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\title{Math 2135: Linear Algebra}
\author{Assignment 4}
\date{Solutions}

\begin{document}
\maketitle


\begin{enumerate}
  \item {\bf Let $V$ and $W$ be vector spaces, and let $T : V \to W$ be any function. The \emph{graph} of $T$ is defined to be the subset $\operatorname{graph}(T) \subset V \times W$ defined by
    $$\operatorname{graph}(T) := \{(v,Tv) \st v \in V\}.$$
    Prove that $T$ is linear if and only if $\operatorname{graph}(T)$ is a vector subspace of $V \times W$.}
    
        We first remark that $(0,T0) \in \operatorname{graph}(T)$, and so $\operatorname{graph}(T)$ is nonempty.

    We will use the fact that the function $\operatorname{graph}(T) \to V$ given by sending $(v,Tv) \mapsto v$ is a bijection. In other words:
    $$ \fbox{$(v,w) \in \operatorname{graph}(T)$ if and only if $w = Tv$.}$$
    
    Suppose that $(v_1,Tv_1)$ and $(v_2,Tv_2)$ are two points in $\operatorname{graph}(T)$. Then their sum is $(v_1,Tv_1) + (v_2,Tv_2) = (v_1 +v_2, Tv_1 + Tv_2)$. Taking $v := v_1 +v_2$ and $w := Tv_1 + Tv_2$ in the boxed statement above, we see that this sum $(v,w)$ is in $\operatorname{graph}(T)$ if and only if $w = Tv_1 + Tv_2 = Tv = T(v_1 + v_2)$.
    Asking this to hold for all $v_1,v_2$, we thus find that $\operatorname{graph}(T)$ is closed under addition if and only if $T$ is additive.
    
    Now suppose that $(v,Tv)$ is some point in $\operatorname{graph}(T)$, and $\lambda \in \bF$ is some scalar. Then $\lambda (v,Tv) = (\lambda v, \lambda Tv)$. Again applying the boxed statement, we see that this point is in $\operatorname{graph}(T)$ if and only if $\lambda Tv = T\lambda v$.
    Asking this to hold for all $v,\lambda$, we thus find that $\operatorname{graph}(T)$ is closed under scalar multiplication if and only if $T$ is homogeneous.
    
    Together, we see that the two conditions (additivity and homogeneity) for $T$ to be linear match the two conditions (closure under addition and scalar multiplication) for the nonempty(!)\ set $\operatorname{graph}(T)$ to be a vector subspace of $V \times W$.
  \item {\bf Let $V_1,V_2,W$ be vector spaces.}
  \begin{enumerate}
    \item {\bf Find an isomorphism between $\cL(W,V_1\times V_2)$ and $\cL(W,V_1) \times \cL(W,V_2)$.}
        
  Given any function $T : W \to V_1 \times V_2$, linear or not, define $T_1 : W \to V_1$ to be the function whose value on $w \in W$ is the first coordinate of $T(w)$, and let $T_2 : W \to V_2$ be the function whose value is the second coordinate. In other words, $T_1,T_2$ are defined by the formula
  $$ \fbox{$T(w) = (T_1(w),T_2(w)).$}$$
  In the other direction, given any functions $T_1 : W \to V_1$ and $T_2 : W \to V_2$, linear or not, the same formula defines a function $T : W \to V_1 \times V_2$.
  
  We claim that if $T$, $T_1$, and $T_2$ are related in this way, then $T$ is linear if and only if $T_1$ and $T_2$ are both linear. This claim follows simply from the way that vector addition and scalar multiplication are defined in $V_1 \times V_2$. Indeed, given $w,w' \in W$ and $\lambda \in \bF$, we see that
  \begin{gather*}
    T(\lambda w+w') = (T_1(\lambda w+w'),T_2(\lambda w+w')), \qquad \lambda T(w)+T(w') = (\lambda T_1(w) + T_1(w'), \lambda T_2(w)+T_2(w')),
  \end{gather*}
  but linearity of $T$ is the statement that the above left-hand sides are equal for all $w,w'$, whereas the statement that the above right-hand sides are equal for all $w,w'$ is precisely linearity of both $T_1$ and $T_2$. Another way to see this is to note that $T_1,T_2$ are nothing but the compositions of $T$ with the projection maps $\pi_1 : V_1 \times V_2 \to V_1$ and $\pi_2 : V_1 \times V_2$ defined by $\pi_i(v_1,v_2) = v_i$, and that these projection maps are linear, and that the composition of linear maps is linear.
  
