8
$\begingroup$

The space $M_n:=M_n(\mathbb{C})$ of complex $n\times n$ matrices has the structure of a finite-dimensional complex vector space.

The space of Hermitian matrices forms a cone in this vector space $M_n$. (By a Hermitian matrix, I mean just a Hermitian symmetric matrix. I make no restrictions on being positive-definiteness.) Suppose you are given finitely many Hermitian matrices $A_1,\ldots, A_k$. The conical hull $con(A_1,\ldots, A_k)$ consists of all positive linear combinations of $A_1,\ldots, A_k$.

Can we find some conditions on $A_1,\ldots, A_k$ such that the conical hull $con(A_1,\ldots, A_k)$ contains at least one unitary matrix? I hope this is not too vague for MathOverflow.

Edit: Let me add more background, as requested by Suvrit. Suppose I have linear forms $p_1,\ldots, p_k$, i.e. homogeneous degree 1 polynomials in the variables $x_1,\ldots, x_n$. I am given real coefficients $c_1,\ldots, c_k$, which are either $1$ or $-1$. Taking linear combinations gives me a quadratic form $c_1 |p_1|^2+\cdots + c_k|p_k|^2$. I want to know when this quadratic form is a positive-definite form in the variables $x_1,\ldots, x_n$.

The norms $|p_t|^2=\sum a_{ij}x_i\bar{x_j}$ are polynomials whose coefficients $(a_{ij})$ form a Hermitian matrix. Using this correspondence, I rephrased the question in terms of matrices, hoping to gain geometric intuition from the convex geometry.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ A useless necessary condition is that there should exist $\alpha_i \ge 0$ (for $1\le i \le k$) such that $1 \le \sum_i \alpha_i \|A_i\|$, where $\|\cdot\|$ is any subordinate norm of $A_i$. This question seems interesting, but perhaps some more background might be helpful? $\endgroup$
    – Suvrit
    Dec 16, 2010 at 10:08
  • $\begingroup$ Colin, a question based on your edit: you write $|p_t|^2 = x^\ast A x$, which already assumes that $A$ is positive definite, not just Hermitian? Or am I missing something in your notation? $\endgroup$
    – Suvrit
    Mar 4, 2013 at 2:10
  • $\begingroup$ any progress on this so far Colin? $\endgroup$
    – Suvrit
    Jul 25, 2013 at 2:06

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.