Given a SVD $A = U \Sigma V^T$, for the column vectors in $U$ - does it matter what order they are in? Does the order affect the order of the columns in $V$ in any way?
If $A$ does not have full rank, then there will be columns in $V$ that's in $\ker A$ and some columns in $U$ maybe be in kernel of $A^T$, right? Does the ordering of these vectors in $U$ affect the ones in $V$? I don't think so, right?
but this tells us that, because the Euclidean tilings are square (4,4), triangle (6,3), and hexagon (3,6), those are the three integer-sidelength rectangles with area equal to perimeter
and the five (integer sidelength) rectangles with area less than perimeter correspond to the Platonic solids
@LibertarianFeudalistBot there could be many. maybe your question is messy and no one wants to read it. Or if your question is written well, maybe your problem is just hard. Or maybe its just your luck that no one clicked on your question. It happens.
There could be other reasons I'm sure
you can try to see if this list helps. You can also post your question here and someone might click
Given the gram matrix $ \begin{bmatrix} 1&0\\ 0&-1 \end{bmatrix}$ which is nothing but the one dimensional minkowski metrik. the question states, find bases in $\mathbb{R^2} \times \mathbb{R^2}$ And considering the "unknown" bilinearform $b$ which belongs to this matrix. The question states, find basis $(v_1,v_2): \forall \: i \; b(v_i,v_i) > 0 ,b(v_i,v_i)<0, b(v_i,v_i)=0 $ Thus we conclude, that the definitness of a bilinear can not be examined n a basis.
so this is what i meant last night, especially, the last comment.
Leslie, do you have good things to say to remove my confusion, i am considering opening a question about this. Since i am still deeply bothered by this, and i have no grasp over it. and i am not sure i understand what you mean.. why is a diagonal matrix anydifferent from other matrix to another base, they are all isomorph.
IE i cant write the matrix back in none diagonal form.. i dont understand......
well that's not going to work at all, you need to pick a definition
if there's no starting point, there's no explanation except, go around and read all the various sources together... that's time consuming and we can't really help
Also quick interruption, the determenant is none basis dependent right? but what happens if i choose a base with reversed orientation, will i get a minus sign in the determinant?
maybe an analogy would help. copper's analogy was about the determinant. it's a function of matrix entries. but it's also invariant under a change of basis.
be careful there, "orientation" itself is often defined in terms of determinants.
if {v_1, v_2} is an ordered basis of R^2, and M is the matrix of a linear transformation on R^2 with respect to that basis, and N is the matrix of the same linear transformation with respect to the basis {v_2, v_1}, then det(M) = det(N)
well stepping away from 'volume' measurement, the determinant it's also a function of matrix entries, right? and matrix entries generally do depend on the choice of basis
but the determinant doesn't
it's maybe surprising that it doesn't, but it doesn't
Alright so if i understand you, returning to our example
Are you saying, that the bilinear form to the matrix i given, has an intrinsic properity of definitness (or none for that matter) that is given the actual naked explicit form of b, can be calculated, and this is for all basis the same?
Okay so the REVERSE engineering the defeitness from the GRAM matrix is not possible, however, if we have explicit form, then the definitness is unique and stays so nomatter change of vision (basis )
the numbers b(v_1,v_1) and b(v_2,v_2) are the diagonal entries of the matrix of b with respect to {v_1,v_2}. you might call the full matrix the gram matrix with respect to {v_1, v_2}
those numbers are not the entire gram matrix
i might regret this, but stepping one level of generality away, the point of the exercise is that the diagonal entries of the gram matrix of b are not, in general, enough to tell you whether b is positive definite, or negative definite, or some other mix
Leslie, is it not that the digonalized matrix is equal to the undiagonalized * some transformation matrices, you mean this operation messes with our result?