  Thus we see that the boxed formula above, read in its two directions, defines a bijective correspondence between $\cL(W,V_1\times V_2)$ and $\cL(W,V_1) \times \cL(W,V_2)$. It remains to check that this correspondence is linear. In other words, given $T = (T_1,T_2)$ and $T' = (T_1',T_2')$ and $\lambda \in \bF$, we must check that, for every $w \in W$,
  $$ (\lambda T+T')(w) = ((\lambda T_1+T_1')(w), (\lambda T_2+T_2')(w)).$$
  This follows from unpacking --- for example, the left-hand side is $\lambda Tw + T'w$ --- and using the definition of the vector space structure on $V_1 \times V_2$.
    
    \item {\bf Find an isomorphism between $\cL(V_1 \times V_2,W)$ and $\cL(V_1,W) \times \cL(V_2,W)$.}
    
    Note that $(v_1,v_2) = (v_1,0) + (0,v_2)$ as vectors in $V_1 \times V_2$, for any vectors $v_1 \in V_1$ and $v_2 \in V_2$. Now suppose that $T : V_1\times V_2 \to W$ is any linear map. Then 
    $$T(v_1,v_2) = T(v_1,0) + T(0,v_2).$$ Define functions $T_1 : V_1 \to W$ and $T_2 : V_2 \to W$ by $T_1(v_1) = T(v_1,0)$ and $T_2(v_2) = T(0,v_2)$. Then the displayed equation equation is equivalent to
    $$ \fbox{$T(v_1,v_2) = T_1(v_1) + T_2(v_2)$.}$$
    If $T$ is linear, then so are $T_1,T_2$, since these are nothing but the compositions of $T$ with the inclusions $\iota_1 : V_1 \to V_1\times V_2$ and $\iota_2 : V_2 \to V_1\times V_2$ defined by $\iota_1(v_1) = (v_1,0)$ and $\iota_2(v_2) = (0,v_2)$. On the other hand, suppose that $T_1 : V_1 \to W$ and $T_2 : V_2 \to W$ are any linear maps, and define a function $T : V_1 \times V_2 \to W$ by the boxed equation. Then $T$ will be linear:
    \begin{multline*}
     T(\lambda(v_1,v_2) + (v_1',v_2')) = T(\lambda v_1+v_1',\lambda v_2+v_2') = T_1(\lambda v_1+v_1') + T_2(\lambda v_2+v_2') \\ = \lambda T_1 v_1 + T_1 v_1' + \lambda T_2 v_2 + T_2 v_2' = \lambda T(v_1,v_2) + T(v_1',v_2')
     \end{multline*}
    Thus we see that the boxed equation defines a bijection between $\cL(V_1 \times V_2,W)$ and $\cL(V_1,W) \times \cL(V_2,W)$. It is straightforward to show that this bijection is linear. Indeed, one must show that if $T,T_1,T_2$ are as in the boxed equation, and if $T',T_1',T_2'$ are, then for any $\lambda$, 
    $$ (\lambda T+T')(v_1,v_2) = (\lambda T_1+T_1')(v_1) + (\lambda T_2+T_2')(v_2)$$
    which can be checked by expanding both sides.    
  \end{enumerate}
  \item {\bf Let $V$ be a vector space over $\bF$. Prove that a nonempty subset $X \subset V$ is an affine subspace (for some vector subspace $U \subset  V$) if and only if, for any $v,w \in X$ and for any $\lambda \in \bF$, $\lambda v + (1-\lambda)w \in X$.}
  
  Note that the statement $\lambda v + (1-\lambda)w \in X$ can be rewritten as
  $$ w + \lambda(v-w) \in X.$$
  
  If $X$ is an affine subset for $U$, then for any $v,w \in X$ and for any $\lambda \in \bF$, we have $v-w \in U$ and so $\lambda(v-w) \in U$ and so $w + \lambda(v-w) \in X$. This proves the ``only if'' direction. In other words, we have shown that if $X$ is an affine subspace, then $\lambda v + (1-\lambda)w \in X$ for all $v,w \in X$.
  
