Let's suppose A is normal matrix. I know that the eigenvectors of normal matrix and its conjugate are the same, and that for the fixed eigenvector the eigenvalues are conjugate. Let's fix some $e_1$ normal eigenvector $(||e_1||=1)$, so $Ae_1= λ_1 e_1$ and $A^*e_1=\bar{λ_1}e_1 $. $$ S_1 = span\{e_1\} $$ $$ T_1 = \{x : (x,e_1)=0\} $$ How can I show that there is an eigenvector $e_2$ which belongs to $T_1$?
P.S. I know that if $x∈T_1$ then $Ax∈T_1,A^*x∈T_1$, because $$ (Ax,e_1)=(x,A^*e_1)=(x,\bar{λ_1}e_1)={λ_1}(x,e_1)=0 $$
So $L_1$ subspace is $A$-invariant, due to the last thing that you prove. And it is a famous theorem about linear operators, that every linear operator has at least one eigenvalue, so there exists some $e_2$ eigenvector which belongs to $T_1$. Here you can find it page 5.