# proof of theorem for normal matrices

1) ($A^{H}=g(A)\rightarrow A$ is normal)

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

 $AA^{H}=Ag(A)=A\sum_{i=0}^{n-1}a_{i}A^{i}=\sum_{i=0}^{n-1}a_{i}AA^{i}=\sum_{i=0% }^{n-1}a_{i}A^{i}A=\left(\sum_{i=0}^{n-1}a_{i}A^{i}\right)A=g(A)A=A^{H}A$

which shows $A$ to be normal.

2) ($A$ is normal $\rightarrow A^{H}=g(A)$)

Let $\lambda_{1},\lambda_{2},\ldots,\lambda_{r}$ , $1\leq r\leq n$ be the distinct eigenvalues of A, and let $\Lambda$ $=$ $diag\{\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\}$; then it’s possible to find a $(r-1)$-degree polynomial $g(t)$ such that $g(\lambda_{i})=\lambda_{i}^{\ast}$ $1\leq i\leq r$, solving the $r\times r$ linear Vandermonde system:

 $\begin{bmatrix}1&\lambda_{1}&\lambda_{1}^{2}&\cdots&\lambda_{1}^{r-1}\\ 1&\lambda_{2}&\lambda_{2}^{2}&\cdots&\lambda_{2}^{r-1}\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ 1&\lambda_{r-1}&\lambda_{r-1}^{2}&\cdots&\lambda_{r-1}^{r-1}\\ 1&\lambda_{r}&\lambda_{r}^{2}&\cdots&\lambda_{r}^{r-1}\end{bmatrix}\begin{% bmatrix}a_{0}\\ a_{1}\\ a_{2}\\ \vdots\\ a_{r-1}\end{bmatrix}=\begin{bmatrix}\lambda_{1}^{\ast}\\ \lambda_{2}^{\ast}\\ \lambda_{3}^{\ast}\\ \vdots\\ \lambda_{r}^{\ast}\end{bmatrix}$

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

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

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

 $A=UTU^{H}$

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

Let’s evaluate $g(A)$.

 $g(A)=g(U\Lambda U^{H})=\sum_{i=0}^{r-1}a_{i}(U\Lambda U^{H})^{i}$

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

 $(U\Lambda U^{H})^{i}=\overset{i\ times}{\overbrace{U\Lambda U^{H}U\Lambda U^{H% }U\Lambda U^{H}\cdots U\Lambda U^{H}}}=U\Lambda^{i}U^{H}$

and so

 $\displaystyle g(A)$ $\displaystyle=$ $\displaystyle\sum_{i=0}^{r-1}a_{i}(U\Lambda^{i}U^{H})$ $\displaystyle=$ $\displaystyle U\left(\sum_{i=0}^{r-1}a_{i}\Lambda^{i}\right)U^{H}$ $\displaystyle=$ $\displaystyle Ug(\Lambda)U^{H}$ $\displaystyle=$ $\displaystyle U\Lambda^{\ast}U^{H}$ $\displaystyle=$ $\displaystyle U\Lambda^{H}U^{H}$ $\displaystyle=$ $\displaystyle(U\Lambda U^{H})^{H}=A^{H}$

which is the thesis.

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

Example:

Let $A=\frac{1}{2}\begin{bmatrix}1+j&-1-j\\ 1+j&1+j\end{bmatrix}$ (which is easily checked to be normal),

with  $U$ $=\frac{1}{\sqrt{2}}\begin{bmatrix}1&-j\\ j&-1\end{bmatrix}$. Then $\sigma(A)=\{1,j\}$ and the Vandermonde system is

 $\begin{bmatrix}1&1\\ 1&j\end{bmatrix}\begin{bmatrix}a_{0}\\ a_{1}\end{bmatrix}=\begin{bmatrix}1\\ -j\end{bmatrix}$

from which we find

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

A simple calculation yields

 $g(A)=(1-j)I+jA=\frac{1}{2}\begin{bmatrix}1-j&1-j\\ -1+j&1-j\end{bmatrix}=A^{H}$
Title proof of theorem for normal matrices ProofOfTheoremForNormalMatrices 2013-03-22 15:36:36 2013-03-22 15:36:36 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 16 Andrea Ambrosio (7332) Proof msc 15A21