Measure, Integration & Real Analysis, 2021a

Section 10C Compact Operators 321
Our next result states that if T is a compact operator and α ̸= 0, then null(T − αI)
and null(T ∗ − αI) have the same dimension (denoted dim). This result about
the dimensions of spaces of eigenvectors is easier to prove in finite dimensions.
Specifically, suppose S is an operator on a finite-dimensional Hilbert space V (you
can think of S = T − αI). Then
dim null S = dim V − dim range S = dim(range S) ⊥ = dim null S ∗ ,
where the justification for each step should be familiar to you from finite-dimensional
linear algebra. This finite-dimensional proof does not work in infinite dimensions
because the expression dim V − dim range S could be of the form ∞ − ∞.
Although the dimensions of the two null spaces in the result below are the same,
even in finite dimensions the two null spaces are not necessarily equal to each other
(but we do have equality of the two null spaces when T is normal; see 10.56).
Note that both dimensions in the result below are finite (by 10.82 and 10.73).
10.91 null spaces of T − αI and T ∗ − αI have same dimensions
Suppose T is a compact operator on a Hilbert space and α ∈ F with α ̸= 0. Then
Proof
dim null(T − αI) =dim null(T ∗ − αI).
Suppose dim null(T − αI) < dim null(T ∗ − αI). Because null(T ∗ − αI)
equals ( range(T − αI) ) ⊥ , there is a bounded injective linear map
R : null(T − αI) → ( range(T − αI) ) ⊥
that is not surjective. Let V denote the Hilbert space on which T operates, and let P
be the orthogonal projection of V onto null(T − αI). Define a linear map S : V → V
by
S = T + RP.
Because RP is a bounded operator with finite-dimensional range, S is compact. Also,
S − αI =(T − αI)+RP.
Every element of range(T − αI) is orthogonal to every element of range RP.
Suppose f ∈ V and (S − αI) f = 0. The equation above shows that (T − αI) f = 0
and RPf = 0. Because f ∈ null(T − αI), we see that Pf = f , which then implies
that Rf = RPf = 0, which then implies that f = 0 (because R is injective). Hence
S − αI is injective.
However, because R maps onto a proper subset of ( range(T − αI) ) ⊥ , we see that
S − αI is not surjective, which contradicts the equivalence of (b) and (c) in 10.85. This
contradiction means the assumption that dim null(T − αI) < dim null(T ∗ − αI)
was false. Hence we have proved that
10.92 dim null(T − αI) ≥ dim null(T ∗ − αI)
for every compact operator T and every α ∈ F \{0}.
Now apply the conclusion of the previous paragraph to T ∗ (which is compact by
10.73) and α, getting 10.92 with the inequality reversed, completing the proof.
Measure, Integration & Real Analysis, by Sheldon Axler