proof of theorem for normal matrices

1) ($A^H=g(A)\to A$ is normal)

Keeping in mind that every matrix commutes with its own powers, let’s compute

$$A{A}^H=Ag(A)=A\sum _{i=0}^{n-1}{a}_i{A}^i=\sum _{i=0}^{n-1}{a}_{i}A{A}^i=\sum _{i=0}^{n-1}{a}_i{A}^{i}A=\left(\sum _{i=0}^{n-1}{a}_i{A}^i\right)A=g(A)A=A^{H}A$$

which shows $A$ to be normal.

2) ($A$ is normal $\to A^H=g(A)$)

Let ${\lambda }_1,{\lambda }_2,\mathrm{\dots },{\lambda }_r$ , $1\le r\le n$ be the distinct eigenvalues of A, and let $\mathrm{\Lambda }$ $=$ $diag\{{\lambda }_1,{\lambda }_2,\mathrm{\dots },{\lambda }_n\}$; then it’s possible tofind a $(r-1)$-degree polynomial $g(t)$ such that $g({\lambda }_i)={\lambda }_i^{\ast }$ $1\le i\le r$, solving the $r\times r$ linear Vandermondesystem:

Since these $r$ eigenvalues are distinct, the Vandermonde matrix is fullrank, and the linear system admits a unique solution; so a $(r-1)$-degreepolynomial $g(t)$ can be found such that $g({\lambda }_i)={\lambda }_i^{\ast }$ $1\le i\le r$ and therefore $g({\lambda }_i)={\lambda }_i^{\ast }$ $1\le i\le n$. Writing these equations in matrix form, we have

$$g(\mathrm{\Lambda })={\mathrm{\Lambda }}^{\ast }$$

By Schur’s decomposition theorem, a unitary matrix $U$ and an uppertriangular matrix $T$ exist such that

$$A=UT{U}^H$$

and since $A$ is normal we have $T=\mathrm{\Lambda }$.

Let’s evaluate $g(A)$.

$$g(A)=g(U\mathrm{\Lambda }{U}^H)=\sum _{i=0}^{r-1}{a}_i{(U\mathrm{\Lambda }{U}^H)}^i$$

But, keeping in mind that $U^{H}U=I$,

$${(U\mathrm{\Lambda }{U}^H)}^i=\stackrel{itimes}{\overbrace{U\mathrm{\Lambda }{U}^{H}U\mathrm{\Lambda }{U}^{H}U\mathrm{\Lambda }{U}^H\mathrm{\cdots }U\mathrm{\Lambda }{U}^H}}=U{\mathrm{\Lambda }}^i{U}^H$$

and so

$g(A)$$=$${\displaystyle \sum _{i=0}^{r-1}}{a}_i(U{\mathrm{\Lambda }}^i{U}^H)$

$=$$U\left({\displaystyle \sum _{i=0}^{r-1}}{a}_i{\mathrm{\Lambda }}^i\right)U^H$

$=$$Ug(\mathrm{\Lambda })U^H$

$=$$U{\mathrm{\Lambda }}^{\ast }{U}^H$

$=$$U{\mathrm{\Lambda }}^H{U}^H$

$=$${(U\mathrm{\Lambda }{U}^H)}^H=A^H$

which is the thesis.

Remark: note that this is a constructive proof, giving explicitly a way tofind $g(t)$ polynomial by solving Vandermonde system in the eigenvalues.

Example:

Let (which is easily checked to be normal),

with  $U$ . Then $\sigma (A)=\{1,j\}$ and the Vandermonde system is

from which we find

$$g(t)=(1-j)+jt$$

A simple calculation yields