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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?

Rodrigo de Azevedo
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dead_space
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2 Answers2

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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.

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

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

Some useful decompositions aiding the other answer.

Balaji sb
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    They are not the tightest bounds! Counter example: $A=\begin{bmatrix}1 & 10 \ 0 & 1\end{bmatrix}$. Its numerical radius (see https://en.wikipedia.org/wiki/Numerical_range) is ~8.57 but its norm is ~10. There is no vector $x$ exists such that $x^T A x = s_\max(A)$. You can only say that there is a vector such that $\lVert A x \rVert = s_\max(A)$. – obareey Mar 28 '24 at 08:12
  • @obareey i have updated my answer. Please take a look. – Balaji sb Mar 29 '24 at 22:45
  • But $(B_1 + i B_2)^H = B_1 - i B_2$, so $B_1 + i B_2$ is not Hermitian. – obareey Mar 30 '24 at 09:32