#linear-algebra
wintry steppe
now... next example i dont understand what it means
just that the norm is what ever is l phi(0) l ???
so this is not a norm, cuz phi (0) could be != 0
?????? or am i mad
like, i could take phi as phi(x) = x
phi(0) = 0 and phi is not the 0 function
lmao you're right
that's a funny question 
the function phi(x) = x is a counterexample to the first condition of a norm
okok, ez
last one is
can i use the first example somehow here?
or since it has something else it has nothing to do?
you might be able to use the first example
okey, but... i dont know how to do the inf norm of a function
mmm
it probably means the maximum of |2phi| on [0, 1]
oh okey, then... i proved first one is a norm
and inf norm is a norm
is there any property that sais the sum of 2 norms is a norm?
or i have to prove this?
you could try to prove that
i recommend it
okey
do i need to prove inf norm is a norm?
no, right?
like, it is a norm by definition
sonic osprey
yes
nocturne jewel
wintry steppe
mmm
im not sure about this, let me paint
something is wrong for sure
i replaced phi and idk for a and b
actually, i believe the second line equals should be <= too, cuz there is another triangular inequality there
sterile pebble
$f=x_1+x_2+e^(x_1+x_2) and g=cos(x_1+x_2) why cant i compose f o g?$
stoic pythonBOT
πeteer
wintry steppe
https://gyazo.com/edbcaf411a687da4460f49398d74f714
@wintry steppe could u check this please? >///<
wintry steppe
or someone else?
dire thunder
why you think |a+b|=|a|+|b|
wintry steppe
yeah hard on paint
dire thunder
also, what is under integral?
dire thunder
i mean dt disappears and i do not understand what is under integral sign now
wintry steppe
yeah
to write faster
it is a(t)
and b(t)
and dt should appear
(l a l + l b l) is under the integral
i just splited the l a + b l into l al + l b l
stoic pythonBOT
Commander Vimes
wintry steppe
the red paint
is the definition of my norm
i am proving it if it is a norm or not
dire thunder
ah ok
tawdry bramble
If $\overline{0}=\overline{x}$ is the only solution to $A\overline{x}=\overline{0}$, what does this mean?
stoic pythonBOT
glomswamp
tawdry bramble
Oh nice.
I've not got to that yet
I've actually never heard of injective or surjective lol
Or bijective
Oh ok
What you said makes a lot more sense now
Interesting, I was asking because I was trying to find the equivalence class of the zero vector of this equation.
lavish jewel
only if the matrix is square
tawdry bramble
The middle vector is xyz and the solution should be ab
I was putting in zeros to try and figure it out
tawdry bramble
Lol I know, but it's still interesting
lavish jewel
well, it's surjective
not injective tho
dat null space
your assumption there that z = 0 is an interesting toy scenario, but in general, z is just some parameter t
the rest of it is correct. you need y = x = 0
but no matter what z is, it is multiplied by zero
this has infinitely many solutions
tawdry bramble
My original problem was to find the equivalence class containing the zero vector.
And that the class had a specific name in linear algebra.
tawdry bramble
It's for part c
I thought I'd start by plugging in the zero vector since it ought to be in its own equivalence class
Then the solution is a=b=0, so I figured I ought to find the vectors who can solve that and I think it's all vectors where x=y=0
Like you were saying
lavish jewel
right
tawdry bramble
I thought about null space but idk. Could it have something to do with linear dependence/independence?
lavish jewel
surely
there is a null space if at least one column is linearly dependent to another
this is what gives you nonzero solutions to the linear combination of the columns of the matrix
and the coefficients of the linear combination are precisely the components of the vector (x,y,z)
so there are nontrivial (x,y,z) that, when multiplied by A, yield a zero vector as long as there are linearly dependent columns
the x,y,z that satisfy this are said to be in the null space of the matrix
i.e. the nontrivial solutions x of Ax = 0 are the set of vectors in the null space of A
x = 0 is trivially in there
and if there are lin dep columns in A, then you have (infinitely) many more
tawdry bramble
In my case, the null space is every vector (0,0,t)?
lavish jewel
yea
tawdry bramble
Which would be the equivalence class containing 0?
lavish jewel
that's what i would say, but i would ask mirza or someone else to double check really quick
as i'm not all that familiar with fancy math words π
lavish jewel
@wintry steppe halp, double check this pl0x
tawdry bramble
:3
lavish jewel
anyway
reading that again, all a b have infinitely many solutions
there is a particular solution in terms of x y
and then infinitely many more by adding some (0,0,t)
w = v + (0,0,t) is the equivalence class
and the reason is the same we already gave
Aw = A(v + (0,0,t)) = Av + A(0,0,t) = Av + 0
so any w of the form w = v + (0,0,t) behaves the same way as just v when transformed by A
tawdry bramble
Man, I already submitted my work. That would've looked extra spicy on there.
lavish jewel
aha, that's the elitist mathematician bullshit i was missing
i just wasn't sure how exactly they wanted the answer to be given, thanks for your input!
hollow finch
what am i missing?
$p(TAT^{-1})=Tp(A)T^{-1}$
stoic pythonBOT
nix
hollow finch
since the exponential function can be represented as a taylor series, the same principle applies
wintry steppe
Im super confused what the difference between parametric, symetric, vector and scalar equations is
could someone explain it
this is my teachers notes but i dont understand it
wintry sphinx
okay the "vector equation" and "parametric equation" terminology here is probably the same thing to a lot of math people
as for the others, it just depends on what the equation looks like
wintry sphinx
t is a parameter
it generally takes on all real numbers as a value
Consider a parametric equation of a line in a plane: r = [1, 1]t
What I'm saying is that if I let t take on all values in the real numbers, then I'll get all of the points that this line contains
stoic pythonBOT
Saccharine
wintry sphinx
A line is just a set of points
no [1,1] is not an interval
ambiguous notation
So do you agree that a line is just a set of points?
