Every Hermitian matrix has an eigenvalue, without FTA?
It is true that every Hermitian matrix* has an eigenvalue. In particular, every real symmetric matrix has an eigenvalue. This follows immediately from the fundamental theorem of algebra even in the real case: the characteristic polynomial has at least one (a priori complex) root; that eigenvalue is real since the matrix is Hermitian.
Question: Is there a proof that doesn't rely on the fundamental theorem of algebra (FTA)?
I've looked at 4 different proofs of this version of the spectral theorem, all of which use the above fact and the FTA: Wikipedia's; PlanetMath's (apparently; it actually just states the above result without reference, saying it's "well-known"); a linear algebra book of mine (which similarly just assumes the existence of such an eigenvalue); and a random blog's.
In a similar vein, can the above fact be used to deduce FTA? If so that would give a nice intuitive reason why the proof of the above seems like it has to run through FTA.
*That is, n by n matrix A over R or C such that A* = A where * denotes the (conjugate) transpose.
Perhaps I didn't make myself clear. It's completely obvious to me why every Hermitian matrix has real eigenvalues; a proof of that fact is not at all what I was looking for.
Moreover it is not true that every matrix has eigenvalues. For instance, the real matrix
[1 1]
[-1 1]
has no (real) eigenvalues [of course it's basically trivial that a matrix has eigenvalues in its algebraic closure, but that's not the point].
For real matrices of odd dimension, the existence of at least one eigenvalue follows from the intermediate value theorem, since then the characteristic polynomial has real coefficients and is of odd degree. So, in this case, the fundamental theorem is unnecessary. My question is, are there clever ways of dispensing with the fundamental theorem similarly in other cases? And if not, can the general existence of an eigenvalue be used to prove the fundamental theorem?
@gianlino: Wonderful, thank you! I don't read French, but I was able to decipher enough to get it. Considering the quadratic was very clever.
For the reverse direction (which I doubt can be made to work, but which still interests me some), you say there aren't enough characteristic polynomials. Why? Naively, there are ~n^2 / 2 degrees of freedom in choosing an n by n Hermitian matrix, whereas only n degrees are needed to choose an arbitrary degree n polynomial.
Figuring out which matrix corresponds to a given arbitrary polynomial is probably unworkable, though.
Yes, of course--I had even noted that these matrices would only give polynomials with real roots a few days ago when I posted this question, but it slipped my mind. Still, restricting the fundamental theorem to polynomials without complex roots gives a non-trivial result, I believe. Extending Hermitian to normal matrices would also entirely overcome the difficulty.