dust
5rows 4cols
slim bridge
ya
wait so
how do i do these types of matrix multiplication?
ive never done ones with different sizes lol
dusky epoch
exactly the same as you normally would...
gray dust
the $(i,j)$ entry of the product of two matrices $A$ and $B$ is the dot product of the $i$th row of $A$ and the $j$th column of $B$
stoic pythonBOT
RokettoJanpu:
slim bridge
ah ok
so for this question i can ignore the other rows in matrix A
and just do matrix multiplication for first row of A and the each column of B
spiral sonnet
So I'm a bit confused on how to find w here. How would I go about finding w?
slim bridge
@wintry steppe awesome thanks man
gray dust
hint, play around with $T(v_1)=T(v_2)$ to get an idea of what a possible choice for $w$ can be, also keeping in mind the properties of linear maps
stoic pythonBOT
RokettoJanpu:
spiral sonnet
I was thinking about T(v_1 + w) = T(v_2 + w). I don't know what T(v_1) computes to so I'm not exactly sure how to solve for w
gray dust
you can't know what T(v_1) is. the q says "for ANY linear map T on R^3 to R^3"
try rearranging T(v_1)=T(v_2) into the form T(...)=0
gray dust
from what i said, whatcha thinking now?
spiral sonnet
T(v_1 - v_2) = 0. When I saw that I thought oh that's pretty similar my idea v_1 + w = v_2.
gray dust
nice there you go
slim bridge
@wintry steppe hey man i just got the answer back
um the answer just looks like row 1 of matrix A...?
am i just stupid? Im so confused
dusky epoch
that's because it IS the first row of A
slim bridge
but didnt the question ask for the first row of Matrix AB?
shadow drift
@slim bridge
So basically the question is asking:
Find row 1 of A*B, then
Find the vB, which equals row 1 of A*B
Since you already have the answer anyway:
Row 1 of A*B is (-8, 1, -9, 36), u can check urself
When finding v, we know its a row vector meaning its dimensions must be: 1xN
B's dimensions are 2x4, so for vB to be valid, v must be 1x2 (Since vB will be (1x2)(2x4))
So we know v must be 1x2.
Which means you have to find something like this:
stoic pythonBOT
Rebase:
shadow drift
You can use matrix multiplication and solve for (a, b).
So:
2a + 0b = -8
2a = -8
a = -8/2 = -4
a = -4
And with b:
0a - 1b = 1
-1b = 1
b = -1
So you get (-4, -1)
In the bigger picture this shows that to make vB row n of matrix AB,
v will be row n of A.
@slim bridge
jagged spindle
So i need to learn this but I don't even know what that means in my language so no idea where to look, tried to translate it as good as I could.
- Orthonormal basis is set of vectors that are all orthogonal to each other right?
2)So should I just find a 3rd vector (e3) that will be orthogonal to both e1 and e2?
gray dust
an orthonormal basis for R^3 is a set, call it S, of vectors picked out from R^3 that satisfies the following:
- first off S needs to be a basis, ie S is linearly independent and span(S)=R^3
- the vectors in S are orthogonal to each other, ie the inner product of one vector with another is 0
- each vector is a unit vector, ie has a norm of 1
to complete the basis given to you, you'd need to find a vector e_3 that is not only orthogonal to both e_1 & e_2 but is also a unit vector
quaint sun
The third point of having a norm of 1 means that the vectors or orthonormal assuming they are orthogonal already
jagged spindle
sure, thanks
gray dust
no prob, good luck
placid oracle
is the some of two eigenvectors corresponding to the same eigenvalue lambda of matrix A always an eigenvector? because ik if theyre different values then it doesnt always sum to an eigenfector
gray dust
you can show this yourself by going through the defn of eigenvalue/vector
placid oracle
im pretty sure it is
gray dust
proof by confidence, that's a new one
placid oracle
you have got to be a very cool person
quartz compass
placid oracle
thats why i said it /:
gray dust
by rank and by merit
quartz compass
it's not too bad to write out
u and v are two eigenvectors
A is the matrix
is w=u+v an eigenvector?
show that it satisfies Aw = lambda w or not by walking through it
round palm
Hey guys, out of curiousity, if a matrix has unique eigenvalues, does that make the eigenvectors unique?
quartz compass
you should write it out and try to figure it out yourself first
round palm
kind of didnt know where to start tbh
wintry steppe
how do u prove stuff like this?
like how would u start
not asking for the solution
thanks in advance!!