Consider the line y = x as an example
I'm sure you know what that looks like
We can write it in set-builder notation as
${(t, t): t \in \mathbb R}$
stoic pythonBOT
Saccharine
wintry sphinx
that's basically the most natural way to write it in set-builder notation, right?
no...
this represents the set ${(1, 1), (2,2), (3, 3), (2.5, 2.5), ...}$
stoic pythonBOT
Saccharine
wary lily
bc x=y
wintry sphinx
do you not see why ${(t, t): t \in \mathbb R} = {(x, y): x, y \in \mathbb R, x = y}$
stoic pythonBOT
Saccharine
wintry sphinx
let's do it as an example
and generalize
So one other choice to represent this is
${(1, 1)t: t \in \mathbb R}$
stoic pythonBOT
Saccharine
wintry sphinx
where we treat points as vectors in R2
in this case, yes, but in general, it is just some dummy variable that goes over all of R
when we write something like "the line given by r = (1, 1)t," what we really mean is ${(1, 1)t, t \in \mathbb R}$
stoic pythonBOT
Saccharine
wary lily
yes, r1 is a point on the line, t a scalar multiplier and m the direction vector
wintry sphinx
yes, in general lines can be written in the form
${\mathbf r_0 + t \mathbf\gamma: t \in \mathbb R}$
stoic pythonBOT
Saccharine
wintry sphinx
you can specify a line by a single point and its direction, yes?
wary lily
you can either have two points or a point and the direction of the line
note that two points can give you the direction too
yes
Saccharine, if this is over, can you explain to me the difference between Symmetric- and Scalar equation?
they are apparently the same, as I see it
just rewritten for convenience
do you know the point slope formula for the line equation?
$y-y_1 = m(x-x_1)$
stoic pythonBOT
az
wary lily
now note that $m = \frac{y_2 - y_1}{x_2 - x_1}$
stoic pythonBOT
az
wary lily
which is the relation between the changes in y to changes in x
yes
this change in y and x is your direction vector
now you can set $a = y_2 - y_1 = b$ and $x_2 - x_1 = a$ and get the Symmetric Equation
stoic pythonBOT
az
wary lily
the Symmetric Equation is like this by default
your teacher just multiplied both sides by b and a
and moved everything to LHS
if the exercises are from published books you can find most solutions on Slader
I don't know
tame mural
Conventionally, if M is a matrix and someone writes M_2, are they referring to rows or columns?
Asking for conventionality
No it's not the same thing
gritty swift
hey when finding the singular value decomposition, how do you know which vectors to use? at first I thought there was only 1 orthonormal eigenbasis but after trying to implement it in code I realize there are multiple (if x1, x2 orthogonal, x1 and -x2 are also orthogonal). how do you pick the right ones?
here's my code if it helps
def eig(A):
"Get sorted eigenvectors and eigenvalues"
evals, evecs = np.linalg.eig(A)
idx = evals.argsort()[::-1]
return evals[idx], evecs[:,idx]
def svd(A):
"Find SVD (A = USV^T)"
eigenvalues, V = eig(A.T @ A)
S = np.sqrt(np.diag(eigenvalues))
_, U = eig(A @ A.T)
# note: eigenvalues for AA^T same as A^TA
return U, S, V
heres a case where it fails
tldr: how do you pick the right eigenvectors when computing the svd? aren't there multiple orthonormal eigenbases?
sonic osprey
Is there a right one?
lavish jewel
any rotation of the basis vectors yields another suitable basis
gritty swift
hmm maybe my code is just inconsistent then
lavish jewel
just pick the first one you get lol
gritty swift
yeah i tried to do that and it diden't work though (see the image) prolly i'm just doing something wrong though
lavish jewel
why'd you sort them like that
what are you trying to do
btw, what you're doing won't always work
there is in general no relationship between eigen and singular values of a matrix
unless the matrix is symmetric
the problem was you got abs(singular values) because you squared (A^T A) and then took the square root
the s-vals were negative, prolly
lavish jewel
it's the natural guess π
gritty swift
it's the natural guess π
so just check negative and positive sqrt to find the right one? i guess that works
lavish jewel
the only relationship between the eig and sing vals of A is: the eigenvalues of A^T A and A A^T are the singular values of A SQUARED
so you unfortunately lose that sign along the way. maybe guessing, as you said
gritty swift
ok cool thanks!
lavish jewel
moonlight, do you know what the parametric equation of a line looks like?