  For the ``if'' direction (i.e.\ if $\lambda v + (1-\lambda)w \in X$ for all $v,w\in X$, then $X$ is an affine subspace), we  use that $X$ is nonempty to fix a basepoint $x \in X$. Set $U := X - x$ to be the set of vectors of the form $v-x$ where $v \in X$. Then $X = x+U$, and it suffices to show that $U$ is a vector subspace. Note that $0 = x-x \in U$.
  
  To this end, we first argue that $U$ is closed under scalar multiplication. Let $u = v-x \in U$. We want to show that $\lambda u = \lambda(v-x) = \lambda v - \lambda x$ is also in $U$. This is equivalent to showing that $\lambda v - \lambda x + x \in X$, which follows (by setting $w=x$) from the assumption about $X$.
  
  The only thing remaining to prove is that $U$ is closed under vector addition. In other words, we assume that we have $u_1,u_2 \in U$ and we want to show that $u_1+u_2 \in U$. But $U$ is closed under scalar multiplication, so $u_1+u_2 \in U$ if and only if $\frac12(u_1+u_2) \in U$. It is this latter statement that we will prove. By definition of $U$, we know that $u_1 +x$ and $u_2 +x$ are both in $X$, and our goal is to prove that $\frac12(u_1+u_2)+x \in X$. But taking $v = u_1+x$ and $w = u_2 + x$ and $\lambda = \frac12$ in the assumed property of $X$ gives exactly this goal.
  
  
  \item {\bf Let $U_1,U_2 \subset V$ be vector subspaces. Suppose that $X_1 \subset V$ is an affine subspace for $U_1$ and that $X_2 \subset V$ is an affine subspace for $U_2$. Prove that $X_1 \cap X_2$ is either empty or an affine subspace for $U_1 \cap U_2$.}
  
  In other words, we wish to show the following two facts:
  \begin{enumerate}
  \item If $x \in X_1 \cap X_2$ and $u \in U_1 \cap U_2$, then $x + u \in X_1 \cap X_2$.
  \item If $x,y \in X_1 \cap X_2$, then $x-y \in U_1 \cap U_2$.
  \end{enumerate}
  
  But if $x \in X_1 \cap X_2$ and $u \in U_1 \cap U_2$, then certainly $x \in X_1$ and $u \in U_1$, and since $X_1$ is affine for $U_1$, then certainly $x+u \in X_1$. By the same token, certainly $x \in X_2$ and $u \in U_2$ and so $x+u \in U_2$. But if $x+u$ is in both $U_1$ and $U_2$, then it is in $U_1 \cap U_2$. This establishes fact 1.
  
  Fact 2 is similar. If $x,y \in X_1 \cap X_2$, then certainly they are both in $X_1$, and so their difference $x-y$ is in $U_1$. But also certainly they are both in $X_2$, and so their difference is in $U_2$. But if $x-y$ is in both $U_1$ and $U_2$, then it is in $U_1 \cap U_2$.
  
  
  \item {\bf Let $U \subset V$ be a vector subspace, let $\iota : U \to V$ denote the ``identity'' map $\iota(u) = u$,  
   and let $\pi : V \to V/U$ the quotient map. Let $W$ be another vector space, and consider the maps
  \begin{gather*}\circ\iota : \cL(V, W) \to \cL(U,W), \; T \mapsto T\circ\iota, \\ \circ \pi : \cL(V/U ,W) \to \cL(V,W), \; S \mapsto S \circ \pi.\end{gather*}
  Finally, let $X \subset \cL(V,W)$ denote the subset
  $$ X := \{T \in \cL(V,W) \st \ker(T) \supseteq U\}.$$}
  \begin{enumerate}
    \item {\bf Show that $\circ\iota$ and $\circ \pi$ are linear, and that $X$ is a vector subspace.}
    
    For the linearity, we will prove the following more general statement. Let $S : U \to V$ denote any linear map. Then $\circ S : \cL(V,W) \to \cL(U,W)$ is linear. This statement, applied to $S=\iota$, gives the first linearity, and applied to $S=\pi$ (with $V$ replaced by $V/U$ and $U$ replaced by $V$) gives the second linearity.
    
