\documentclass[reqno,12pt]{amsart}
\usepackage{amscd,amssymb}
\usepackage[all]{xy}
\DeclareMathOperator{\image}{Im}
\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\Mat}{Mat}
\newcommand{\compose}{\circ}
\newcommand{\set}[1]{\left\{#1\right\}}
\newcommand{\bR}{{\mathbb R}}
\newcommand{\ra}{\rightarrow}
\newcommand{\from}{\colon}
\title{Math 23a Problem Set 4}
\dedicatory{due October 22th}

\begin{document}
\maketitle
\begin{enumerate}
\item Let $L,M$ be finite-dimensional vector spaces. Show that
$\dim(L\oplus M)= \dim(L)+\dim(M)$.

\item Let $V$ be a finite-dimensional vector space, $L,M\subset V$
subspaces of $V$. Consider the map $F\colon L\oplus M\rightarrow V$ given by
$F(l,m) =l+m,l\in L,m\in M$.

\begin{enumerate}
\item Check that the map $F$ is linear.

\item Show that $F$ is an isomorphism of vector spaces [that is $F$ is
one-to-one and onto] if and only if we can
decompose any vector $v\in V$ as a sum $v=l+m,l\in L,m\in M$ and such a
decomposition is unique.

\item Show that $F$ is an isomorphism of vector spaces if and only if
$\dim(V)=\dim(L)+\dim(M)$ and $L\cap M=\set{0}$.

\end{enumerate}

\item Let $L,M$ be finite-dimensional vector spaces and $\Hom(L,M)$ be the
set of linear maps $F$ from $L$ to $M$.

\begin{enumerate}
\item Define the structure of a vector space on the set $V=\Hom(L,M)$,
that is, define two operations $+\from V \times V \to V$, $\cdot\from
\bR \times V \to V$ and show that with these as vector addition and
scalar multiplication, $\Hom(L,M)$ is a vector space.

\item Let $\set{w_1,\dots,w_l}\in L, \set{v_1,\dots,v_m}\in M$ be bases of $L$ and
$M$, and  $\set{w^1,\dots,w^l}, \set{v^1,\dots,v^m}$ be the dual bases
of $L^*$ and $M^*$. To any linear map  $F$ from $L$ to $M$ we
associate an $l\times m$ matrix $A(F)$ such that  $A(F)_{i,j}=v^j
(F(w_i)), 1\leq i\leq l, 1\leq j \leq m$. Show that the map $\Hom(L,M)\to
\Mat(l,m)$ is an isomorphism of vector spaces.

\item Prove that $\dim(\Hom(L,M))=\dim(L)\dim(M)$.
\end{enumerate}

\item Let $L,M$ be vector spaces. To any linear map $F\from L\to M$
and a linear functional $\lambda \in M^*$ 
we associate a function $F^* (\lambda)=F^*_\lambda\from L \to \bR$
by
$$F^* (\lambda)(v)=\lambda (F(v)), \forall v\in L$$ 

\newpage

This can be summarized in the following commutative diagram:

\begin{figure}[h]
\ \xymatrix{
 L\ar[r]^{F}\ar@{-->}[dr]_{F^*_\lambda} & M\ar[d]^{\lambda}\\
             & \bR
}
\end{figure}

Here's how to read a commutative diagram: it's just a summary of
all the maps you have sitting around, with solid arrows for the maps
you start with and dashed arrows for the maps you get from the
existing maps.
It is called ``commutative'' because it doesn't matter which path you
take from one point to another; it's the same map. Thus,
$F^*_\lambda=\lambda\compose F$, since they both start at $L$ and end in $\bR$.

Nils finds these diagrams to be really pretty and much easier to
understand than formulas.

\begin{enumerate}
\item Show that the function $F^*_\lambda\from L \to \bR$ is linear.

\item Thus, $F^*$ defines a map $F^*\from M^* \to L^*$. Show that this
map is linear.

\item Let $\set{w_1,\dots,w_l}\in L,\set{v_1,\dots,v_m}\in M$ be bases of $L$ and
$M$,  and $\set{w^1,\dots,w^l},\set{v^1,\dots,v^m}$ be the dual bases
of $L^*$ and $M^*$. Show that for any linear map $F\from L\to M$ the
matrix $A(F^*)$ is equal to the transpose of the matrix $A(F)$.
\end{enumerate}

\item Let $P_2$ be the linear space of polynomials $p(x)$ of degree $\leq
2$. Consider the linear functionals $\lambda_i, 1\leq i\leq 3$ on $P_2$
given by $\lambda_i (p(x))=p(i)$.

\begin{enumerate}
\item Show that $\lambda_i, 1\leq i\leq 3$ is a basis of the space $P_2^*$.

\item Find the dual basis $\lambda_i^*, 1\leq i\leq 3$ of the space
$(P_2^*)^*=P_2$.
\end{enumerate}

\end{enumerate}
\end{document}