almost, be careful with the signs
yeap
it's what you just did
points in 2D are of the form (x,y)
and you just found that x = -3 -3t and y = 1-2t
yeah
you can see this as "all the points (x,y) that lie on the line passing through P with direction d"
you can use a dot product to find the cosine of the angle
nocturne jewel
if 2 lines are parallel, they have the same direction vector
lavish jewel
nocturne jewel
(and the set of points dont match, ie the position vector of 1 line cant be a point on the other)
cause then they're coincidental lines
yes
[-1,2]^T
or row vectors if that's what your teacher uses
or just [-1,2]
lavish jewel
c[-1,2] for any real c, really
nocturne jewel
any R c except 0
lavish jewel
right so (2,-4) is c(-1,2) with c = -2
nocturne jewel
there's a correct option
lavish jewel
so it works
nocturne jewel
-2[-1,2]=[2,-4]
You can also just note that 1 entry is negative and one is positive, so whatever answer it is, it has to have 1 neg and 1 pos
and scalars*
finite geyser
Hello all π , i'm trying to change some transform matrices between coordinate systems (basically unreal engine (left-handed z-up) to opencv (right-handed)). The red arrows are whats is expected while the 3d plot is the result externally (wrong). What should i do to change between systems
[
{
"Transform": [
[ -0.95091152191162109, -0.017452528700232506, 0.30897003412246704, 50 ],
[ 0.016598338261246681, -0.9998476505279541, -0.005393133033066988, 0 ],
[ 0.3090171217918396, -2.7939677238464355e-09, 0.95105648040771484, 50 ],
[ 0, 0, 0, 1 ]
]
},
{
"Transform": [
[ 1, -0, 0, -50 ],
[ 0, 1, -0, 0 ],
[ 0, 0, 1, 50 ],
[ 0, 0, 0, 1 ]
]
},
{
"Transform": [
[ -0.77051305770874023, -0.63742417097091675, 0, 0 ],
[ 0.63742417097091675, -0.77051305770874023, -0, -50 ],
[ 0, 0, 1, 50 ],
[ 0, 0, 0, 1 ]
]
},
{
"Transform": [
[ 8.6426734924316406e-06, 0.24871107935905457, -0.96857762336730957, 0 ],
[ 0.80901980400085449, 0.56931018829345703, 0.14619439840316772, 50 ],
[ 0.58778119087219238, -0.78359979391098022, -0.20120728015899658, 20 ],
[ 0, 0, 0, 1 ]
]
}
]
Matrices in question
nocturne jewel
symmetric form is isolate for the parameter and set them equal
gritty swift
so you unfortunately lose that sign along the way. maybe guessing, as you said
I figured out something better then guessing, $A = USV^T \implies U^TAV = S$ :D
stoic pythonBOT
uli
gritty swift
(since U and V are orthonormal and they must be invertible since A^TA is symmetric so it has a full set of eigenvectors)
lavish jewel
i don't follow what you did
gritty swift
multiply the right by V and multiply the left by U^T
lavish jewel
ah
yeah well, you just diagonalized the matrix
that's the same as just doing the SVD straight up
so yeah, that works
idk about that one, sorry
idk what symmetric equations are
humble oak
hello just wondering if this is true,
T:V -> W is an isomorphism <=> V and W are isomorphic
sonic osprey
Is this not the definition of isomorphic?
How have you defined what it means for V and W to be isomorphic?
@humble oak
humble oak
my question was bad, it is most definitely the definition of isomorphic i'm just dumb :c
sonic osprey
nah you're good
finite geyser
Hello all π , i'm trying to change some transform matrices between coordinate s...
^ checking if anyone has any feedback in regards to my previous question
drowsy wadi
Anyone know how to find the covariance of multivariate guassian
wintry steppe
idk how to translate it but, it sais something like: Considering M 2x2 the vector space made by 2x2 matrixes. Guess its dimension and a base
hoary osprey
try finding a basis
gray dust
it may help to recall the usual rules of adding/scaling in that space before finding a basis
wintry steppe
try finding a basis
yeah but this is what i mean
[[1, 0], [0, 1]]
is the canonic base (?) idk if canonic exists, just the usual one
But i have 4 params on the matrix: a b c d
then i am also asked if some formulas are a norm of that vector space or not
like, the first one is a norm, and the second one isnt (cuz a matrix like [[-1, 0], [1, 0]], with that norm) measures 0, and the matrix is not the matrix 0
But idk about the dimension
wintry steppe
it may help to recall the usual rules of adding/scaling in that space before fin...
wdym?
hello
is anyone available to help w linear algebra 2 work
kind of lost
if you have f(x,v) where x is a non empty set and v is any vector space then does f have to be a linear transformation for f(x,v) to be a vector space?
sonic osprey
What
wintry steppe
if f is a set of all functions that f: x -> v
sorry i'll copy paste the question i wrote it p badly
sonic osprey
please don't use the same letter for different things
wintry steppe
Let X be any nonempty set and let V be any vector space. Let F(X,V) be the set of all functions f:XβV. Explain how to define the operations of addition and scalar multiplication on F(X,V) so that F(X,V) is a vector space under these operations
so i'm thinking that the operations of addition and scalar multiplication are the same as for linear transformations right
f(x+v) = f(x) + f(v)
and f(cx) = cf(x)
sonic osprey
No that doesn't make any sense
X is just a set
It doesn't make sense to add two elements of x together
or to scale x by a scalar
wintry steppe
i see, thank you. I'll look into sets i didn't see them mentioned in the lectures
i'm kind of confused
whats the difference between a nonempty set of vectors and a vector space
wintry steppe
its just a set
limber sierra
sets alone dont have any inherent structure - they're just a collection of elements
wintry steppe
i copied the question word for word here
limber sierra
vector spaces are what you get when you take a set and imbue it with vector space structure
if you JUST have a set, addition and multiplication dont make sense
that's the whole point of vector spaces
they make addition and multiplication make sense
wintry steppe
i see so how would a function turn a nonempty set into a vector space?
