and the true solution to the problem seems to be beta_hat, so that y = X beta_hat
#linear-algebra
lavish jewel
so when beta = beta_hat, || y - X beta || = 0
though in general the argmin is not unique here
i guess you'll see that later
this whole problem is formulated like poop :x
wintry steppe
so when beta = beta_hat, |\| y - X beta || = 0
here tho wouldn't it be y-xBeta_hat?
or would it be beta
lavish jewel
doesn't matter because beta = beta_hat
lavish jewel
but yes, throughout the problem, one of beta or beta_hat is the true solution and the other is a variable
and they are not consistent with which is which
wintry steppe
yea that gave me some troubles with expanding :/
i guess a follow up question would be, do u know any good resources to help understand the argument relating the normal eqns for a given X and y training set to minimizing the associated RSS
lavish jewel
yep
that one doesn't have all the explanations though
it assumes you already know why those things are true
wintry steppe
oh hmm
yea im just learning this for the first time so i need help understanding those concepts
lavish jewel
oh nvm i mixed it up with another
boyd shows up in a couple of these and some are higher level than others
this one is fine
the problem of minimizing the 2-norm can easily be shown to be convex
and strictly so depending on the rank of the matrix involved
wintry steppe
do u know which chapter / section to look at? in 1.3 it just has this
lavish jewel
you'll probably have to read the whole thing from the start
and if that doesn't help, you'll need to read something else first
try this
i wrote these for undergrad engineering students
wintry steppe
oh wow awesome ty!
void meteor
Can anyone briefly explain the idea of column pivoting in QR algorithms? Do I reorganize a matrix A first so that the columns are ordered by largest norms in descending order, and then orthonormalize it with an algorithm?
Or does each iteration need the A matrix reorganized by norm size before going through the orthonormalizing algorithm
wintry steppe
for this ik its a hyperplane but is the normal bector 7 -3 1 2 7 a row or column vector?
also is the 7 as the constant on the rhs included in the vector?
1)column
2)no, think about the dimensions, it's 4 and if you add the seven it becomes 5.
ah ok
cool tysm
wintry steppe
when it asks for the normal vector and the equation of the hyperplane, one that ive already gotten into X^TB = c form, what does that mean?
so for example this is my problem and this is my work
and like idk what it means by normal vector
btw @woven zephyr sorry to ping u if this isnt allowed but this is the same one from yesterday that we discussed
woven zephyr
sorry, i'm a bit busy today - someone else can answer, or you can put it up in a help channel
wintry steppe
no worries thank u!
lavish jewel
xT b = c is the equation of a hyperplane, and b is the normal vector
you should google normal vector and hyperplane
topaz sierra
dim lotus
does d (x, y ) = | min(x − y) | satisfy the triangle inequality?
if so, how can I show it?
quartz compass
seems like your question is missing context, what set is this defined on, or what is the min of?
dim lotus
seems like your question is missing context, what set is this defined on, or wha...
x and y are any real-valued vectors and min is a function that takes a real-valued vector and return the smallest element in that vector
wintry steppe
you should google normal vector and hyperplane
thanks for the explanation, i appreciate it!
additionally, the way this is a euclidean ball's formula, what would it be for a sphere?
halcyon spindle
https://cdn.discordapp.com/attachments/495064710278938662/938623653216653372/Scr...
Replace <= with =.
rugged plaza
If I have a matrix A(4x4) and the system AX = 0, how can I find a base after finding the solution of the system?
zinc timber
all the LI solutions will for a basis of your Null(A)
rugged plaza
Thank you 🙂
spiral osprey
ayo if I post a picture can someone tell me if I messed up or not
I assume x5 is free and I just solve
My answers end up being basically all free
X1=t
X2=2.5t
X3=2t
X4=t
X5= free t -> t is all real numbers
K I did it right
disregardo
stable kindle
is there an easy way to do this
i can laplace expand but god damn i have better things to do with my life
stable kindle
it still seems long after row operations
proper cradle
hope it will help
proper cradle
just make max entries zero in one column using row operation
stable kindle
just make max entries zero in one column using row operation
?
oh wait yeah i can do better
proper cradle
like R3-->R3-R2, R4-->R4-R3
proper cradle
then change R3<-->R1 then det(A) will be -det(A)
spiral osprey
Can someone explain to me how X6 lowest value is not 0?
dusky epoch
$x_4 = x_6 - 70$ and presumably $x_4$ must be nonnegative?
stoic pythonBOT
Ann
dusky epoch
@spiral osprey
deep oasis
can someone explain this part here to me? I don't understand how they are equal since AB is not equal to BA for matrices?
tranquil steeple
The matrices in a trace of a product can be switched without changing the result:
deep oasis
thanks a lot!
analog apex
restive raft
That doesn’t look like linear algebra mate, maybe #precalculus
wintry steppe
hello all very cool maths people
I'm making a package made for lin alg on python
what operations and functions do you think people would benefit from
lavish jewel
fast multiplication of multilevel symmetric/hermitian toeplitz matrices
wintry steppe
I'm just a beginner to matrices and stuff
@lavish jewel explain what that is
I'm 3rd week of my 12 week course
I'm interested
lavish jewel
ah then don't worry about it 😛
are you making this as a requirement for your class/to learn, or to actually make something useful
cuz if it's the latter, just don't
wintry steppe
just something to put on pypi packages for my own fun
going to making the vector and matrix operations in python
it's mostly to learn
I know how to make a function which tells you if the function is a linear transformation but I'm not good enough at numpy to do that 🗿
dusky epoch
i can see at least one hole
your lemma 1 claims $\int_0^{\pi} \cos(nt) \cos(kt) \dd{t} = 0$ for all natural $n, k$ but this is not true as stated --- did you mean to also require $n \neq k$?
stoic pythonBOT
Ann
wintry steppe
Yes sorry, it is meant for k = 1,2,...,n-1
dusky epoch
very well...
the sentence immediately after the lemma also reads weird and/or doesnt make much sense grammatically
are you trying to show that {y_n} is the result of gram-schmidting your {x_n}?
if so then you would do good to say "we show that applying the gram-schmidt process to {x_n} yields {y_n}" or something to that effect
nocturne jewel
also subset not \in after the lemma
valid cypress
I gonna start linear algebra today
wintry steppe
if so then you would do good to say "we show that applying the gram-schmidt proc...
