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Elements of the set $P$ of positive hermitian $n×n$ matrices over complex numbers have some special properties:

(i) they are closed under sum,

(ii) they are closed under multiplication by positive scalars,

(iii) spectrum of every matrix is positive, (all eigenvalues are nonnegative, and not all are equal to 0),

(iv) $P+-P+iP+-iP=M_n(\mathbb{C})$.

Does any other subset of matrix algebra $M_n(\mathbb{C})$ satisfy these properties except for $tPt^{-1}$, where $t$ is an invertible element in $M_n(\mathbb{C})$?

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  • $\begingroup$ $X^*AX \ge 0$ for all $X$ if $A \ge 0$. $\endgroup$
    – Suvrit
    Commented Oct 29, 2011 at 13:07
  • $\begingroup$ 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 $P$. $\endgroup$
    – spelas
    Commented Oct 29, 2011 at 13:54
  • $\begingroup$ ah, ok. i did not read (iv) at all :-) $\endgroup$
    – Suvrit
    Commented Oct 29, 2011 at 14:21
  • $\begingroup$ 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 $\mathcal U + -\mathcal U + i \mathcal U + -i \mathcal U + \mathcal L + -\mathcal L + i \mathcal L + -i \mathcal L = M_n(\mathbb C)$. $\endgroup$
    – Jack Poulson
    Commented Oct 29, 2011 at 19:48

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I think I recall seeing this question in a Halmos book on linear algebra, either "Finite Dimensional Vector Spaces" or the "Linear Algebra Problem Book", but I don't remember which, and I don't have them on hand.

Here are some subsets which satisfy 3 out of 4 conditions:

Jack Poulson already mentioned upper triangular matrices, which only violate (iv).

The set of all Hermitian matrices only violates (iii).

The set of Hermitian matrices $P_r$, where all eigenvalues are greater than some positive real $r$ is closed under addition — but not positive scaling — and every matrix can be written as an element of $P_r+(−P_r)+iP_r + (−iP_r)$. This set is a strict subset of $P$, and any element of $P \setminus P_r$ is not contained in $tP_rt^{-1}$ for any invertible $t \in M_n(\mathbb{C})$ (consider diagonalization).

The set of non-diagonalizable matrices with real, non-negative eigenvalues satisfies everything but (i). For $M_2$ explictly, consider matrices of the form $$ A = \left[ {\begin{array}{cc} r_1 & z \\\ c\bar{z} & r_2 \\\ \end{array} } \right] $$ where $r_1$, $r_2$ are real, $r_1 + r_2 > 0$, $z \neq 0$, and $c = -\left(\frac{r_1 - r_2}{2|z|}\right)^2$. Then $A$ has one repeated eigenvalue, $\frac{r_1 + r_2}{2}$, and one linearly independent eigenvector $(z, \frac{r_2-r_1}{2})$. The set of all such matrices satisfies (ii), (iii), (iv), and is not conjugate to $P$ — since everything in $P$ is diagonalizable — but is not closed under addition.

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  • $\begingroup$ I cannot find the question in Halmos books. There were some other useful information. Thank you. For case $n=2$, might Mathematica be able to compute this? $\endgroup$
    – spelas
    Commented Oct 30, 2011 at 16:02

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