  For an operator $A$ on a complex Banach space $X$ and a closed subspace
$M\subseteq X$, the local commutant of $A$ at $M$ is the set $C(A;M)$ of all
operators $T$ on $X$ such that $TAx=ATx$ for every $x\in M$. It is clear that $
C(A;M)$ is a closed linear space of operators, however it is not an algebra, in
general. For a given $A$, we show that $C(A;M)$ is an algebra if and only if
the largest subspace $M_A$ such that $C(A;M)=C(A;M_A)$ is invariant for every
operator in $C(A;M)$. We say that these are ultrainvariant subspaces of $A$.
For several types of operators we prove that there exist non-trivial
ultrainvariant subspaces. For a normal operator on a Hilbert space, every
hyperinvariant subspace is ultrainvariant. On the other hand, the lattice of
all ultrainvariant subspaces of a non-zero nilpotent operator can be strictly
smaller than the lattice of all hyperinvariant subspaces.
