# How to prove the condition number $\kappa(P) = n$?

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 $$$$B=\begin{bmatrix} & & & & -n \\ \frac{1}{2} & & & & \\ & \frac{2}{3} & & & \\ & & \ddots & & \\ & & & \frac{n-1}{n} & \end{bmatrix}_{n \times n}.$$$$ However, $$B$$ is not diagonal. I have obtained the eigendecomposition of $$B$$ as follow $$$$BP = P\Lambda,$$$$ where $$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 $$$$AXP - XBP = CP \Rightarrow A(XP) - (XP) \Lambda = CP.$$$$ Thus, I want to estimate the condition number of $$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., $$$$\kappa(P)=\left\|P^{-1}\right\|\|P\| = \frac{\sigma_{\max}(P)}{\sigma_{\min}(P)} = n.$$$$ But I don't know how to prove $$\kappa(P)=n$$ yet.

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

## 1 Answer

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 $$D$$.)