# Proof of normal matrix criterion

1. Jul 30, 2008

### TTob

I need to proof:
A in normal matrix if and only if $$trace(A^*A)=|t_1|^2+...|t_n|^2$$
where $$t_1,...,t_n$$ are the characteristic roots of A.

I have a problem only with the second direction:
$$trace(A^*A)=|t_1|^2+...|t_n|^2$$ --> A is normal.

can you help me ?

2. Jul 30, 2008

### morphism

Presumably A is a complex matrix, so we can put it into Schur form, i.e. we can find a unitary matrix Q and an upper triangular matrix T such that A=Q-1TQ. Notice that trace(A*A)=trace(T*T). So if A is not normal, then <blank>.

3. Jul 31, 2008

### TTob

thanks.

note $$T=[\alpha_{ij}]$$
so $$trace(T^*T)=trace(A^*A)=|t_1|^2+...|t_n|^2$$
hence
$$\sum_{\substack{ 0\leq\i\leq n \\ 0\leq\j\leq n }} |\alpha_{ij}|^2 =|t_1|^2+...|t_n|^2$$
because of the eigenvalues are the diagonal entries of T we have
$$\sum_{\substack{ 0\leq\i\leq n \\ 0\leq\j\leq n \\ j\ne i }} |\alpha_{ij}|^2 =0$$
hence for $$i\ne j$$ we have $$\alpha_{ij}=0$$

so T is diagonal matrix. A=Q^-1TQ and hence A is normal.