# Characterisation of positive matrices [closed]

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 $$X= [\xi_i \overline{\xi}_j]_{i,j}$$

• The important thing to remember, if you choose to post a question you intend to answer, is to post a high-quality question with context: why this topic is of interest to you, your level in math, source of the question, etc. writing it as though you have no answer, and trying to help provide as many details as you can, inviting others for their input, e.g.), If you then answer that well written question I see no problems in your posting such a question, and then answering. May 5 at 13:58
• But you failed to post a high-quality question. This can be edited, from you, to address the lacking context, such as what I describe above. May 5 at 13:59

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 $$\overline{\xi_i}$$.

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.

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 $$\xi_j$$.

• Can you elaborate how the spectral theorem gives this?
– user839372
May 4 at 23:06
• I have added some more detail. Let me know if that's enough for you. May 5 at 0:58