Let $V$ inner product space from finite dimension under $\mathbb C$ and $T: V\to V$ linear transformation so that $T^2=\frac{T+T^*}{2}.$
I proved that is normal and I need also to prove that $T^2-T=0$.
I'd love your help with this.
Thanks!
Let $V$ inner product space from finite dimension under $\mathbb C$ and $T: V\to V$ linear transformation so that $T^2=\frac{T+T^*}{2}.$
I proved that is normal and I need also to prove that $T^2-T=0$.
I'd love your help with this.
Thanks!
Since $T$ is normal, both $T$ and $T^*$ are semi-simple (diagonalizable) in the same basis (they commute). Moreover, if $v$ is an eigenvector of $T$ associated with eigenvalue $\lambda$, then $v$ is an eigenvector of $T^*$ associated with eigenvalue $\bar{\lambda}$. This implies that w.r.t the diagonalizing basis $B$ we have $[T]_B=\rm{diag}(\lambda_1,...,\lambda_n)$ and $[T^*]_B=\rm{diag}(\bar{\lambda_1},...,\bar{\lambda_n})$.
Now use the equality $T^2=\frac{1}{2}(T+T^*)$: it holds also in base $B$, so for any eigenvalue $\lambda$ of $T$ you should have $\lambda^2=\frac{1}{2}(\lambda+\bar{\lambda})$. By representing $\lambda=a+ib$ and substituting, we have $a^2=a, b=0$. It follows that $\lambda\in\{0,1\}$. This implies that $T^2=T$.