Let $T \in Hom(ℂ^3)$, $T$ a normal operator and $T(1,1,1)=(2,2,2)$. If $(x,y,z) \in ker(T)$, then $x+y+z=0$.
I was trying to solve this question, one of my attempts was using the fact that if $T$ is a normal operator, then eigenvectors of $T$ corresponding to distinct eigenvalues are orthogonal. In this case: $T(1,1,1)=2 \cdot (1,1,1)$, thus $2$ is an eigenvalue associated to the eigenvector $(1,1,1)$.
Another property is that if $T$ is a normal operator, then $ker(T)=ker(T^*)$, and I was using other properties but, unfortunately I can't find the answer, anyone can help me?.
If $T(1,1,1)=(2,2,2)$, then $(1,1,1)$ is an eigenvector of $T$, with eigenvalue $2$. If $(x,y,z)\in\ker T$, then either $(x,y,z)=0$ or $(x,y,z)$ is another eigenvector of $T$, with eigenvalue $0$. If $(x,y,z)=0$, then clearly $x+y+z=0$. Otherwise, since $T$ is normal and since $2\ne0$, $(1,1,1)$ and $(x,y,z)$ are orthogonal, which means that, again, $x+y+z=0$.