the function would have to make the set fulfill these two conditions right
fickle skiff
in your problem you want to define addition and scalar multiplication on F(X, V), i.e., you want to tell us how to add 2 functions f, g: X -> V
wintry steppe
so if x1 and x2 are vectors in X
to make it into a vectors space i would have to make it so
fickle skiff
you don't want to make X into a vector space here
you're looking at F(X, V) instead
sonic osprey
I mean you can add two functions
If it helps, maybe think of functions from R to R
(f + g)(x) = f(x) + g(x)
wintry steppe
ohh i see
i'm slow haha
so then would it be correct to say that the operations should be defined as (f+g)(x,v) = f(x,v) + g(x,v) and f(c(x,v)) = cf(x,v)
for it to be a vector space
sonic osprey
These are functions to X to V
they take in an element of X and return an element of V
So it's just f(x) and g(x)
but yes thats the right idea
wintry steppe
oh i see
thank you i get it now, i had the complete wrong idea lol
if channel is free... could someone help me? Q.Q
if its lin alg 1 i can maybe
wintry steppe
idk how to translate it but, it sais something like: Considering M 2x2 the vecto...
here*
idk
why do i need more than 1 matrix?
ah true
cuz my vector space if of matrixes
with mine i cant get the matrix 0 1 00
ok ok
so
the basis calle e
?
Like C00
????
dimension is 1 i think
mmm
the vector space C00
is (1,0,0,0,...), (0,1,0,0,0...)
and so on
it is infinity
but nvm, a basis would be 4 matrix with 1 on each position?
stoic pythonBOT
mirzathecutiepie
stoic pythonBOT
mirzathecutiepie
wintry steppe
no
wintry steppe
i meant this
so my space has dim 4, right? cuz yeah, it is a basis
yeah, canonical
sure, ty
wintry steppe
like, the first one is a norm, and the second one isnt (cuz a matrix like [[-1, ...
and am i right here?
forget that XD
wind pasture
Give examples of two DISTINCT bases for the vector space of polynomials with degree less than or equal to 3
wintry steppe
XD
wind pasture
can i just write 1, at and 1, at, bt^2 where a and b are constants?
are they distinct?
nocturne jewel
They arent bases for $\mathbb{R}[t]_{\leq 3}$
stoic pythonBOT
moshill1
wind pasture
ok
changing constants in front makes the bases different?
why is what i said wrong?
stable kindle
you missed t^2 in the first base
nocturne jewel
Pretty sure you're also missing t^3.. since it's deg less than or equal to 3
~~
nocturne jewel
yes, bases are sets which span and are lin indep
I originally mentioned they said deg less than or equal to 3
4:39pm, not an excuse S M H \j
brisk fractal
yes
non zero determinant implies that A is an isomorphism, so A is surjective. the column space must be equal to the codomain
tame mural
For matrices, how do you distinguish between an automorphism and an isomorphism? Or is the distinction purely theoretical?
tame mural
Huh! I thought automorphisms are <precisely> invertible endomorphisms
quartz compass
sorry idk why I said that lol
tame mural
its cool I'm trying to polish my definitions too
tame mural
ah well if you have a lead, I'll take it
robust pond
@rose cairn can you think of a way to represent a 2x2 matrix squared
maybe use that to make a system of equations
use classic algebra techniques on it
or matlab
robust pond
say i wrote $\begin{pmatrix} a & b \ c & d \end{pmatrix}$
stoic pythonBOT
jan Niku
rose cairn
we can square that
robust pond
what do you get
idk i guess its just clear you'd use this to make a system of equations
4 equations, 4 unknowns
so you can definitely draw a conclusion with a little work about how many solutions there are
rose cairn
4
rose cairn
wait what
robust pond
what do you get when you square that matrix
rose cairn
yup
robust pond
a^2+bc=96, ...
robust pond
π
rose cairn
There is no matrix that exist
tame mural
For matrices, how do you distinguish between an automorphism and an isomorphism? Or is the distinction purely theoretical?
wintry steppe
isomorphism = invertible linear map between two vector spaces
automorphism = invertible linear map from a vector space to itself
i guess they do coincide for matrices, since the domain and codomain must then be F^n (F the field from which your matrix pulls its scalars, and n the dimension of the matrix)
but on the level of linear maps you can't really say that anymore
if you wanted to, it'd be dependent on a choice of basis
the same linear isomorphism V -> W of n-dimensional vector spaces over a field F could give two different-looking automorphisms of F^n (they'd be the same up to a change in basis)
sharp idol
There is no matrix that exist
why do you say that?
rose cairn
Because when I solved the system of equations for the unknown a b c and d and squared the matrix
It didnβt equal that of x^2
dire thunder
well you have matrix and there is an isomorphism between matrices and linear mappings
but speaking more seriously
they just show that TS is also linear mapping if T and S are both linear
i mean see that you have
T(y_1,y_2) = (6y_1+7y_2, 8y)1+9y_2)
where (y_1,y_2)=S(x_1,x_2)=(x_1+2x_2, 3x_1+5x_2)
thus you have
TS(x_1,x_2)
that is, S maps (x_1,x_2) to (y_1, y_2) which is mapped to (z_1,z_2) by S
wintry steppe
I see so it makes a little triangle like the diagrams if you put them next to each other
dire thunder
yes
wintry steppe
But what this is is a proof to show it is a linear transformation
dire thunder
well it is like with all function compositions if you are familiar with that but here you have particular case
wintry steppe
I don't understand what would make it not a linear tranformation in this case
dire thunder
But what this is is a proof to show it is a linear transformation
just show that it mets defn of linear transformation
wintry steppe
if that makes sense
Yes they also do the definition after
but they said this was the "brute force" method
dire thunder
well e.g. if 0 would be mapped to not 0 it would be not linear mapping
dire thunder
but they said this was the "brute force" method
i mean they prolly shown before that for each matrix there is associated linear mapping
and they got matrix in the end
thus they got linear map
wintry steppe
here let me post the question itself
dire thunder
what i mean that by finding matrix they've found an explicit formula for your mapping
dire thunder
Oh I see so if they did not find a formula it would not be linear?