Thanks
dusky epoch
plus you will also need to assume y_k is what it needs to be for 1 ≤ k ≤ n
wintry steppe
also subset not \in after the lemma
good point I will change it
nocturne jewel
Also personally as a reader, I feel like a lot is omitted
wintry steppe
plus you will also need to assume y_k is what it needs to be for **1 ≤ k ≤ n**
So I should switch to strong induction?
wintry steppe
Also personally as a reader, I feel like a lot is omitted
Like what?
dusky epoch
yes, strong induction is more appropriate here
zinc timber
(proof by "I already know it's true")
dusky epoch
gram-schmidt by its nature is a strong-inductive process, if i may be so informal in my descriptions
dusky epoch
if so then you would do good to say "we show that applying the gram-schmidt proc...
i think this is one of the things mosh was referring to
nocturne jewel
why does your normalization of y_0 have cos(t) in it?
wintry steppe
I kind of forgot strong induction since I haven't used it in a long time... should I do something like "assume that y_n is true for k in {1,...,n} for some n in N"?
nocturne jewel
when y_0 is the constant vector
wintry steppe
That's a typo
nocturne jewel
You also haven't shown any vectors are normalized that I can see
faint dune
we have determined the relative error |dx|/|x| ~ 10^(s-k), I cant explain what they mean with losing s decimals.
dusky epoch
a relative error on the order of 10^-k means you have k decimal places of precision
faint dune
Ah, they mean then losing s decimals depending on the norm you use.
but the overall relative error is 10^(s-k)
tired fossil
hey guys does anyone know how to approach this
lavish jewel
use the definition of linear independence and linear dependence
tired fossil
So x cannot be written as a linear combination of u,v
lavish jewel
you can probably start with {u,v,x} and use a contradiction working backwards
tired fossil
So u,v, x is dependent then x is in the span {u,v}
lavish jewel
mhm
uvx dependent, this means au + bv + cx = 0 -> rearrange into x being in the span of u and v, contradiction
maybe?
zinc timber
wym maybe
lavish jewel
it means i'm eating and could be making mistakes 😌
the rest of the time i'm not eating, but the mistakes could still be there
dusky epoch
some details will need to be filled in
{u,v,x} dependent => exist a, b, c not all zero s.t. au + bv + cx = 0 => given the independence of {u,v}, which of the coefficients can be zero and which can't?
dusky epoch
"c wouldn't"?
tired fossil
so when you rearrange, it pasically shows the independence of {u,v}
yes, i mean a, b=0
c/=0
dusky epoch
that... sounds a little off
i would have rather said c cannot be zero as that would force a and b to be zero as well by independence of {u,v}, hence the equation is equivalent to x = (-a/c)u + (-b/c)v which contradicts x ∉ span{u,v}
indigo stirrup
It would be helpful if I can get help on this problem. "Let vectors u and v be in R^n, vector b be in R^m, and A be an m x n matrix. If there exist scalars c and d such that c(A times vector u) + d(A time vector v) = vector b, then vector b is in the span of the columns of A, but is not necessarily in Span(u, v)." I have to answer either true or false, but I just don't know how to approach it. I first thought that any vector that is in Span(u,v) must be in R^n. If n and m are not equal, b is not in R^n, and therefore, not in the span of {u,v}. Would the above statement be true then? Thanks
lavish jewel
seems true
you can rearrange this as A(cu + dv) = b, then let x = cu + dv so that Ax = b, which means b is a linear combination of the columns of A as desired. then, as you said, since m \neq n in general, b is in general not in the span of {u,v}
vague berry
do adjacency matrixes double the entrance value for loops? does it happen in directed and non directed graphs?
I cant understand adjacency matrixes potencies...any good video tutorial?? I cant find nothing.
what is the translation for adjacency matrixes potencies?
burnt hearth
"Determine the value(s) of h such that the
matrix is the augmented matrix of a consistent linear system"
answer is "all h" but why??
nocturne jewel
cause you never have 0 0 * for non-zero * as a row in your row reduced matrix
a consistent system is one with solutions, doesn't matter how many solutions (ie 1 or inf many)
@burnt hearth
burnt hearth
so in my own words since the last row does not contain a pivot column, i can have any h?
(last row in the rref matrix)
burnt hearth
ok i got it now thanks!
stoic pythonBOT
lewis
Let $V$ be a finite-dimensional vector space over $\mathbb{K}$. Show that (U^\perp)^\top \subset U$ where we define for $M \subset C$ that $M^\perp := \{f \in V^* \mid f(m) \text{ for all } m \in M\}$ and for $S \subset V^*$ define $S^\top := \{ v \in V \mid s(v) = 0 \text{ for all } s \in S\}$. Note that $M^\perp$ is a subspace of $V^*$ and $M^\top$ is a subspace of $V$.
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<inserted text>
$
l.55 ...tor space over $\mathbb{K}$. Show that (U^
\perp)^\top \subset U$ whe...
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.
Preview: Tightpage -1310720 -1310720 1310720 1310720
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stoic pythonBOT
lewis
viscid lagoon
f(m) = 0*
halcyon spindle
Hmm is my approach to this question ok. I have for an element in $f \in P_3$ we have $f(x) = a_3x^3 + a_2x^2 + a_1x + a_0$ where $a_j \in F$. Writing out the linear combination of that system we have $(x^3+2x)\lambda_1 + (x^2+x+1)\lambda_2 + (x^3+5)\lambda_3 = \(\lambda_1 + \lambda_2)x_3 + \lambda_3x^2 + (2\lambda_1 + \lambda_2)x + (\lambda_2 + 5\lambda_3)$. \So I thought I just needed to analyze the coefficient matrix of this following system \ $\lambda_1 + \lambda_2 = a_3 \ \lambda_3 = a_2 \ 2\lambda_1 + \lambda_2 = a_1 \ \lambda_2 + 5\lambda_3 = a_0$.
stoic pythonBOT
Plegasus
desert palm
My bad
quartz compass
Hmm is my approach to this question ok. I have for an element in $f \in P_3$ we ...
no I think that's too much work
P_3 is 4 dimensional and has a basis 1,x,x^2,x^3
so there's no chance 3 vectors can span it
halcyon spindle
You right, I literally just went through this theorem about it in this section. Thanks.
halcyon spindle
no I think that's too much work
In this case where we trying to determine if 4 polynomials in P_3 would generate P_3 would the approach I gave work?
grim bridge
the same argument works in this case but you have to be careful
basically if you prove the four vectors span V, they are a basis since the space is 4-dimensional
hardy inlet
so i basically just gotta add the dimention argument?
nocturne jewel
dim(V) = 4 since the basis has 4 vectors, so it is sufficient to show any other set of 4 vectors from V spans or are independent iirc
hardy inlet
and thanks for the help on #1 i'll try and type up a solution
can someone please check this explanation to a different problem
jade karma
If my orthogonal complement has a basis of only B={1,1,-1} , and I know to use gram schmits or w/o does that mean that the ortoghonal basis is the same as B ?
hardy inlet
What is the easiest way to prove these are bases?
wintry steppe
First you have to realize that u only have one vector, and it’ll span only 1 dim, u need to span 3 sum and in order to do that u got to know what are the two vector that’ll complete to basis
So u need 2 more linearly independent vectors that are also independent with b1
Always look at the standard basis
And then you can see that you have those
Also
Another option
0
0
1
hardy inlet
I don't have 1 vector
I have 3 sets of 3 vectors
and the span of each set is R3
i just need to show
someone, is this sufficient evidence that its a basis
nocturne jewel
someone, is this sufficient evidence that its a basis
no
cause you've told me a vector depends on its entries
hardy inlet
?
nocturne jewel
your statement doesn't say anything about a basis
hardy inlet
I've shown every vector in R3 can be written as a combination of v1 v2 and v3
2.42 says
Suppose $V$ is a fdvs. Then every spanning list of vectors in $V$ with length $\operatorname{dim}V$ is a basic of $V$
nocturne jewel
you haven't though
stoic pythonBOT
MattDog_222
fringe fjord
But your v1 and v2 and v3 depend on the vector you want to express ....
nocturne jewel
you've written the canonical basis fancily
what you want is to find a,b c in R such that $[x,y,z]=av_1+bv_2+cv_3$
stoic pythonBOT
Mosh
hardy inlet
ok so $(x,y,z) = \frac{1}{2}v_1 + \frac{3}{2} v_2 + \frac{1}{2} v_3$
stoic pythonBOT
MattDog_222
nocturne jewel
ok, but what are v1,v2,v3?
hardy inlet
nocturne jewel
ok, so you want to show they span?
hardy inlet
What is the easiest way to prove these are bases?
did u see the op
nocturne jewel
No, I saw your absurd statement
nocturne jewel
ok, so you want to show they span?
