Exercise 3.5.1.
Show that the dual basis is indeed a basis for \(V^*\text{.}\)
Section 3.5 Project: dual basis
Let \(V\) be a vector space over \(\R\text{.}\) (That is, scalars are real numbers, rather than, say, complex.) A linear transformation \(\phi:V\to \R\) is called a linear functional.
Here are some examples of linear functionals:
- The map \(\phi:\R^3\to \R\) given by \(\phi(x,y,z) = 3x-2y+5z\text{.}\)
- The evaluation map \(ev_a:P_n(\R)\to \R\) given by \(ev_a(p) = p(a)\text{.}\) (For example, \(ev_2(3-4x+5x^2) = 2-4(2)+5(2^2) = 14\text{.}\))
- The map \(\phi:\mathcal{C}[a,b]\to \R\) given by \(\phi(f) = \int_a^b f(x)\,dx\text{,}\) where \(\mathcal{C}[a,b]\) denotes the space of all continuous functions on \([a,b]\text{.}\)
Note that for any vector spaces \(V,W\text{,}\) the set \(\mathcal{L}(V,W)\) of linear transformations from \(V\) to \(W\) is itself a vector space, if we define
\begin{equation*}
(S+T)(v) = S(v)+T(v),\quad \text{ and } \quad (kT)(v)=k(T(v))\text{.}
\end{equation*}
In particular, given a vector space \(V\text{,}\) we denote the set of all linear functionals on \(V\) by \(V^*=\mathcal{L}(V,\R)\text{,}\) and call this the dual space of \(V\text{.}\)
We make the following observations:
- If \(\dim V=n\) and \(\dim W=m\text{,}\) then \(\mathcal{L}(V,W)\) is isomorphic to the space \(M_{mn}\) of \(m\times n\) matrices, so it has dimension \(mn\text{.}\)
- Since \(\dim \R=1\text{,}\) if \(V\) is finite-dimensional, then \(V^*=\mathcal{L}(V,\R)\) has dimension \(1n=n\text{.}\)
- Since \(\dim V^*=\dim V\text{,}\) \(V\) and \(V^*\) are isomorphic.
Here is a basic example that is intended as a guide to your intuition regarding dual spaces. Take \(V = \R^3\text{.}\) Given any \(v\in V\text{,}\) define a map \(\phi_{v}:V\to \R\) by \(\phi_{v}(w) = v\dotp w\) (the usual dot product).
One way to think about this: if we write \(v\in V\) as a column vector \(\bbm v_1\\v_2\\v_3\ebm\text{,}\) then we can identify \(\phi_{v}\) with \(v^T\text{,}\) where the action is via multiplication:
\begin{equation*}
\phi_{v}(w) = \bbm v_1\amp v_2\amp v_3\ebm\bbm w_1\\w_2\\w_3\ebm = v_1w_1+v_2w_2+v_3w_3\text{.}
\end{equation*}
It turns out that this example can be generalized, but the definition of \(\phi_v\) involves the dot product, which is particular to \(\R^n\text{.}\)
There is a generalization of the dot product, known as an inner product. (See Chapter 10 of Nicholson, for example.) On any inner product space, we can associate each vector \(v\in V\) to a linear functional \(\phi_v\) using the procedure above.
Another way to work concretely with dual vectors (without the need for inner products) is to define things in terms of a basis.
Given a basis \(\{v_1,v_2,\ldots, v_n\}\) of \(V\text{,}\) we define the corresponding dual basis \(\{\phi_1,\phi_2,\ldots, \phi_n\}\) of \(V^*\) by
\begin{equation*}
\phi_i(v_j) = \begin{cases} 1, \amp \text{ if } i=j\\ 0, \amp \text{ if } i\neq j\end{cases}\text{.}
\end{equation*}
Note that each \(\phi_j\) is well-defined, since any linear transformation can be defined by giving its values on a basis.
For the standard basis on \(\R^n\text{,}\) note that the corresponding dual basis functionals are given by
\begin{equation*}
\phi_j(x_1,x_2,\ldots, x_n) = x_j\text{.}
\end{equation*}
That is, these are the coordinate functions on \(\R^n\text{.}\)
Next, let \(V\) and \(W\) be vector spaces, and let \(T:V\to W\) be a linear transformation. For any such \(T\text{,}\) we can define the dual map \(T^*:W^*\to V^*\) by \(T^*(\phi) = \phi\circ T\) for each \(\phi\in W^*\text{.}\)
Exercise 3.5.2.
Confirm that (a) \(T^*(\phi)\) does indeed define an element of \(V^*\text{;}\) that is, a linear map from \(V\) to \(\R\text{,}\) and (b) that \(T^*\) is linear.
