If a linear map is a
bijection then it is called a linear isomorphism. In the case where $V=W$, a linear map is called a (linear) endomorphism. Sometimes the term linear operator refers to this case,^{
[1]} but the term "linear operator" can have different meanings for different conventions: for example, it can be used to emphasize that $V$ and $W$ are
real vector spaces (not necessarily with $V=W$),^{[
citation needed]} or it can be used to emphasize that $V$ is a
function space, which is a common convention in
functional analysis.^{
[2]} Sometimes the term linear function has the same meaning as linear map, while in
analysis it does not.
A linear map from V to W always maps the origin of V to the origin of W. Moreover, it maps
linear subspaces in V onto linear subspaces in W (possibly of a lower
dimension);^{
[3]} for example, it maps a
plane through the
origin in V to either a plane through the origin in W, a
line through the origin in W, or just the origin in W. Linear maps can often be represented as
matrices, and simple examples include
rotation and reflection linear transformations.
Let $V$ and $W$ be vector spaces over the same
field$K$.
A function $f:V\to W$ is said to be a linear map if for any two vectors ${\textstyle \mathbf {u} ,\mathbf {v} \in V}$ and any scalar $c\in K$ the following two conditions are satisfied:
Homogeneity of degree 1 / operation of scalar multiplication
$f(c\mathbf {u} )=cf(\mathbf {u} )$
Thus, a linear map is said to be operation preserving. In other words, it does not matter whether the linear map is applied before (the right hand sides of the above examples) or after (the left hand sides of the examples) the operations of addition and scalar multiplication.
By
the associativity of the addition operation denoted as +, for any vectors ${\textstyle \mathbf {u} _{1},\ldots ,\mathbf {u} _{n}\in V}$ and scalars ${\textstyle c_{1},\ldots ,c_{n}\in K,}$ the following equality holds:^{
[4]}^{
[5]}
Denoting the zero elements of the vector spaces $V$ and $W$ by ${\textstyle \mathbf {0} _{V}}$ and ${\textstyle \mathbf {0} _{W}}$ respectively, it follows that ${\textstyle f(\mathbf {0} _{V})=\mathbf {0} _{W}.}$ Let $c=0$ and ${\textstyle \mathbf {v} \in V}$ in the equation for homogeneity of degree 1:
A linear map $V\to K$ with $K$ viewed as a one-dimensional vector space over itself is called a
linear functional.^{
[6]}
These statements generalize to any left-module ${\textstyle {}_{R}M}$ over a ring $R$ without modification, and to any right-module upon reversing of the scalar multiplication.
Examples
A prototypical example that gives linear maps their name is a function $f:\mathbb {R} \to \mathbb {R} :x\mapsto cx$, of which the
graph is a line through the origin.^{
[7]}
More generally, any
homothety${\textstyle \mathbf {v} \mapsto c\mathbf {v} }$ where $c$ centered in the origin of a vector space is a linear map.
The zero map ${\textstyle \mathbf {x} \mapsto \mathbf {0} }$ between two vector spaces (over the same
field) is linear.
The
identity map on any module is a linear operator.
For real numbers, the map ${\textstyle x\mapsto x^{2}}$ is not linear.
For real numbers, the map ${\textstyle x\mapsto x+1}$ is not linear (but is an
affine transformation).
If $A$ is a $m\times n$real matrix, then $A$ defines a linear map from $\mathbb {R} ^{n}$ to $\mathbb {R} ^{m}$ by sending a
column vector$\mathbf {x} \in \mathbb {R} ^{n}$ to the column vector $A\mathbf {x} \in \mathbb {R} ^{m}$. Conversely, any linear map between
finite-dimensional vector spaces can be represented in this manner; see the
§ Matrices, below.
If ${\textstyle f:V\to W}$ is an
isometry between real
normed spaces such that ${\textstyle f(0)=0}$ then $f$ is a linear map. This result is not necessarily true for complex normed space.^{
[8]}
Differentiation defines a linear map from the space of all differentiable functions to the space of all functions. It also defines a linear operator on the space of all
smooth functions (a linear operator is a linear endomorphism, that is, a linear map with the same
domain and
codomain). An example is
An indefinite
integral (or
antiderivative) with a fixed integration starting point defines a linear map from the space of all real-valued integrable functions on $\mathbb {R}$ to the space of all real-valued, differentiable functions on $\mathbb {R}$. Without a fixed starting point, the antiderivative maps to the
quotient space of the differentiable functions by the linear space of constant functions.
