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Linear forms

A linear form or a linear functional on a vector space $V$ over a field $F$ is a linear map $h: V\to F$.

  • The space $\Hom(V,F)$ of linear forms is a vector space called the dual vector space to $V$. DF use the notation $V^{\ast}$ for $\Hom(V,F)$.

Suppose that $S$ is a basis for $V$ (not necessarily finite). For each $s\in S$, define a linear map $s^{\ast}: V\to F$ by saying that $s^{\ast}(s)=1$ and $s^{\ast}(x)=0$ for any other $x\in S$; then extending $s^{\ast}$ by linearity to all of $V$. Note that the linear map $s^{\ast}$ depends on all of $S$, not just on $s$ itself.

If $V$ is finite dimensional, and $s_1,\ldots, s_n$ is a basis for $V$, then $s_1^{\ast},\ldots, s_n^{\ast}$ is a basis for $V^{\ast}$ called the dual basis. To show that it spans $V^{\ast}$, let $f$ be any linear form and compute $f(s_{i})$. Then \(f=\sum f(s_{i})s_{i}^{\ast}\) since the right side agrees with $f$ on the basis $s_{i}$. The $s_{i}^{\ast}$ are linearly independent since if \(\sum a_{i}s_{i}^{\ast}=0\) Then \((\sum a_{i}s_{i}^{\ast})(s_{i})=0=a_{i}\) for all $i$.

If $V$ is infinite dimensional and $S$ is a basis, you can still construct elements $s^{\ast}$ dual to the elements of $S$ and they are linearly independent. However they won’t span.

Dual transformations

Suppose $L:V\to W$ is a linear map. Then there is a “dual map” $L^{\ast}:W^{\ast}\to V^{\ast}$ defined abstractly by setting $((L^*)(f))(v)=f(L(v))$.

Proposition: Given $L:V\to W$ and $G:W\to H$, we have \((G\circ L)^{\ast}=L^{\ast}\circ G^{\ast}\).

Proof: $(G\circ L)^{\ast}$ is a linear map from $H^{\ast}\to V^{\ast}$. If $h^{\ast}\in H^{\ast}$, we have

\[(G\circ L)^{\ast}(h^{\ast})(v)=h(G\circ L)(v)=(h\circ G\circ L)(v).\]

On the other hand, \(L^{\ast}\circ G^{\ast}(h^{\ast})=L^{\ast}(h\circ G) = (h\circ G\circ L)\) so the two things are the same.

Proposition: If A is a finite basis for $V$ and $B$ is a finite basis for $W$ then let $A^{\ast}$ and $B^{\ast}$ be the corresponding dual bases. Then \([L^\ast]_{B^{\ast}}^{A^{\ast}}\) is the transpose of $[L]_{A}^{B}$.

Proof: Consider $b_i^{\ast}$ in $B^{\ast}$. Then \(L^{\ast}(b_{j}^{\ast})(a_{i})=b_{j}^{\ast}(L(a_i))\) which is the coefficient of $b_{j}$ in the expansion of $L(a_{i})$. This is by definition the entry $x_{ji}$ in the matrix of $L$ relative to the bases $A$ and $B$.

On the other hand, if we write \(L^{\ast}(b_{j}^{\ast})=\sum y_{ij}a_{i}^{\ast}\) where $y_{ij}$ are the matrix entries of $L^{\ast}$ relative to the dual bases, then we see that $x_{ji}=y_{ij}$. In other words, the matrix entries for $L^{\ast}$ are those for $L$, but with rows and columns interchanged.

Let $H\subset V$ be a subspace. Then any linear form on $V$ restricts to one on $H$, so there is a map $V^{\ast}\to H^{\ast}$. This map is surjective since any linear form on $H$ extends to one on $V$. It’s kernel is the set of linear forms on $V$ that vanish on $H$; this is called the “annihilator of $H$”.

Corollary: The row and column ranks of a matrix coincide.

Proof: Let $L:V\to W$ and $L^{\ast}:W^{\ast}\to V^{\ast}$ be a linear map and its dual. $L$ gives an isomorphism from $V/K$ to the image $H$ of $L$ in $W$ where $K$ is the kernel of $L$. So we can view $L:V/K\to H$.

The dual transform $L^{\ast}$ takes a linear form on $W$ and makes it one on $V$ by the formula $L^{\ast}(f)(v)=f(L(v))$. Since $L(v)\in H\subset W$, $L^{\ast}(f)$ is determined by its values on $H$ and the kernel of $L^{\ast}$ is the annihilator of $H$. Therefore the image of $L^{\ast}$ is $H^{\ast}$. But $H^{\ast}$ and $H$ have the same dimension, so the rank of $L^{\ast}$ and the rank of $L$ are the same. However, the rank of $L^{\ast}$ is the column rank of its matrix representation, which is the row rank of the matrix of $L$.

Some remarks on analysis

In very rough terms, the Riesz Representation theorem says that if $X$ is a compact hausdorff space and C(X) are the continuous functions on X, then any(positive) continuous linear form $a: C(X)\to R$ there is a unique Borel measure $\mu$ satisfying some extra properties such that $a(f)=\int f(x)d\mu$.

Roughly speaking, continuous linear forms are the same as measures.

Continuity is essential here; the space of continuous linear forms is MUCH SMALLER than the space of all linear forms. Here the topology on $C(X)$ is the metric topology given by the sup norm.

“Functional Analysis” is the study of possible topologies on vector spaces and their relationship to spaces of linear forms.

The double dual

The “Double dual” space $V^{\ast\ast}=\Hom(\Hom(V,F),F)$ is another vector space. Notice that if $V$ is of dimension $n$ then $V^{\ast\ast}$ is also of dimension $n$. But the relationship is closer than that.

Given $v$ in $V$, define $e_{v}$ in $V^{\ast\ast}$ by $e_{v}(f)=f(v)$. This is called the “evaluation map”.

If $V$ is finite dimensional, $e$ is an isomorphism.

In general, the evaluation map is injective but far from surjective.