Numerical Range

The numerical range of an operator $A$ is the set $W(A)$ of all complex numbers of the form $\left\langle Af, f \right\rangle$, where $f$ varies over all vectors on the unit sphere (that is, $\left\| f\right\| = 1$, not $\left\| f\right\| \leq 1$).

Recall the expression of a projection $P$ of rank $1$ in terms of a unit vector $f$ in its range: $Pg = \left\langle g, f \right\rangle f$ for all $g$. If $A$ is an arbitrary operator, then $PAP$ is an operator of rank $1$, and therefore a finite-dimensional concept such as trace makes sense for it; since $Pf = f$, the value of that trace is $\operatorname {tr} PAP = \left\langle APf, Pf \right\rangle = \left\langle Af, f \right\rangle$. These remarks can be summarized as follows: $W(A)$ is equal to the set of all complex numbers of the form $\operatorname {tr} PAP$, where $P$ varies over all projections of rank $1$. Replace $1$ by an arbitrary positive integer $k$, and obtain the $k$-numerical range of $A$, in symbols $W_k(A)$: it is the set of all complex numbers of the form $\operatorname {tr} PAP$, where $P$ varies over all projections of rank $k$. The ordinary numerical range is the $k$-numerical range with $k=1$.

Solution 212 exhibited a compact operator $A$ such that $W(A) = (0,1]$; that is, $0$ is in the closure of $W(A)$ but not in $W(A)$ itself.

The trivial corollary that asserts that if $A = B+iC$, with $B$ and $C$ Hermitian, then $\operatorname {spec} A \subset \overline{W(B)} + i\overline{W(C)}$ is the Bendixson--Hirsch theorem.

If $A$ is a quasinilpotent operator, then, by Problem 214, $0 \in \overline{W(A)}$. By Solution 212, the set $W(A)$ may fail to be closed, so that from $0 \in \overline{W(A)}$ it does not follow that $0 \in W(A)$. Observe that any such example is a solution of Problem 212.

The convex hull of a set $M$, in symbols $\operatorname {conv} M$, is by definition the smallest convex set that includes $M$; in other words, $\operatorname {conv} M$ is the intersection of all the convex sets that include $M$. It is a non-trivial fact of finite-dimensional Euclidean geometry that the convex hull of a compact set is closed; the convex hull of a compact set is the intersection of all the closed half planes that include it. The closure of a convex set is convex.

As an application: if $\left|\lambda \right| = 1$ and

then $A$ is unitary, and therefore normal, so the result implies that its numerical range is the convex hull of its eigenvalues (the cube roots of $\lambda$) --- a triangle with those eigenvalues as vertices. Likewise, the numerical range of a direct sum is the convex hull of the numerical ranges of its summands; applying this to the direct sum of $\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ and $(1)$ recaptures the domed-cone shape of the numerical range of $\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$.

The numerical range, like the spectrum, associates a set with each operator. There is a closely related numerical function $w$, called the numerical radius, defined by $w(A) = \sup \left\{ \left|\lambda \right| : \lambda \in W(A)\right\}$ (cf.the definition of spectral radius $r(A)$, Problem 88). It is easy to prove that $w$ is a norm: $w(A) \geq 0$, and $w(A) = 0$ if and only if $A = 0$; $w(\alpha A) = \left|\alpha \right| \cdot w(A)$ for each scalar $\alpha$; and $w(A+B) \leq w(A) + w(B)$. This norm is equivalent to the ordinary operator norm: $\tfrac {1}{2}\left\| A\right\| \leq w(A) \leq \left\| A\right\|$. The norm $w$ has many other pleasant properties; thus, for instance, $w(A^{*}) = w(A)$, $w(A^{*}A) = \left\| A\right\|^2$, and $w$ is unitarily invariant, in the sense that $w(U^{*}AU) = w(A)$ whenever $U$ is unitary.

Since $\operatorname {spec} A \subset \overline{W(A)}$ (Problem 214), there is an easy inequality between the spectral radius and the numerical radius: $r(A) \leq w(A)$. The existence of quasinilpotent (or, for that matter, nilpotent) operators shows that nothing like the reverse inequality could be true.

