Decomposition of a $3 \times 3$ matrix

Question

Given a matrix

$$B = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$$

what are necessary and sufficient conditions on a $3 \times 3$ real matrix $A$ such that there is a solution $R$ to the equation

$$A = R^T B R$$

where $R$ is $2 \times 3$ and has full rank?

A few observations I've made so far are that, if the decomposition exists then $A$ has to be symmetric and the rank of $A$ is $2$ (i.e. zero is an eigenvalue). I get the feeling that I need to incorporate the entries of $B$ into my conditions somehow.

Edit: Direct computation with a concrete example shows that the diagonals of $B$ aren't the eigenvalues of $A$.

Any tips or hints would be appreciated.

Answer 1

Here are a couple of hints.

Firstly, recall that any real symmetric matrix $A$ can be diagonalized by an orthogonal matrix, $A=O^TDO$. So we should determine which diagonal matrices can be obtained, and any symmetric matrix with the same eigenvalues can also be obtained.

Secondly, if $$ R=\begin{bmatrix}a&0&0\\0&b&0\end{bmatrix} $$ then $$ A=\begin{bmatrix}a^2&0&0\\0&-b^2&0\\0&0&0\end{bmatrix} $$ so we can obtain any symmetric matrix with one positive, one negative and one zero eigenvalue.