Let $V$ be a finite dimensional vector $k$-space with inner product and $T\in \mathcal{L}(V)$. Prove that the operator $S=T^{*}T+I$ is invertible.
I have drafted:
Using the Euclidean Inner Product,
$$\left \langle x,Sx \right \rangle=\left \langle x,x \right \rangle+\left \langle x,T^{*}Tx \right \rangle=\left \langle x,x \right \rangle+\left \langle Tx,Tx \right \rangle=\left \| x \right \|^{2}+\left \| Tx \right \|^{2}\geq \left \| x \right \|^{2}.$$
Therefore, $Sx=0\implies ||x||=0\implies x=0$. That is, $S$ is a linear operator on a finite-dimensional space and $\ker{S}=\{0\}$. By the Rank Nullity Theorem, $S$ is invertible.
Am I wrong?
Best Answer
Yes it is correct.
Another way to establish this result is by using eigenvalues:
$T^*T$ being self adjoint $(T^*T)^*=T^*T$ has real eigenvalues $\lambda_k \ge 0$.
Therefore $T^*T+I$ has real eigenvalues $\mu_i:=\lambda_k+1 \ge 1 > 0$,
therefore is invertible (indeed its eigenvalues are $1/\mu_i$, all non-zero).