why
it would
oh wait
i mean they could do defn based proof
or find formula and do proof on it or just use that it is matrix
wintry steppe
Yes after this they do a definition based proof
Still very confused on how that first so called brute force method shows it is linear
soft burrow
without context, no idea
sudden nacelle
Abb is the covariance matrix for xb
lower case are vectors
I found this
but in this one the linear term is bTx
in the example above it's xT b
dusky epoch
$b^Tx = x^Tb$ does it not
stoic pythonBOT
Ann
sudden nacelle
why
how does that work
i think you're right
but i'm not sure why
oh
it's dot product
nvm ty
dusky epoch
yes
sudden nacelle
does it work for covariance matrix too
dusky epoch
?
sudden nacelle
does xT sigma^-1 u = x sigma^-1 uT
dusky epoch
you mean $x^T \Sigma^{-1} u = u^T \Sigma^{-1} x$?
stoic pythonBOT
Ann
sudden nacelle
yep
dusky epoch
if Sigma is symmetric then it should work
sudden nacelle
so it works for all symmetric matrices
thanks i've got one more question
where did the second term of this come from
I tried expanding the first term from the top
but i couldn't get to the second term
dusky epoch
wow what a mess
i have no idea and i'm not going to spend precious mental energy on this rn
wintry steppe
@drifting night here
native rampart
You are probably better off asking in the physics server
drifting night
lavish jewel
do you know how to do a linear regression?
also, depending on how you do it, this is a multivariable calculus question instead of linalg
sudden nacelle
how do i show a function is well defined
tame mural
how do i show a function is well defined
a function is well defined if it has a mapping from every input
limber sierra
a unique mapping, in the sense that its impossible for f(x) to equal, say, both 5 and 7 for the same x
[unless 5 = 7]
somewhat more precisely: a relation f: X -> Y is a well-defined function if:
- for every x in X, f(x) = y for some y in Y
- for every x, x' in X, x = x' implies f(x) = f(x')
unkempt light
Guys question, when a system has more unknowns than equations, then the system has infinitely many solutions. Does that only apply to homogeneous linear systems or also regular linear systems?
dusky epoch
it is possible for a system with more unknowns than equations to have no solutions at all
but if it does have at least one, then it has infinitely many
very simple example: $\begin{cases} x_1 + x_2 + x_3 + x_4 + x_5 = 2 \ x_1 + x_2 + x_3 + x_4 + x_5 = 7 \ 4x_1 - 7x_4 + x_5 = 0 \end{cases}$
stoic pythonBOT
Ann
unkempt light
So it's better to solve and see when the system is homogeneous.
Thanks for the help!
lavish jewel
remember you can write the line given a point p and a direction d
they chose p = (2,4)
how do you think you could come up with a direction d?
wintry steppe
hi. How can i check if a norm is a norm on an infinity space?
actually, on the space made by all polynomials of 1 variable?
i have to check if this is a norm https://gyazo.com/c8eaeb12dc4cdc41e4c2f3fdacfe9207
which i think it is, but idk how to prove for infinity spaces
ot this space isnt actually infinity?
dire thunder
i mean defn of norm does not differ for finite and infinite dim spaces
wintry steppe
But for polynomials, to prove the triangular inequality, if i suppose A until n, and B until m (as degrees), the n>m, i do the sum until n and then i do the norm of B with the coeficients from n-m to m?
dire thunder
it should have the same props for all spaces
wintry steppe
i mean defn of norm does not differ for finite and infinite dim spaces
yes yes, i know, i was struggling with how to to the sum if it was inifity
dire thunder
it is not infinity
wintry steppe
i know ^^'
dire thunder
But for polynomials, to prove the triangular inequality, if i suppose A until ``...
ye
take two polynoms
WLOG assume they have the same deg
otherwise just extend smallest one by zeros
wintry steppe
what does wlog mean?
dire thunder
without loss of generality
wintry steppe
ah okey
dire thunder
otherwise just extend smallest one by zeros
i mean because of that
wintry steppe
i mean because of that
oh true
i could do that too
thanks
and to say something is not a norm, with an example is enough right?
restive moon
can someone help with a linear question?
wintry steppe
ask
dire thunder
and to say something is not a norm, with an example is enough right?
i mean yes if you manage to find counterxample you are done
restive moon
what does this mean
dire thunder
it means that you should find formula for B^n
dire thunder
try apply defn of B^n
wintry steppe
it is asking how will ur matrix look like if u do B^2, how will it look like if u do B^3
and so on
restive moon
hmm okay
wintry steppe
i mean yes if you manage to find counterxample you are done
https://gyazo.com/b3b0960d0e141031cf025be43b61ce91 this norm is the inifinity norm for polymonials?
dire thunder
this is limit of p-norm as p to infty
wintry steppe
stoic pythonBOT
Commander Vimes
$$\left( \sum_{i=0}^{n-1} |a_i|^p\right)^{\frac{1}{p}} = \norm{x}_p$$
Commander Vimes
wintry steppe
does this match my norm? like, i dont see any sum on mine, nor exponents
wintry steppe
still, this one would be the sum of all the coeficients
dire thunder
their modules*
wintry steppe
well, yes
wintry steppe
https://gyazo.com/b3b0960d0e141031cf025be43b61ce91 this norm is the inifinity no...