.
hardy inlet
well i want to show they're a basis but i should need to show they span to show they're a basis
nocturne jewel
$[x,y,z]=a[-1,1,2]+b[1,0,0],c[0,1,0]$
stoic pythonBOT
Mosh
nocturne jewel
so $x=-a+b \ y=a+c \ z=2a$
stoic pythonBOT
Mosh
nocturne jewel
thus you can find a b and c as functions of x, y and z
hardy inlet
$a = z/2 \ c = y - z/2 \ b = x + z/2$
nocturne jewel
yes
stoic pythonBOT
MattDog_222
nocturne jewel
and for any (x,y,z) in R^3, those functions are well defined (they're polynomials)
so scalars exist uniquely for all vectors in R^3
QED
hardy inlet
so like? (v3 should be 0,1,0 not what in the picture)
wintry steppe
quick question
so for this problem
thats the solution
but does the result depend on the assumption that the elements of y_hat sum to one?
i think it does
thin wing
Yes, otherwise the largest y_i might not satisfy min_k||t_k-ŷ||. For example, say ŷ=(1,2), then min_k||t_k-ŷ|| would say 1 is the largest element of ŷ, when we know 2 is.
wintry steppe
🙂
burnt hearth
guys i dont understand
if I got 0=0 it means its a free variable right? so why does this seem contradicting??
for example, if i got a free variable then the top picture says i cant assume the linear system is consistent?
but the bottom says i can, like i dont understand??
thin wing
Well [ 0 0 ... 0 | b] with b nonzero is a paradox, agreed?
half ice
"it" means "its" a free variable
What is "it" and "it" here?
The top picture is trying to say that "has a free variable" and "consistent" aren't directly related
burnt hearth
for Matrix A i got the last row 0=0 after row reducing
and i assumed "0=0" meant the linear system was consistent
nocturne jewel
You can have as many free variables or pivot variables as you want. If you have a row of the form 0 0 0 ... 0 * for nonzero *, you instantly have an inconsistent system
half ice
You don't need to assume, the blue box says exactly when a matrix is consistent
No row [0 0 0 ... b] = consistent
burnt hearth
No row [0 0 0 ... b] = consistent
exactly i got no row of [0 0 0 ...b] which means consistent
right?/
let me upload my work
to show you guys
light arrow
Can someone explain to me how to express fractions as decimals
burnt hearth
wait so am i right
nocturne jewel
you can see that my last row is a free variable
again, that doesn't mean the system is consistent
however if you kept getting to RREF you'd see no row causes problems
hence it's consistent
burnt hearth
so when its in REF form i can't assume the linear system is consistent, but when its RREF i can?
nocturne jewel
if it's in RREF you 100% know if you have a problem row
burnt hearth
so im right
thin wing
Your justification is inaccurate as I have said.
That's why I'm concerned if he knows the significance of a "consistent" linear system. In other words why an inconsistent linear system forms a paradox.
thin wing
That's why I'm concerned if he knows the significance of a "consistent" linear s...
Let's say you have the case [ 0 0 0 | b ], then that means there exists x,y,z such that 0x+0y+0z=b. In this case, if the system was consistent, then b would have to be 0, since 0x+0y+0z=0 for all x,y,z. If the system was inconsistent then it'd be saying there exists an x,y,z such that 0x+0y+0z=b≠0, which is a paradox.
dull latch
is R^2 2d and R^3 3d and so on?
thin wing
is R^2 2d and R^3 3d and so on?
Yeah.
nocturne jewel
R^2 is 2D space (the xy plane) and R^3 is 3D space, yes
R^n is more generally called Coordinate or Euclidean space
dull latch
👍
burnt hearth
Let's say you have the case [ 0 0 0 | b ], then that means there exists x,y,z su...
i see
thin wing
Let's say you have the case [ 0 0 0 | b ], then that means there exists x,y,z su...
Note that even if we have one row where we think it's consistent, there could still be another row that isn't. This is why a free variable doesn't automatically imply consistency, there could still be another row that isn't consistent. The best way to tell a system is consistent is by getting RREF and verifying that all the rows are consistent.
burnt hearth
ok thank you my mistake was that i didnt go to RREF form bc i thought i could stop at REF. ill add this to my notes. again thanks guys for helping
dull latch
Just to make sure I'm on the right track, v_1, v_2, v_3, and b are all linearly independent correct? (this is in R^2)
nocturne jewel
Maybe
looks like some linear combination of v1 and v2 could make b
actually no, more than 2 vectors so automatically a dependent set
dull latch
right, so {v_1, v_2}, {v_1, v_3}, {v_3, v_2} would all be linearly independent though right?
dull latch
so for R^n, and combination of would any n + 1 amount of vectors automatically be dependent?
nocturne jewel
yes
hardy inlet
yes
nocturne jewel
since any n size set could constitute a basis if it spans
likewise any set with cardinality less than n will not span
dull latch
makes sense, gotcha
How does that work for determining the potential solution(s) for
would there be many solutions? if so, why?
nocturne jewel
yes there would be many solutions
x_1=0, x2,x3 whatever they need to be
x_2=0, x1,x3 whatever
x_3=0, x1,x2 whatever
since they pairwise form bases
dull latch
so bc there exists a greater than n amount of vectors that when put into their n-tuples for R^n are linearly independent, there are many solutions?
dull latch
👍
thin wing
so bc there exists a greater than n amount of vectors that when put into their n...
The first line is from the fact that {vᵢ | i∈{1,2,3}} is not linearly indepedent (It's missing the ∃a₁,a₂∈F:, but hopefully you get the point). Hold up, a couple of mistakes..
dull latch
gotcha 
thin wing
This should be the correct version.
dull latch
ah, I believe I get it, ty
hardy inlet
so since v1 through v4 are linearly independent and dimV=4, U subspace V with v1 v2 in U but not v3 v4. dimU = 2 and {v1,v2} is a l.i. list who's length matches the dimention, therefore its a basis?
vital girder
Let $V$ be a vector space and $U,W$ be subspaces of $V$ s.th. $U\bigoplus W=V$. Given $v=u_1+w_1=u_2+w_2$, where $u_1, u_2 \in U, w_1, w_2 \in W, v \in V$, does this imply that $u_1=u_2$ and $w_1=w_2$?
stoic pythonBOT
person2709505
thin wing
so since v1 through v4 are linearly independent and dimV=4, U subspace V with v...
