-1
$\begingroup$

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?

  • 0
    $A\mathbf{x}$ as in an all real vector? Or did you mean $\mathbf{x}^\mathrm{T}A\mathbf{x}$ as a real scalar?2012-10-27
  • 0
    @EuYu it means all entries of $Ax$ are real.2012-10-27
  • 0
    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.2012-10-27

2 Answers 2

2

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

  • 0
    ... and $A$ can be any matrix, not necessarily symmetric.2012-10-27
  • 0
    ...as long as the entries are real :-).2012-10-27
  • 0
    Thanks, so for any matrix $A$, if $Ax$ and $x$ is real, then $A$ is real.2012-10-27
1

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.