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\title{MAT 247 \\ Assignment 9 \\ Due Tuesday April 5}


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\maketitle
\begin{enumerate}
\item Let $ V = \mathbb{Q} $ (i.e. the 1-dimensional vector space over the field of rational numbers).  Let $ H $ be a non-zero bilinear form on $ V $.
\begin{enumerate}
\item Show that we can find a basis $ v $ for $ V $ such that $ [H]_v = \bigl[n \bigr] $, where $ n$ is a square-free integer (a non-zero integer which is not divisible by the square of an integer bigger than 1).
\item Show that such an $ n $ is unique (in the same sense that the signature of symmetric bilinear form on a real vector space is unique). 
\end{enumerate}

\item
Let $ V $ be a vector space and let $ \Omega$ be a symplectic form on $ V$.  Let $ L, M $ be two Lagrangian subspaces of $ V$ such that $$ L \cap M = 0.$$  
\begin{enumerate}
\item Show that $ V = L \oplus M $.
\item Define a linear map $ T: M \rightarrow L^* $ by setting $ T(v)$ to be the linear functional defined by $ (T(v))(w) = \Omega(v,w) $.  Prove that $ T $ is an isomorphism.
\item Use the above results to find a symplectic basis $$ q_1, \dots, q_n, p_1, \dots, p_n $$ for $ V $ such that $ q_1, \dots, q_n \in L $ and $ p_1, \dots, p_n \in M $.
\end{enumerate}


\item
Let $ V, \langle, \rangle $ be a complex inner product space.  We can regard $ V $ as a real vector space by just considering scalar multiplication by elements of $ \mathbb{R} $.  Define a real bilinear form $ \Omega $ on $ V $ by $$ \Omega(v,w) = Re (\langle v, iw \rangle), $$ where $ Re $ denotes the real part of a complex number.
\begin{enumerate}
\item Show that $ \Omega $ is a real symplectic form on $ V $.
\item Let $ v_1, \dots, v_n $ be a orthonormal basis for $ V $ (regarded as a complex inner product space).  Show that $$ v_1, \dots, v_n, iv_1, \dots, iv_n $$ is a symplectic basis for $ V $ (regarded as a real vector space with symplectic form $ \Omega$).
\end{enumerate}

\item
Recall that if $ T : V \rightarrow W $ is a linear map, then there is a linear map $T^* : W^* \rightarrow V^* $ which is defined by $$ (T^*(\beta))(v) = \beta(Tv),$$ for $ \beta \in W^* $ and $ v \in V $.  Applying this reasoning twice, we see that there is a linear map $ (T^*)^* : (V^*)^* \rightarrow (W^*)^* $.

In class, for each vector space $ V $, we defined the isomorphism $ \psi_V : V \rightarrow (V^*)^* $ by setting $ \psi_V(v) $ to be the linear functional on $ V^* $ defined by $$ (\psi_V(v))(\alpha) = \alpha(v).$$
for each $ \alpha \in V^* $.

Prove that for any two vector spaces $ V, W $ and any linear map $ T : V \rightarrow W $, we have $ (T^*)^* \psi_V = \psi_W T $. 

(This is the sense in which $ \psi $ is ``natural''.)
\end{enumerate}

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