you absolutely can tell what is going on with b's definiteness if you have b(v_1,v_1), b(v_2,v_1), b(v_1,v_2), and b(v_2,v_2) - i.e. the full gram matrix. but that's not what this exercise is focused on. it's looking only at the diagonal entries
mad: you generally can't tell how a matrix M will diagonalize if you only know the diagonal entries of M
if i said, i have some matrix, and it's diagonalizable, and it has all zeros down the diagonal, how does it diagonalize, the answer is "who knows?" there's not enough information. it's partial information about the eigenvalues but not enough to say what they are
similarly if all i tell you about a bilinear form is the diagonal entries of its matrix with respect to some basis, you can't determine the eigenvalues of that matrix from that, and so you can't determine if the form is positive definite or singular or what its signature is
the exercise above seems to be oriented toward leading you to see it in a concrete example
"oh wow, with the same simple bilinear form, if i change what basis i use, i can make the diagonal of its gram matrix look like all sorts of things." -> leading to the message "i guess the diagonal of the gram matrix with respect to some random basis won't be enough to tell me much about the form"
Alright leslie, i guess thats somewhat satisfying, thanks!. i will look further into it, i seem to not find alot of search results in english, am i naming this wrongly? how do you call it?
Is there a name for the following equation? I want to be able to look it up to be able to characterise it correctly:
$n(x,y,z)=d\cdot(1-x^2/a^2-y^2/b^2-z^2/c^2)$, where $a,b,c,d$ are constants.
At first I thought it was an inverted parabola, since if it was missing a dimension it would be one, then I thought maybe it's a type of ellipsoid, but n varies. So now I am at a loss of what to look up in order to understand it properly.
In the Taylor series $x^0=1$ for all x is not 0 but why do we say $0^0=1$? I know it makes sense for $x^x$ when x approaches 0. It still bothers me if it is okay to do this.
it's just a useful convention. there are reasons unrelated to that limit to adopt the convention. it makes it easy to state the binomial theorem without qualification, for example
and other expressions involving polynomials. i would include taylor series in that
@leslietownes Thanks for mentioning this. So far it's been better than the webpage I found (which appears to be a forum post anyway) at visualising the equation.
user: the level set n(x,y,z) = k seems to be classifiable in terms of the sign of k/d - 1. if it's 0, you get a point. if it's negative, you get an ellipsoid that can be rewritten in 'standard form' by rescaling a, b, and c. and if it's positive, no solutions.
as for visualizing those ellipsoids, i dunno. they all appear to be similar to one another, with the relative scale of the axes established by a, b, c. and as k goes to d from below, they get smaller and go to a point.
why is there notation for the resolvent R(x, T) when $(x-T)^{-1}$ is perfectly fine, and as a bonus no one needs to guess what your convention for the resolvent is ((T-x) vs (x-T))
calvin: you're welcome to use that. some people find that stuff like the resolvent identity and other identities become easier to grok if you have a name for that.
e.g. if A and B are not going to commute you still might want to recognize R(z,A) and R(z,B) when you see them even if you are generally not comfortable with formally inverting more complicated expressions in A and B.
not: if you would like to be able to write a_0 + a_1 x + a_2 x^2 as sum_{k=0..2} a_k x^k then you need to adopt the convention that 0^0 is 1. or you can treat the constant term separately and not adopt the convention. it doesn't really matter.
same with the binomial formula (a + b)^n = sum k=0..n C(n,k) a^k b^{n-k} which looks great although if you are uncomfy with interpreting a^0 and b^0 in general then maybe you would have to write both the k = 0 and k = n terms separately.
but if you adopt the convention then you don't have to treat those things separately.
just pointing out that you might want to do that even if you have never heard of limits, or don't care about what the limit of x^x is as x goes to 0. in adopting the convention you can just be adopting a notational convention, not deciding what limits are.
"0^0" in calculus class is what people call an 'indeterminate form,' i.e. generally knowing that f goes to 0 and g goes to 0 as x goes to c isn't enough to tell you what f^g goes to as x goes to c. so even the fact that x^x goes to 0 as x goes to 0 is something of a happy coincidence.
does someone know about marty's theorem in complex analysis?
We have the following form. Let F be a family of meromorphic functions on \Omega where \Omega is open and connected. then F is spherically normal iff F^#=\{f^#: f\in F\} is locally bounded. We denote by f^# the sperical derivative of f.
