\documentclass{amsart}
\usepackage{amsmath,amssymb}

\DeclareMathOperator{\im}{Im}

\newcommand{\R}{{\mathbb R}}

\title{Math 23A SOLUTION SET 4}
\author{Matthew G. Lee}

\begin{document}
\maketitle
\begin{enumerate}

\item
Let $\{w_1,w_2, \dots ,w_l \} \in L$ be a basis for $L$, and
$\{v_1,v_2, \dots ,v_m \}$ be a basis for $M$.

Claim:
$\{(w_1,0),(w_2,0),\dots,(w_l,0),(0,v_1),(0,v_2),\dots,(0,v_m)\}$
is a basis for $L \oplus M$.

Proof: We will first prove that
$\{(w_1,0),(w_2,0),\dots,(w_l,0),(0,v_1),(0,v_2),\dots,(0,v_m)\}$
spans $L \oplus M$.  Any element $w$ of $L$ can be written as
${c_1}{w_1}+\dots+{c_l}{w_l}$ for some $c_1,\dots,c_l \in \R$. Any
element $v$ of $M$ can be written as ${d_1}{v_1}+\dots+{d_m}{v_m}$
for some $d_1,\dots,d_m \in \R$.  So any element $(w,v)$ of $L
\oplus M$ can be written as
$$({c_1}{w_1}+\dots+{c_l}{w_l},{d_1}{v_1}+\dots+{d_m}{v_m})$$
$$={c_1}{(w_1,0)}+\dots+{c_l}{(w_l,0)}+
{d_1}{(0,v_1)}+\dots+{d_m}{(0,v_m)}$$ for
$c_1,\dots,c_l,d_1,\dots,d_m \in \R$. So
$\{(w_1,0),(w_2,0),\dots,(w_l,0),(0,v_1),(0,v_2),\dots,(0,v_m)\}$
spans $L \oplus M$.

Next we will prove that
$\{(w_1,0),(w_2,0),\dots,(w_l,0),(0,v_1),(0,v_2),\dots,(0,v_m)\}$
is linearly independent.  Suppose
$${c_1}{(w_1,0)}+\dots+{c_l}{(w_l,0)}+
{d_1}{(0,v_1)}+\dots+{d_m}{(0,v_m)}=(0,0)$$ for some
$c_1,\dots,c_l,d_1,\dots,d_m \in \R$.  Then
$${c_1}{w_1}+\dots+{c_l}{w_l}=0.$$

Since the $w_i$ are linearly independent, $$c_1=\dots=c_l=0$$


Also, $${d_1}{v_1}+\dots+{d_m}{v_m}=0$$

Since the $v_j$ are linearly independent, $$d_1=\dots=d_m=0$$

Since $$c_1=\dots=c_l=d_1=\dots=d_m=0,$$
$\{(w_1,0),(w_2,0),\dots,(w_l,0),(0,v_1),(0,v_2),\dots,(0,v_m)\}$
is linearly independent.

Since it spans $L \oplus M$ and is linearly independent,
$$\{(w_1,0),(w_2,0),\dots,(w_l,0),(0,v_1),(0,v_2),\dots,(0,v_m)\}$$
is a basis for $L \oplus M$.
\item
\begin{enumerate}
\item
Let $l_1, l_2 \in L, m_1, m_2 \in M, a \in \R$.  Then

$$F((l_1 ,m_1)+(l_2, m_2)) = F(l_1 + l_2,m_1 + m_2)$$
$$= l_1 + l_2 + m_1 + m_2$$
$$ = F(l_1,m_1)+F(l_2,m_2).$$

Also, $$F(a(l_1, m_1))=F({a}l_1,{a}m_1)$$
$$={a}l_1+{a}m_1$$
$$=a(l_1+m_1)$$
$$=a(F(l_1,m_1)).$$
\item
We need to show that $F$ is an isomorphism if and only if we can
decompose any vector $v \in V$ into a unique sum $v = l + m$ for
some $l \in L, m \in M$.

Suppose $F$ is an isomorphism.  Let $v$ be an arbitrary element of
$V$.  Since $F$ is surjective, there exists $(l,m) \in L \oplus M$
such that $F(l,m)=v$.  So $l+m=v$.

Now we will prove that this decomposition is unique.  Suppose that
for some $l_1,l_2 \in L$ and $m_1, m_2 \in M, l_1+m_1=v=l_2+m_2$.
Then $F(l_1,m_1)=v=F(l_2,m_2)$. Because $F$ is injective, we must
have $(l_1,m_1)=(l_2,m_2)$, and so $l_1=l_2$ and $m_1=m_2$.