    To this end, we first show that $\circ S$ is additive. Let $T_1,T_2 \in \cL(V,W)$. Then $T_1 + T_2$ is the function which sends $v \mapsto (T_1+T_2)(v) := T_1v + T_2v$. Its composition with $S$ is thus the function which sends $u \mapsto T_1Su + T_2Su$. But this is precisely the function $T_1S + T_2S$. This confirms that $\circ S$ is additive. Homogeneity is confirmed similarly: let $T \in \cL(V,W)$ and $\lambda \in \bF$. Then $\lambda T$ is the function $v \mapsto \lambda T v$, and so $(\lambda T)\circ S$ is the function $u \mapsto \lambda T S u$, which is precisely the function $\lambda (T\circ S)$.
    
    We now show that $X$ is a vector space. To do so we unpack the condition ``$\ker(T) \supseteq U$'': it is equivalent to saying
    $$ \fbox{$ Tu = 0 \quad \forall u \in U. $}$$
    Now, $X$ is nonempty becuase the zero map $T=0$ satisfies this condition. Suppose that $T_1,T_2 \in X$. In other words, $T_1u=T_2u = 0$ for all $u\in U$. But then $(T_1+T_2)u = T_1u + T_2u = 0+0=0$, so $T_1+T_2 \in X$. And if $T\in X$ and $\lambda \in \bF$, then $(\lambda T)u = \lambda Tu = \lambda 0 = 0$ for all $u \in U$, so $\lambda T \in X$.
    
    \item {\bf Show that $\circ \pi$ is injective.}
    
    A linear map is injective if and only if its kernel is zero. So the question is equivalent to showing that $\ker(\circ \pi) = \{0\}$. Unpacked, this is equivalent to showing that if $S : V/U \to W$ satisfies $S\circ \pi : V \to W$, then $S=0$. Well, $\pi$ is the function that sends $v \mapsto v+U$, and so
     $S\circ \pi$ is the function which sends $v \in V$ to $S(v+U)$. So saying that this is identically the zero function is saying that $S(v+U) = 0$ for every $v+U \in V/U$. But every element of $V/U$ is of the form $v+U$, and so $S$ vanishes identically, i.e.\ it is the zero function.
     
    \item {\bf Show that $\im(\circ \pi) = X$. Conclude that $\cL(V/U, W) \cong X \subset \cL(V,W)$.}
    
    We wish to show that $T \in X \subset \cL(V,W)$ if and only if there exists an $S \in \cL(V/U, W)$ such that $S\circ \pi = T$. Unpacked, we wish to show that the following conditions on $T : V \to W$ are equivalent:
    \begin{enumerate}
      \item $Tu = 0$ for all $u \in U$.
      \item There exists $S : V/U \to W$ such that $Tv = S(v+U)$ for all $v \in V$.
    \end{enumerate}
    
    Well, if ii.\ holds, then $Tu = S(u+U) = S(0+U) = T0 = 0$, so i.\ holds. 
    
    On the other hand, suppose that i.\ holds. Then recall that there is a linear map $\bar{T} : V/U \to W$ defined by the formula $\bar{T}(v+U) = Tv$. Indeed, for this formula to define a function, we need to know that if $v+U = v'+U$, then $Tv = Tv'$ (so that the formula really does return a well-defined output), but this is equivalent to the statement that $T(v-v') = 0$, which follows from the fact that $v-v' \in U$ and that $T$ vanishes on $U$. Now take $S := \bar{T}$; then ii.\ holds.
    
    The conclusion $\cL(V/U, W) \cong X \subset \cL(V,W)$ follows from the fact that $\circ \pi$ is injective by (b), and so it provides an isomorphism between its domain $\cL(V/U,W)$ and its image $X$.
        
    \item {\bf Show that $\ker(\circ\iota) = X$.}
    
    By definition, $\ker(\circ\iota) \subset \cL(V,W)$ is the set of $T : V \to W$ such that $T\circ\iota = 0$, or equivalently such that $0 = T\iota u = Tu$ for all $u \in U$. In other words, $u \in \ker T$ if $u \in U$, or in other words $U \subseteq \ker T$. So this is simply the definition of $X$.
    
