Suppose A is Non Singular and $\lambda_1,.......,\lambda_n$ are roots of the characteristic polynomial of A

Note : These are Operators on the Vector Space of square n by n Matrices.

$L : X \rightarrow AXA^T$

$S : X \rightarrow AXA^{-1}$

Show that L has Eigenvalues $\lambda_i *\lambda_j$ where $i,j = 1,...n$

Show that S has Eigenvalues $\lambda_i \; / \; \lambda_j$ where $ i,j = 1,...n$

  • $\begingroup$ Is that $(A)(X^T)(A)$ or $(A)(X)(^TA)$? $\endgroup$ – Marc van Leeuwen Mar 9 '15 at 13:24
  • $\begingroup$ My book technically says $(A)(X^T)(A)$ but it's possible that's a typo. $\endgroup$ – Exc Mar 9 '15 at 15:53
  • $\begingroup$ Edit : I've corrected the question. $\endgroup$ – Exc Mar 9 '15 at 22:51

Take $v_i$ an eigenvector associated with the eigenvalue $\lambda_i$. For $V_{i,j}=v_iv_j^T$ (this is an $nxn$ matrix. One has


When $A$ is invertible $\lambda_i\neq 0$ and is an eigenvalue of $A^T$ take $w_i$ a corresponding eigenvector. Then $(A^{-1})^{-1}w_i=\frac{1}{\lambda_i}w_i$. Let's compute for $W_{i,j}$


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  • $\begingroup$ @ marwalix , I see that Exc is not an eagle because you show only an inclusion. Moreover, you assume that $L:X\rightarrow AXA^T$ ; is really the OP's question ? $\endgroup$ – user91684 Mar 9 '15 at 10:21
  • $\begingroup$ I don't think one can assume $A$ is diagonalisable (in case some of the $\lambda_i$ coincide). Therefore an argument based on eigenvectors only does not seem sufficient. $\endgroup$ – Marc van Leeuwen Mar 9 '15 at 13:28
  • $\begingroup$ I did not assume that $A$ is diagonalisable, I just proved that if $\lambda_i$ are eigenvalues of $A$ then $\lambda_i\lambda_j$ are eigenvalues of $L$ (not necessarily all) $\endgroup$ – marwalix Mar 9 '15 at 13:40
  • $\begingroup$ I'm confused, surely $L(v_i v_j^T) = A v_j v_i^T A$? The question has $L(X) = A X^T X$, not $L(X) = A X A^T$. $\endgroup$ – copper.hat Mar 9 '15 at 14:36
  • $\begingroup$ @copper.hat you mean $A(X^T)A$ in that case I think it works with an adaptation by taking $v_iw_j$ with $v_i$ and $w_j$ eigenvector of $A$ and $A^T$ respectively or is it the other way round $\endgroup$ – marwalix Mar 9 '15 at 14:46

@ Exc , more generally, consider the function $f:X\in M_n(\mathbb{C})\rightarrow AXB^T$ where the spectra of $A,B$ are $(\lambda_i)_i,(\mu_i)_i$. If we stack a square complex matrix row by row, then $f=A\otimes B$ is a $n^2\times n^2$ matrix.

cf. http://en.wikipedia.org/wiki/Kronecker_product

The key is: if $P,Q$ are invertible, then $A\otimes B$ and $(PAP^{-1})\otimes (QBQ^{-1})$ are similar. Thus we may assume that $A,B$ are upper triangular ; then $A\otimes B$ is upper triangular and its diagonal is composed of the $n^2$ products $\lambda_i\mu_j$.

Example: $\begin{pmatrix}a&b\\0&c\end{pmatrix}\otimes \begin{pmatrix}d&e\\0&f\end{pmatrix}=\begin{pmatrix}ad&ae&*&*\\0&af&*&*\\0&0&cd&ce\\0&0&0&cf\end{pmatrix}$

PS: To calculate the spectrum of $g:X\rightarrow AX^TB$ is more complicated. (Consider $g\circ g$, ...)

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