Conversely, suppose we can decompose any vector $v \in V$ into a
unique sum $v = l + m$ for some $l \in L, m \in M$.  Then there
exists a unique ordered pair $(l,m), l \in L, m \in M$, such that
$v = l + m$. Since $l+m=F(l,m)$ for all $l \in L$ and $m \in M$,
there exists a unique ordered pair $(l,m) \in L \oplus M$ such
that $F(l,m)=v$.

\item
Suppose that F is an isomorphism of vector spaces.  Then ker $F =
\{0\}$ and Img $F = V$.  By the Rank-Nullity Theorem, $$
\mbox{dim} V = 0 + \mbox{dim}(L \oplus M).$$  So dim($V$)= dim($L
\oplus M$) = dim($L$) + dim ($M$).

Suppose, for the sake of contradiction, that $L \cap M \neq
\{0\}$. Then there exists $x$ such that $x \in L$ and $x \in M$.
Since $M$ is a vector space, $-x \in M$.  So $F(x,-x)=0=F(0,0)$.
So F is not one-to-one, so F is not an isomorphism.  From this
contradiction we can conclude that $L \cap M = \{0\}$.

Conversely, suppose dim($V$) = dim($L$) + dim ($M$) and $L \cap M
= \{0\}$.  Suppose $F(l_1,m_1)=F(l_2,m_2)$ for $l_1,l_2 \in L$ and
$m_1, m_2 \in M$.  Then $l_1+m_1=l_2+m_2$, so $l_1-l_2 = m_2-m_1
\in L \cap M$.  So $l_1-l_2=0=m_1-m_2$, and therefore $l_1=l_2$
and $m_1=m_2$, so $(l_1,m_1)=(l_2,m_2)$ and $F$ is one-to-one.

Since $F$ is one-to-one, ker $F = \{0\}$ and dim(ker $F$) = 0. So,
by the Rank-Nullity Theorem and using the result of Problem 1,
dim(Img $F$) = dim($L \oplus M$) = dim($L$) + dim ($M$) =
dim($V$).  So any basis for Img $F$ is a set of dim($V$) linearly
independent vectors in $V$ and thus a basis for $V$.  So Img $F$ =
$V$ and $F$ is onto.  Since we know that $F$ is linear from part
(a), $F$ is an isomorphism.

\end {enumerate}

\item
\begin {enumerate}
\item
Let $F, G \in V=$Hom$(L,M)$ and $c \in \R$. Define $+\colon V
\times V \rightarrow V$ by $(F+G)(w)=F(w)+G(w)$ for all $w \in W$.
Define $\cdot\colon \R \times V \rightarrow V$ by
$(cR)(w)=c(R(w))$ for all $w \in W$.

We will now verify that, with these operations as vector addition
and scalar multiplication, Hom$(L,M)$ is a vector space.
Hom$(L,M)$ is closed under vector addition and scalar
multiplication by our definition of the domain and range of these
functions.  We then need to verify that Hom$(L,M)$ satisfies the
eight axioms of a vector space given in Hubbard \& Hubbard, p.
189. I will verify the first one, and you should be able to verify
the rest similarly.

(1) Additive identity.  Define $Z \in$ Hom($L,M$) to be the
function such that $Z(w)=0$ for all $w \in W$.  Then, for any $F$
in Hom($L,M$), $(Z+F)(w)=Z(w)+F(w)=0+F(w)=F(w)$, so $Z+F=F$.

\item
To show that the map $A\colon $ Hom($L,M$) $\rightarrow$
Mat($l,m$) is an isomorphism, we need to check that it is linear,
that it is onto, and that it is one-to-one.  Please note that $A$
is not what we normally think of as the matrix associated with a
linear transformation.  Each element $A(F)_{i,j}$ is the
coefficient $c_j$ when we express $F(w_i)$ as the linear
combination $c_1v_1+c_2v_2+\dots+c_mv_m$.  $A(F)$ is actually the
transpose of the $m$-by-$l$ matrix we normally think of as
associated with the linear transformation $F$.

Proof of linearity: $$A(F+G)_{i,j}=v^j ((F+G)(w_i))$$ $$=v^j
(F(w_i)+G(w_i)$$ $$=v^j (F(w_i)) + v^j (G(w_i))$$ $$=A(F)_{ij}
+A(G)_{i,j} .$$ So $A(F+G)=A(F)+A(G)$.

$$A(cF)_{i,j}=v^j ((cF) w_i)$$ $$=v^j (c(F(w_i)))$$ $$=c v^j
(F(w_i))$$ $$=c(A(F)_{i,j}).$$ So $A(cF)=cA(F)$.