If $V$ and $W$ are finite-dimensional vector spaces over a field F, of respective dimensions m and n, then the function that maps linear maps ${\textstyle f:V\to W}$ to n × m matrices in the way described in
§ Matrices (below) is a linear map, and even a
linear isomorphism.
The
expected value of a
random variable (which is in fact a function, and as such a element of a vector space) is linear, as for random variables $X$ and $Y$ we have $E[X+Y]=E[X]+E[Y]$ and $E[aX]=aE[X]$, but the
variance of a random variable is not linear.
The function ${\textstyle f:\mathbb {R} ^{2}\to \mathbb {R} ^{2}}$ with ${\textstyle f(x,y)=(2x,y)}$ is a linear map. This function scales the ${\textstyle x}$ component of a vector by the factor ${\textstyle 2}$.
The function ${\textstyle f(x,y)=(2x,y)}$ is additive: It doesn't matter whether vectors are first added and then mapped or whether they are mapped and finally added: ${\textstyle f(\mathbf {a} +\mathbf {b} )=f(\mathbf {a} )+f(\mathbf {b} )}$
The function ${\textstyle f(x,y)=(2x,y)}$ is homogeneous: It doesn't matter whether a vector is first scaled and then mapped or first mapped and then scaled: ${\textstyle f(\lambda \mathbf {a} )=\lambda f(\mathbf {a} )}$
Linear extensions
Often, a linear map is constructed by defining it on a subset of a vector space and then extending by linearity to the
linear span of the domain.
Suppose $X$ and $Y$ are vector spaces and $f:S\to Y$ is a
function defined on some subset $S\subseteq X.$
Then a linear extension of $f$ to $X,$ if it exists, is a linear map $F:X\to Y$ defined on $X$ that
extends$f$^{
[note 1]} (meaning that $F(s)=f(s)$ for all $s\in S$) and takes its values from the codomain of $f.$^{
[9]}
When the subset $S$ is a vector subspace of $X$ then a ($Y$-valued) linear extension of $f$ to all of $X$ is guaranteed to exist if (and only if) $f:S\to Y$ is a linear map.^{
[9]} In particular, if $f$ has a linear extension to $\operatorname {span} S,$ then it has a linear extension to all of $X.$
The map $f:S\to Y$ can be extended to a linear map $F:\operatorname {span} S\to Y$ if and only if whenever $n>0$ is an integer, $c_{1},\ldots ,c_{n}$ are scalars, and $s_{1},\ldots ,s_{n}\in S$ are vectors such that $0=c_{1}s_{1}+\cdots +c_{n}s_{n},$ then necessarily $0=c_{1}f\left(s_{1}\right)+\cdots +c_{n}f\left(s_{n}\right).$^{
[10]}
If a linear extension of $f:S\to Y$ exists then the linear extension $F:\operatorname {span} S\to Y$ is unique and
holds for all $n,c_{1},\ldots ,c_{n},$ and $s_{1},\ldots ,s_{n}$ as above.^{
[10]}
If $S$ is linearly independent then every function $f:S\to Y$ into any vector space has a linear extension to a (linear) map $\;\operatorname {span} S\to Y$ (the converse is also true).
For example, if $X=\mathbb {R} ^{2}$ and $Y=\mathbb {R}$ then the assignment $(1,0)\to -1$ and $(0,1)\to 2$ can be linearly extended from the linearly independent set of vectors $S:=\{(1,0),(0,1)\}$ to a linear map on $\operatorname {span} \{(1,0),(0,1)\}=\mathbb {R} ^{2}.$ The unique linear extension $F:\mathbb {R} ^{2}\to \mathbb {R}$ is the map that sends $(x,y)=x(1,0)+y(0,1)\in \mathbb {R} ^{2}$ to
$F(x,y)=x(-1)+y(2)=-x+2y.$
Every (scalar-valued)
linear functional$f$ defined on a
vector subspace of a real or complex vector space $X$ has a linear extension to all of $X.$
Indeed, the
Hahn–Banach dominated extension theorem even guarantees that when this linear functional $f$ is dominated by some given
seminorm$p:X\to \mathbb {R}$ (meaning that $|f(m)|\leq p(m)$ holds for all $m$ in the domain of $f$) then there exists a linear extension to $X$ that is also dominated by $p.$
If $V$ and $W$ are
finite-dimensional vector spaces and a
basis is defined for each vector space, then every linear map from $V$ to $W$ can be represented by a
matrix.^{
[11]} This is useful because it allows concrete calculations. Matrices yield examples of linear maps: if $A$ is a real $m\times n$ matrix, then $f(\mathbf {x} )=A\mathbf {x}$ describes a linear map $\mathbb {R} ^{n}\to \mathbb {R} ^{m}$ (see
Euclidean space).
