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Oct 30, 2011 at 8:26 answer added Zack Wolske timeline score: 1
Oct 29, 2011 at 19:48 comment added Jack Poulson The set of upper (lower) triangular matrices with non-negative diagonals satisfies (i), (ii), and (iii) trivially since the eigenvalues lie on the diagonal. If we call the set of such upper triangular matrices $\mathcal U$, and the set of such lower triangular matrices $\mathcal L$, then we have a variant of (iv) which is $\mathcal U + -\mathcal U + i \mathcal U + -i \mathcal U + \mathcal L + -\mathcal L + i \mathcal L + -i \mathcal L = M_n(\mathbb C)$.
Oct 29, 2011 at 14:21 comment added Suvrit ah, ok. i did not read (iv) at all :-)
Oct 29, 2011 at 13:54 comment added spelas But $x^∗ax$ is also hermitian matrix if $a$ is. So $x^∗Px⊂P$, and $x^∗M_n(\mathbb{C})x=M_n(\mathbb{C})$ iff $x$ is invertible. So $x^∗Px$ either does not satisfy (iv) or equals $P$.
Oct 29, 2011 at 13:07 comment added Suvrit $X^*AX \ge 0$ for all $X$ if $A \ge 0$.
Oct 29, 2011 at 12:01 history asked spelas CC BY-SA 3.0