proof of normal matrix criterion

[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 ?

[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>.

[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.