Exercise 6.4.18

We get that $T^{\text{∗}}T$ is positive semidefinite by Exercise 6.4.17(a). Similarly, we get the same result for $T{T}^{\text{∗}}$.

and so $T(x)=0$. If $x\in N(T)$, we have $T^{\text{∗}}T(x)=T^{\text{∗}}(0)=0$.

Now we get that $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->(T^{\text{∗}}T)=\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->(T)$ and $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->(T{T}^{\text{∗}})=\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->(T^{\text{∗}})$ since $T^{\text{∗∗}}=T^{\text{∗}}$. Also, we have $\mathrm{rank}<!--\; FUNCTION\; APPLICATION\; -->(T)=\mathrm{rank}<!--\; FUNCTION\; APPLICATION\; -->(T^{\text{∗}})$ by the fact

for some orthonormal basis $\beta$. Finally by Dimension Theorem we get the result