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\begin{document}
\begin{center}
{\bf Homework 1}\\
(due Wednesday, Jan 25)
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\vspace*{.3in}


Let $V$ be a finite dimensional (say $\dim V = n$) vector space over a field $\F$.  On the last day of fall term I talked about the idea of the dual space of $V$.  I made a lot of assertions during that lecture, but since time was short, I didn't prove most of them.  Your job on this homework will be to supply the missing proofs.  As you'll see fairly quickly, this is a very formal kind of thing that mostly amounts to correctly applying definitions.  Still it's tricky to get the details right and trickier still to say it all nicely.  If nothing else, it serves as a good review of some of the basic concepts (bases, linear transformations, etc) from last semester.  In the interest of full disclosure, I'll point out that you can get all your proofs from Treil's chapter 8.  But really, how much fun would that be? 

Recall that a \emph{linear functional} (also called a \emph{covector} in many contexts) on $V$ is a linear transformation $\vv':V\to \F$.  I encourage you to think of $\vv'$ as a way of `measuring' (i.e. assigning scalar values to) vectors $\vv\in V$.  Since linear functionals on $V$ may be added to each other and multiplied by scalars, the collection of all linear functionals on $V$ forms a vector space $V'$ called \emph{the dual space} of $V$.

A quick (and hopefully helpful) final remark: what takes getting used to here is that elements $\vv'\in V$ are on the one hand vectors, that can be added to each other and multiplied by scalars; but they are also functions that take vectors $\vv\in V$ and turn them into scalars $\vv'(\vv) \in \F$.  In particular, whenever you have an equality $\vv' = \tilde \vv'$ involving elements of of $V'$, it is equality in the sense of functions; i.e. it means that $\vv'(\vv) = \tilde\vv'(\vv)$ for every $\vv\in V$.  

\begin{prob}
Let $\cB = \{\vv_1,\dots,\vv_n\}$ be a basis for $V$.  Explain why for each $1\leq j\leq n$ there is a unique linear functional $\vv'_j\in V'$ such that $\vv'_j(\vv_k) = 1$ if $j=k$ and $\vv'_j(\vv_k) = 0$ otherwise.  Show that the set $\cB' = \{\vv'_1,\dots,\vv'_n\}\subset V'$ is a basis for $V'$.  This is called the \emph{dual basis} for $V$ relative to $\cB$.
\end{prob}

\begin{prob} 
\begin{enumerate}
 \item Explain why there is a unique linear transformation $\phi:V\to V'$ such that $\phi(\vv_j) = \vv'_j$ for $1\leq j\leq n$ and that $T$ is in fact an isomorphism. 
 \item A crucial point is that the isomorphism $\phi$ depends very much on which basis you use.  Show this by taking $V = \R^2$ and comparing the isomorphism $\phi_\cE$ that you get from the standard basis with the isomorphism $\phi_{\cB}$ that you get from the basis $\cB = \{(1,0),(1,1)\}$.
\end{enumerate}
\end{prob}
 
\begin{prob} 
Now let $U$ be another vector space over $\F$ with basis $\cC = \{\vu_1,\dots,\vu_m\}$.  Suppose that $T:U\to V$ is a linear transformation.  Given $\vv'\in V'$, we define the \emph{pullback} $T'\vv':U\to\F$ of $\vv'$ by $T'\vv'(\vu) = \vv'\circ T(\vu).$  
\begin{enumerate}
\item Explain why $T'\vv'$ is linear (i.e. $T'\vv'\in U'$) and $T':V'\to U'$ is a linear transformation (the \emph{dual transformation} of course).
\item Let $\cB$ and $\cC$ be bases for $U$ and $V$ and $A = [T]_{\cB\cC}$ be the corresponding matrix for $T$.  Show that the matrix for $T'$ relative to the dual bases $\cB'$ and $\cC'$ is given by
$$
[T']_{\cC'\cB'} = A^T.
$$
\end{enumerate}
\end{prob}

\begin{prob} Finally, consider the case when $\F = \R$ and $U$ and $V$ are real inner product spaces.  Observe that each vector $\vv\in V$ gives rise to a linear functional $*\vv\in V'$ given by the formula $*\vv(\vw) = \pair{\vw}{\vv}$.  
\begin{enumerate}
\item Prove that the function $*:V \to V'$ is an isomorphism. (Hint: you know that $\dim V = \dim V'$, so once you show that $*$ is linear, you need only show further that $*$ is injective rather than bijective. So one advantage of having an inner product space over just a plain vector space is that the inner product gives us a `canonical' (i.e. basis independent) isomorphism $*:V\to V'$.  In fact, one way you can prove $*$ is an isomorphism is by showing that $* = \phi_{\cB}$ for any orthonormal basis $\cB\subset V$.
\item Suppose that $U$ is also an inner product space over $\R$ and $T:U\to V$ is linear.  Write $*_U$ and $*_V$ to distinguish the canonical isomorphisms between $U$ and $V$ and their dual spaces.  Show that the adjoint transformation $T^*:V\to U$ is given by 
$$
T^* = *_U^{-1}\circ T' \circ *_V.
$$
\item How does this change if $U$ and $V$ are \emph{complex} inner product spaces?
\end{enumerate}
\end{prob}






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