    \item {\bf Assuming that $V$ is finite-dimensional, show that $\circ\iota$ is surjective. (In fact, $\circ\iota$ is surjective even if $V$ is infinite-dimensional, but proving this requires some set-theoretic results that go beyond the class.) Conclude that $\cL(U,W) \cong \cL(V,W)/X$.}
    
    We must show that if $S : U \to W$ is any linear map, then $S = T\circ \iota$ for some linear map $T : V \to W$. In other words, we are given the values of $T$ on $U$ and must extend it to all of $V$. Using finite-dimensionality, let's assume that $\dim V = n$ and $\dim U = m \leq n$. Pick a basis $v_1,\dots,v_m$ for $U$, and extend it to a basis $v_1,\dots,v_m,v_{m+1},\dots,v_n$ for $V$. Now define
    $$ Tv_i := \begin{cases} Sv_i, & i\leq m, \\ 0, & i > m. \end{cases}$$
    Remember that for any basis of $V$, any linear map $V \to W$ is uniquely determined by its values on that basis. Thus there is a unique linear map $T : V \to W$ which takes these values. It is the following: for any $v \in V$, write $v = \sum_{i=1}^n \alpha_i v_i$, and then $Tv = \sum_{i=1}^n \alpha_i Tv_i$.
    
    We claim that $T\circ \iota = S$, or in other words that $Tu = Su$ if $u \in U$. Well, suppose that $u \in U$, and write $u = \sum_{i=1}^n \alpha_i v_i$. Since $v_1,\dots,v_n$ is a basis for $V$, there is a unique such expression for $v$. Because $v_1,\dots,v_m$ is a basis for $U$, we can write $u$ as a linear combination of just the first $m$ basis vectors, i.e.\ we don't need the last $n-m$. Since the sum $u = \sum_{i=1}^n \alpha_i v_i$ uniquely determines the $\alpha_i$, we conclude that $\alpha_{m+1}=\dots=\alpha_n = 0$. Thus $Tv = \sum_{i=1}^n \alpha_i Tv_i = \sum_{i=1}^m \alpha_i Tv_i + \sum_{i=m+1}^n \alpha_i Tv_i = \sum_{i=1}^m \alpha_i Tv_i + \sum_{i=m+1}^n 0 Tv_i = \sum_{i=1}^m \alpha_i Tv_i + 0 = Su$.
        
        This establishes the surjectivity of $\circ \iota : \cL(V,W) \to \cL(U,W)$. The conclusion follows because, by (d), $\ker(\circ\iota) = X$.
  \end{enumerate}
  {\bf To summarize these results: $\cL(-,W)$ takes subs to quotients and quotients to subs.}
  \item {\bf Let $V_1,V_2,W$ be vector spaces. A function $T : V_1 \times V_2 \to W$ is called \emph{bilinear} if for any $v_1 \in V_1$, the function $T(v_1,-) : V_2 \to W$ defined by $v_2 \mapsto f(v_1,v_2)$ is linear, and also for any $v_2 \in V_2$ the function $T(-,v_2) : V_1 \to W$ defined by $v_1 \mapsto T(v_1,v_2)$ is linear. Let $\cB\cL(V_1,V_2;W)$ denote the set of bilinear functions $V_1 \times V_2 \to W$.}
  \begin{enumerate}
    \item {\bf When $V_1 = V_2 = W = \bR$, show that the addition function $A(v_1,v_2) := v_1 + v_2$ is linear but not bilinear, whereas the multiplication function $M(v_1,v_2) := v_1 v_2$ is bilinear but not linear.}
    
    The statement that $A$ is linear unpacks to:
    \begin{description}
      \item[additivity] $A((v_1,v_2)+(v_1',v_2')) = A(v_1,v_2) + A(v_1',v_2')$.  But the left-hand side is $A(v_1+v_1',v_2+v_2') = v_1+v_1'+v_2+v_2'$ and the right-hand side is $v_1+v_2+v_1'+v_2'$, which agree by commutativity of addition.
      \item[homogeneity] $A(\lambda(v_1,v_2)) = \lambda A(v_1,v_2)$. But the left-hand side if $A(\lambda v_1,\lambda v_2) = \lambda v_1 + \lambda v_2$, and the right-hand side is $\lambda(v_1+v_2)$, which agree by the distributivity law.
    \end{description}
    $A$ is not bilinear. If it were, then for any choice of $v_1$, the function $A(v_1,-)$ would be linear. So let's choose $v_1 = 2$ to test this. Then we are asking whether $A(2,-) : v \mapsto 2+v$ is a linear map. It isn't, for example it is inhomogeneous ($A(2,\lambda v) = 2+\lambda v \neq \lambda (2+v)$).
    