Proof that A is onto: Let $B \in$ Mat($l,m$).  To define a linear
map, it is sufficient to define its values on a basis.  So we
define $F\colon L \rightarrow M$ by $F(w_i)= \sum_{j=1}^{m}
b_{i,j} v_j$. Then $A(F)_{ij}= v^j (F(w_i)) = v^j (\sum_{j=1}^{m}
b_{i,j} v_j)=b_{i,j}$.  So $A(F)=B$.  So $F$ is onto.

Proof that A is one-to-one: It suffices to show that ker $A$ =
\{0\}.  Suppose that $F \in $ker$ A$.  Then $A(F)=0$.  So, for all
$1 \leq i \leq l$ and $1 \leq j \leq m$, $v^j (F(w_i))=0$.  Since
${v_1,\dots,v_m}$ is a basis for $M$, for all $i$ we can write
$F(w_i)=c_{i,1} v_1 + \dots c_{i,m} v_m$.  Then, for all $i$ and
$j$, $0 = v^j (F(w_i)) = c_{i,j}$.  So, for all $i$, $F(w_i)=0$.
Since the value of $F$ at each basis vector is 0, $F(w)=0$ for all
$w \in W$ and $F = 0$.  So ker $A$ = \{0\}, so $A$ is one-to-one.

So A is an isomorphism.

\item
Using the Rank-Nullity Theorem on $A$,
$$\mbox{dim(Hom(}L,M\mbox{))=dim(ker }A\mbox{)+dim(Img }A).$$
Since A is an isomorphism,
$$\mbox{dim(Hom(}L,M\mbox{))=0+dim(Mat(}l,m\mbox{))=}lm\mbox{=dim(}L\mbox{)dim(}M).$$


\end {enumerate}

\item
\begin {enumerate}
\item
For this part, we need to show that for any linear map $F\colon L
\rightarrow M$ and linear functional $\lambda \in M^*,$ and for
any vectors $v_1,v_2 \in L$ and scalar $c \in \R$, $$F^*(\lambda
)(v_1+v_2) = F^*(\lambda)(v_1) + F^*(\lambda) (v_2)$$ and
$$F^*(\lambda ) (cv_1) = c F^*(\lambda) (v_1).$$

We have
$$F^*(\lambda)(v_1+v_2)=\lambda(F(v_1+v_2)) $$
$$=\lambda(F(v_1)+F(v_2)) $$
$$=\lambda(F(v_1))+\lambda(F(v_2)) $$
$$=F^*(\lambda)(v_1)+F^*(\lambda)(v_2).$$

Also, $$F^*(\lambda)(cv_1)=\lambda(F(cv_1))$$
$$=\lambda(cF(v_1))$$ $$=c \lambda (F(v_1)) $$
$$=cF^*(\lambda)(v_1).$$


\item
For this part, we need to show that for any linear map $F\colon L
\rightarrow M$ and linear functionals $\lambda_1$ and $\lambda_2
\in M^*$, and for any scalar $c$ $\in \R$,
$$F^*(\lambda_1+\lambda_2)=F^*(\lambda_1)+F^*(\lambda_2)$$ and
$$F^*(c \lambda_1) = cF^*(\lambda_1)$$.

Note that we are comparing functions in $L^*$.  To show that two
functions in $L^*$ are equal, it suffices to show that at all $v
\in L$, the two functions evaluate to the same value.

Let $v \in L$.  Then
$$F^*(\lambda_1+\lambda_2)(v)=(\lambda_1+\lambda_2)(F(v)),$$
which by definition of addition in $M^*$ is equal to
$$\lambda_1 (F(v))+\lambda_2 (F(v))$$
$$=F^*(\lambda_1)(v)+F^*(\lambda_2)(v).$$

So $F^*(\lambda_1+\lambda_2)=F^*(\lambda_1)+F^*(\lambda_2)$.  Also,

$$F^*(c \lambda_1)(v)=(c \lambda_1)(F(v))$$
$$=c \lambda_1 (F(v))$$
$$=c F^*(\lambda_1)(v).$$

So $F^*(c \lambda_1) = cF^*(\lambda_1)$.  So the map $F^*\colon
M^* \rightarrow L^*$ is linear.

\item
The solution to this problem is very short, but many of you who
attended my office hours were confused with by the notation.  One
must be careful to apply the definition of $A$ very carefully in
writing out $A(F^*)$. $A(F)_{i,j}$ is defined to be $v^j(F(w_i))$,
where $v^j$ is the $j$-th vector in the dual basis of the range,
and $w_i$ is the $i$-th vector in the basis of the domain.  Since
$F^*\colon M^* \rightarrow L^*$, the $j$-th vector in the dual
basis of the range of $F^*$ is actually the basis vector dual to
$w^j$. This turns out to be the element of $L^{**}$ representing
evaluation at $w_j$. And the $i$-th vector in the basis of the
domain is the element $v^i$ of $M^*$.  So we have $A(F^*)_{i,j}=$
the evaluation at $w_j$ of $F^*(v^i) = F^*(v^i)(w_j) =
v^i(F(w_j))=A(F)_{j,i}$. So $A(F^*)=A^T(F)$.

