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@monoidaltransform Obviously not for 1 — take Y to be a disjoint union of X and something which is not a topological manifold. Yes for 2: smooth structure is a local structure, and you just transport it via local inverse of f.

Oh, well, sorry, I was somewhat blind and assumed that b was something else

Yep, exactly

I'd better say ellipsoids.

You will find extremal point this way (which can be a saddle, i. e. maximal in some directions and minimal in other), and then you want to understand whether it's maximum or minimum. It's done either by Hessian method or by direct comparing of values

that's your a and b

analogously, but for positive-definite hessian, you'll find minimum

we can restrict to standard unit sphere because x^tAx/x^tx is scale-invariant

(actually, local maximum, but in this case it'll be global)

if at some point where differential is zero Hessian is negative definite, it's a maximum

then you want to compute differential of this function

Look at unit sphere for standard norm; you have real-valued function x^tAx on it

Hermitian n by n matrices always have n linearly independent eigenvectors

ah, that's because the theorem in form you cite cannot apply verbatim to non-Hermitian case

can you point at some points in theorem which are not completely clear to you?

and a is greater than 2 — which is value of x^tAx/x^tx on x = (1, 1)

I've shown that your computation (which was absent in your post — just some "... = ..." of a and b is certainly wrong

if matrix is Hermitian, then you can bound this value by eigenvalues.

Well, Rayleigh theorem says: take any vector and divide x^tAx by x^tx

Multiplication by some number (which comes from action of base field) is represented by diagonal matrix with that number in all diagonal entries.

Do you know what a vector space is?

You correctly found eigenvalues; but if matrix has all eigenvalues equal to something (1 in your case) doesn't mean that it is multiplication by 1 on all vectors.

I can try to explain you what's it about here to not clump comments.

$x$'s in statement of theorem are arbitrary

What that whatever-his-name theorem says: take any matrix $A$ and construct quadratic form $q_A$ on $\Bbb C^n$ from it in obvious fashion, $q_A (x) = |x^t A x|$. Then, if $A$ Hermitian, when you restrict it to sphere $\{q_{Id}(x) = 1\}$, maximum of $q_A$ on that whole sphere — not on eigenvectors — will be equal to maximal eigenvalue of $A$, and minimum equal to minimal eigenvalue. There's also part about intersections of this sphere with some subspaces, but it's not very relevant. Does it make sense?

I've quoted your mistake.