Let $\{\mathbf {v} _{1},\ldots ,\mathbf {v} _{n}\}$ be a basis for $V$. Then every vector $\mathbf {v} \in V$ is uniquely determined by the coefficients $c_{1},\ldots ,c_{n}$ in the field $\mathbb {R}$:
which implies that the function f is entirely determined by the vectors $f(\mathbf {v} _{1}),\ldots ,f(\mathbf {v} _{n})$. Now let $\{\mathbf {w} _{1},\ldots ,\mathbf {w} _{m}\}$ be a basis for $W$. Then we can represent each vector $f(\mathbf {v} _{j})$ as
Thus, the function $f$ is entirely determined by the values of $a_{ij}$. If we put these values into an $m\times n$ matrix $M$, then we can conveniently use it to compute the vector output of $f$ for any vector in $V$. To get $M$, every column $j$ of $M$ is a vector
corresponding to $f(\mathbf {v} _{j})$ as defined above. To define it more clearly, for some column $j$ that corresponds to the mapping $f(\mathbf {v} _{j})$,
where $M$ is the matrix of $f$. In other words, every column $j=1,\ldots ,n$ has a corresponding vector $f(\mathbf {v} _{j})$ whose coordinates $a_{1j},\cdots ,a_{mj}$ are the elements of column $j$. A single linear map may be represented by many matrices. This is because the values of the elements of a matrix depend on the bases chosen.
The matrices of a linear transformation can be represented visually:
Matrix for ${\textstyle T}$ relative to ${\textstyle B}$: ${\textstyle A}$
Matrix for ${\textstyle T}$ relative to ${\textstyle B'}$: ${\textstyle A'}$
Transition matrix from ${\textstyle B'}$ to ${\textstyle B}$: ${\textstyle P}$
Transition matrix from ${\textstyle B}$ to ${\textstyle B'}$: ${\textstyle P^{-1}}$
The relationship between matrices in a linear transformation
Such that starting in the bottom left corner ${\textstyle \left[\mathbf {v} \right]_{B'}}$ and looking for the bottom right corner ${\textstyle \left[T\left(\mathbf {v} \right)\right]_{B'}}$, one would left-multiply—that is, ${\textstyle A'\left[\mathbf {v} \right]_{B'}=\left[T\left(\mathbf {v} \right)\right]_{B'}}$. The equivalent method would be the "longer" method going clockwise from the same point such that ${\textstyle \left[\mathbf {v} \right]_{B'}}$ is left-multiplied with ${\textstyle P^{-1}AP}$, or ${\textstyle P^{-1}AP\left[\mathbf {v} \right]_{B'}=\left[T\left(\mathbf {v} \right)\right]_{B'}}$.
Examples in two dimensions
In two-
dimensional space R^{2} linear maps are described by 2 × 2
matrices. These are some examples:
The composition of linear maps is linear: if $f:V\to W$ and ${\textstyle g:W\to Z}$ are linear, then so is their
composition${\textstyle g\circ f:V\to Z}$. It follows from this that the
class of all vector spaces over a given field K, together with K-linear maps as
morphisms, forms a
category.
The
inverse of a linear map, when defined, is again a linear map.
If ${\textstyle f_{1}:V\to W}$ and ${\textstyle f_{2}:V\to W}$ are linear, then so is their
pointwise sum $f_{1}+f_{2}$, which is defined by $(f_{1}+f_{2})(\mathbf {x} )=f_{1}(\mathbf {x} )+f_{2}(\mathbf {x} )$.
If ${\textstyle f:V\to W}$ is linear and ${\textstyle \alpha }$ is an element of the ground field ${\textstyle K}$, then the map ${\textstyle \alpha f}$, defined by ${\textstyle (\alpha f)(\mathbf {x} )=\alpha (f(\mathbf {x} ))}$, is also linear.