\end {enumerate}

\item
\begin {enumerate}
\item
To show that ${\lambda_1, \lambda_2, \lambda_3}$ form a basis for
$P_2^*$, it suffices to show that these three vectors are linearly
independent.  This is because dim($P_2^*$)=dim($P_2$)=3, and any
three vectors in a space of dimension three are linearly
independent.

Suppose $$a_1 \lambda_1 + a_2 \lambda_2 + a_3 \lambda_3 = 0$$ for
$a_1=a_2=a_3 \in \R$.  Then, for all $p(x) = c_0 + c_1 x + c_2
x^2$, we must have $$a_1 \lambda_1(p(x)) + a_2 \lambda_2(p(x)) +
a_3 \lambda_3(p(x)) = 0.$$ By the definition of the $\lambda_i$,
$$a_1 p(1) + a_2 p(2) + a_3 p(3) = 0.$$ So $$a_1 (c_0+c_1+c_2) +
a_2 (c_0+2c_1+4c_2) + a_3 (c_0 + 3c_1 + 9c_2) = 0.$$ So
$$(a_1+a_2+a_3) c_0 + (a_1 + 2a_2 + 3 a_3) c_1 + (a_1 + 4a_2 +
9a_3) c_3 = 0.$$

Since this equation has to hold for all values of the $c_i$, we
must have $$a_1+a_2+a_3=0,$$ $$a_1+2a_2+3a_3=0,$$
$$a_1+4a_2+9a_3=0.$$

The only solutions to this system of equations are
$a_1=a_2=a_3=0$.  So ${\lambda_1, \lambda_2, \lambda_3}$ are
linearly independent and hence a basis for $P_2^*$.

\item
We are now asked to find the dual basis $\lambda_i^*, 1\leq i\leq
3$, of the space $(P_2^*)^*=P_2$.

First, a review of fundamental concepts:

An element $p$ of $P_2$ is a polynomial in $x$ of degree $\leq 2$.

An element $\lambda$ of $P_2^*$ is a linear functional on $P_2$.
We showed above that all elements of $P_2^*$ can be written as a
linear combination of "evaluation at 1," "evaluation at 2," and
"evaluation at 3."

An element $\lambda^*$ of $P_2^{**}$ is a linear functional on
$P_2^*$. Any element of $P_2^{**}$ can be expressed as evaluation
at a polynomial member of $P_2$, and in this way we can identify
$P_2^{**}$ with $P_2$.

Now the search is on:

The dual basis $\lambda_i^*$ of $\lambda_i$ must satisfy the
condition $\lambda_i^*(\lambda_j)=1$ for $i=j$ and
$\lambda_i^*(\lambda_j)=0$ for $i \neq j$.

We will now look for polynomials $p_1$, $p_2$, and $p_3$ such that
evaluation at $p_i$ satisfies the condition for $\lambda_i^*$.  So
we need $\lambda_j(p_i)=1$ for $i=j$ and $\lambda_j(p_i)=0$ for $i
\neq j$.

Notice that these are exactly the same equations that we would
have if we started out trying to solve for the basis ${p_1, p_2,
p_3}$ of $P_2$ that the $\lambda_i$ are the dual of.  In general,
the dual of the dual of a basis consists of the linear functionals
representing evaluation at the original basis vectors.

I will solve for $p_1$, and the other solutions are similar.  We
need a polynomial of degree $\leq 2$ such that $p_1(1)=1,
p_1(2)=0, p_1(3)=0$.  One could write out these equations in terms
of the coefficients of $p_1$ and solve three equations in three
unknowns.  Or simply note that since 2 and 3 are roots of $p_1$,
$x-2$ and $x-3$ must be factors of $p_1$.  And to get $p_1(1)=1$,
we let $p_1=\frac{(x-2)(x-3)}{(1-2)(1-3)} = \frac{1}{2}x^2 -
\frac{5}{2}x + 3$.  Similarly, $p_2=-x^2+4x-3$ and $p_3=
\frac{1}{2}x^2 - \frac{3}{2}x + 1$. And each $\lambda_i^*$ is the
function representing evaluation at the respective $p_i$.
\end{enumerate}
\end{enumerate}




\end{document}
