What are the bounds on $x^TAx$ when the matrix $A$ has complex eigenvalues with positive real parts?

Question

We all are familiar with the fact that

$$ \lambda_{\min}(A)\|x\|^2 \leq x^TAx \leq \lambda_{\max}(A)\|x\|^2 $$

where $x \in \mathbb{R}^n$ and $A \in \mathbb{R}^{n \times n}$ happens to be a positive definite matrix. But what will be the equivalent identity when $A \in \mathbb{R}^{n\times n}$ is a matrix with complex eigenvalues where the real part is positive? What will be the upper and lower bound in that case?

Answer 1

The theorem you stated is $Rayleigh-Ritz$ theorem which is valid for Hermitian matrices(if $A=A^H$.) So in general for any matrix $A$ - it can be written as $A = \frac{1}{2}(A+A^H) + \frac{1}{2}(A-A^H).$ Substitute this A back in $x^HAx \to$ this is equivalent to $\frac{1}{2}x^H(A+A^H)x$. Now you can bound the eigen values of this $A'$ hermitian matrix.

Answer 2

$$\frac{1}{2}(A + A^H) = \frac{1}{4}\left((A+I)^H(A+I) - (A-I)^H(A-I) \right)$$

$$\frac{-i}{2}(A - A^H) = \frac{1}{4}\left((A+iI)^H(A+iI) - (A-iI)^H(A-iI) \right)$$

Some useful decompositions aiding the other answer.