Thus the set ${\textstyle {\mathcal {L}}(V,W)}$ of linear maps from ${\textstyle V}$ to ${\textstyle W}$ itself forms a vector space over ${\textstyle K}$,^{
[12]} sometimes denoted ${\textstyle \operatorname {Hom} (V,W)}$.^{
[13]} Furthermore, in the case that ${\textstyle V=W}$, this vector space, denoted ${\textstyle \operatorname {End} (V)}$, is an
associative algebra under
composition of maps, since the composition of two linear maps is again a linear map, and the composition of maps is always associative. This case is discussed in more detail below.
Given again the finite-dimensional case, if bases have been chosen, then the composition of linear maps corresponds to the
matrix multiplication, the addition of linear maps corresponds to the
matrix addition, and the multiplication of linear maps with scalars corresponds to the multiplication of matrices with scalars.
A linear transformation ${\textstyle f:V\to V}$ is an
endomorphism of ${\textstyle V}$; the set of all such endomorphisms ${\textstyle \operatorname {End} (V)}$ together with addition, composition and scalar multiplication as defined above forms an
associative algebra with identity element over the field ${\textstyle K}$ (and in particular a
ring). The multiplicative identity element of this algebra is the
identity map${\textstyle \operatorname {id} :V\to V}$.
An endomorphism of ${\textstyle V}$ that is also an
isomorphism is called an
automorphism of ${\textstyle V}$. The composition of two automorphisms is again an automorphism, and the set of all automorphisms of ${\textstyle V}$ forms a
group, the
automorphism group of ${\textstyle V}$ which is denoted by ${\textstyle \operatorname {Aut} (V)}$ or ${\textstyle \operatorname {GL} (V)}$. Since the automorphisms are precisely those
endomorphisms which possess inverses under composition, ${\textstyle \operatorname {Aut} (V)}$ is the group of
units in the ring ${\textstyle \operatorname {End} (V)}$.
If ${\textstyle V}$ has finite dimension ${\textstyle n}$, then ${\textstyle \operatorname {End} (V)}$ is
isomorphic to the
associative algebra of all ${\textstyle n\times n}$ matrices with entries in ${\textstyle K}$. The automorphism group of ${\textstyle V}$ is
isomorphic to the
general linear group${\textstyle \operatorname {GL} (n,K)}$ of all ${\textstyle n\times n}$ invertible matrices with entries in ${\textstyle K}$.
${\textstyle \ker(f)}$ is a
subspace of ${\textstyle V}$ and ${\textstyle \operatorname {im} (f)}$ is a subspace of ${\textstyle W}$. The following
dimension formula is known as the
rank–nullity theorem:^{
[14]}
The number ${\textstyle \dim(\operatorname {im} (f))}$ is also called the
rank of ${\textstyle f}$ and written as ${\textstyle \operatorname {rank} (f)}$, or sometimes, ${\textstyle \rho (f)}$;^{
[15]}^{
[16]} the number ${\textstyle \dim(\ker(f))}$ is called the
nullity of ${\textstyle f}$ and written as ${\textstyle \operatorname {null} (f)}$ or ${\textstyle \nu (f)}$.^{
[15]}^{
[16]} If ${\textstyle V}$ and ${\textstyle W}$ are finite-dimensional, bases have been chosen and ${\textstyle f}$ is represented by the matrix ${\textstyle A}$, then the rank and nullity of ${\textstyle f}$ are equal to the rank and nullity of the matrix ${\textstyle A}$, respectively.
This is the dual notion to the kernel: just as the kernel is a subspace of the domain, the co-kernel is a
quotient space of the target. Formally, one has the
exact sequence
These can be interpreted thus: given a linear equation f(v) = w to solve,
the kernel is the space of solutions to the homogeneous equation f(v) = 0, and its dimension is the number of
degrees of freedom in the space of solutions, if it is not empty;
the co-kernel is the space of
constraints that the solutions must satisfy, and its dimension is the maximal number of independent constraints.
The dimension of the co-kernel and the dimension of the image (the rank) add up to the dimension of the target space. For finite dimensions, this means that the dimension of the quotient space W/f(V) is the dimension of the target space minus the dimension of the image.
As a simple example, consider the map f: R^{2} → R^{2}, given by f(x, y) = (0, y). Then for an equation f(x, y) = (a, b) to have a solution, we must have a = 0 (one constraint), and in that case the solution space is (x, b) or equivalently stated, (0, b) + (x, 0), (one degree of freedom). The kernel may be expressed as the subspace (x, 0) < V: the value of x is the freedom in a solution – while the cokernel may be expressed via the map W → R, ${\textstyle (a,b)\mapsto (a)}$: given a vector (a, b), the value of a is the obstruction to there being a solution.
