Product of symmetric matrices equals zero matrix

Question

Let $X \in \mathbb{R}^{n \times n}$ and $Z \in \mathbb{R}^{n \times n}$ be symmetric matrices.

Assume that $Q=\left[q_{1}, \ldots, q_{r}\right]$ is an $n \times r$ matrix whose columns form an orthonormal basis of

$$ \operatorname{ker} X := \{ v \in \mathbb{R}^{n} \mid Xv = 0 \} \,. $$

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I am working on proving the following claim:

If

$$ X Z = 0_{n,n} $$

then there exists a symmetrix matrix $Y \in \mathbb{R}^{r \times r}$ such that

$$ Z=Q Y Q^{\top} \,. $$

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This claim is part of a lemma from a paper I am currently reading. In this paper the above claim is proved by the following:

$X Z=0_{n,n}$ implies that

$$ \operatorname{range} Z \subset \operatorname{ker} X=\operatorname{span}\left\{q_{1}, \ldots, q_{r}\right\} \,, $$

where $\operatorname{range} Z := \{ u \in \mathbb{R}^{n} \mid \exists v \in \mathbb{R}^{n} \, \text{ s.t. } \, Zv = u \}$.

This implies that

\begin{equation} Z \in \operatorname{span}(\{q_{i} q_{j}^{\top}+q_{j} q_{i}^{\top} \mid i, j=1, \ldots, r\})=\left\{Q Y Q^{\top} \mid Y \text{ is a symmetric } r \times r \text{ matrix}\right\} \end{equation}

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I am not sure how to proceed from here. Any help is much appreciated.

Answer 1

Assume that $Z=[z_1,\dots,z_n]$ with column vectors $z_j$. The assumption says that $Xz_j=0$ for each $j$. Thus $z_j\in \ker(X)$, which means that $z_j$ is a linear combination of $q_1,\dots, q_r$. Thus there exists a vector $a_j\in M_{r\times 1}(\mathbb{R})$ such that $$z_j=Qa_j.$$ Set $A=[a_1,\dots,a_n]\in M_{r\times n}(\mathbb{R})$, we get $$Z=QA.$$ This implies that $Z^t=A^tQ^t$. Since $Z=Z^t$, we get $XA^tQ^t=0$. Note that $Q^tQ=I_r$. Thus $XA^t=0$. In particular, each column vector in $A^t$ is in $\ker(X)$. Repeat the above process, it is easy to find a matrix $Y\in M_{r\times r}(\mathbb{R})$ such that $A^t=QY^t.$ Thus $A=YQ^t$ and $$Z=QA=QYQ^t.$$ Since $Z=Z^t$, we have $QY^tQ^t=QYQ^t$. Using $Q^tQ=I_r$ again, we can obtain that $Y=Y^t$.