but i was asking of this
the teacher miss wrote 2 |
dire thunder
**Commander Vimes**
```latex
$$\left( \sum_{i=0}^{n-1} |a_i|^p\right)^{\frac{1}{...
it is what you get letting p approach infty
well you can just define it to be such norm
wintry steppe
it is what you get letting p approach infty
sorry, i dont see it
dire thunder
but also it is limit of p-norm as p to infty
wintry steppe
on the p-norm i have a sum. Here i have just the max
dire thunder
sorry, i dont see it
assume e.g that |a_0| is max. then factor a_0 and see that other terms go to 0
royal relic
Why are non leading variables considered free variables when there are more unknowns than equations?
I just know that non leading variables are free variables, idk why tho
Could someone please explain
lavish jewel
the non-leading ones end up being defined by rows that contain only zeros
that corresponds to 0 = 0, which is always true
the variable corresponding to that row (when reduced) can then take any value, as it does not change the fact that 0 = 0
the leading ones, as you call them, do participate in equations that are not all 0, so they must satisfy some specific relationship
royal relic
the non-leading ones end up being defined by rows that contain only zeros
oh
that makes much more sense
thank you
sharp idol
I have a question. With normal absolute values, is there a way to show diagonally dominant matrices do not have a determinant of 0?
Note: I am only interested in a method that uses determinants alone. I know it can be done in an easier way.
blissful vault
am i misunderstanding something
or for c)
Q inverse Q = identity matrix
so T beta = T alpha?
sharp idol
No no, keep in mind matrix multiplication is NOT commutative.
So $A^{-1}BA\neq BA^{-1}A=B$
stoic pythonBOT
dackid
wintry steppe
pallid rampart
You mean M_n(R)
blissful vault
ah ok π€£
wintry steppe
You mean M_n(R)
i don't work with things that aren't lie groups
don't "M_n(R) is Lie(GL(n, R))" me
sharp idol
I see you have a preference
desert light
hey guys, I need some help with this question
desert light
ummm i know you can do the cross product of a vector, not sure about a scalar
sharp idol
Absolutely not. Vector multiplication, cross or dot products, is with vectors only
In other words, the expression you have is absolutely meaningless
desert light
alright so cross product - dot product x a vector is not possible?
desert light
oh alright thanks!
sharp idol
You bet.
brisk fractal
it's the l-p norm
you can think of l-p(S) as the space of functions that have bounded total sum of outputs on S
and it turns out that every hilbert space is isomorphic to some l-2(S) for some set S
that would be a banach space
although yeah hilbert spaces are also complete
(well not necessarily, just usually)
no, just different
hilbert spaces are infinite-dimensional inner product spaces with a complete orthonormal system
to be technical
prehilbert spaces are not complete sorry
if anything I hate hilbert spaces
they're necessary but kind of annoying to work with because of their tendency to be infinite dimensional
banach spaces are just complete normed vector spaces
maybe I am unqualified to speak on this
complete in the case of hilbert spaces is that the orthonormal system has a dense span in the hilbert space if I remember correctly, which implies analytic forms of completion
it's weirdly complicated
the closure of the span of the "basis" is the entire space
closure is the set of limit points
I need to some exercises again on this stuff too lol, my memory is very foggy
stoic pythonBOT
slimvesus
coral sage
does the fact <x, Ax> = <A^T x, x> have a name? not a very easy thing to google
YES! thank you @wintry steppe I was looking for the adjoint
couldn't remember that word for the life of me
compact relic
How do I show that the given set W is a vector space?
what I attempt is turning it into Ax = 0 form and make the argument that it is a subspace of R^2
[2 -1 0 -3 [x1 [0
0 1 -1 -2 ] x2 0 ]
x3 =
x4 ]
gray dust
@compact relic looking at Ax=0, x has 4 entries, so ker(A) is a subspace of R^4, not R^2. also rewriting W as ker(A) is cheating a bit since we're taking for granted that ker(A) is a subspace. we can show W is a vector space by the vector space axioms, but noting W is a subset of R^4, our work can be cut down if we instead just show W is a subspace of R^4, which has easy-to-check criteria
humble oak
am i tripping, when i normalize this i need to multiply the matrix by sqrt(3/2)
right?
my textbook is multiplying by sqrt(2/3)
stable kindle
matrix?
stable kindle
the magnitude is sqrt(3/2), so to normalise you divide by sqrt(3/2), ie. multiply by sqrt(2/3)
yes? no?
humble oak
ahhhh yes
wary lily
yes
humble oak
i was missing
the the whole fraction
i forgot it was 1/magnitude
thank you very much
blissful vault
when they say devine v0 = 0
is it asking me to define it, or is it saying that i can assume that v0 = 0
blissful vault
ok thanks
plain saffronBOT
humble oak
in this video, the prof shows that P = zero vector to prove that the linear transformation T is 1-1, why is this enough to show T is 1-1?
empty copper
It's a well-known theorem that a linear transformation between vector spaces is injective if and only if its kernel is trivial
wintry steppe
can anyone help me with this?
Find a basis {p(x),q(x)} for the vector space {f(x)βP3[x]β£fβ²(7)=f(1)} where P3[x] is the vector space of polynomials in x with degree less than 3.
is there an efficient way to do this or do i just guess and check
i have one possible equation x^3 - 3x^2 + b = 0
but i'm pretty sure i'm on the wrong track
humble oak
thanks @empty copper
empty copper
np
trim vortex
can anyone help me with this?
