1
$\begingroup$

I recently have been considering the Sylvester matrix equation given by $AX-XB = C$, where the rank of $C$ is lower than $X$, and $A$ is a diagonal matrix, and $B$ is defined as follow \begin{equation} B=\begin{bmatrix} & & & & -n \\ \frac{1}{2} & & & & \\ & \frac{2}{3} & & & \\ & & \ddots & & \\ & & & \frac{n-1}{n} & \end{bmatrix}_{n \times n}. \end{equation} However, $B$ is not diagonal. I have obtained the eigendecomposition of $B$ as follow \begin{equation} BP = P\Lambda, \end{equation} where $P = [p_1\ p_2 \ \cdots \ p_n]$ is a square matrix whose columns are the $n$ linearly independent eigenvectors of $B$ and $\|p_i\| = 1$, and $\Lambda$ is a diagonal matrix where each diagonal element $\Lambda_{ii}$ is the eigenvalue associated with the $i$-th column of $P$, and $\Lambda_{ii}$ is the $n$ (shifted) roots of unity, i.e, $\Lambda_{ii}^n + 1 = 0$.

Then, it follows that \begin{equation} AXP - XBP = CP \Rightarrow A(XP) - (XP) \Lambda = CP. \end{equation} Thus, I want to estimate the condition number of $P$ to compare the singular value of $X$ and $XP$.

Through computer simulation, I have found that the condition number of $P$ is $n$, i.e., \begin{equation} \kappa(P)=\left\|P^{-1}\right\|\|P\| = \frac{\sigma_{\max}(P)}{\sigma_{\min}(P)} = n. \end{equation} But I don't know how to prove $\kappa(P)=n$ yet.

$\endgroup$
1
  • $\begingroup$ $B^*B = {\rm diag}\{n^2,(1/2)^2,\cdots, ((n-1)/n)^2\}$, where $B^*$ is the complex conjugate of $B$. $\endgroup$
    – xiuhua
    Jan 18, 2022 at 12:04

1 Answer 1

0
$\begingroup$

Note that $B=wDCD^{-1}$ where $w=\exp(\frac{i\pi}{n}),\,D=\operatorname{diag}(1,\frac{w}{2},\ldots,\frac{w^{n-1}}{n})$ and $$ C=\pmatrix{&&&&1\\ 1\\ &1\\ &&\ddots\\ &&&1}. $$ It is well known that $C=UD^2U^\ast$ where $U$ is a unitary DFT matrix. (See Wikipedia for instance.) Therefore $B=P_0\Lambda P_0^{-1}$, where $P_0=DU$ and $\Lambda=wD^2$. Since $U$ is a DFT matrix, the moduli of all its elements are equal to $\frac{1}{\sqrt{n}}$. It follows that all columns of $P_0=DU$ have the same norm, namely, $r=\sqrt{\sum_{k=1}^n\frac{1}{nk^2}}$.

Now, since all eigenvalues of $\Lambda$ (or $B$) are distinct, its eigenspaces are one-dimensional. So, if $B$ has a diagonalisation $P\Lambda P^{-1}$ in which all columns of $P$ are unit vectors, we must have $P=\frac{1}{r}P_0S=\frac{1}{r}DUS$ for some unitary diagonal matrix $S$. It follows that the condition number of $P$ is identical to the condition number of $D$, which is $n$. (We actually know more: the singular values of $P$ are $\frac{1}{rk}$ for $k=1,2,\ldots,n$, but it suffices to conclude using the condition number of $D$.)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .