# Product of symmetric matrices equals zero matrix

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

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

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

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

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

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

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

• Please don't change the symbols in your question in a way that renders the existing answer unreadable. Commented Jun 14, 2022 at 8:12

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 $$Y=Y^t$$.