An example illustrating the infinite-dimensional case is afforded by the map f: R^{∞} → R^{∞}, ${\textstyle \left\{a_{n}\right\}\mapsto \left\{b_{n}\right\}}$ with b_{1} = 0 and b_{n + 1} = a_{n} for n > 0. Its image consists of all sequences with first element 0, and thus its cokernel consists of the classes of sequences with identical first element. Thus, whereas its kernel has dimension 0 (it maps only the zero sequence to the zero sequence), its co-kernel has dimension 1. Since the domain and the target space are the same, the rank and the dimension of the kernel add up to the same
sum as the rank and the dimension of the co-kernel (${\textstyle \aleph _{0}+0=\aleph _{0}+1}$), but in the infinite-dimensional case it cannot be inferred that the kernel and the co-kernel of an
endomorphism have the same dimension (0 ≠ 1). The reverse situation obtains for the map h: R^{∞} → R^{∞}, ${\textstyle \left\{a_{n}\right\}\mapsto \left\{c_{n}\right\}}$ with c_{n} = a_{n + 1}. Its image is the entire target space, and hence its co-kernel has dimension 0, but since it maps all sequences in which only the first element is non-zero to the zero sequence, its kernel has dimension 1.
Index
For a linear operator with finite-dimensional kernel and co-kernel, one may define index as:
namely the degrees of freedom minus the number of constraints.
For a transformation between finite-dimensional vector spaces, this is just the difference dim(V) − dim(W), by rank–nullity. This gives an indication of how many solutions or how many constraints one has: if mapping from a larger space to a smaller one, the map may be onto, and thus will have degrees of freedom even without constraints. Conversely, if mapping from a smaller space to a larger one, the map cannot be onto, and thus one will have constraints even without degrees of freedom.
Algebraic classifications of linear transformations
No classification of linear maps could be exhaustive. The following incomplete list enumerates some important classifications that do not require any additional structure on the vector space.
Let V and W denote vector spaces over a field F and let T: V → W be a linear map.
Monomorphism
T is said to be injective or a monomorphism if any of the following equivalent conditions are true:
T is
monic or left-cancellable, which is to say, for any vector space U and any pair of linear maps R: U → V and S: U → V, the equation TR = TS implies R = S.
T is
left-invertible, which is to say there exists a linear map S: W → V such that ST is the
identity map on V.
Epimorphism
T is said to be surjective or an epimorphism if any of the following equivalent conditions are true:
T is
epic or right-cancellable, which is to say, for any vector space U and any pair of linear maps R: W → U and S: W → U, the equation RT = ST implies R = S.
T is
right-invertible, which is to say there exists a linear map S: W → V such that TS is the
identity map on W.
Isomorphism
T is said to be an isomorphism if it is both left- and right-invertible. This is equivalent to T being both one-to-one and onto (a
bijection of sets) or also to T being both epic and monic, and so being a
bimorphism.
If T: V → V is an endomorphism, then:
If, for some positive integer n, the n-th iterate of T, T^{n}, is identically zero, then T is said to be
nilpotent.
Given a linear map which is an
endomorphism whose matrix is A, in the basis B of the space it transforms vector coordinates [u] as [v] = A[u]. As vectors change with the inverse of B (vectors are
contravariant) its inverse transformation is [v] = B[v'].
An example of an unbounded, hence discontinuous, linear transformation is differentiation on the space of smooth functions equipped with the supremum norm (a function with small values can have a derivative with large values, while the derivative of 0 is 0). For a specific example, sin(nx)/n converges to 0, but its derivative cos(nx) does not, so differentiation is not continuous at 0 (and by a variation of this argument, it is not continuous anywhere).
Applications
A specific application of linear maps is for
geometric transformations, such as those performed in
computer graphics, where the translation, rotation and scaling of 2D or 3D objects is performed by the use of a
transformation matrix. Linear mappings also are used as a mechanism for describing change: for example in calculus correspond to derivatives; or in relativity, used as a device to keep track of the local transformations of reference frames.
^"Linear transformations of V into V are often called linear operators on V."
