Characterisation of positive matrices

Question

I want to know if the following statement is true. A simple yes or no is enough.

Let $X\in M_n(\mathbb{C})$ be a positive matrix (thus positive eigenvalues and self-adjoint). Does there exist a vector $\xi \in \mathbb{C}^n$ such that $$X= [\xi_i \overline{\xi}_j]_{i,j}$$

Answer 1

You have $X=\xi\,\xi^*$. This is obviously rank-one, since $X\eta=(\xi^*\eta)\,\xi$ for all $\eta$, so the range is the one-dimensional subspace $\mathbb C\,\xi$.

The Spectral Theorem gives you that any positive $n\times n$ matrix $X$ has an orthonormal basis of eigenvectors $\eta_1,\ldots,\eta_n$. So we have $X\eta_j=\lambda_j\,\eta_j$. Since $X$ is positive, $$ \lambda_j=\langle T\eta_j,\eta_j\rangle\geq0. $$ Now let $\xi_j=\lambda_j^{1/2}\,\eta_j$. Then $$ (\xi_j\xi_j^*)\xi_j=(\xi_j^*\xi_j)\,\xi_j=\lambda_j\,\xi_j. $$ Thus $$ X=\sum_{j=1}^n\xi_j\,\xi_j^*, $$ with $\xi_1,\ldots,\xi_n$ pairwise orthogonal. It follows that $\xi_1,\ldots,\xi_n$ are a basis of eigenvectors for $X$, and $\|\xi_j\|^2=\xi_j^*\xi_j$ is the eigenvalue for $\xi_j$.

Answer 2

Hint The matrix $[\xi_i \overline{\xi}_j]_{i,j}$ has rank $\leq 1$.

You can see that by observing that $\mbox{Row} i$ multiplied by $\overline{\xi_j}$ is the same as $\mbox{Row} j$ multiplied by $\overline{\xi_i}$.

Answer 3

Self-answering my question: assume $$I_n = [\xi_i \overline{\xi_j}]$$

Then $\xi_i \overline{\xi_j}=\delta_{i,j}$ and hence $\xi_1 \xi_2 = 0$ but also $|\xi_1|^2 = |\xi_2|^2 = 1$ which is a contradiction.