fossil wagon
I don't understand how is projection of a vector onto another related to their linear combination 
wintry steppe
this is not a vector
fossil wagon
oh wait I wasn't referring to your question, sorry for the misunderstanding
wintry steppe
ohbsorry!
viscid kernel
What is an “ordered” basis ?
tardy vault
Yo am I having a stroke i don't know how to approach this. Is it not just the eigenvector?
steady fiber
so that norm is one where you take the value of maximum magnitude
so ||A||_infty would be 8
since that's the magnitude of the max value of A
if one eigenvalue is 8
then yes, x could be just the eigenvector for the eigenvalue of 8
but if there is no such eigenvalue
you have to find a vector so that the maximum value of an entry in Ax is 8 times the maximum value of an entry of x
could probably arrive there with a tad bit of trial and error
slim bridge
@shadow drift hey thanks for the explaination :D
from what i read it seems in a simple explanation the question is basically asking.
Find vB, so that vB = AB (so v = A lol)
with the 'trick' where u have to work with first rows of vectors.
I get basically everything you wrote out its a really clear explanation :D
But Im just wondering how you figured out the dimensions of the vector v
is it simply the fact that for matrix multiplication you take the rows of the first matrix, and the columns of the 2nd matrix?
and since the rows of the 'first row' of AB, is just 1. and rows of B = 2, therefore its a 1x2 matrix?
steady fiber
that is not linear algebra
oblique spire
Can someone help me with this pls?
strong bison
What would $\grad v^{T}(Ax - b)$ equal, with respect to the x vector?
steady fiber
is v a column vector
if it is, then yes
it would
you can expand it out and try the gradient definition directly
strong bison
turns out it was a row vector and thus fucking up my code (e.g. I needed A^(t) v)
thank you for the help!!
shadow drift
is it simply the fact that for matrix multiplication you take the rows of the first matrix, and the columns of the 2nd matrix?
@slim bridge Yep, so its basically the rule of of Matrix Multiplication where the first matrices length of columns must be equal to the rows of the other matrix.
Basically when you write the dimension of two matrices down.
e.g A = 2x5, B = 5x3
For A*B to be valid, you can look at A and B's dimensions:
2x5 and 5x3
and check if the two middle numbers (in this case 5) are same, otherwise it would be invalid/impossible to multiply.
And the result A*B = 2x3 (the two outer numbers)
long blade
hello
i have just started learning vector spaces
and am not that sure
this question was my first question in my hw
and i have no clue
on what how to even begin
dusky epoch
do you have the axioms of a vector space in front of you?
dusky epoch
check 'em all one by one
long blade
but
dusky epoch
if the set you're checking fails even one axiom, write it off as not a vector space and move on
if your set meets all the axioms, conclude that it is a vector space.
you get a bunch for free since you're told to use the "usual" operations, i.e. the most obvious operations of addition and scaling
dusky epoch
what are you confused about
long blade
about the axioms
dusky epoch
which ones
long blade
like
hmm
a3
for example
u + v = v+ u
isnt this obvious
a1 is u+ v is an element of V
dusky epoch
well
long blade
what does this even mean
dusky epoch
commutativity of addition is obvious in your case since you're using the usual addition operations rather than something wacky
same with associativity
long blade
can u go through with me the axioms for b)
the polynomial
one
and checking if it fails or not
dusky epoch
ok
can i have your list of axioms
so that i can adhere to your book's wording
to make it easier for you
dusky epoch
ok
great
so
here our V is the set of all real polynomials with positive coefficients
for example, x^2 + 7x + 232 is an element of V, while 10x^8 - x^5 - 22x^2 + 1 is not
this isn't about any axioms. i just wanna make sure you understand what set we're working with
does this make sense to you so far
long blade
yeah
dusky epoch
if your polynomial has at least one negative coefficient then it's not an element of V
dusky epoch
yeah ok so
axiom 1:
the sum of any two elements of V is itself a member of V
for our case, the sum of two polynomials with positive coefficients is itself a polynomial with positive coefficients.
is this true?
long blade
yes
dusky epoch
brilliant
axioms 2 and 3 should be obvious because they follow directly from addition on real numbers having this very same property
long blade
yeah
long blade
it fails
dusky epoch
yup
dusky epoch
yeah so no need to continue further
dusky epoch
it ain't a vector space
dusky epoch
for the very same reason axiom 5 fails, axiom 6 fails too, since you can pick -1 (or any negative number really) as your k