Since v1 through v4 forms a basis, they are L.I. Since v1 through v4 is L.I., any subset is also L.I, thus v1 and v2 are L.I. Since dim U = 2 = |{v1, v2}| with v1 and v2 being L.I., v1 and v2 spans U and finally form a basis over U.
zinc timber
in case of ⊕, yes
zinc timber
**person2709505**
.
vital girder
Ok, thanks
hardy inlet
Since v1 through v4 forms a basis, they are L.I. Since v1 through v4 is L.I., an...
i have found out that it is false
let v1 to v4 be the trivial basis in R^4, then (image)
dimU = 3
do these explanations look good?
zinc timber
👍
hardy inlet
Let $V = \mathcal{P}(\mathbf{R})$ be the vector space of all polynomials with real coefficients. If $p$ is any polynomial, let $Tp$ be the polynomial defined by $(Tp)(x) = p(x+1) - p(x)$. Show that $T$ is a linear transformation.
stoic pythonBOT
MattDog_222
zinc timber
show T(ap+bq)=aT(p)+bT(q)
hardy inlet
so like for a concrete example if $p=x^2 + x$ then $(Tp)(x) = (x+1)^2 + (x+1) - (x^2 + x)$?
stoic pythonBOT
MattDog_222
zinc timber
yes
hardy inlet
show T(ap+bq)=aT(p)+bT(q)
so like i did that in abstract algebra on homomorphism stuff, but idk how to apply that here
ik in class when we literally just started talking about it, we did scalar and addition separately, so does this work for addition?
zinc timber
no, you can't do then unless T is linear which you are proving
why not write T(f+g)(x)=(f+g)(x+1)?
hardy inlet
because $T(f+g)(x)=(f+g)(x+1) - (f+g)(x)$
stoic pythonBOT
MattDog_222
hardy inlet
or are you saying I should solve T(f+g)(x) singly, then solve T(f) + T(g) singly, and compare the results?
bc thats why i was using ?=
zinc timber
T(f+g)(x)=Tf(x)+Tg(x) is what you are supposed to prove
you assumed that true in yr first line
zinc timber
why not write T(f+g)(x)=(f+g)(x+1)?
try this instead
zinc timber
yes
hardy inlet
is this a correct start and do i need to go further?
lavish jewel
seems ok
hardy inlet
idk if i did the lambda correct
dusky epoch
yes
hardy inlet
ok thanks
moving onto this problem, would the strategy be to find the set of polynomials with roots at 1 and 3, then add the constant function?
i think this is a good start, but i remember something in class sorta where (x-1)²(x-3) and (x-1)(x-3)² are redundant to have both, why?
zinc timber
W1 says p(0)=0 means x is a factor of elements on W1. for W2, similarly (x-1) is a factor of elements of W2.
hardy inlet
i think this is a good start, but i remember something in class sorta where (x-1...
silly goose, just multiply the first one by x repeatedly
past yew
W1 says p(0)=0 means x is a factor of elements on W1. for W2, similarly (x-1) is...
Can u share detail answer in written format
hardy inlet
if the dimension of two vector spaces are equal, does that mean the vector spaces are equal?
right now I have this work for finding and extending the basis
trying to show oplus equal tho and idk if im going in the right direction
wintry steppe
if the dimension of two vector spaces are equal, does that mean the vector space...
if one's a subset of the other
hardy inlet
so is this legal
lavish jewel
if U and W are subspaces of W
hardy inlet
$W = \mathcal{P}_4(\mathbb{R}) \ U = \operatorname{span}({1,, (x-1)(x-3),, x(x-1)(x-3),, x^2(x-1)(x-3)}) \ W = \operatorname{span}(x)$
stoic pythonBOT
MattDog_222
hardy inlet
so i added
this isn't fun. Do i just show that the linear mapping holds for arbitrary a when a= 0, and then one by one for b and c?
lavish jewel
you can show it component-wise
tranquil cloak
is the RREF of a square matrix equal to it's inverse?
lavish jewel
no
hardy inlet
so the ← direction would be, let a=b=c=0 then prove its linear. and the → direction is 'component wise'
I think i can handle ← on my own now, but how would i do the forward direction. i'm not really sure what u mean
tranquil cloak
I thought that in order to calculate the inverse of a square matrix, we write the matrix in [ A | I ] form, then row reduce it?
lavish jewel
yes
and then the rref of A is simply an identity matrix
the inverse is in the other block
in the forward direction, you look at each component and show that the components don't satisfy the definition of linearity
hmm
yeah
start with T linear, so it satisfies the def of linearity, and show that a,b,c=0 for this to hold
otherwise it violates the def of linearity
hardy inlet
is the result a vector?
like can I write it as a column vector since its so large left to right
lavish jewel
sure
lavish jewel
multiply it out?
hardy inlet
lavish jewel
looks ok
hardy inlet
so what did u mean about doing this component wise
lavish jewel
that even if 1 of the 3 components of the vectors violates the definition of linearity, the whole transformation isn't linear
so all 3 of them must satisfy it independently
hardy inlet
oh so thats why theres an a in the 1st, b in the 2nd, c in the 3rd
so u have to show all 3 violated
lavish jewel
so now you have 3 small problems, each of which must be linear
lavish jewel
so u have to show all 3 violated
no, all 3 satisfied
if even just 1 is violated, then the whole transformation is not linear
hardy inlet
but we're not asking to show its nonlinear
we're asked to show its linear iff a=b=c=0
lavish jewel
yes
ah i understand your wording now
you mean it's violated it a or b or c neq 0
then yes
hardy inlet
my friend said the teacher said theres an easier way but we dont know what it is
lavish jewel
yeah so
my gut instinct was that having a nonzero constant added makes this an affine transformation
you just say that and you're done
hardy inlet
so i've shown for the first component
ok so i've shown under addition a = b = c = 0, do I then need to show under scalar?
lavish jewel
it's enough to show that one of the two requires it
you already showed the opposite direction in the previous step, so this already satisfies linearity
lavish jewel
the easier method was what i said earlier
the expressions are nonzero scalars in general, so any of a or b or c neq 0 means it's affine and therefore not linear
and that's all
hardy inlet
is this a better explanation of why it suffices
like it makes sense since if scalar allowed a = 0 or 1 for example, that the or doesn't even matter since that value wouldn't hold under condition
hardy inlet
better yet, lets say scalar multiplication had no restrictions
it wouldn't matter because addition has restrictions that are an AND condition
but if scalar didn't hold for a=b=c=0 then we'd know from proving the reverse direction
this is all i put on the backwards direction
idk how to construct something like this. i tried (x,y) → x + 1 but thats neither
and (x,y)→2x holds
hardy inlet
i took inspiration from the internet to find and change up a counterexample, but idk how to construct more
do u just use some nonlinear operator
zinc timber
that does not work as √a²=|a| not a
hardy inlet
damn the example i looked at had cube root. makes sense
but does this example work for complex the other way around
zinc timber
what other way around?
