Does $x^T |A| x \geq 0 $ hold?

Question

Problem statement:

Let $|A|$ be defined as $|A| \equiv X |\Lambda|X^{-1}$, where $|\Lambda|$ holds the absolute values of the eigenvalues of $A$ along the diagonal and $X$ is a matrix whose columns hold the eigenvectors of $A$.

Does the following hold for any arbitrary vector $\mathbf{x}$?

$\mathbf{x}^T |A|\mathbf{x} \geq 0$

Solution attempt:

\begin{align} \mathbf{x}^T |A|\mathbf{x} &= \mathbf{x}^T X |\Lambda|X^{-1} \mathbf{x} \\[6pt] &= (X^T \mathbf{x})^T|\Lambda| (X^{-1} \mathbf{x})\end{align}

So does $\mathbf{x}^T |A|\mathbf{x} \geq 0$ hold if and only if $X^T=X^{-1}$?

Yes, you need $X^T=X^{-1}$, which is equivalent to the orthogonality of a matrix. See unitary matrix to know more. — GNUSupporter 8964民主女神地下教會, Dec 13 '17 at 19:31
@user1551 it is common to take $|A| = \sqrt{A^*A}$, even in the case where $A$ is not Hermitian. — Ben Grossmann, Dec 13 '17 at 20:46

score 1 · Answer 1 · 2017-12-27T12:55:57.213

If a non-symmetric matrix $B$ is $\geq 0$, then their eigenvalues have a $\geq 0$ real part.

On the other hand, the matrix $\begin{pmatrix}1&10\\0&2\end{pmatrix}$ is not $\geq 0$.

The NS condition is $B+B^T\geq 0$.

EDIT. I come back to the question asked by the OP. According to above,

$|A|\geq 0$ iff $X|\Lambda|X^{-1}+X^{-T}|\Lambda|X^T\geq 0$ iff $X^T(X|\Lambda|X^{-1}+X^{-T}|\Lambda|X^T)X\geq 0$

iff $V=U|\Lambda|+|\Lambda|U\geq 0$ where $U=X^TX$ is symmetric $>0$.

Note that, if the $|\lambda_i|$ are distinct and close to $1$ (for example), then $|A|\geq 0$. In particular, the "condition $X$ is orthogonal" is not necessary.

Does $x^T |A| x \geq 0 $ hold?

1 Answers1