= Solution
Put $T=A-zI$. We use the following closed-range lemma:
$$
0\notin\operatorname{Sp}_{\rm ess}(T^*T)
\quad\Longleftrightarrow\quad
T\text{ is upper semi-Fredholm}.
$$
Indeed,
$$
\ker(T^*T)=\ker T.
$$
Away from this kernel, zero is separated from the spectrum of the positive self-adjoint operator $T^*T$ exactly when
$$
\|Tx\|\geq c\|x\|,
\qquad x\perp\ker T,
$$
for some $c>0$. This is equivalent to closed range. Zero is then absent from the essential spectrum exactly when $\dim\ker T<\infty$. Similarly,
$$
0\notin\operatorname{Sp}_{\rm ess}(TT^*)
\quad\Longleftrightarrow\quad
T\text{ is lower semi-Fredholm},
$$
because $\ker T^*=(\operatorname{ran}T)^\perp$ measures the cokernel.
Applying the lecture definitions of the three <essential spectra of a closed operator> now gives
$$
\boxed{
\operatorname{Sp}_{\rm ess,1}(A)
=\left\{z:
0\in\operatorname{Sp}_{\rm ess}(T^*T)
\cap\operatorname{Sp}_{\rm ess}(TT^*)\right\}},
$$
$$
\boxed{
\operatorname{Sp}_{\rm ess,2}(A)
=\left\{z:
0\in\operatorname{Sp}_{\rm ess}(T^*T)\right\}},
$$
and
$$
\boxed{
\operatorname{Sp}_{\rm ess,3}(A)
=\left\{z:
0\in\operatorname{Sp}_{\rm ess}(T^*T)
\cup\operatorname{Sp}_{\rm ess}(TT^*)\right\}}.
$$
If $A$ is normal, so is $T$. The spectral theorem gives
$$
T^*T=|T|^2,
$$
and maps the spectral mass of $T$ at $0$ exactly to the spectral mass of $T^*T$ at $0$. Thus zero is isolated with finite multiplicity for $T^*T$ exactly when $z$ is an isolated eigenvalue of finite multiplicity for $A$. Consequently
$$
\boxed{
\operatorname{Sp}_{\rm d}(A)
=\{z:0\in\operatorname{Sp}_{\rm d}(T^*T)\}}.
$$
Normality is essential. Let $Ue_j=e_{j+1}$ be the unilateral shift and take $A=U^*$. Its spectrum is the closed unit disk, so $0$ is not in the <discrete spectrum>. But
$$
A^*A=UU^*=I-P_{\{e_1\}},
$$
whose zero eigenvalue is isolated and simple. Hence $0\in\operatorname{Sp}_{\rm d}(A^*A)$ while $0\notin\operatorname{Sp}_{\rm d}(A)$.
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