In a Hilbert space $X$, $T \in B(X)$, $|\lambda|=\lVert T \rVert$. Prove that $Im(\lambda I - T)+Ker(\lambda I-T)$ is dense in $X$

Question

Let $X$ be a Hilbert space, $T\in B(X)$ and $\lambda$ be a scalar such that $|\lambda|=\lVert T \rVert$. Prove that $Im(\lambda I - T)+Ker(\lambda I-T)$ is dense in $X$.

Since $Ker(\lambda I -T)=Im(\bar{\lambda}I-T^*)^\perp$ where $T^*$ is the adjoint of $T$, we need to show $Im(\lambda I-T)+Im(\bar{\lambda} I-T^*)^\perp$ is dense. My attempt is to assume it is not dense. Hence we can find something outside its closure. However, I am kind of stuck at this point.

I have proved that if $T^*x = \bar{\lambda}x$, then $Tx = \lambda x$. I don't know if this helps.

Answer 1

Suppose that $x$ is orthogonal to this space, and let's also normalize, $\|x\|=1$. Then, in particular, $x\in R(T-\lambda)^{\perp}=N(T^*-\overline{\lambda})$, so $$ \overline{\lambda} \langle x, Tx \rangle = \langle x, TT^* x\rangle = \|T^*x\|^2 = |\lambda|^2 $$ and thus $\langle x, Tx \rangle =\lambda$. On the other hand, $$ |\langle x, Tx \rangle | \le \|Tx\| \le |\lambda| \quad( = |\langle x, Tx \rangle |) . $$ Equality in the Cauchy-Schwarz inequality means that the vectors are linearly dependent, so $Tx = cx$, and only $c=\lambda$ works here.

So, in conclusion $x\in N(T-\lambda)$, but we also assumed that $x$ is orthogonal to this space, so $x=0$. (This whole argument assumes that $\lambda\not= 0$; of course, the claim is trivial if $T=0$.)