\documentclass{amsart}

\newtheorem{prob}{Problem}
\newtheorem{hard}{Hard Form}
\renewcommand{\L}{\mathcal L}
\newcommand{\F}{\mathbb F}
\newcommand{\R}{\mathbb R}

\title{Problem Set \#4}
\author{Due: November 15th}

\begin{document}
\maketitle

This set will be run a little differently. Many of the problems will have a regular version and a ``hard  form''. You can choose which ones you want to do (though the harder one subsumes the easier). Additionally: hard ones are worth 15 total points (so you can earn up to 5 extra credit points).

\subsection*{The Interplay between $\L(U,V)$ and $V\otimes U^\ast$}

\begin{prob}
Show for $U=V=\R^2$ that $V\otimes U^\ast\to\L(U,V)$ described in class is an isomorphism. Show also that when we identify $U^\ast$ with row vectors and $V$ with column vectors that the isomorphism is ``matrix multiplication'':
\[\vec{v}\otimes\vec{u}^t\mapsto \vec{v}\cdot\vec{u}^t.
\]
\end{prob}

\begin{hard}
Show that for general finite dimensional $U$ and $V$,
\[
V\otimes U^\ast\to\L(U,V)
\]
is given by $\bar{v}\otimes f\mapsto f(-)\cdot\vec{v}$ and hence is also ``matrix multiplication''.
\end{hard}

\begin{prob}
Given a basis $\{\vec{u}_1,\dots,\vec{u}_n\}$ of $U$ and $\{\vec{v}_1,\dots,\vec{v}_m\}$ of $V$, we get a basis $\{\vec{v}_i\delta_{\vec{u}_j}\}$ of $\L(U,V)$ by specifying
\[
\vec{v}_i\delta_{\vec{u}_j}(\vec{u}_k)=\begin{cases}
\vec{v}_i & j=k \\
0 & \text{otherwise.}
\end{cases}
\]

Show that under the map 
\[
\L(U,V)\otimes\L(U',V')\to\L(U\otimes U',V\otimes V')
\]
described in class, we have
\[
(\vec{v}_i\delta_{\vec{u}_j})\otimes(\vec{v}'_k\delta_{\vec{u}_\ell'})\mapsto (\vec{v}_i\otimes\vec{v}_k')\delta_{\vec{u}_j\otimes\vec{u}_\ell'}.
\]
\end{prob}

\subsection*{Playing with $\L\big(U,\L(V,W)\big)\cong\L(U\otimes V,W)$}
For the next few parts, Let $f\in\L\big(U,\L(V,W)\big)$ (so this is a linear transformation on $U$ with values in the linear transformations from $V$ to $W$.

Define $\hat{f}\colon U\times V\to W$ by
\[
\hat{f}(\vec{u},\vec{v})=f(\vec{u})(\vec{v}).
\]

\begin{prob}
Show that $\hat{f}$ is bilinear.
\end{prob}

Now let $L\colon U\otimes V\to W$ be linear. Let $f_L\colon U\to\L(V,W)$ be defined by
\[
f_L(\vec{u})(\vec{v})=L(\vec{u}\otimes\vec{v}).
\]

\begin{prob}
Show that if $L_f$ is the map $U\otimes V\to W$ associated to the bilinear map $\hat{f}$, then the linear transformation $f_{L_f}$ is $f$.
\end{prob}

\begin{prob}
Show that $L_{f_L}$ is $L$. Conclude that the two maps are inverses and thus the two spaces are [naturally] isomorphic.
\end{prob}

\begin{hard}
Show that $\L(U,\L(V,-))$ and $\L(U\otimes V,-)$ are naturally isomorphic by showing that they represent the same collection of objects (naturally in the $-$ variable). In other words, show that (a) they give the same collection of objects for each value of $W$, and (b) given a map $W\to W'$, the identifications are compatible with the map.
\end{hard}

\subsection*{Determinants}
This problem completes the proof of the ``coordinate free'' form of the determinant discussed in class.

\begin{prob}
Verify for $n=2$ or $3$, and $V=\mathbb R^n$, and for $L\colon V\to V$ that
$\Lambda^n(L)$ is multiplication by $|[L]_{\mathcal B}|$ for some (and therefore any) basis $\mathcal B$ of $V$.
\end{prob}

\begin{hard}
Show for a general $n$-dimensional vector space $V$ over $\F$ and $L\colon V\to V$ that
$\Lambda^n(L)$ is multiplication by $|[L]_{\mathcal B}|$ for some (and again, for any) basis $\mathcal B$.
\end{hard}

\subsection*{Invariant Form of the Trace}
The vector space $\F$ has a distinguished basis vector $1$. We've seen that this gives an isomorphism
\[
\L(\F,W)\cong W.
\]
We can apply this to $W=\L(V,V)=V\otimes V^\ast$. There is a distinguished element in this vector space: the identity map $V\to V$. Thus we get a map
\[
\F\to V\otimes V^\ast
\]
that sends $1$ to the identity. We call this map ``coevaluation''.

\begin{prob}
Let $V=\mathbb R^3$. Describe the image of $1$ under this map as an element of $V\otimes V^\ast$ and also as an element of $\L(V,V)=M_{3}(\R)$.
\end{prob}

\begin{hard}
Repeat this for $V=\F^n$.
\end{hard}

\begin{prob}
Now consider the map $V\times V^\ast\to \F$ given by $(\vec{v},f)\mapsto f(\vec{v})$. Show this is bilinear.

Let $ev\colon V\otimes V^\ast\to\F$ be the corresponding linear map, and compute for $V=\R^3$ what this map does to a general element (viewed as a $3\times 3$-matrix).
\end{prob}

\begin{hard}
Repeat this for an arbitrary finite dimensional $V$ and $\F$. Conclude that the composite
\[
\F\to V\otimes V^\ast\to \F
\]
is the dimension and is independent of choice of basis.
\end{hard}

Since $V\otimes V^\ast$ is the same thing as $\L(V,V)$, given any $T\in \L(V,V)$ we get a map $T_\ast\colon V\otimes V^\ast\to V\otimes V^\ast$ given by composing with $T$.

\begin{prob}
In the $V=\R^2$ case, after identifying $V\otimes V^\ast$ with $M_2(\R)$, show that this amounts to matrix multiplication.
\end{prob}

Now we can define the trace:
$Tr(T)$ is the composite \[
\F\xrightarrow{coev} V\otimes V^\ast\xrightarrow{T_\ast} V\otimes V^\ast\xrightarrow{ev}\F.
\]
Since this is a linear map $\F\to\F$ it amounts to multiplication by a number. We will also denote this number $Tr(L)$.

\begin{hard}
After choosing a basis, show this coincides with the usual notion of trace.
\end{hard}




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