outer goblet
if i have to use the lagrange interpolation formula to construct a polynomial, from a set of points, do i need to first sort the point so that i start with the lowerst x value point and end with biggest?
zinc timber
that's not necessary but a nice thing to do
outer goblet
ait ait
zinc timber
contradiction
zinc timber
yes
also show that $(A_k^{-1}\cdots A_1^{-1})(A_1\cdots A_k) = I$ for complete argument
stoic pythonBOT
zinc timber
(though it's not necessary)
junior bane
contradiction
So we say
if A1 A2... Ak is invertible,
suppose one of the Ai is not invertible?
zinc timber
yes
well no
say A is not invertible, then show that at least one of Ai is not invertible
junior bane
say A is not invertible, then show that at least one of Ai is not invertible
wait is that proof by contraposition
outer goblet
junior bane
or contradiction
outer goblet
any help on how to write that out?
zinc timber
both
junior bane
both<:catKing:812386361297862747>
i still cant think of the link tho
outer goblet
all is came to is to write
a_1=a_3+a_4
a_2=a_2
a_3=a_1-a_4
a_4=a_1-a_3
a_5=a_5
but idk what more to do
junior bane
jus cuz A1...Ak is not invertible, how do we link it to at least one of Ai is not invertible
zinc timber
if at least one of the Ai is non-invertible then A cannot be invertible
outer goblet
nvm i did it
thin wing
I believe this works if you want the straight forward way (as opposed to proof by contradiction). See if you can spot the mistake, lol. Next post should be correct.
thin wing
~~I believe this works if you want the straight forward way (as opposed to proof...
Sorry, here's the corrected edit.
magic light
Looking for understanding where I went wrong. Homework says that this function isn't a linear transformation but it seems that it is? Field is C.
T(z) = cz for c in C
so I took
X = a + bi
Y = c + di
c = f + gi some constant in C
T(X + Y) = T(a + bi + c + di) = (a + bi + c + di)(f + gi) = (a + bi)(f+gi) + (c+di)(f+gi) = T(a + bi) + T(c + gi) = T(X) + T(Y) - closed for addition
for K = c + di
kT(X) = (c+di)(f+gi)(a+bi) = (f+gi)(ac + adi + bic + bidi) = T(ac + adi + bic + bidi) = T((c+di)(a+bi)) = T(kX) - closed for multiplication
so T is linear, no?
$$T(z) = cz, C \in C, X=a+bi, Y=c+ di, c = f + gi$$
stoic pythonBOT
blackmamba[they/them]
magic light
$$T(X + Y) = T(a + bi + c + di) = (a + bi + c + di)(f + gi)$$
$$ = (a + bi)(f+gi) + (c+di)(f+gi) = T(a + bi) + T(c + gi) = T(X) + T(Y)$$
$$kT(X) = (c+di)(f+gi)(a+bi) = (f+gi)(ac + adi + bic + bidi) = $$
$$T(ac + adi + bic + bidi) = T((c+di)(a+bi)) = T(kX)$$
magic light
<@&286206848099549185>
wintry steppe
can you post the specific homework problem as a screenshot?
that function is definitely linear...
thin wing
Looking for understanding where I went wrong. Homework says that this function i...
Yeah it should be linear (picture from examples of linear transformations). It's a case of rotation-dilation, which is linear. There's either a misunderstanding of the question or a typo.
magic light
can you post the specific homework problem as a screenshot?
"find if the function is linear, and if so find the kernel and image of T"(or f in this case)
From the answers:
c is a constant, T is the representatino of it
he says it's not linear, no explanation.
maybe it's because I took X from C like a constant instead of a vectoral representation?
like instead of X = a + bi, X= [a, b]?
magic light
<:stare:826920038623346729>
he said because R^2 is isomorphic to C then he changed all functions to T:R^2->R^2 but that still woulnd't change anything I guess
wintry steppe
well an isomorphism should preserve linearity
so that's indeed not changing anything
hmmm
yeah i don't know
throw eggs at his house or something
they're isomorphic as both R and C vector spaces, too, so i have no idea what your prof is getting at
magic light
probably some error, idk.
I absolutely hate this course because the professor is a stickler to the most annoying technicalities. Like, you didn't write V is a vector space in some question then you don't get any points etc
@wintry steppe if DimV = DimW does that mean V=W?
because I'm a bit confused on that part
because then R^2=C?
nocturne jewel
<@456226577798135808> if DimV = DimW does that mean V=W?
no
R^2 and C are just isomorphic
$\varphi([a,b]^T)=a+bi$ is a bijective mapping $\varphi:\mathbb{R}^2\to\mathbb{C}$
stoic pythonBOT
Mosh
nocturne jewel
hence isomorphic
wintry steppe
if one's a subset of the other
refer to
without an inclusion of some kind, all you can say is that they are isomorphic
wintry steppe
I absolutely hate this course because the professor is a stickler to the most an...
now i'm remembering something
i had a real analysis course with a prof like that
3rd year course, so you wouldn't expect the prof and his TA to be total pedants
but here i am losing marks for giving a picture proof that, say, the topologist's sine circle is path connected
all you can really do is deal with it and leave a scathing review at the end
thin wing
<@456226577798135808> if DimV = DimW does that mean V=W?
I think a counter-example is that of a sphere (surface of a ball) and a plane.
wintry steppe
still not a vector space
thin wing
So like.. There's V=W, and there's (V,+,‧)=(W,⊕,⊙)..
wintry steppe
the fibers are, but they're two dimensional vector spaces
thin wing
still not a vector space
Oh yeah I agree, I keep thinking bundles are VS. That's why I crossed it out, lol.
thin wing
So like.. There's V=W, and there's (V,+,‧)=(W,⊕,⊙)..
So you could have two vectors spaces with the same dimension, but a different defined set and addition/scalar-multiplication operations.
thin wing
So you could have two vectors spaces with the same dimension, but a different de...
Finally.. That argument works, right?
wintry steppe
it works
gray dust
So you could have two vectors spaces with the same dimension, but a different de...
@thin wing this describes isomorphic spaces, and there are many examples
eg R^2 & C as spaces over R
thin wing
<@!171055699479101450> this describes isomorphic spaces, and there are many exam...
