I have a symmetric positive-definite matrix $A\in R_{n\times n}$.
Its eigenvectors $e_i$ are an orthonormal basis of $R_n$.
Are the $n^2$ matrices $[e_i\,e_j^T]$ a basis of $R_{n\times n}$?
I have noticed an interesting property: $[e_i\,e_j^T]$ commute with $A$ iff $\lambda_i=\lambda_j$
If $A$ does not commute with $[e_a\,e_b^T]$ and $[e_c\,e_d^T]$, can it commute with $\left( [e_a\,e_b^T]+[e_c\,e_d^T] \right)$?
Would the matrices $[\bar{e}_i\,\bar{e}_j^T]$ (eigenvectors with the same eigenvalue) generate the space of all matrices that commute with $A$?
Best Answer
First, consider the case in which the $e_i$ are the standard basis vectors. Or equivalently, take $A$ to be the matrix of the transformation relative to the basis of eigenvectors $e_1,\dots,e_n$. With that, $A$ must be a diagonal matrix. Write $$ A = \operatorname{diag}(\overbrace{\lambda_1,\cdots,\lambda_1}^{m_1}, \cdots, \overbrace{\lambda_k,\cdots,\lambda_k}^{m_k}) $$ That is, $$ A = \pmatrix{\lambda_1 I_{m_1} \\ & \ddots \\ && \lambda_k I_{m_k}} $$ where $I_m$ is the identity matrix of size $m$. $A$ commutes with all conformally partitioned block-diagonal matrices, that is, all matrices of the form $$ B = \pmatrix{B_1\\& \ddots \\ && B_k} $$ where $B_i$ is $m_i \times m_i$.