A given linear function: $f(x) = \bar{A}x$ can be mapped as a generalized Sylvester equation as follows:

$f(X) = AX + CXB$, where $x = \rm vec{(X)}$ (column stacking).

The eigenvalues $\lambda_i$, $\mu_j$ and $\gamma_k$ of matrices $A\in\mathbb{R}^{n\times n}$, $\mathbb{B}^{m\times m}$ and $\mathbb{C}^{n\times n}$, respectively, are known, where $i=1,\dots,n$, $j=1,\dots,m$ and $k=1,\dots,n$.

Is it possible to know the eigenvalues of $A$ directly from the Sylvester equation? That is, using knowledge of $\lambda_i$, $\mu_j$ and $\gamma_k$.

If it helps, I know that the eigenvalues of $F(X)=CXB$ are $\mu_j\gamma_k$, and the eigenvalues of $F(X) = CX+XB$ are $\mu_j + \gamma_k$.

Also, the Sylvester equation can be rewritten as the following operations with Kronecker products:

$f(x) = (I_n\otimes A + B^T\otimes C)x$,

where $I_n\in\mathbb{R}^{n\times n}$ is the identity matrix.

If it is not possible to know the eigenvalues, does restricting $C$ to a matrix whose eigenvalues are only $0$s and $1$s helps (except for the identity matrix)?


Your dimensions are not consistent. Assume $A,C\in M_{m,m},B\in M_{n,n}$ and $X\in M_{m,n}$

Unfortunately, the functions $g:X\rightarrow AX$ and $h:X\rightarrow CXB$ do not commute except if $AC=CA$. If $AC\not= CA$, then we cannot say anything.

If $AC=CA$, then $A,C$ are simultaneously triangularizable and there are orderings of the spectra of $A,C$ s.t. $diagonal(P^{-1}AP)=(a_i),diagonal(P^{-1}CP)=(c_i)$. Moreover, $g,h$ are simult. triang. s.t.

$(*)$ $diagonal(p^{-1}gp)=(a_i)$ $n$ times each and $diagonal(p^{-1}hp)=(c_ib_j)$ with some orderings.

On the other hand, since $spectrum(AC)=(a_ic_i)$, we deduce that $spectrum(g\circ h)=(a_ic_ib_j)$, what gives the orderings of the above diagonals $(*)$.

Finally $spectrum(f)=(a_i+c_ib_j)$, a set with $mn$ elements

Notice that there is an imposed correspondence $a_i<->c_i$.

  • $\begingroup$ Indeed, I had a typo, $C$ was supposed to be of the same dimension of $A$. Thank you for your answer. $\endgroup$
    – maracas
    May 4 '18 at 20:09

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