Exercise 3.5.3.
Let \(V=P(\R)\) be the space of all polynomials, and let \(D:V\to V\) be the derivative transformation \(D(p(x))=p'(x)\text{.}\) Let \(\phi:V\to \R\) be the linear functional defined by \(\phi(p(x)) = \int_0^1 p(x)\,dx\text{.}\)
What is the linear functional \(D^*(\phi)\text{?}\)
Exercise 3.5.4.
Show that dual maps satisfy the following properties: for any \(S,T\in \mathcal{L}(V,W)\) and \(k\in \R\text{,}\)
- \(\displaystyle (S+T)^* = S^*+T^*\)
- \(\displaystyle (kS)^* = kS^*\)
- \(\displaystyle (ST)^* = T^*S^*\)
In item Item 3.5.4.3, assume \(S\in \mathcal{L}(V,W)\) and \(T\in \mathcal{L}(U,V)\text{.}\) (Reminder: the notation \(ST\) is sometimes referred to as the “product” of \(S\) and \(T\text{,}\) in analogy with matrices, but actually represents the composition \(S\circ T\text{.}\))
We have one topic remaining in relation to dual spaces: determining the kernel and image of a dual map \(T^*\) (in terms of the kernel and image of \(T\)). Let \(V\) be a vector space, and let \(U\) be a subspace of \(V\text{.}\) Any such subspace determines an important subspace of \(V^*\text{:}\) the annihilator of \(U\text{,}\) denoted by \(U^0\) and defined by
\begin{equation*}
U^0 = \{\phi\in V^* \,|\, \phi(u)=0 \text{ for all } u\in U\}\text{.}
\end{equation*}
Exercise 3.5.5.
Determine a basis (in terms of the standard dual basis for \((\R^4)^*\)) for the annihilator \(U^0\) of the subspace \(U\subseteq \R^4\) given by
\begin{equation*}
U = \{(2a+b,3b,a,a-2b)\,|\, a,b\in\R\}\text{.}
\end{equation*}
Hint.
Write \(\phi = c_1\phi_1+c_2\phi_2+c_3\phi_3+c_4\phi_4\text{,}\) and note that each \(\phi_j\) simply gives the \(j\)th component of a vector in \(\R^4\text{.}\)
Here is a fun theorem about annihilators that I won’t ask you to prove.
Theorem 3.5.6.
Let \(V\) be a finite dimensional vector space. For any subspace \(U\) of \(V\text{,}\)
\begin{equation*}
\dim U + \dim U^0 = \dim V\text{.}
\end{equation*}
Here’s an outline of the proof. For any subspace \(U\subseteq V\text{,}\) we can define the inclusion map \(i:U\to V\text{,}\) given by \(i(u)=u\text{.}\) (This is not the identity on \(V\) since it’s only defined on \(U\text{.}\) In particular, it is not onto unless \(U=V\text{,}\) although it is clearly one-to-one.)
Then \(i^*\) is a map from \(V^*\) to \(U^*\text{.}\) Moreover, note that for any \(\phi\in V^*\text{,}\) \(i^*(\phi)\in U^*\) satisfies, for any \(u\in U\text{,}\)
\begin{equation*}
i^*(\phi)(u) = \phi(i(u))=\phi(u)\text{.}
\end{equation*}
Thus, \(\phi\in \ker i^*\) if and only if \(i^*(\phi)=0\text{,}\) which is if and only if \(\phi(u)=0\) for all \(u\in U\text{,}\) which is if and only if \(\phi\in U^0\text{.}\) Therefore, \(\ker i^* = U^0\text{.}\)
By the dimension theorem, we have:
\begin{equation*}
\dim V^* = \dim \ker i^* + \dim \operatorname{im} i^*\text{.}
\end{equation*}
With a bit of work, one can show that \(\operatorname{im} i^* = U^*\text{,}\) and we get the result from the fact that \(\dim V^*=\dim V\) and \(\dim U^* = \dim U\text{.}\)
There are a number of interesting results of this flavour. For example, one can show that a map \(T\) is injective if and only if \(T^*\) is surjective, and vice-versa.
One final, optional task: return to the example of \(\R^n\text{,}\) viewed as column vectors, and consider a matrix transformation \(T_A:\R^n\to \R^m\) given by \(T_A(\vec{x}) = A\vec{x}\) as usual. Viewing \((\R^n)^*\) as row vectors, convince yourself that \((T_A)^* = T_{A^T}\text{;}\) that is, that what we’ve really been talking about all along is just the transpose of a matrix!
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