What about ℤⁿ and ℝⁿ over ℤ? Since ℤⁿ and ℝⁿdon't have a bijective map, they wouldn't be isomorphic. However they have the same dimension of n.
wintry steppe
Z^n isn't a vector space
gray dust
also Z isnt a field
thin wing
Ohhkay gotcha, thanks!
tough veldt
Why do the rows transposed of the RREF of A transpose give basis vectors of im A
$$\textnormal{columns of }(rref(A^T))^T = \textnormal{basis of }\Im A$$
stoic pythonBOT
Shuri2060
tough veldt
haughty berry
$$\textnormal{columns of }(rref(A^T))^T = \textnormal{basis of }\Im A$$
So the image of A is its column space, which is the transpose of the row space of A transpose. So the idea is that you can prove that row the row space of A is the same as the row space as A after you apply some row operation. So the row space of A = row space of RREF(A), and since the (non-zero) vectors of RREF(A) are linearly independent and span the row space, they are a basis
tough veldt
Thanks. Just forgot why this 'hack' works 😅
haughty berry
Np
tough veldt
I also just realised that if you want some vectors that span the image of f = Ax
You just take the columns
The only reason to do the above is to find a basis
haughty berry
Ye
dapper gorge
Given a module M over a ring R, if R is not a division ring, then there is no chance that there is a basis for M, right?
subtle walrus
there are free modules over any ring?
dapper gorge
I'm confused with what you have said
subtle walrus
your question seems to imply that no module over a ring which is not a division ring is free
dapper gorge
My question implies that I think that's possibly the case but I don't know
subtle walrus
R is a free R-module
dapper gorge
R is a ring ?
ok so
there are modules over non-division rings which have a basis?
dapper gorge
yeah true
dapper gorge
there are always modules with a basis
This means that given any ring you can find/construct some module over that ring with a basis?
subtle walrus
if you give me a ring R and some set S, you can construct an R-module with basis S
spice ruin
How is theory of matrices any different from linear algebra?
I have heard it described as linear algebra 2 but this syllabus looks like stuff I have already learned in linear 1
subtle walrus
what do you mean?
well, first of all matrices are only really useful for finite dimensional vector spaces
spice ruin
Is this class just a direct continuation of linear algebra? none of this looks new.
subtle walrus
this looks like a linear algebra class
spice ruin
yea, thats what I thought. this is my theory of matrices syllabus though
subtle walrus
well ok 🤷
i dont think your picture covers infinitely dimensional vector spaces a lot
maybe you did linear algebra only over R and this is more general
its impossible to tell if this will have much new content for you
spice ruin
true. I took linear over a year ago so I cant remember exactly what was covered but this all looks essentially the same to me. The professor said on the first day that we will primarily be focusing on finite dimensional vector spaces in R
Reduced row echelon form go brrrrrr?
subtle walrus
should be a very small part of your picture
if its covered at all
the whole eigenvalue thing is pretty big
and so are canonical forms
spice ruin
that makes sense. I remember we computed eigenvalues/vectors but I dont think we were ever given the context in which they are useful. Maybe we will have to verify axioms for them or something?
subtle walrus
they are useful because they give simple descriptions of linear maps
a matrix is a linear map plus a choice of coordinates (a basis)
so there are multiple matrices that describe the same linear map (with different choice of basis)
but some matrices are simpler than others, so this motivates is to pick a specific basis that makes a matrix very simple
this will motivate the last two chapters
spice ruin
Hmmm, interesting. iirc this book goes all the way to multilinear products
(Schaums outline of linear alg)
I honestly might go to the future chapters so I can have more context for the motivation of nuances in these ones.
subtle walrus
the prerequisite to 3b1b linear algebra and calculus videos is knowing linear algebra and calculus respectively
discussion channels would probably be more appropriate, topic channels are more about strictly mathematical questions
nocturne jewel
I wouldn't say the prereq to Grant's essence series are the topics themselves
subtle walrus
the prerequisite to 3b1b linear algebra and calculus videos is knowing linear al...
but to elaborate on this: the videos are meant as additional tools for your intuition, they do not teach you the topic
i mean you can watch them and enjoy them
but they will not teach you linear algebra in any capacity
so i guess it depends what you want to get out of it, if its just enjoyment or motivation to study the topic that will work well, if you want to learn the topic the videos are not that great
actually they might also work as a supplement really well, while reading a book or learning from another source
wintry steppe
what is the difference between the inner and outer products?
fringe fjord
Inner products produce a scalar. Outer products produce a tensor (or a matrix).
magic light
determinants confuse me a bit, when do I put in a (-) when developing?
like say you develop via the third row, third column
(everything else in this row is 0)
quartz compass
you can imagine making a checkerboard over the matrix that alternates + and -
and those are the signs you put on the terms as you go along that row or column
here's the first image I could find by search engine to show what I mean:
top left is always +, so you can just go from there
quartz compass
cool
magic light
Thanks. i was under the false impression that the first column is always +
so many pointless mistakes. lol
Another basic question, if i have base C = ((2, 1), (7,4)) and I want for example to find [2,10]c then I need to find a,b s.t a(2,1) + b(7,4) = (2, 10) right? or in other words 2a + 7b = 2 & a + 8b = 10?
wintry steppe
Inner products produce a scalar. Outer products produce a tensor (or a matrix).
hmm im a bit confused bc
i have this problem and i got the outer products
but i have no idea how to get the inner products
fringe fjord
Inner products are just dot products.
wintry steppe
i guess my question is
what does the notation x_i^T x_j mean
is it just
find the dot product of
x_1^Tx_1
x_1TX_2
etc etc
and then how do u use those to get XTX? just them all up?
fringe fjord
It's treating x_ 1 as a column vector, that is, a 2×1 matrix. x_1^T is the transpose of that vector and x_1^Tx_1 is an ordinary matrix product.
(Which works out such that it's actually just the dot product).
fringe fjord
2x^Tx2 should be 2.
Oh foo.
I think I've seen this madness before, haven't I?
Boldface x and lightface x means different vectors to this author.
wintry steppe
yea i think so
its baeed on ISL/ESL texts
i guess what am i supposed to do with the inner products to get the
XTX?
fringe fjord
I think the point of the exercise must be for you to sit down and write everything out in explicit coordinates all the way, to see for yourself how the two procedures it sketches work.
fringe fjord
How to format them?
fringe fjord
My suggestion is to set x_1=(a,b), x_2=(c,d), x_3=(e,f), and then write all of bold X, bold x1, bold x2 and all of the various products out explicitly in terms of a,b,c,d,e,f.
wintry steppe
yea exactly ;. im not sure what even its asking for
fringe fjord
Well, that's my suggestion, at least. ¯_(ツ)_/¯
fringe fjord
These are the non-bold x vectors.
defined in the first-sentence of the text you quoted.
wintry steppe
so it would be (2,1) (1, 1) and (-1, -1)
fringe fjord
Yes, but keep them as letters while you're writing things out to get a feel for what's going on.
