0
$\begingroup$

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 $A$?

$\endgroup$
2
$\begingroup$

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 $m_i \times m_i$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.