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.
Best Answer
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 $\mathbb R^{m\times m}$.