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\noindent
\parbox{2.25in}{\bf Math 4010-101}
\hfill {\Large \bf Homework \#4} \hfill
\parbox{2.25in}{\bf \hfill Due: Wed., Feb. $16^\mathrm{th}$, 2022}

%\vspace{0.2in}
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%\noindent {\bf\large Name:} \quad \underline{\large\qquad This is my Name \qquad}

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\noindent
Please remember when submitting any work via email or in person to\dots\\
\force \hfill \fbox{\fbox{\bf\large\color{red}PUT YOUR NAME ON YOUR WORK!}} \hfill \force


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\begin{description}
\item[\#1. A Based Problem] Let $\std = \{f_1,f_2,f_3\}=\{1,x,x^2\}$ be the standard basis for $P_2 =\{ax^2+bx+c \;|\; a,b,c \in \R \}$.\\ \force\hfill Also, let $\alpha = \{g_1,g_2,g_3\}=\{ 1+x^2, 1+2x+x^2, 2+x \}$. 

\begin{enumerate}[(a)]
\item Explain why $\alpha$ is a basis for $P_2$. Then find $[I]_\std^\alpha$.
\item Give explicit formulas for $f_1^*,f_2^*,f_3^*,g_1^*,g_2^*,$ and $g_3^*$. For example, $f_3^*(ax^2+bx+c)=a$. 
\item Compute $[3x^2+2x+1]_\alpha$ using your change of basis matrix found in (a) and then using your formulas for $\alpha^*=\{g_1^*,g_2^*,g_3^*\}$ found in part (b).
\item Let $m(ax^2+bx+c)=-4a+3c$. Find $[m]_{\std^*}$ and $[m]_{\alpha^*}$.
\end{enumerate}

\item[\#2. Completely Annihilated by Math] Let $V$ be a vector space over $\F$ and let $W$ be a subspace of $V$. We define $A(W) = \{ f \in V^* \;|\; f(w)=0 \text{ for all } w\in W\}$. In other words, $f \in A(W)$ if $f(W)=\{0\}$ (i.e., $f$ annihilates all of our subspace $W$).

\begin{enumerate}[(a)]
\item Show that $A(W)$ is a subspace of $V^*$.

{\color{blue}[I'll do this one for you.]

First, the zero functional sends all vectors to zero (the scalar). Thus $0 \in A(W)$ (i.e., the annihilator of $W$ is a non-empty subset of $V^*$). Let $f,g \in A(W)$ and $s \in \F$. Notice that for all $\B w \in W$, we have $(f+g)(\B w)=f(\B w)+g(\B w)=0+0=0$ and $(sf)(\B w)=s\,f(\B w)=s\,0=0$. Thus $f+g,sf \in A(W)$. Therefore, $A(W)$ is a subspace.
}

\item Suppose $U$ is a subspace of $W$. Explain why $A(W) \subseteq A(U)$.

{\color{blue}[And this one too.]

Let $f \in A(W)$. This means that $f(\B w)=0$ for all $\B w \in W$. Suppose $\B u \in U$. Then because $U \subseteq W$ we have $\B u \in W$ and so $f(\B u)=0$. Therefore, $f$ annihilates all of $U$ and thus $f \in A(U)$. Thus $A(W) \subseteq A(U)$. Briefly, if we annihilate all of $W$, then since $U$ is contained in $W$, we certainly annihilate all of $U$.
}

\item Suppose $V=U \oplus W$. Show that $V^* = A(W) \oplus A(U)$. \\
\force \qquad {\it Note:} You need to show that every dual vector is a sum of a dual vector annihilating $W$ and one\\ \force \qquad \qquad \quad annihilating $U$. Also, you need to show that if $f \in A(W) \cap A(U)$ then $f=0$. 

\force \qquad {\it Big hint:} Consider $\pi_U:V \to V$ defined by $\pi(\B u+\B w)=\B u$ where $\B u \in U$ and $\B w \in W$ (this is well defined since every vector in $V$ is a {\it unique} sum of a vector in $U$ and a vector $W$ -- because $V$ is a {\it direct sum} of those spaces). This $\pi_U$ is called a projection onto $U$. It is linear. Likewise, define $\pi_W$. Consider composing $f \in V^*$ with these maps.

\item Let $T:V \to V$ be a linear operator and suppose that $T(W) \subseteq W$ (i.e., $W$ is $T$-invariant). Show that $T^*(A(W)) \subseteq A(W)$ (i.e., $A(W)$ is $T^*$-invariant). 

\force \qquad {\it Recall:} $T^*:V^* \to V^*$ is the transpose of $T$ defined by $T^*(f)=f\circ T$.
\end{enumerate}

\item[\#3. Mapping Stuff] Let $T:V \to W$ be a linear transformation.

\begin{enumerate}[(a)]
\item Let $U$ be a subspace of $V$. Show that $T(U)=\{T(\B u) \;|\; \B u \in U\}$ is a subspace $W$.

\qquad {\it Note:} As a consequence, $T(V)$ (i.e., the range of $T$) is a subspace of $W$.

{\color{blue}[I'll do this one for you.]

Notice that $\B 0 \in U$ (since $U$ is a subspace) and thus $\B 0=T(\B 0) \in T(U)$. Therefore, $T(U)$ is a non-empty subset of $W$. Next, let $\B x,\B y \in T(U)$ and $s \in \F$. This implies that there exists some $\B a,\B b \in U$ such that $T(\B a)=\B x$ and $T(\B b)=\B y$. Therefore, $\B x+\B y=T(\B a)+T(\B b)=T(\B a+\B b) \in T(U)$ since $\B a+\B b \in U$. Likewise, $s\,\B x=s\,T(\B a)=T(s\,\B a) \in T(U)$ since $s\,\B a \in U$. Thus $T(U)$ is a subspace.
}

\item Let $U$ be a subspace of $W$. Show that $T^{-1}(U) = \{ \B v \in V \;|\; T(\B v) \in U\}$ is a subspace of $V$.

\qquad {\it Note:} As a consequence, $T^{-1}(\{\B 0\}) = \mathrm{ker}(T)$ is a subspace of $V$.

\qquad {\it Hint:} $\B v \in T^{-1}(U)$ if and only if $T(\B v) \in U$. This proof {\it should} be easier than the one in part (a).

\item {}[The Second Isomorphism Theorem] Let $U_1$ and $U_2$ be subspaces of $V$.\\ \force \hfill Show that $(U_1+U_2)/U_2 \cong U_1/(U_1 \cap U_2)$. 

\qquad {\it Hint:} Consider the map $\B v \mapsto \B v+U_2$ from $U_1$ to $(U_1+U_2)/U_2$. Apply the First Isomorphism Theorem.

\end{enumerate}

\end{description}

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