Reading a math text, I found, with no proof given, the following assertion.
Suppose $A$ is a real $n \times n$ matrix, and suppose the real part of its spectrum lies between $a$ and $b$; i.e., the maximum real part of the eigenvalues of $A$ is $b$, and the minimun real part of the eigenvalues of $A$ is $a$. Then, for every real n-vector $x$ we have
$a ||x||^2 \leq (Ax,x) \leq b||x||^2$ , where $||.||$ is the 2-norm.
Has anyone heard of this? Could you prove it or provide a source where it is proven? Otherwise, could you disprove it?
I find it a little hard to believe, since it would imply, e.g., that in order to see if a non hermitian matrix satisfies $x^t A x > 0 \; \forall x \neq 0$ we can just check if all the eigenvalues of $A$ have positive real part, and I have never heard such a thing (it is said in a variety of sources that for that, one have to look at $A+A^t$ and apply the well known criteria for positive definiteness).
Edit:
This doesn't hold if A is not diagonalizable. For example, it isn't true for $A = [-2, 1; 0, -2]$. So, now I ask, can it be shown that it always holds for diagonalizable matrices?
This is false, it does not hold disregarding wether the matrix is diagonalizable or not. For a non diagonalizable matrix, take $A=[-2,1;0, -2]$. For a diagonalizable matrix, take $A=[-1,0; 3, -0.2]$.
Edit: the textbook I mencioned in the question was incomplete. The correct theorem is that there exist a basis for which this holds. For source, go to "Differential equations, dynamical systems and linear algebra" (Hirsch, Smale). It's the lemma at the beginning of chapter 7 (bottom of page 145). Also, the hypotesis should be read as "...suppose the real part of its spectrum lies strictly between a and b...".