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\noindent
\parbox{2.25in}{\bf Math 5230}
\hfill {\Large \bf Homework \#8} \hfill
\parbox{2.25in}{\bf \hfill Due: Wed., Oct. 30${}^\mathbf{th}$\!, 2024}

\vspace{0.2in}

\begin{description}
\item[\#1 Ultimately a Based Problem] Let $T:V \to W$ be a linear transformation between vector spaces $V$ and $W$ (both over the field $\mathbb{F}$).

\vspace{0.05in}

\fbox{Let $S$ be a linearly independent subset of $V$ and let $T$ be one-to-one. Show $T(S)$ is linearly independent.}

\vspace{0.05in}

{\it Note:} We have $T\left(\sum\limits_{i=1}^\ell c_i\B v_i \right) = \sum\limits_{i=1}^\ell c_i T(\B v_i)$ for any $c_1,\dots,c_\ell \in \mathbb{F}$ and $\B v_1,\dots,\B v_\ell \in V$. Thus the image of any linear combination of elements of $V$ is a linear combination of the images of those elements. Therefore, given any subset $S \subseteq V$, we have $T\left(\mathrm{span}(S)\right) = \mathrm{span}\left(T(S)\right)$. Consequently, if $S$ spans $V$, then $T(S)$ spans $T(V)$. Thus if $T$ is onto, then a spanning set for $V$ maps to a spanning set for $W$.

\vspace{0.05in}

Putting this together with the above homework problem, we get that \fbox{isomorphisms map bases to bases}.

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\item[\#2 Concrete Quotient] Let $W = \left\{ \begin{bmatrix} a+b+4c & 2a+b+3c \\ 3a+b+2c & 4a+b+c \end{bmatrix} \middle| a,b,c \in \mathbb{R} \right\}$. Give a quick justification for why $W$ is a {\it subspace} of $\mathbb{R}^{2 \times 2}$. Then find a basis for $W$ and a basis for $\diagquotient{\mathbb{R}^{2 \times 2}}{W}$. 

     
\item[\#3 Abstract Quotient] Let $U$ and $W$ be subspaces of some vector space $V$ (over a field $\mathbb{F}$). 
   \begin{enumerate}[(a)]
   \item Prove the Second (or Diamond) Isomorphism Theorem:  $\diagquotient{W}{U \cap W} \cong \diagquotient{U+W}{U}$.
   
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   {\it Hint:} Consider $\varphi:W \to \diagquotient{U+W}{U}$ defined by $\varphi(\B w) = \B w + U$ and apply the First Isomorphism Theorem.
   
   
   \item Give a relationship among the dimensions of $U,W,U+W,$ and $U\cap W$ determined by the above theorem
   \item What can we say in the special case when $U+W = U \oplus W$?
   \end{enumerate}
   
\item[\#4 Concrete Dual]  Let $\alpha = \{ (1,0,0), (1,-1,0), (2,0,1) \}$. 
\begin{enumerate}[(a)]
\item Explain why $\alpha$ is a basis for $\mathbb{R}^3$. Then find $\alpha^*$ for $(\mathbb{R}^3)^*$ (i.e. find the basis dual to $\alpha$).
\item Explain why $f \in (\mathbb{R}^3)^*$ where $f(x,y,z)=3x+2y+z$. Then write $f$ as a linear combination of $\alpha^*$ elements (i.e., find its $\alpha^*$-coordinates).
\end{enumerate}

\end{description}

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\item[\#5. Completely Annihilated] 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}
   
