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For a Hermitian nonnegative-definite matrix $A$, if $Ax$ is always real for any real vector $x$, can we conclude that $A$ is also real?

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$A\mathbf{x}$ as in an all real vector? Or did you mean $\mathbf{x}^\mathrm{T}A\mathbf{x}$ as a real scalar? –  EuYu Oct 27 '12 at 2:57
    
@EuYu it means all entries of $Ax$ are real. –  chaohuang Oct 27 '12 at 3:03
    
by symmetric, did you try to say $A=A^{T}$ or $A=A^{H}$, because for complex matrices, $A=A^{T}$ and $A=A^{H}$ imply totally different things. –  dineshdileep Oct 27 '12 at 3:10

2 Answers 2

up vote 2 down vote accepted

Since $e_i^T A e_j = [A]_{i,j}$, and $Ax$ is real for real $x$, then $A$ must be real.

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... and $A$ can be any matrix, not necessarily symmetric. –  wj32 Oct 27 '12 at 3:10
    
...as long as the entries are real :-). –  copper.hat Oct 27 '12 at 3:13
    
Thanks, so for any matrix $A$, if $Ax$ and $x$ is real, then $A$ is real. –  chaohuang Oct 27 '12 at 3:23

Imagine if $A$ had some non-real entries. Then pick the column which has those non-real entries, say it is the i'th column. $A e_i$ is equal to the i'th column of $A$, so it has non-real entries.

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