 Solution

Put $T=A-zI$. We use the following closed-range lemma:Indeed,Away from this kernel, zero is separated from the spectrum of the positive self-adjoint operator $T^{\ast }T$ exactly whenfor some $c>0$. This is equivalent to closed range. Zero is then absent from the essential spectrum exactly when $\dim \ker T<\infty$. Similarly,because $\ker T^{\ast }=(\mathrm{ran}T{)}^{\perp }$ measures the cokernel.

Applying the lecture definitions of the three essential spectra of a closed operator now givesand

If $A$ is normal, so is $T$. The spectral theorem givesand maps the spectral mass of $T$ at $0$ exactly to the spectral mass of $T^{\ast }T$ at $0$. Thus zero is isolated with finite multiplicity for $T^{\ast }T$ exactly when $z$ is an isolated eigenvalue of finite multiplicity for $A$. Consequently

Normality is essential. Let $U{e}_j=e_{j+1}$ be the unilateral shift and take $A=U^{\ast }$. Its spectrum is the closed unit disk, so $0$ is not in the discrete spectrum. Butwhose zero eigenvalue is isolated and simple. Hence $0\in {\mathrm{Sp}}_{\mathrm{d}}(A^{\ast }A)$ while $0\notin {\mathrm{Sp}}_{\mathrm{d}}(A)$.