Consider the following result which I recently came across in a research paper in my area (Signal Processing)

Let $X$ be a $N\times N$ positive semidefinite (psd) matrix whose rank is $r$. Let $A$ be any symmetric $N\times N$ matrix. Then, there exist a set of vectors $x_1,\dots,x_r$ such that \begin{align} X & = \sum_{i=1}^{r}x_ix_i^T \\ x_i^TAx_i &= \frac{\mbox{trace}\{AX\}}{r},~~~\forall i \end{align}

The following is the proof for it which I can't verify.

**Proof:**
Consider the following step-wise procedure whose inputs are $X$ and $A$.

$~~$0.$~~$ Inputs are $X$ (given $X\geq 0$, $rank(X)=r$) and $A$ (symmetric).

Decompose $X=RR^T$.

Generate the eigen decomposition $R^TAR=U\Lambda U^T$.

Let $h$ be any $N\times 1$ vector such that $\lvert h_i\rvert=1$ (each entry of $h$). Generate the vector $x_1$ and matrix $X_1$ as \begin{align} x_1&=\frac{1}{\sqrt{r}}RUh \\ X_1&=X-x_1x_1^T \end{align}

Outputs are $X_1$ and $x_1$.

The paper then claims that

- $X_1$ is psd and has rank $r-1$
- $x_1^TAx_1=\frac{1}{r}trace(AX)$

**While I am able to verify the second claim, am not able to verify the first one? How is it true?** If this can be done, the rest of the proof is straight forward. I am looking for a rigorous proof.

Read this if you are interested to know where this proof heads. Now do the stepwise algorithm earlier with inputs $X_1$ and $A$ to get $x_2$ and $X_2$ such that \begin{align}X_2&=X_1-x_2x_2^T\\&=X-x_1x_1^T-x_2x_2^T\end{align} and $$x_2^TAx_2=\frac{1}{r-1}trace(AX_1)=\frac{1}{r}trace(AX)$$ Then the result of the paper is that you can do this procedure $r$ times and get a rank-one decomposition $$X=\sum_{i=1}^{r}x_ix_i^T$$ with the property $$x_i^TAx_i\,=\,\frac{1}{r}trace(AX),~\forall i$$for any given psd X with rank $r$ and any symmetrix $A$.