I mean, take a sheet of paper and literally write down $x_1=\begin{bmatrix}a\b\end{bmatrix}$ $x_2=\begin{bmatrix}c\d\end{bmatrix}$ $x_3=\begin{bmatrix}e\f\end{bmatrix}$ $\mathbf X=\begin{bmatrix}a&b\c&d\e&f\end{bmatrix}$ $\mathbf x_1=\begin{bmatrix}a\c\e\end{bmatrix}$ and so on for all the various matrices and vectors and products of the same in the exercise.
stoic pythonBOT
Troposphere
fringe fjord
I'm not seeing any a, b, c, d, e, f on that paper, but if you understand what's happening, that's good.
wintry steppe
yea i think i got it!
with the same problem, it asks to do this, what does this even mean?
i found beta_hat
thorugh doing
(XTX)^-1 * XTY
gray dust
do u know what a linear combo is
wintry steppe
do u know what a linear combo is
yeah i do
gray dust
so whats ur confusion
wintry steppe
i guess like i know what it is conceptually but not how to actually do it
gray dust
find scalars $a,b$ where $\mbf{\hat y}=a\mbf x_1+b\mbf x_2$
stoic pythonBOT
RokabeJintaro
wintry steppe
but here what are x_1 and x_2
wintry steppe
ah ok ok
and then to find a and b
couldnt it be like anything technically?
x_1 = 2 1 -1
and x_2 = 1 1 -1
y_hat = 2 .5 -.5
gray dust
valid values of a,b depend on the vectors
gray dust
the number of sols depends on the vectors
wintry steppe
no ik lol
but im saying
like this is what i have rn
so i just have to solve for a and b right?
gray dust
yes
wintry steppe
but isnt it possible that theres more than one solution
gray dust
the number of sols depends on the vectors
depending on what x1,x2,y are there can be no sol, 1 sol, or many sols
fringe fjord
In this case there happens to be 1.
gray dust
but the question asks u to write y as a linear combo so i presume at least 1 sol exists
nocturne jewel
what's the space with basis {1,sinx,cosx,...} called? Just Hilbert space?
Or Fourier space cause it's used more in Fourier
gray dust
what's the space with basis {1,sinx,cosx,...} called? Just Hilbert space?
hilbert is a type of space, not a specific space
nocturne jewel
ye that's what I figured
gray dust
that set is a basis of the space of square integrable functions on [-pi,pi]
nocturne jewel
$\int_{-\pi}^\pi f^2(x)\dd{x}<\infty$?
stoic pythonBOT
Mosh
gray dust
yes if the space is over R
nocturne jewel
yeah, trying to write an explanation for why sines and cosines are orthogonal, but wasn't sure what space the Inner would be over
gray dust
u dont need any of this to just talk about orthogonality
the key is just smart integration
keen sierra
Probably more appropriate in #prealg-and-algebra
zealous jetty
cross(Av,Ax), A is operator, v,x are vector, can i somehow factor out A
wintry sphinx
for rotation matrices I guess you could
wintry steppe
what's the space with basis {1,sinx,cosx,...} called? Just Hilbert space?
Its called the space of polynomials
dusky epoch
the space of trigonometric polynomials
wintry steppe
the space of _trigonometric_ polynomials
same thing up to isomorphism
tho the isomorphism isnt unique
so maybe your answer is better cuz of that, Ann
dusky epoch
same thing up to isomorphism
tho the isomorphism isnt unique
i mean in linear algebra especially you generally dont treat things as the same just because theyre isomorphic (ie have the same dimension)
so yeah
magic light
Another basic question, if i have base C = ((2, 1), (7,4)) and I want for exampl...
anyone?
subtle gust
Hey
Just a quick question
If we have a system of 4 equations in 3 unknowns
And the last row is all zeros
Do we just ignore it and assume it's a 3×3 matrix? 🌞
frail timber
hello. Quick question. Can anyone explain why this is not an internal product, if that is how you call it in english.
wintry steppe
Do we just ignore it and assume it's a 3×3 matrix? 🌞
as long as the right side is 0 yes? cuz then the last row means 0=0.
wintry steppe
hello. Quick question. Can anyone explain why this is not an internal product, i...
Do you know the axioms for inner products?
lavish jewel
as carla points out, one of them is "positive definiteness"
you can check that this operation is not positive definite
frail timber
as carla points out, one of them is "positive definiteness"
Is is because its inner product can be 0, but u2 and v2 can still be different than 0?
lavish jewel
mhm
frail timber
what I have here for "positive definiteness" is for <u,u>=0 => u=0 . Does it work for <u,v>=0 => u=0 and v=0 as well?
frail timber
oh, ok. That makes sense. thx
subtle gust
as long as the right side is 0 yes? cuz then the last row means 0=0.
But doesn't this mean that the system has infinitely many solutions
lavish jewel
i think carla meant for the last row only
but even if the last column would all be 0, the number of solutions depends on the rank of the matrix
subtle gust
i think carla meant for the last row only
Yeah that's what i mean
Doesn't a row of zeros imply an infinite number of solutions
lavish jewel
no
subtle gust
Oh ...
lavish jewel
especially here that your matrix is tall (4 x 3)
even if you have a row of all 0s, the matrix can still be rank 3
if the 3 columns are lin indep, the null space is trivial
subtle gust
I don't think we've reached that part yet ....
You're using stuff that i'm not really familiar with
We only reached determinants
In class i mean
subtle gust
It's not...
lavish jewel
not invertible
lavish jewel
right, that does mean it's not invertible
but doesn't immediately betray a nontrivial null space (infinitely many sols)
subtle gust
Well doesn't a matrix being not invertible mean that a homogeneous system doesn't only have the trivial sol ?
subtle gust
Cuz it's like only two cases for homogeneous systems
lavish jewel
this is a counter example
subtle gust
Oh so if a A is not invertible
lavish jewel
in this case, Ax = 0 only has the trivial solution
subtle gust
Ax=0 can still only have the trivial sol
lavish jewel
invertibility is not related to the size of the null space, not directly
lavish jewel
yes
subtle gust
But wouldn't that imply that A is invertible ........
subtle gust
So the logic statement doesn't go both ways ..
lavish jewel
right
subtle gust
Invertible -> only the trival sol
lavish jewel
if the matrix were square, things would be different
but this matrix is rectangular
subtle gust
Only the trivial sol != matrix is invertible
lavish jewel
right
subtle gust
Yeah i got it
lavish jewel
at best it implies injectivity
versed parrot
invertibility is not related to the size of the null space, not directly
it isn't? if the Nullspace is just 0 then isn't it always invertible?
versed parrot
oh cause it maps from R^3 to R^4
and not all vectors in R^4 are in the column space, right
queen snow
hey, have a question
is a line determined by some point T and vector(2,2) the same as the line determined by that same point T and vector(3,3)?
in other words what im asking is if you are given 2 lines with the determined by the same vectors but different points and you have to compute a plane between them, since they have the same vector the cross product will be 0, so my question is can i just subtract 1 from each component of one vector and than do the cross product since the vectors stay parallel?
queen snow
Thanks a lot
Just one more question
Lets say i have a vector(1,2,0) after i subtract 1 i get (0,1,-1), than if i write it in a form of the determinant the cross product doesnt equal to 0 any more
Which in my mind makes sense because they are parallel so the cross product should be a vector vertical to vector(1,2,0) and (0,1,-1)
ember kraken
quasi vale
Okay so first of all, cross product is only defined for 3 dimensional vectors. I misread your statement at first. Second of all, cross product of two parallel vectors is the zero vector.
@queen snow
And no, it isn't nec essary that you get parallel vectors if you subtract 1 from each component
In your case, (1,2,0) and (0,1,-1) are not parallel
wintry steppe
Okay so first of all, cross product is only defined for 3 dimensional vectors. I...
