(I have no idea what definition of tensor product you're working with, but you should be able to sort it out, if necessary, by using $W \cong (W^*)^*$.)
Right, for a simple tensor, yes, and then extending bilinearly.
Hmm... more generally, then, still assuming $V$ and $W$ are finite-dimensional real vector spaces, does $V\otimes W$ have a "nice" interpretation, too? Or do I have to use the fact that $V\cong V^{**}$ to make sense of it?
hi @Semiclassic. You can think of it as a map (acting on elements of $V^*\times W^*$), or you can think of it as a "bivector" (especially when $V=W$). You can think of it as vectors in $V$ that are "twisted" by vectors in $W$. This will make more sense when you get further in certain fields of math. :)
Studiosus commented on my answer, told me a source about how to find a manifold that has no affine torsion-free connections that are locally symmetric.
@tylerl-uxai I could parse that sentence a year ago. I started learning calculus about six, seven years ago. One could take a less circuitous route than I did, if they wanted.
@tylerl-uxai I don't see what its transcendence has to do with anything. All exponential are equivalent up to scaling of the input parameter, and exp is naturally normalized. Presumably neuroscientists considerations yield a heuristic differential equation with the logistic function as it's solution.
3 is an exact value for 3, while 2.7123123123 is not an exact value for e. but what is your point? the purpose of these values is not to have exact values, but to approximate real-world phenomena. I don't see you criticizing the usage of pi because it's transcendental. Would you seriously claim we should use 3 instead for the area of a unit circle because it's exact whereas 3.14 is not an exact value for pi?
I need to show that $\mathbb S^n$ is orientable using the fact that if $M=M_1\cup M_2$ is a manifold, where $M_1$ and $M_2$ are open, orientable submanifolds such that $M_1\cap M_2$ is connected, then $M$ is orientable.
My idea is to say that the set of all open balls in $\mathbb R^{n+1}$ each intersected with $\mathbb S^n$ forms a cover of $\mathbb S^n$ comprising open, orientable submanifolds with connected intersections.
To use the fact to prove $\Bbb S^n$ is orientable, you need to split $\Bbb S^n$ into two orientable open submanifolds whose intersection is connected no?
like, say, M_1 = S^n minus north pole and M_2 = S^n minus south pole
well, that is essentially what I said, but each region needs to be a little bit over the equator to be a covering, and for simplicity we might as well go all the way to the poles
See attempt below
I am interested in effective computations in finding approximate spectral decompositions in some suitable format.
Let $A: H \rightarrow H$ be a Hermitian operator on an $n-$dimensional Hilbert space $H$ with the spectrum $\{\lambda_1, ... \lambda_m\}, m \leq n$. Then, $A$ can ...
@Balarka If you do not mind can i ask you something about curl. I have been introduced to it in electrodynamics. The book says that it is extremely useful in physics. But mathematically why do we care about the curl of a vector say $h$
so curvature is how fast the direction if the normal to the tangent plane of the curve changes, and curl is the part of this which is orthogonal to the direction of the curve.
I am still unsure what that means. I agree curvature is how fast the normal of the curve changes direction, that's fine. But what do you mean part of it being "orthogonal to the direction of the curve"? I assume by direction of the curve you mean the tangent vector of the curve, but normal of the curve by defn is orthogonal to the tangent vector.
now fix a normal to the tangent plane of unit length, and see it as a vector starting at the chosen point on the curve
@BalarkaSen Isn't tangent plane the natural thing to start with in order to get the normal to the curve? I might be misremembering, as it has been like 10 years
I don't know. To me normal is unit vector orthogonal to the tangent vector of the curve, which is the most natural thing to start with IMO (it's the image of the derivative). Curves are 1-dimensional, so tangent lines should be more natural than planes.
anyway, we can see the curvature as a measure of how fast the end-point of this normal moves. And the curl is how fast it moves in the direction orthogonal to the direction of the curve (still along the tangent plane)
@BalarkaSen The tangent vector of the curve has an entire plane of orthogonals
@BalarkaSen But the important thing is that we have a distinguished plane with a normal vector
and we can intuitively see the curvature as a measure of how the end-point of the normal moves around (well, how fast it does so), while the curl tells us how fast it moves if we only consider the direction of movement orthogonal to the tangent vector of the curve
Wait. Let me get this straight. In the tangent plane there exists this vector which gives me direction of the curve and and how fast the normal to the tangent plane moves along the direction of the curve is the curl. Right?
