linear algebra, higher linear algebra
(…)
A $K$-linear map (also $K$-linear function, $K$-linear operator, or $K$-linear transformation) is a morphism in $K$-Vect (or $K$-Mod), that is a homomorphism of vector spaces (or modules). Often one suppresses mention of the field (or commutative ring or rig) $K$.
In elementary terms, a $K$-linear map between $K$-linear spaces $V$ and $W$ is a function $T\colon V \to W$ such that
for $x$ and $y$ elements of $V$ and $r$ an element of $K$. (It is an easy exercise that this one identity is enough to ensure that $T$ preserves all linear combinations.)
The morphisms between topological vector spaces are of course the continuous linear maps. Between Banach spaces (including of course Hilbert spaces), these are the same as the bounded linear maps, so they're often called bounded operators (with linearity tacitly assumed).
In this context, linear operators are more general; they are (in general) only partial functions. However, we still require the domain of the partial function to be a linear subspace, after which the definition above applies. Because one typically restricts attention to complete spaces, the densely-defined operator?s (where the domain is a dense subspace) are the most general needed. To specify that the domain of a linear operator $T\colon V \to W$ is all of $V$, one may use a non-‘operator’ term, such as linear mapping.
Notice that we do not require partially-defined linear operators to be continuous; see unbounded operator. However, we have the theorem that any densely-defined, continuous, linear map $T\colon V \to W$, with $W$ complete and Hausdorff, extends uniquely to all of $V$. Thus one typically assumes that a continuous (or bounded) linear operator $T$ is defined on all of $V$ while an arbitrary linear operator $T$ is defined only on a dense subspace of $V$.
In elementary mathematics (at least as taught in the United States, perhaps elsewhere?), the term ‘linear function’ is usually used more generally, for an affine map (but still between vector spaces); in this same context, the term ‘linear transformation’ is often used instead for specifically linear maps. (Another difference at this level is that ‘linear functions’ are usually scalar-valued, while ‘linear transformations’ are usually vector-valued.)
In operator theory?, one sometimes distinguishes ‘linear maps’ (defined everywhere, but not necessarily continuous in general) from ‘linear operators’ (partially defined in general, but assumed to be defined everywhere if continuous between complete Hausdorff spaces). There is also a tendency for ‘operator’ to be used only for (possibly partial) endomorphisms, that is $T\colon V \to V$; then operators may be composed, giving rise to an operator algebra. If $V$ is a function space, then an endomorphism of $V$ is an operator in the sense of higher-order logic; the more general meaning of ‘linear operator’ is abstracted from this.
piecewise linear function?
Last revised on October 19, 2022 at 07:28:04. See the history of this page for a list of all contributions to it.