# Bounds for inner product of $Ax$ and $x$

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?

• I don't think this is true, but I'm having a bit of difficulty proving it. My understanding was that the quadratic form corresponding to a non-Hermitian matrix is in general not even bounded below.
– Ian
Commented Aug 19, 2015 at 23:20
• Is it possible that the author intended to only consider diagonalizable matrices (perhaps diagonalizable over $\mathbb{C}$)? This seems more plausible in that context. Commented Aug 19, 2015 at 23:35
• If $x_1,x_2$ are unit eigenvectors then $(x_1+x_2)^T A (x_1+x_2)=(x_1+x_2)^T (\lambda_1 x_1 + \lambda_2 x_2) = \lambda_1 + \lambda_2 + (\lambda_1 + \lambda_2) \cos(\theta)$.
– Ian
Commented Aug 20, 2015 at 0:44
• it is true if $A$ is normal. Probably also only if. Aside: If you divide by $\|x\|^2$, you obtain the Raleigh quotient. The eigenvalues is contained in the range of Raleigh quotients. There is a bunch theory about eigenvalue approximation by Raleigh quotient. Commented Aug 20, 2015 at 0:53
• @Omnomnomnom like I said no counter example :D Commented Aug 20, 2015 at 3:16

## 1 Answer

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...".

• why so complicated? a non diagonalizable example $[0, 1; 0, 0]$. Commented Aug 20, 2015 at 0:56
• a diagonalizable counter example $[0, 1 ; 0, 1]$. Commented Aug 20, 2015 at 1:21
• The desired condition is normality Commented Aug 20, 2015 at 1:47