Rudin 1976, p. 207
^Let V and W be two real vector spaces. A mapping a from V into W Is called a 'linear mapping' or 'linear transformation' or 'linear operator' [...] from V into W, if ${\textstyle a(\mathbf {u} +\mathbf {v} )=a\mathbf {u} +a\mathbf {v} }$ for all ${\textstyle \mathbf {u} ,\mathbf {v} \in V}$, ${\textstyle a(\lambda \mathbf {u} )=\lambda a\mathbf {u} }$ for all $\mathbf {u} \in V$ and all real λ.
Bronshtein & Semendyayev 2004, p. 316
^Rudin 1991, p. 14 Here are some properties of linear mappings ${\textstyle \Lambda :X\to Y}$ whose proofs are so easy that we omit them; it is assumed that ${\textstyle A\subset X}$ and ${\textstyle B\subset Y}$:
${\textstyle \Lambda 0=0.}$
If A is a subspace (or a
convex set, or a
balanced set) the same is true of ${\textstyle \Lambda (A)}$
If B is a subspace (or a convex set, or a balanced set) the same is true of ${\textstyle \Lambda ^{-1}(B)}$
is a subspace of X, called the null space of ${\textstyle \Lambda }$.
^Rudin 1991, p. 14. Suppose now that X and Y are vector spaces over the same scalar field. A mapping ${\textstyle \Lambda :X\to Y}$ is said to be linear if ${\textstyle \Lambda (\alpha \mathbf {x} +\beta \mathbf {y} )=\alpha \Lambda \mathbf {x} +\beta \Lambda \mathbf {y} }$ for all ${\textstyle \mathbf {x} ,\mathbf {y} \in X}$ and all scalars ${\textstyle \alpha }$ and ${\textstyle \beta }$. Note that one often writes ${\textstyle \Lambda \mathbf {x} }$, rather than ${\textstyle \Lambda (\mathbf {x} )}$, when ${\textstyle \Lambda }$ is linear.
^Rudin 1976, p. 206. A mapping A of a vector space X into a vector space Y is said to be a linear transformation if: ${\textstyle A\left(\mathbf {x} _{1}+\mathbf {x} _{2}\right)=A\mathbf {x} _{1}+A\mathbf {x} _{2},\ A(c\mathbf {x} )=cA\mathbf {x} }$ for all ${\textstyle \mathbf {x} ,\mathbf {x} _{1},\mathbf {x} _{2}\in X}$ and all scalars c. Note that one often writes ${\textstyle A\mathbf {x} }$ instead of ${\textstyle A(\mathbf {x} )}$ if A is linear.
^Rudin 1991, p. 14.
Linear mappings of X onto its scalar field are called linear functionals.
^Rudin 1976, p. 210
Suppose ${\textstyle \left\{\mathbf {x} _{1},\ldots ,\mathbf {x} _{n}\right\}}$ and ${\textstyle \left\{\mathbf {y} _{1},\ldots ,\mathbf {y} _{m}\right\}}$ are bases of vector spaces X and Y, respectively. Then every ${\textstyle A\in L(X,Y)}$ determines a set of numbers ${\textstyle a_{i,j}}$ such that
Observe that the coordinates ${\textstyle a_{i,j}}$ of the vector ${\textstyle A\mathbf {x} _{j}}$ (with respect to the basis ${\textstyle \{\mathbf {y} _{1},\ldots ,\mathbf {y} _{m}\}}$) appear in the j^{th} column of ${\textstyle [A]}$. The vectors ${\textstyle A\mathbf {x} _{j}}$ are therefore sometimes called the column vectors of ${\textstyle [A]}$. With this terminology, the range of Ais spanned by the column vectors of ${\textstyle [A]}$.
^Nistor, Victor (2001) [1994],
"Index theory", Encyclopedia of Mathematics,
EMS Press: "The main question in index theory is to provide index formulas for classes of Fredholm operators ... Index theory has become a subject on its own only after M. F. Atiyah and I. Singer published their index theorems"
^Rudin 1991, p. 15
1.18 TheoremLet ${\textstyle \Lambda }$ be a linear functional on a topological vector space X. Assume ${\textstyle \Lambda \mathbf {x} \neq 0}$ for some ${\textstyle \mathbf {x} \in X}$. Then each of the following four properties implies the other three:
${\textstyle \Lambda }$ is continuous
The null space ${\textstyle N(\Lambda )}$ is closed.
${\textstyle N(\Lambda )}$ is not dense in X.
${\textstyle \Lambda }$ is bounded in some neighbourhood V of 0.
^One map $F$ is said to
extend another map $f$ if when $f$ is defined at a point $s,$ then so is $F$ and $F(s)=f(s).$