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This is a good example of math being presented in an unnecessarily confusing way. Values lambda_i(A) are introduced, but then in the formula also values lambda
by auggierose 7y ago
This is a good example of math being presented in an unnecessarily confusing way.
Values lambda_i(A) are introduced, but then in the formula also values lambda_i(M) are referenced for some M which is not A.
Basically, we have no idea what lambda_i(M) might mean from the statement of the theorem. Is the order lambda_1(M), lambda_2(M) ... related to the order lambda_1(A), lambda_2(A), ... How many lambda_i(M) exist? And so on.
Maybe this is all clear if you deal with minor matrices all the time, but when I see a statement like that I am not even interested in its proof anymore.
- banachtarski 7y agoTerry is being abundantly clear. Why are you trying to correlate indices in two separate product series?
- auggierose 7y agoHe is not being "abundantly clear". He doesn't give a definition of lambda_i(M) for the various M appearing in the theorem that are not A.
- _hardwaregeek 7y agoDoesn't he say it here: where {M_j} is the {n-1 \times n-1} Hermitian matrix formed by deleting the {j^{th}} row and column from {A}.
- auggierose 7y agoYes, he says what M_j is, but he does not say what lambda_i(M_j) is.
- tomrod 7y agoOkay, Donald Knuth might be upset that Terry failed to introduce the notion of a spectrum space in a Matrix Analysis paper to an audience that should know about it already. If I tell you f(x) = y, do you get upset that I don't define the domain, range, codomain, notion of a function, and so forth? Or if I tell you that https:// https:// prepends a good portion of URLs on the world-wide web, would you subject it to the same scrutiny? I hope note -- to require such would be unnecessarily limiting. My $0.02.
- auggierose 7y agoIf you tell me that thing about https:// https://, it's not math, it's just a conversation, so why would I care? If you tell me that f(5) = 10, and then you tell me later on that now I can compute g(x) = f(x) - 4, for all x, from that knowledge, then I surely would wonder what you are talking about.
- banachtarski 7y agoJust read your original opening statement. I think what everyone is telling you is that you were being unnecessarily hyperbolic towards an author that writes for a mathematical argument. If it offends you so much as you've indicated that you just "moved on," then so be it.
- auggierose 7y agoI still stand by my opening statement. Now, after the discussion here I understand what the theorem statement is saying. That's a lot of work I had to do to understand the theorem statement, and it would have been unnecessary, had the theorem been stated "syntactically" correctly. Now why wasn't the theorem stated syntactically correct? If we had to state it in a mechanised theorem prover like Isabelle then we would extract the definition of the eigenvalues from the statement into a separate definition, like: For a hermitian n x n matrix H, let lambda_1(A), ..., lambda_n(A) be the eigenvalues of H. But of course, here we assume an order of the eigenvalues, and we don't really want that, because it is immaterial to what follows. So we rather have to write it like that: Definition: For a matrix H, let L(H) be the set of its eigenvalues. Theorem: If H is a n x n Hermitian matrix, then |L(H)| = n. But, of course, this is not true, as |L(H)| < n is a possibility. So you would have to account for that somehow as well in your definition / notation. So it is probably better to go back instead again, and explain the notation as follows: Notation: For a Hermitian n x n matrix H, let lambda_1(H), ..., lambda_n(H) be its eigenvalues (in an arbitrary order). Theorem: What Tao said. The notation could still not be written in Isabelle like that, but at least it has proper scoping now.
- auggierose 7y ago
- chombier 7y agoHe says that eigenvalues of A are the lambda_i(A), so it's pretty obvious that lambda_i(M_j) is the i-th eigen value of M_j. What might not be obvious is the ordering of the eigenvalues, which must be the same for the formula to make sense.
- auggierose 7y agoWell, you see, the ordering of the eigenvalues doesn't really matter. What, that's not obvious to you from the statement?
- FabHK 7y agoNo, the ordering of the eigenvalues of M doesn't matter, because we take their product (or rather, the product of their difference from a fixed number), and multiplication commutes. It's maybe not obvious, and as such I can understand auggierose's complaint to an extent, though it was maybe unnecessarily harshly phrased.
- chombier 7y agoYou're right, my bad.
- debbiedowner 7y agoI think you know that lambda_i(M_j) is the ith eigenvalue of M_j. Order them how you like it won't change the identity.
- _hardwaregeek 7y agoActually...yeah the two orderings are related by the Cauchy Interlace Theorem: https://www.semanticscholar.org/paper/Cauchy%27s-Interlace-Theorem-for-Eigenvalues-of-Hwang/68b2f12f71940481380503651adc8306d2d70a7e https://www.semanticscholar.org/paper/Cauchy%27s-Interlace-T....
- debbiedowner 7y agoIn that formula the order of the eigenvalues doesn't matter...
- auggierose 7y agoThe order of the eigenvalues does not really matter in that formula, no. Thank you for decoding that for me. Still, how many lambda_i(M_j) are there? From the formula I am led to believe there are n-1 such eigenvalues of M_j.
- banachtarski 7y agoM is hermitian and an (n-1) x (n-1) matrix so that would be reasonable.
- kmill 7y agoThey're expecting you to know the spectral theorem for Hermitian matrices. Since M_j is obtained by striking out the jth row and jth column, it too is Hermitian, so it has has n-1 real eigenvalues (possibly with multiplicity). The notation $\lambda_i(M)$ ought to refer to the $i$th eigenvalue of M in sorted order to be unambiguous, but this does not matter for the proof.
- tomrod 7y agoYes, it is written for people who deal with minor matricies all the time. Academic papers are incremental materializations -- it is up to the reader to bring or build the requisite base understanding.
- auggierose 7y agoWell, I am sure the meat of the theorem is just in those eigenvalues of the minors. Which are not even defined in the statement of the theorem. That's not academic. That's just bad style.
- snakeboy 7y agoHe also didn't define eigenvalue in the theorem statement, but you don't seem to be criticizing that as bad style. Part of the art of mathematical exposition is to walk the line between under and over explaining your process based on your audience. If the intended audience is academic mathematicians or even just people with a math degree, then the notation is probably familiar. I imagine that's the target audience here.
- auggierose 7y agoI can now say exactly what I don't like about the theorem statement: That the definition of the eigenvalue notation was not well-scoped. This made it much harder for me to understand what is going on, and has nothing to do with the actual semantic meaning of eigenvalues. Caring about proper scoping is also something mathematicians usually do not, making everything more complicated than it has to be in the process.
- liuyao 7y agoCome on, if I defined f(x)=x^2, and later used f(y) for y=x/2 or something, do you have problem with that?