I don't see your point. The vector you chose is not an eigenvector of $A$ since its second element has to be zero based on my proof (if I made a mistake there please let me know).

I've quoted your mistake.

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?

It seems that the theorem you are referring to is precisely the Rayleigh's Theorem I wrote. But, still you haven't explained why your choice of $\mathbf{x}$ (which has to be in the span $S$ of the orthonormal eigenvectors) is valid. For, any $\mathbf{x}=[x_1\ x_2]^T$ to be an eigenvector I have shown that $x_2=0$, by solving the system $A\mathbf{x}=\lambda\mathbf{x}$. So, how can you choose $\mathbf{x}=[1\ 1]^T$?

$x$'s in statement of theorem are arbitrary

Then can you check if the theorem you are talking about is the Rayleigh's Theorem I wrote? Also do you think that my calculations of the quantities I denoted as $a$ and $b$ are correct? If not, can you show me how I can calculate them?

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

OK, I am here

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.

What is not a multiplication by 1 on all vectors?

Also, I want to show that the theorem does not necessarily hold for non-Hermitian matrices. So, I need one suitable vector x to show this

Do you know what a vector space is?

Of course

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

OK, I understand this, then what?

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

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

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

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

I see what you are saying but I am not convinced it is correct. The theorem says that this vector has to belong in S which is the span of the orthonormal basis of eigenvectors. In our case S is just one vector of the form [x_1 0], for example one can choose S={[1 0]}. My point is that you cannot choose whatever vector you want.

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

Hermitian n by n matrices always have n linearly independent eigenvectors

I see, then I understand your counterexample. Can you show me how to compute quantities a and b?

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

then you want to compute differential of this function

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

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

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

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

that's your a and b

OK, so for a, I did the following. Let x=(y, z). Then, I found that x^TAx/x^Tx=...=(y+z)^2/(y^2+z^2). Now, I think that if set the gradient of this quantity to zero, I will find the maximum. Is this true?

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

I'd better say ellipsoids.

So what you are saying is that after I find the gradient and the points that it evaluates to zero, then I will compute the Hessian for these points. At the point where the Hessian is negative definite, it will be the maximum i.e. the quantity a and the point where the Hessian is positive definite it will be the minimum i.e. the quantity b.

Yep, exactly

OK, one more thing. Quantity b is the real part of the maximum of xAx/xx where x is in general complex (this is by definition of b). So are you sure I can find b in the above way?

I meant x^*Ax/x^*x, where the asterisk denotes the Hermitian operation

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

Please see the original post. b is defined there

Alright, so in our case a=b