Wrong, it can also be defined for 0,1 or 7 dimensions
queen snow
Ah
Yeah thanks a lot after graphing it out in geogebra i understand it much better
i couldnt wrap my head around how two planes can be vertical on a third plane but not parallel with each other
thanks a lot for ur help @quasi vale
versed parrot
Wrong, it can also be defined for 0,1 or 7 dimensions
well with geometric algebra it can be defined for more
dusky epoch
ok we
well*
in part b, have you written out the normal equations being referred to?
dusky epoch
right
well, it should follow immediately from that, should it not
because you have $X^T(Xw - y) = 0$
stoic pythonBOT
Ann
dusky epoch
subtract X^Ty from both sides and factor out X^T lol
raw stag
and we want to show that the error vector, ⃗e = ⃗y − X ⃗w∗
, is orthogonal to the columns of X
okay, one sec
raw stag
ohh okay
so all I have to show is that X^T(Xw-y) = 0 and that proves the orthogonality of the error vector to the columns of X?
dusky epoch
so all I have to show is that X^T(Xw-y) = 0
i mean, yes? but also that's literally just your normal equations rewritten with a tiny bit of matrix algebra
dusky epoch
sure
youll need to recall that passage i screenshotted and pasted here about a matrix equation encoding orthogonality to multiple vectors at once,
and also know the structure of the design matrix for a prediction rule of the kind shown in part b
raw stag
okay
dusky epoch
part (a) should also serve to prime you into the right observation
raw stag
okay so for part a, I got that it would just be equal to the sum of all the elements in b
dusky epoch
yes, that's correct.
raw stag
okay cool. So let's see what I can do from here..
dusky epoch
it will be important that the prediction rule has an intercept term.
raw stag
okay, so we have the following facts. Our prediction rule is this:
H(⃗x) = w0 + w1x
(1) + w2x
(2) + ... + wdx
(d)
and we have that 1^Tb is the sum of all elements in b
and if A^Tb = 0, that b is orthogonal to all the column vectors of A
im just a bit confused on the design matrix part
dusky epoch
math formulas and copy-paste do not mix well.
anyway, the point i was trying to make is that the leftmost column of X will actually be all 1's, precisely because of that intercept term.
raw stag
hmm okay, the leftmost column of X in that error equation?
dusky epoch
yes
...idk if i can continue this
definitely don't have the energy rn to do part d and its subparts
raw stag
hey, so sorry, ill b back in 1 min loll
but can u show me how ro do part c? im not really getting it
wintry steppe
tldr the k nearest neighbors method is just a way of averaging things out right?
like the closeness metric can be seen as euclidean distance and the formula for regression is 1/k(summation of y's )
isnt it then in essence an average?
wintry steppe
$\begin{bmatrix}
\alpha & 1 & 1 & 1 & 1 \ 1 &\alpha &1 &1 &1\ 1 &1 &\alpha &1 &1 \ 1 &1 &1 &\alpha &1 \ \end{bmatrix}$
stoic pythonBOT
mate
wintry steppe
so this is a system of linear equations (alpha * x1 + x2 + x3 + x4 = 1 and so on)
could somebody help solve this? im confused
$\alpha \in \bR$
stoic pythonBOT
mate
teal grotto
could somebody help solve this? im confused
what is the problem to be solved?
teal grotto
all good
raw stag
its problem 2 parts c and d
nocturne jewel
its problem 2 parts c and d
Post a screenshot of the questions, rather than something that needs to be downloaded
raw stag
ohh okay sorry, I didn't know it couldnt be downloaded
teal grotto
oh, i was asking someone else. i dont know how to do your problem
nocturne jewel
ohh okay sorry, I didn't know it couldnt be downloaded
it can be downloaded, however no one in their right mind would download a random file off Discord
I haven't seen your question, cause I'm trying to get you to post it
raw stag
oh my bad, that thought didn't even occur to me sorry
okay i will post screenshots right now, one second
I will have to post multiple screenshots so you get the entire context of the problem but only parts c and d need to be solved haha
wintry steppe
what is the problem to be solved?
i have a system of linear equations with $\alpha \in \bR$. have to find solutions (depending on $\alpha$)
stoic pythonBOT
mate
teal grotto
solutions to what equation?
wintry steppe
solutions to what equation?
$\alpha \cdot x_1 + x_2 + x_3 + x_4 = 1$
stoic pythonBOT
mate
teal grotto
oh oh, my bad
wintry steppe
and so on, look at the matrix
teal grotto
i couldnt tell from the matrix, it didnt look augmented
wintry steppe
yeah sorry i dont know latex rules well.. will learn it after my exams finish
im dying atm lol
raw stag
@nocturne jewel any chance you know my problem?
wintry steppe
like i know how to solve similar problems but alpha is everywhere and thats why im stuck
nocturne jewel
nope
raw stag
oh man
wintry steppe
it becomes too complicated
raw stag
@wintry steppe any chance you can help me after csquared?
wintry steppe
<@456226577798135808> any chance you can help me after csquared?
sorry i am a beginner
raw stag
ahh okay, no worries
wintry steppe
just started doing matrices
raw stag
gotchya
<@&286206848099549185>
need someone to help me with a linear algebra questoin
plsssss
ripe shale
if T(v) = cv then v is the eigenvectors and c is its associated eigenvalues. If TA = cA, where T, A are matrices then what can we tell about A and c
zinc timber
if T(v) = cv then v is the eigenvectors and c is its associated eigenvalues. If ...
( v ≠ 0)
zinc timber
didn't get the question,
ripe shale
I was just wondering what property a matrix like A would have
zinc timber
let's say you fix your basis, then [v] will be an eigen vector of the matrix A=[T] with eigen value c
wintry steppe
Let $T: V_2 \to V_2$ map each point with polar coordinates $(r,\theta)$ onto the point with polar coordinates $(2r,\theta)$. Also, $T$ maps the origin onto itself. Show that $T$ is linear.
stoic pythonBOT
n/c
wintry steppe
I'm having some trouble with this, because if we define $T(r,\theta) = (2r,\theta) = 2r(\cos \theta,\sin \theta)$ then I don't see how we get linearity.
stoic pythonBOT
n/c
wintry steppe
Because $T(a,\theta) + T(b,\phi) = 2a(\cos\theta,\sin\theta) + 2b(\cos\phi,\sin\phi)$, but this is not equal to $T(a + b,\theta + \phi)$ for all values of phi, theta...
stoic pythonBOT
n/c
wintry steppe
Because look at this
,w a * sin(theta) + bsin(phi) = 2(a+b)*(cos(theta)cos(phi) - sin(theta)sin(phi))
dusky epoch
isnt T just the map x |-> 2x
linearity should be obvious unless you explicitly refuse to allow yourself to make this observation
wintry steppe
But I should be able to do it in terms of polar coordinates, shouldn't I?
dusky epoch
well addition in polar coordinates is somewhat hairy to express
it's not just addition of radii and angles
dusky epoch
and there's no reason to force yourself into viewing stuff through an inconvenient lens just for the sake of "practicing dealing with polar coords" or whatever