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		\title{Weekly Assignment 2}
		
		\author{WRITE YOUR NAME HERE\\
		MATH 117: Advanced Linear Algebra}
		
		\maketitle
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	Some hints/comments for this assignment may be written in the footnotes. See the \href{https://people.ucsc.edu/~jbreland/teaching/SM22_MATH117/weekly_assignments.html}{weekly assignment webpage} for due dates, templates, and assignment description.
	Make sure to justify any claims you make. You may not appeal to any results that we have not discussed in class. 
	
	
\begin{enumerate}[\bfseries 1.]
	
		


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		\item Prove Proposition 2.1.7: Let $V$ and $W$ be vector spaces over $F$ and let $L:V\to W$ be a linear map. Then
		\begin{enumerate}[(a)]
			\item $\im(L)$ is a subspace of $W$; and
			
			\item $\ker(L)$ is a subspace of $V$.
		\end{enumerate}
		
		\begin{proof} Write your proof here.
	
		\end{proof}

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		\item Prove Theorem 2.5.1: Suppose that $V,W$ are vector spaces over $F$ and $L,K:V\to W$ are linear maps. Then $\alpha L + \beta K$ is a linear map for all $\alpha,\beta \in F$.\footnote{Since the zero map is linear, this implies that $\mathrm{Hom}(V,W)$ is a subspace of $W^V$.}


		\begin{proof} Write your proof here.
	
		\end{proof}
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\item Let $V_1,V_2$ be vector spaces over $F$. The direct sum $V_1\oplus V_2$ comes with linear maps 
\[
\iota_1:V_1 \to V_1\oplus V_2,\, v_1 \mapsto (v_1,0) \quad \text{and} \quad \iota_2:V_2\to V_1\oplus V_2, \, v_2 \mapsto (0,v_2).
\]
Let $Z$ be any other vector space and let $L_1:V_1 \to Z$ and $L_2:V_2 \to Z$ be any other linear maps. Prove that there is a unique linear map $L:V_1\oplus V_2 \to Z$ with the property that $L\circ \iota_1 = L_1$ and $L\circ \iota_2 = L_2$. See Remark~\ref{rmk1}.

\begin{proof} Write your proof here.
	
\end{proof}
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\item Let $V_1,V_2,W_1,W_2$ be vector spaces over $F$ and let $L_1:V_1\to W_1$ and $L_2:V_2\to W_2$ be linear maps. 
\begin{enumerate}[(a)]
	\item Use the Universal Property of the Direct Sum (see Remark~\ref{rmk1}) to show that there is a unique linear map
	\[
	L_1\oplus L_2:V_1\oplus V_2 \to W_1\oplus W_2
	\]
	satisfying $(L_1\oplus L_2)(v_1,v_2) = (L_1(v_1),L_2(v_2))$. 
	
	\item Suppose additionally that $V_1,V_2,W_1,W_2$ are finite-dimensional with ordered bases $B_1=(a_1,\ldots,a_k)$, $B_2=(b_1,\ldots,b_l)$, $C_1 = (c_1,\ldots,c_m)$, and $C_2=(d_1,\ldots,d_n)$, respectively.
	\begin{enumerate}[(i)]
		\item Prove that $B:=((a_1,0),\ldots,(a_k,0),(0,b_1),\ldots,(0,b_l))$ is a basis for $V_1 \oplus V_2$. Similarly, $C:= ((c_1,0),\ldots,(c_l,0),(0,d_1),\ldots,(0,d_n))$ is a basis for $W_1\oplus W_2$.
		
		\item Prove that the matrix for $L_1\oplus L_2$ with respect to $B$ and $C$ has the following block diagonal form:
		\[
		[L_1\oplus L_2]_B^C = \begin{pmatrix}[L_1]_{B_1}^{C_1} & 0\\ 0 & [L_2]_{B_2}^{C_2}
		\end{pmatrix}.
		\]
	\end{enumerate}
\end{enumerate}

\begin{proof} Write your proof here.
	
\end{proof}
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\end{enumerate}

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\begin{remark}\label{rmk1} In other words, $L$ is the unique linear map making the following diagram commute
	% https://q.uiver.app/?q=WzAsNCxbMSwxLCJWXFxvcGx1cyBXIl0sWzAsMCwiViJdLFswLDIsIlciXSxbMywxLCJaIl0sWzEsMCwiXFxpb3RhX1YiLDJdLFsyLDAsIlxcaW90YV9XIl0sWzEsMywiTF9WIiwwLHsiY3VydmUiOi0yfV0sWzIsMywiTF9XIiwyLHsiY3VydmUiOjJ9XSxbMCwzLCJcXGV4aXN0cyAhIEwiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=
	\[\begin{tikzcd}
		V_1 \\
		& {V_1\oplus V_2} && Z \\
		V_2
		\arrow["{\iota_1}"', from=1-1, to=2-2]
		\arrow["{\iota_2}", from=3-1, to=2-2]
		\arrow["{L_1}", curve={height=-12pt}, from=1-1, to=2-4]
		\arrow["{L_1}"', curve={height=12pt}, from=3-1, to=2-4]
		\arrow["{\exists ! L}", dashed, from=2-2, to=2-4]
	\end{tikzcd}\]
	The property described in Problem 1 is usually called the Universal Property of the Direct Product. It provides a precise answer to the question ``How do I define a linear map out of the direct sum of two vector spaces?". One simply defines linear maps $L_1$ and $L_2$ as in the statement. Your proof will provide the recipe for constructing the desired linear map $L$. Moreover, it establishes a bijection of sets
	% https://q.uiver.app/?q=WzAsNCxbMCwwLCJIb20oVlxcb3BsdXMgVyxaKSJdLFsxLDAsIkhvbShWLFopXFxvcGx1cyBIb20oVyxaKSJdLFswLDEsIkwiXSxbMSwxLCIoTFxcY2lyY1xcaW90YV9WLExcXGNpcmNcXGlvdGFfVykiXSxbMCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJhcnJvd2hlYWQifX19XSxbMiwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn19fV1d
	\[\begin{tikzcd}
		{\Hom_F(V_1\oplus V_2,Z)} & {\Hom_F(V_1,Z)\oplus \Hom_F(V_2,Z)} \\
		L & {(L\circ\iota_1,L\circ\iota_2)}
		\arrow[tail reversed, from=1-1, to=1-2]
		\arrow[maps to, from=2-1, to=2-2]
	\end{tikzcd}\]
	However, notice that the domain and codomain are actually vector spaces. It can be shown that this bijection is a linear map, i.e., a vector space isomorphism.
\end{remark}


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