But now I asked myself don't we need some more information to speak about local boundedness. So I mean where is ist locally bounded? Would't it make more sense if we say locally bounded in \Omega?
Let $F=\{z^n\}$ be the set of complex functions and I want to show the following 3 claims:
$F$ is normal on $\Bbb{D}$
$F$ is normal on $\Bbb{C}\setminus \Bbb{\bar D}$
$F$ is not normal on $|z|<2$
In the lecture our prof gave us the hint to use Marty's theorem. So let me quickly introduce our ...
So about this polar topology, I switched to absolute polars instead because I've noticed that they only really consider balanced sets anyway. So that's that. I think they just had absolute polars in mind from the start.
If there is some counter-example where using real polars doesn't generate a Hausdorff locally compact TVS then feel free to share
(I believe the greatest issue here is proving whetever the multiplication is continuous)
why do people write $\frac{\delta F}{\delta \rho}(\rho)$ for the first variation of a functional $F$ at $\rho$, like why not just $\frac{\delta F}{\delta \rho}$
Let $Y \subseteq X$ be (discrete?) spaces, and let $i_X : X \to \beta X$ be the map associated to the Stone-Cech compactification of $X$. Is it true that $\beta Y$ can be identified with $\overline{i_X(Y)}$?
It seems like it should and it's just about chasing definitions down and the universal property, but I just wanted to cech (pun intended).
Hello if we have a function $f$ defined on some space $V$, and $V$ is a direct sum of others, how would you express f on the direct sum? $V= \bigoplus M_i$
Say something like $f_{\big| \bigoplus M_i}= f_{\big|M_1} + ... ?$
if we defined the restriction to be the identity outside.. and then use composition.. maybe it works., problem would be when you reach the function where the point is not ouside.. then you just get a new element, this new element might be in one of the sets. if our function goes into the same set, so we have a problem.
Define $f_i$ as functions indexed by $i\in I$.
Is it ok to say that $\bigoplus_{i\in I}f_i = f=(f_1,f_2,f_3,......)$?
I understand that the direct sum is often used in vector spaces or algebra structures. Is the usage here in functions common? I have never seen this kind of usage before, though.
Well if you have $f:A\to C$ and $g:B\to C$ you can define $h:A\oplus B\to C$ by $h=(f,g)$, but if $f:A\to B$ and $g:A\to C$ you can define $h:A\to B\oplus C$ by $h=f+g$
Man i swear to god i just wanted to prove this statement and i stopped at this question i want to calculate the gram matrix of a bilin form $f$ inductively on sub spaces, whose direct sum produce the vector space. so i am trying to get to the result where i can say, the left matrix is equal to the direct sum of matrices on the right
Because, wikipedia is telling me thats the answer lol!!
Now, let's do something weird: let's flip the colors in the top-right quadrant. We get this image. If we focus on each color individually, we see that black looks like this and red looks like this. They're each wacky-looking single spirals now!
So the union of two double spirals equals the union of two wacky single spirals!
the main problem was i want to show the matrix of a bilin form, has a spesific construction, for that, the proof i am doing right now, which lead me to this point, is i can write the vector space, as a direct sum of smaller ones, and i know how the matrix of the form looks on those smaller ones. so according to the formula that if weh ave matrices then the direct sum of those, are hte matrices on a diagonal of a bigger one, which is actually the final form of the matrix of te function.
However in order to reach the last result, i need to show that the matrix on the left, is truly the one on the right, so this means, the matrix of the function on the direct sum, will equal the direct sum of the matrices.
Get it ?
So if we denot A as the matrix for f, i want to show that $A = A_1 \oplus A_2 ....$, where as the $A_i$ are the restrictions on the smaller subspaces whose addition result the main vector space
Now i have seen the final result, and it is actually TRUE, that $A$ will have that form, however i am stuck at the part where i write the function as constraint on the subspaces, and dont know how to procceed..
I'm telling you you have to look at what the actual mapping is and see if breaking it into pieces is valid. Certainly if $V=V_1\oplus V_2$ and $f\colon V\to W$ is linear, then $f(v_1+v_2) = f(v_1)+f(v_2)$. There is no more to say.
