  It is well known that for a single bounded operator $A_0$ on a Hilbert
$\mathfrak{H}$, if $\mathfrak{M}\subset \mathfrak{H}$ is hyperinvariant for
$A_0$, then the spectrum of $A_0|_{\mathfrak{M}}$ is contained in the spectrum
of $A_0$. In this note, we modify an example of Taylor to prove the following.
There exist a quadruple $A=(A_1,A_2,A_3,A_4)$ of commuting bounded Hilbert
space operators and a hyperinvariant subspace $\mathfrak{X}_1$ for $A$ such
that the Taylor joint spectrum of $A$ restricted to $\mathfrak{X}_1$ is a not a
subset of the Taylor joint spectrum of $A$.
