Can symmetric rank two matrices be written as $WW^{\top}$?

Question

Suppose $A$ is a symmetric matrix of rank two. Then $A$ can be written as $A=vu^{\top}+zw^{\top}$. I could also write $A=UV^{\top}$, where $U$ and $V$ are $n \times 2$ matrices of rank two. I'd like to know if it's posible to express $A$ as $$A=WW^{\top},$$ for some $n \times 2$ rank two matrix $W$, given that $A$ is symmetric. I know that $A=UV^{\top}=(UV^{\top})^{\top}=VU^{\top}$, but don't know how to proceed from this point. Ideas?

Answer 1

If you are dealing with real matrices, it is not generally possible to express any symmetric matrix $A$ as $WW^T$, since the latter form implicitly invokes positive semi-definiteness and $A$ may not confine to such a structure. An example is $$ A=\begin{bmatrix} 1&0&0\\ 0&-1&0\\ 0&0&0 \end{bmatrix} $$ which is a rank-$2$ indefinite matrix, where no proper choice of $W$ is available such that $A=WW^T$. But, if $A$ is symmetric and positive semi-definite, Schur decomposition of $A$ yields $$ A=UDU^T $$ where $U$ is unitary and $D$ is diagonal with non-negative diagonal entries. If $A$ is rank-$2$, then $$ D=\begin{bmatrix} \hat D&0_{2\times n-2}\\ 0_{n-2\times 2}&0_{n-2\times n-2} \end{bmatrix} $$ where $$ \hat D=\begin{bmatrix} d_1&0\\ 0&d_2\\ \end{bmatrix}. $$ Now define $$ S=\begin{bmatrix} P_{2\times 2}\\ 0_{n-2\times 2} \end{bmatrix} $$ where $$ P=\begin{bmatrix} \sqrt{d_1}&0\\ 0&\sqrt{d_2}\\ \end{bmatrix}. $$ and let $W=US$. You obtain what you wanted to prove. A similar approach cane be used for any general rank, symmetric positive semi-definite matrix $\blacksquare$