@Albas No, the curl is the part orthogonal to the direction of the curve
so in some sense we always expect to have the normal move at least a bit in the direction of the curve, as long as the curvature is not zero. But it might only move in this direction (think of the curve of sin embedded in 3-space)
I am still having issues with the direction of a curve at a point. The direction if a curve at a point is given by the tangent vector at that point . Now curl is part orthogonal to the direction of the curve. That means the curl is orthogonal to the tangent vector at that point
In a nutshell, you look at a curve $\gamma$, it's unit normal $\mathbf{N}$ at $x_0 \in \gamma$, and curl measures how much $\mathbf{N}$ moves forward (as we go from $x_0$ to $x_0 + h$) towards the direction of the tangent vector at $x_0$, yeah, @Tobias?
How do I determine which columns of the projection matrix to use? I don't have a concrete matrix, I only have a formula $P_i:=\frac{1}{2\pi i}\oint\limits_{\mathcal{C}(c_i, \varepsilon)} (A-z I)^{-1}\,dz$ that project onto an m-dimensional subspace
The MathJax/LaTeX rendering seems to be broken for question titles in the "Related" section. Example from Show that 1 + $\lambda$ is an eigenvalue of $I + A$:
(Observed with Safari and Chrome on OS X, Firefox on Windows.)
Btw, is this $P_i:=\frac{1}{2\pi i}\oint\limits_{\mathcal{C}(c_i, \varepsilon)} (A-z I)^{-1}\,dz$ really a projector? How to show it. Suppose, we know that A has m eigenvalues within the circle C(ci, eps)
@ValerySaharov if $A$ is diagonalisable with $B$ st $BAB^{-1}$ is diagonal, check out what $B^{-1}(A-zI)^{-1}B$ looks like. If you calculate the relevant residues you should see that the integral becomes a projector onto the eigenspaces of $A$ with eigenvalues inside of the contour.
In the case that $A$ is not diagonalisable look at Jordan form instead, I can't see it without pen and paper but maybe the same arguments will go through
See attempt below
I am interested in effective computations in finding approximate spectral decompositions in some suitable format.
Let $A: H \rightarrow H$ be a Hermitian operator on an $n-$dimensional Hilbert space $H$ with the spectrum $\{\lambda_1, ... \lambda_m\}, m \leq n$. Then, $A$ can ...
I think I misunderstand, when checking to see if a general statement is true, do you care about what theorems you use to prove its truth? Because what I wrote is a way of seeing that the formula indeed gives projectors.
In your question you also write that you cannot decide what multiplicity roots have, but this is easy. Just take derivatives of the polynomial, if the first $n$ derivatives have a root at $x$ then the original polynomial has a root at $x$ of multiplicity at least $n$. If the $n+1$ derivative doesn't have a root at $x$ then the multiplicty is exactly $n$
hmm actually that last thing may not be true sorry, if it has root $x$ of multiplicity $n$ then the first $n$ derivatives have a root $x$, but the converse could be false.
also isn't matrix inversion more expensive computationally than eigenspace decomposition?
If $A$ is diagonalisable, these things give the projectors you want. If $A$ is not diagonalisable, consider a jordan block $A=I \lambda + n$, ($n$ nilpotent) then $(A-zI)^{-1}=((\lambda-z)I +n)^{-1}=\sum_{k=0}^\infty (-1)^k \frac{n^k}{(\lambda -z)^{k+1}}$.
