For any positive semi-definite Hermitian matrix $A$ of rank $r\geq1$ and $\text{tr}(A)=1$, is it always possible to find matrix $P$ so that $$r A=P^\dagger P,$$ where $P$ is consisted of row vectors $p_i$ satisfying $\text{norm}(p_i)=1$.

My attempt:

I tested the simplest case that $A=\text{diag}(\alpha,1-\alpha)$, where $\alpha \in \mathbb{R}$. It turned out that $$ P=\begin{pmatrix} \sqrt{\alpha} & \sqrt{1-\alpha} \\ \sqrt{\alpha} & -\sqrt{1-\alpha} \\ \end{pmatrix} $$ satisfied $$ \begin{pmatrix} \sqrt{\alpha} & \sqrt{1-\alpha} \\ \sqrt{\alpha} & -\sqrt{1-\alpha} \\ \end{pmatrix}^\dagger \begin{pmatrix} \sqrt{\alpha} & \sqrt{1-\alpha} \\ \sqrt{\alpha} & -\sqrt{1-\alpha} \\ \end{pmatrix} =2\begin{pmatrix} \alpha & 0 \\ 0 & 1-\alpha \\ \end{pmatrix}. $$ Note that $P$ may not be unique since $$ Q=\begin{pmatrix} -\sqrt{\alpha} & \sqrt{1-\alpha} \\ -\sqrt{\alpha} & -\sqrt{1-\alpha} \\ \end{pmatrix} $$ also do the job. In fact, for this simplest $A$, there are $8$ different $P$'s.

Besides, I also tested that this result stands for the general two-by-two matrix case.

I have checked various matrix decomposition but I still do not have any clue. Any idea is welcomed. Thanks.

Edit: $A$ is positive semi-definite matrix.

I think this problem is related to transform a matrix into one with constant diagonal elements.

  • $\begingroup$ Do you assume proved the main theorem of spectral decomposition of hermitian matrices? Any hermitian matrix A can be written $A = UDU^*$ with U unitary and D diagonal and real. Or is it the question? $\endgroup$ – Edouard Berthe Nov 23 '16 at 10:55
  • $\begingroup$ @Edouardb I do know the spectral decomposition of hermitian matrices and tried proving the problem starting from this theorem. However, I failed to connect them in further derivation. $\endgroup$ – Louis Yu Nov 23 '16 at 11:09
  • $\begingroup$ I think you should have a look at the Cholesky decomposition of a hermitian matrix. If $A$ is only semi-definite, the decomposition still exists if you allow non-zero element on the diagonal of the Cholesky matrix, this may be the purpose of multiplying $A$ by $r$ (because the trace is invariant by similitude). $\endgroup$ – Edouard Berthe Nov 23 '16 at 11:32
  • $\begingroup$ @Edouardb Thank you for the information about Cholesky decomposition. I will look into it since it seems useful. $\endgroup$ – Louis Yu Nov 23 '16 at 12:18
  • $\begingroup$ No worries. And you were right, there is no way to directly apply spectral decomposition. $\endgroup$ – Edouard Berthe Nov 23 '16 at 12:22

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