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 $A$?


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 $m_i \times m_i$.

  • $\begingroup$ I understand that the diagonalized matrix will commute with any block matrices as B. And so, if $M=V\,A\,V^T$ and $N = V\,B\,V^T$, M and N will commute. But can we define a basis for the commutative matrices N? $\endgroup$ – Daniel Cunha Jun 2 '17 at 17:34
  • $\begingroup$ @DanielCunha using matrices with exactly one non-zero entry, construct a basis for the block matrices $B$. Applying the same similarity gets you a basis for the commuting matrices $N$. $\endgroup$ – Omnomnomnom Jun 2 '17 at 17:36

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