The sum eventually terminates because of the nilpotency of $n$. Only the first term contributes to the residue.
But the first term is exactly the projection onto the Jordan block.
So the same argument will go through and the formula will give a projection onto the generalised eigenspaces if the matrix is not actually diagonalisable
@ValerySaharov if you have a hermitian matrix and you know some EV lie in an $\epsilon$ ball around some point, then the integral as an expression will give you the exact projectors onto the corresponding eigenspaces. If you use some algorithm to calculate the integral you will get something as close as you like to the projector in operator norm. If you then write $$\|(A-\sum_i \lambda_i P_i)\|=\|\sum_i (\tilde \lambda_i \tilde P_i - \lambda_i P_i)\|≤\sum_i |\lambda_i| \| \frac{\tilde \lambda_i}{\lambda_i} \tilde P_i - P_i\|$$
@ValerySaharov if you can exactly calculate the integral you will get an exact projector onto the eigenspaces corresponding to the eigenvalues inside the contour (provided no eigenvalues lie on the contour itself)
@ValerySaharov I can't/don't know how without assuming that a diagonal form/jordan normal form exists. But the proof does not need you to know the explicit way in which the matrix you are considering is diagonalised. What I understand is you want a computational method to calculate the eigenspaces, not a computational proof of this expression being the same. I confess here that I have no idea what a "computational proof" is supposed to be though.
@s.harp No. I don't want to find a computational proof nor do I want find a computational method of calculating eigenspaces. I need to find eigenvalues/eigenvectors/eigenspaces in APPROXIMATE format without any reference to exact eigenvectors/eigenspaces. Approximate in the sense as I described in the quesstion. In paritcular, for a vector it's || Av - cv || <= eps
Same idea. You measure how much tensor product functor and the hom functor fails to be exact, in both cases.
@AkivaWeinberger You start with a left or right exact functor, use a injective or projective resolution of objects and apply the functor to get a chain complex and extract out the homologies which turn out to be independent of the resolution.
@MikeMiller This is probably a stupid question, but a chart describes $\alpha$ locally, so how do we pick a chart that is nce for the kernel of $\alpha$?
I'm trying to break the classfication of locally riemannian symmetric spaces to little steps to make it more comprenhesible (and s.t. the technical details can be verified without drowning completely).
The first big step which I find difficult to break to concise litlle pieces is how to get fro...
unclear to me that writing everything in 'the most modern language' actually helps intuition, but sure. de Rham's decomposition theroem is proved in Kobayashi-Nomizu
ok. I'm not sure how important it is but you might want to get some more familiarity with actually working with and thinking about Riemannian manifolds, maybe - I say this because your isometric extension theorem is clearly false
I'm trying to understand why the function f(x)=x^2 for x in the reals does not satisfy the definition of DIScontinuity. I know that it is obviously true because I can prove it is continuous (which is the negation of continuity) but I can't see how I can show the definition of discontinuity breaks down for f(x)=x^2. So why does x^2 NOT satisfy discontinuity at a point c in R?
The question @GridleyQuayle asked was how he could show that a function is not discontinuous, he said he had shown that it was continuous but did not see how to show discontinuity fails. The point is that proof of not (not A) is the same as a proof of A
See attempt below
I am interested in effective computations in finding approximate spectral decompositions in some suitable format.
Let $A: H \rightarrow H$ be a Hermitian operator on an $n-$dimensional Hilbert space $H$ with the spectrum $\{\lambda_1, ... \lambda_m\}, m \leq n$. Then, $A$ can ...
The MathJax/LaTeX rendering seems to be broken for question titles in the "Related" section. Example from Show that 1 + $\lambda$ is an eigenvalue of $I + A$:
(Observed with Safari and Chrome on OS X, Firefox on Windows.)
I'm trying to break the classfication of locally riemannian symmetric spaces to little steps to make it more comprenhesible (and s.t. the technical details can be verified without drowning completely).
The first big step which I find difficult to break to concise litlle pieces is how to get fro...