Find a basis {p(x),q(x)} for the vector space {f(x...
I think if you take a general polynomial with coefficients a,b,c,d, you should be able to find what conditions these must satisfy to be in the vector space. from there, you should be able to find the basis polynomials more easily
wintry steppe
oh ok i'll try that thank you
trim vortex
np
nocturne jewel
Can Gershgorin actually be used to estimate eigenvalues or is it just a region in the complex plane where they exist?
tame mural
What is an expected problem that a student might encounter if you treat vectors and matrices as the same?
nocturne jewel
What is an expected problem that a student might encounter if you treat vectors ...
multiplication
also vectors im assuming you mean R^n vectors?
tame mural
Yeah
or are saying for C^n it doesn't work?
because if you think of vectors as 1 Γ n or n Γ 1 matrices, matrix multiplication will still work
nocturne jewel
right but matrices have 1 type of multiplication, cartesian vectors have 2
trail dirge
V=M3(R), and S is the subspace of all vectors in V such that the trace is zero and the sum of entries in the first row is zero find the dimension of S
Can someone drop some hnits on this?
desert rapids
digital bough
because if you think of vectors as 1 Γ n or n Γ 1 matrices, matrix multiplicatio...
Ive treated it as matrices where the 1xn matrix represent the basis and the nx1 matrix represents the coordinate-matrix.
Or uh, i may have misunderstood the question
lavish jewel
do you mean quatratic forms with symmetric matrices?
blissful vault
i'm not sure if i understand correctly
usually i have to prove that T(u+v) = T(u)+T(v)
and then that T(kv) = kT(v)
but here it seems like T(u)=T(v)
since they use f(2), no matter what x it's always gonna be f(2)
spring pasture
but here it seems like T(u)=T(v)
Actually u have to take f1 and f2
T(u(x))=u(2)
And T(v(x))=v(2)
They are same when u(2)=v(2)
To prove
T(u(x)+v(x))=T((v+u)(x))=(v+u)(2)=v(2)+u(2)=T(v(2))+T(u(2))
tame mural
Is [3] a shorthand for { 0, 1, 2 } ?
dusky epoch
if your book defines it to be
trim moat
@blissful vault be careful to note that T maps from the space of polynomials of degree two or less to matrices in R^{2x2}--not simply to R
I'm assuming your class will allow you to use properties of calculus as axiomatic, so you should be able to find that T is linear by using the fact that differentiation is linear
wary lily
I have $\frac{b}{2} \neq a$ for no solution and $\frac{b}{2} = a$ for infinitely many solutions, but there is no combination of a and b where the system gives exactly one solution bc they are parallel.
stoic pythonBOT
az
unkempt light
Is there any way to know if a matrix is invertible other than the fact that the Determinant should not be 0?
raw sand
Sure there are
If you did gaussian elimination then youd get some form of the identity matrix
or rank(A) = n (nxn matrix)
or the nullity of A is 0
theres plenty more
just google invertible matrix theorem
lavish jewel
there is also the question of what you mean by "invertible"
sometimes matrices are invertible w.r.t. rows or columns, but not both
when the matrix isn't square
lavish jewel
not quite
the pseudoinverse is a "back-projection" onto the row space
it's only equal to the original right-multiplied vector if the null space contains only the 0 vector
but the operation exists regardless
it just leaves you in a subspace of the row space
Ax = b -> x_rowspace = (A^T A)^-1 A^T b, yeah
when x is multiplied with A from the right side
if the null space contains only the 0 vector, this pseudo inverse gets x exactly
and so the matrix is invertible for vectors multiplied to the right of A
even if the null space spans a larger space, you can still do the pseudo inverse
but then x = x_rowspace + x_nullspace
and then it is no longer invertible for x multiplied to the right of A, since you cannot get this x_nullspace back
weak glade
I'll be really thankful if someone could verify this proof. I'm not 100% sure it is right. I mention that i've proved that these sets are posets, just the part with morphism interests me. Thanks in advance
wintry steppe
a polynomial
wintry steppe
well, i just need to calculate its norm 1 and norm inf
norm 1 is n and norm inf is 1?
sonic osprey
yes
wintry steppe
ty
what does it mean "prove both norm are not equivalent"?
like, both are p-norm
right?
sonic osprey
Do you know what it means for norms to be equivalent
sonic osprey
yes
wintry steppe
so for this to be
a and b should be 1/n
why they cant?
cuz 1/n does not belong to my succession right?
sonic osprey
your scalars are probably real numbers or something
wintry steppe
it said they have to belong to E
wintry steppe
Oh
then are both equivalents or not? cuz 1/n is real number, and if a = b = 1/n
then 1 <= 1 <= 1
which is true, so they are equivalents
but problems sais "prove they are not equivalents"
and 1/n is > 0
sonic osprey
The problem is you're only doing that for a single x
you have to show that this is true for all x for the norms to be equivalent
wintry steppe
what? no no no
wintry steppe
it is refering to norm 1 and norm inf with the polynomials from the previous exercice
sonic osprey
Yes I understand
Again, you need to show that those inequalities are true for all x, not just the single P(t) you had in your previous question
wintry steppe
what all x?
sonic osprey
From what I'm understanding, it's not?