Is it okay to rearrange the Fourier series? After learning a bit about how scary it is to rearrange infinite series in calculus I find it disturbing to even think about it. I don't understand how should I prove if it is absolutely convergent.
@Ted Shifrin Hi, in this answer math.stackexchange.com/questions/3894028/… you state that $\frac{\partial\phi}{\partial x} = F(\tfrac yx) -\frac yx F(\tfrac yx) \quad\text{and}\quad \frac{\partial\phi}{\partial y} = F(\tfrac yx)$ but shouldn't it be $\frac{\partial\phi}{\partial x}=F(\frac{y}{x})-\frac{y}{x}\frac{\partial F}{\partial x}$ and $\frac{\partial\phi}{\partial y}=\frac{\partial F}{\partial y}$? What am I missing? Thanks
@lorenzo There certainly were primes missing. I don't know how that happened. But, no, $\partial F/\partial x$ and $\partial F/\partial y$ make no sense. $F$ is a function of a single variable.
As robjohn said, the usual definition is to split the doubly infinite sum into two infinite sums, one with negative terms, one with positive terms. So you "assumed" that.
@TedShifrin I see, so $F'(\frac{y}{x})=\frac{dF(\frac{y}{x})}{dx_1}$ and, for example, $\frac{\partial \phi}{\partial x}=\frac{\partial }{\partial x}\left(xF(\frac{y}{x})\right)=F(\frac{y}{x})+\frac{\partial F}{\partial x}=F(\frac{y}{x})+x\frac{d F}{dx_1}\cdot\frac{\partial x_1}{\partial x}=F(\frac{y}{x})+x\frac{d F}{dx_1}\cdot (-\frac{y}{x^2})=F(\frac{y}{x})-\frac{y}{x}\frac{dF}{dx_1}$ right? (You're welcome, and thank you for the help in understanding the exercise.)
Careful. What you have written makes no sense. Writing $\dfrac{dF}{dx}\big|_{x=a}$ is very different from $\dfrac{dF(a)}{dx}$. The prime notation is better.
You’re still writing $\partial F/\partial x$, which makes mo sense. STOP!
There’s a composition of functions. Break your terrible calculus habits of using the same letter for a function and for its composition with another function. This is horrible single-variable habits!
so if I understand correctly I cannot write $\frac{\partial F}{\partial x}$ because $F$ doesn't depend directly on $x$, so would a notation like $\frac{\partial (F\circ x_1)}{x}$ make more sense?
I seem to be forgetting some basic general topology and I am hoping someone can correct me. Let $X$ be a Hausdorff space and $A$ a subspace. Is it true that $X/A$ is Hausdorff if and only if $A$ is closed?
Write $1\Bbb Q$ for the indicator function of the rationals. Then $1\Bbb Q$ has a symmetric derivative at every rational but not at any irrational
(the symmetric derivative is $\lim (f(x+h)-f(x-h))/(2h)$)
@ShaVuklia It defines it in the next sentence: "The latter is the quotient map identifying the basepoint of $S^1$ with its antipodal point"
(this is the same as Astyx's drawing)
I suppose this is $\theta\to\begin{cases}(2\theta,0),&0<\theta<\pi\\ *,&\theta=0,\pi\\ (2\theta,1),&\pi<\theta<2\pi\end{cases}$, or something like that
Let $F=\{z^n\}$ be the set of complex functions and I want to show the following 3 claims:
$F$ is normal on $\Bbb{D}$
$F$ is normal on $\Bbb{C}\setminus \Bbb{\bar D}$
$F$ is not normal on $|z|<2$
In the lecture our prof gave us the hint to use Marty's theorem. So let me quickly introduce our ...
Define $f_i$ as functions indexed by $i\in I$.
Is it ok to say that $\bigoplus_{i\in I}f_i = f=(f_1,f_2,f_3,......)$?
I understand that the direct sum is often used in vector spaces or algebra structures. Is the usage here in functions common? I have never seen this kind of usage before, though.
