Given a set of orthogonal, symmetric, rank-1 matrices, can the set be completed with additional rank-1 matrices to form a basis for symmetric matrices

Question

Given a set of orthogonal, symmetric, rank-1 matrices, can the set be completed with additional rank-1 matrices to form a basis for symmetric matrices?

For example, consider a symmetric matrix $A\in\mathbb{R}^{m\times m}$, which has a spectral decomposition of $A = V D V^T$ where $V = [v_1,\dots,v_m]$. The set of rank-1 matrices, $S=\{v_i v_i^T\}_{i=1}^m$ is orthonormal with respect to the trace inner product, $\langle A,B\rangle = tr(A^TB)$. However, the space of symmetric matrices has dimension $m(m+1)/2$ and the set $S$ contains only $m$ elements.

I'd like a way to generate the additional $m(m-1)/2$ elements to complete the basis with rank-1 elements that are mutually orthogonal, if possible.

Answer 1

For $m \geq 2$, the answer here is no. The maximal number of mutually orthogonal, rank-1 matrices that can be attained is $m < m^2$.

To see that this is the case, first note that two rank-1 symmetric matrices $A,B$ will satisfy $\langle A,B\rangle = 0$ if and only if $AB = BA = 0$. Indeed, we note that we can necessarily write $A = \pm uu^T$ and $B = \pm vv^T$ for some non-zero column-vectors $u,v$. We find that $\langle A,B \rangle = (u^Tv)^2$, $AB = \pm (u^Tv)uv^T$, and $BA = \pm(u^T v) vu^T$. We thus conclude that the conditions $AB = 0, BA = 0,$ and $\langle A,B \rangle = 0$ are all logically equivalent to $u^Tv = 0$.

Now, suppose for the purpose of contradiction that we have a set $\{A_i\}_{i=1}^{m+1}$ mutually orthogonal rank 1 matrices. This is a set of commuting symmetric matrices, so it is necessarily simultaneously orthogonally diagonalizable. Thus, we have an associated set of diagonal and mutually orthogonal rank-1 matrices $\{D_i\}_{i=1}^{m+1}$, which is impossible.

Answer 2

No, unless $m=1$. If two rank-one symmetric matrices $auu^T$ and $bvv^T$ are orthogonal to each other with respect to the Frobenius inner product, the vectors $u$ and $v$ must be mutually orthogonal too, as $\langle auu^T,bvv^T\rangle_F=ab\operatorname{tr}(uu^Tvv^T)=ab(v^Tu)^2$. Hence there can be no more than $m$ mutually orthogonal rank-one symmetric matrices in $\mathbb R^{m\times m}$.