How is P(t) generic
It's just a single vector in E
sonic osprey
Again, the elements 1,t,t^2, t^3,... don't form a vector space
wintry steppe
then idk what to do
sonic osprey
Figure out what your vector space E is first
wintry steppe
a0 + a1 * t + a2 * t^2 and so on?
sonic osprey
idk read your problem
sonic osprey
Right
wintry steppe
so then it sais, consider P(t) the polynomial succession of SUM(t^k) k=0, n
calculate norm 1 and norm inf
sonic osprey
Sure
sonic osprey
Yes that is true
wintry steppe
okey, and then it sais "prove they are not equivalent"
sonic osprey
right
wintry steppe
and here is where idk what to do
sonic osprey
To show they're not equivalent, you need to show that for any choice of a and b, there's some element x of E that makes the inequality not true
wintry steppe
i cant find any
wintry steppe
i am thinking XD
sonic osprey
If I give you a = 0.5 and b = 1, can you find a vector x that violates the inequality?
sonic osprey
I'm not sure what you mean by that
sonic osprey
like a0 + a1 * t
how is this a counterexample?
wintry steppe
sharp idol
Yep, that generalizes to normed spaces. I know exactly what you are referring to
wintry steppe
how is this a counterexample?
sorry i meant 1+t
norm inf of this is 1, while norm 1 is 2
sonic osprey
Well not quite cause that makes them all equal
sonic osprey
Ah okay I switched the norms
wintry steppe
u can switch them, right?
sonic osprey
Yes, either order is fine
wintry steppe
so this is fine?
sonic osprey
I was thinking about the other order
Yes, but you've only proved a single case
You need to show that
No matter what a and b I give you, then you can find a polynomial that violates the inequality
wintry steppe
isnt 1 counter example enough to prove something?
sonic osprey
Depends on what you're trying to disprove
wintry steppe
like, if u say something works for everything, finding 1 who doesnt is enough, isnt it?
sonic osprey
If you're trying to prove that something holds for all (), then a single counterexample is enough yes
sharp idol
So $||a_0+a_1t||_\infty=\max{a_0,a_1t}$ if I am understanding what is happening correctly.
Alright, I'll let the peep talking after Max carry on. Sorry for the intrusion.
sonic osprey
But what is happening here is that the norms are equivalent if you can find any a,b that makes the norms satisfy the inequality
wintry steppe
lmao zoph got called max
stoic pythonBOT
dackid
sharp idol
lmao zoph got called max
Fixed-ish xp
wintry steppe
If you're trying to prove that something holds for all (), then a single counter...
so for our case, isnt 1 counterexample enough? like, problems sais "prove they are not equivalent". I think showing one that doesnt accomplish it, is enough. i may be wrong tho
sonic osprey
Again, two norms are equivalent if you can find any a,b that makes all x satisfy the inequalities
So you can't disprove this with a single counterexample
sonic osprey
Because maybe there's some other a,b that works
wintry steppe
same but to t^10
wintry steppe
increase the n
wintry steppe
like... they are not equivalents cuz this space is inf-dimensional?
sonic osprey
i mean
its true that all norms are equivalent for finite dimensional spaces
But norms can be equivalent on infinite dimensional spaces too
just consider some norm ||.|| and like 2||.||
wintry steppe
but what exactly do you need to increase it to
i mean, what i need to do is taking n = max(a,b)
sonic osprey
exactly
sonic osprey
uh what
sonic osprey
Not sure what you're saying
wintry steppe
if i can find a and b that a * N1 <= N2 <= b * N1
it means they are equivalents, no?
sonic osprey
But you can't?
The whole thing you've been trying to show
is that for any a,b you take, you can find something that violates that inequality
sonic osprey
You can always find a vector in your vector space that violates the inequality
which is what we were looking for
sharp idol
iirc, Norm inf should be N2
But it sounds like they are saying comparability is not relevant here.
wintry steppe
actually no xD that makes it harder
tame mural
What's a common way to indicate the subset of naturals smaller than x? [x]?
wintry steppe
okey so what he was saying i think is: i have 1+t+t^2
dire thunder
S_x
wintry steppe
norm 1 = 3 and norm inf = 1
stoic pythonBOT
Commander Vimes
wintry steppe
so a * norm inf <= norm 1 <= b * norm inf
i can pick a = 1 and b = 4 and it satisfy the inequality
dire thunder
what are your defns of norms
wintry steppe
but i can go further with my polynomial, and go to t^5
then b will need to increase too, but i can keep making P bigger
is this what u mean@sonic osprey ?
sonic osprey
Again no
wintry steppe
Q.Q
sonic osprey
You don't get to pick different a,b for every different polynomial
Two norms are equivalent if you can find a,b such that the inequality holds for all polynomials
wintry steppe
so the only way a,b hold for all polynomials, is if b = inf, cuz i have inf polynomials, but this isnt valid
this?
sonic osprey
it's not really about having infinite polynomials again
sonic osprey
Again
the point is that for any a,b, you can find a polynomial that violates the inequality
@wintry steppe this is not the right channel
wintry steppe
yes but, i can find find it because i have infinity polynomials. I dont know why u keep saying this is not the reason :/
sonic osprey
Okay
Let's consider two different norms
Consider ||.||_1 and 2||.||_1
these two norms are equivalent even though you have infinitely many polynomials
dire thunder
sonic osprey
@wintry steppe Please read the channel description, this is for undergraduate level linear algebra. Move to #precalculus
So the reason must be something other than "infinitely polynomials"
you need to use some actual facts about the 1 norm and inf norm
You need to use what we said earlier about the polynomials P_n
wintry steppe
well, i dont see it
wintry steppe
these two norms are equivalent even though you have infinitely many polynomials
this is what i dont