If $A$ is normal, then $w(A) = \left\| A\right\|$. Wintner called an operator $A$ with $w(A) = \left\| A\right\|$ normaloid. Another useful (but nameless) property of a normal operator $A$ (Problem 216) is that $\overline{W(A)}$ is the convex hull of $\operatorname {spec} A$; call an (not necessarily normal) operator with this property convexoid. Still another (nameless) property of a normal operator $A$ is that $r(A) = w(A)$; call an operator with this property spectraloid. It is a consequence of Problem 218 that every normaloid operator is spectraloid. It is also true that every convexoid operator is spectraloid: since the closed disc with center $0$ and radius $r(A)$ includes $\operatorname {spec} A$ and is convex, it follows that if $A$ is convexoid, then that disc includes $W(A)$; this implies that $w(A) \leq r(A)$, and hence that $A$ is spectraloid.

The best way to ask the question is in terms of the Hausdorff metric for compact subsets of the plane. To define that metric, write $M + (\varepsilon ) = \left\{ z + \alpha : z \in M, \left|\alpha \right| < \varepsilon \right\}$ for each set $M$ of complex numbers and each positive number $\varepsilon$. In this notation, if $M$ and $N$ are compact sets, the Hausdorff distance $d(M,N)$ between them is the infimum of all positive numbers $\varepsilon$ such that both $M \subset N + (\varepsilon )$ and $N \subset M + (\varepsilon )$.

Since the Hausdorff metric is defined for compact sets, the appropriate function to discuss is $\overline{W}$, not $W$. The only thing that is immediately obvious is that if $\overline{W}$ is continuous with respect to any topology, then so is $w$, and consequently, if $w$ is discontinuous, then so is $\overline{W}$.

The good properties of the numerical range and the numerical radius have to do with convexity and linearity; the relations between the numerical range and the multiplicative properties of operators are less smooth. Thus, for instance, $w$ is certainly not multiplicative, i.e., $w(AB)$ is not always equal to $w(A)w(B)$. (Example with commutative normal operators: if $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$, then $w(A) = w(B) = 1$ and $w(AB) = 0$.) The next best thing would be for $w$ to be submultiplicative ($w(AB) \leq w(A)w(B)$), but that is false too. (Example: if $A = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, then $w(A) = w(B) = \tfrac {1}{2}$ and $w(AB) = 1$.) Since $w(AB) \leq \left\| AB\right\| \leq \left\| A\right\| \cdot \left\| B\right\|$, it follows that for normal operators $w$ is submultiplicative (because if $A$ and $B$ are normal, then $\left\| A\right\| = w(A)$ and $\left\| B\right\| = w(B)$), and for operators in general $w(AB) \leq 4w(A)w(B)$ (because $\left\| A\right\| \leq 2w(A)$ and $\left\| B\right\| \leq 2w(B)$). The example used to show that $w$ is not submultiplicative shows also that the constant $4$ is best possible here.

Commutativity sometimes helps; here it does not. Examples of commutative operators $A$ and $B$ for which $w(AB) > w(A)w(B)$ are a little harder to come by, but they exist. Here is one: with

and $B = A^2$, it is easy to see that $w(A^2) = w(A^3) = \tfrac {1}{2}$. The value of $w(A)$ is slightly harder to compute, but it is not needed; the almost obvious relation $w(A) < 1$ will do. Indeed: $w(AB) = w(A^3) = \tfrac {1}{2} > w(A) \cdot \tfrac {1}{2} = w(A)w(B)$.

The only shred of multiplicative behavior that has not yet been ruled out is the power inequality $w(A^n) \leq (w(A))^n$.

This statement is obviously a consequence of the power inequality. To show that it also implies the power inequality, reason as follows. If $w(A) = 0$, then $A = 0$, and everything is trivial. If $w(A) \neq 0$, then write $B = A/w(A)$, note that $w(B) \leq 1$, use the statement of this problem to infer that $w(B^n) \leq 1$, and conclude that $w(A^n) \leq (w(A))^n$.