Let $V=\mathbb{C^3}$ with the standard inner product. find normal operator $T:V\rightarrow V$ given Ker(T)=span{$(1,1,0),(0,1,1)$} and $f_T=x^2(x-1)$ find $T(x,y,z)$.
So what I know is that I have $$T(1,1,0)=0$$ $$T(0,1,1)=0$$ and I have another eigenvector $\vec{v}$ for $\lambda=1$ such that $T\vec{v}=1\cdot\vec{v}$
I know that T should be normal so $\vec{v}$ should be orthogonal to those vectors in the span. I am not see how does it helps me, what elese I should use here to find $T$ in the efficient way?
$\ker(T)$ is two-dimensional. Find a unit vector $u\perp\ker(T)$. As $T$ is normal, eigenspaces for different eigenvalues are orthogonal to each other. Therefore, $u$ is an eigenvector of $T$. Let $\lambda$ be the corresponding eigenvalue.
Let $v$ and $w$ be two orthonormal vectors in $\ker(T)$. If we put $U=\pmatrix{u&v&w}$, where the coordinates of $u,v$ and $w$ are those with respect to the standard basis, then the matrix representation of $T$ with respect to the standard basis is given by $$ A=U\pmatrix{\lambda\\ &0\\ &&0}U^{-1}=U\pmatrix{\lambda\\ &0\\ &&0}U^\ast=\lambda uu^\ast. $$ Hence in practice we only need to find $u$; there is no need to calculate $v$ and $w$. The value of $\lambda$ can be determined from the given characteristic polynomial of $T$, and $$ T(x,y,z)=\lambda uu^\ast\pmatrix{x\\ y\\ z}. $$