    The statement that $M$ is bilinear unpacks to two statements. First, we are asserting that for fixed $v_1$, the function $M(v_1,-)$ is linear, and second we are asserting that for fixed $v_2$, the function $M(-,v_2)$ is linear. Note that $V_1 = V_2 = \bR$, and so we we will change letters: we are asserting that for $v_1 = \lambda \in \bR$ fixed, the function $M(\lambda,-) : v \mapsto \lambda v$ is linear, and we are asserting that for $v_2 = \lambda \in \bR$, the function $M(-,\lambda) : v \mapsto v\lambda$ is linear. These are the same function, and it is linear by the distributivity and associativity of multiplication.
    
    
    \item {\bf Show that $\cB\cL(V_1,V_2; W)$ is a vector subspace of the set $W^{V_1\times V_2}$ of all functions.}
    
    To see that $\cB\cL(V_1,V_2; W)$ is nonempty, note that the zero map $0(-,-)$ is bilinear. Indeed, if you fix $v_1$, then $0(v_1,-)$ is the zero map, which is linear, and if you fix $v_2$, then $0(-,v_2)$ is the zero map, which is linear.
    
    To see that $\cB\cL(V_1,V_2; W)$ is closed under addition, suppose that $T,T'$ are both bilinear. Then for any $(v_1,v_2) \in V_1 \times V_2$, we have $(T+T')(v_1,v_2) = T(v_1,v_2) + T'(v_1,v_2)$. Now, if we fix $v_1$, we are looking at the function $(T+T')(v_1,-) = T(v_1,-) + T'(v_1,-) : V_2 \to W$, which is a sum of two linear functions and hence linear. Similarly, if we fix $v_2$, then $(T+T')(-,v_2) = T(-,v_2) + T'(-,v_2) : V_1 \to W$ is a sum of two linear functions and hence linear. So $T+T'$ is bilinear.
    
    To see that $\cB\cL(V_1,V_2; W)$ is closed under scalar multiplication, suppose that $T$ is bilinear and $\lambda \in \bF$. Then $\lambda T$ is the function defined by $(\lambda T)(v_1,v_2) = \lambda \, T(v_1,v_2)$. If you fix $v_1$, you get the function $\lambda T(v_1,-) : V_2 \to W$, which is $\lambda$ times the linear function $T(v_1,-) : V_2 \to W$, and hence linear. Similarly, if you fix $v_2$, you get $\lambda T(-,v_2) : V_1 \to W$, which is a multiple of a linear function and hence linear. So $\lambda T$ is bilinear.
    
    \item {\bf Show that $\cB\cL(V_1,V_2; W)$ is isomorphic to $\cL(V_1, \cL(V_2,W))$ and also to $\cL(V_2, \cL(V_1,W))$.}
    
    There is a manifest isomorphism $\cB\cL(V_1,V_2; W) \cong \cB\cL(V_2,V_1; W)$ given by sending $T : V_1\times V_2 \to W$ to the function $T' : V_2 \times V_1 \to W$ defined by $T'(v_2,v_1) = T(v_1,v_2)$. So it suffices to establish an isomorphism $\cB\cL(V_1,V_2; W) \cong \cL(V_1, \cL(V_2,W))$.
    
    Well, suppose that $T \in \cB\cL(V_1,V_2; W)$. Then for any $v_1$, the function $T(v_1,-) : V_2 \to W$ is linear. In other words, $T^\sharp : v_1 \mapsto T(v_1,-)$ is a function $V_1 \to \cL(V_2,W)$. 
    
    We first claim that $T^\sharp$ is linear, i.e.\ it is an element of $\cL(V_1, \cL(V_2,W))$. Unpacked, we want to show the following: given $v_1,v_1' \in V_1$ and $\lambda \in \bF$, we want to show that we have an equality
    $$ T^\sharp(\lambda v_1 + v_1') \overset?= \lambda T^\sharp(v_1) + T^\sharp(v_1')$$
    of elements of $\cL(V_2,W)$. In other words, this should be an equality of functions, and two functions are equal if they have the same values. So we want to show that for every $v_2 \in V_2$,
    $$ T^\sharp( \lambda v_1 + v_1')(v_2) \overset?= \lambda T^\sharp(v_1)(v_2) + T^\sharp(v_1')(v_2)$$
    is an equality in $W$. Except the LHS is $T(\lambda v_1 + v_1',v_2)$ and the right-hand side is $\lambda T(v_1,v_2) + T(v_1', v_2)$, and these are equal because the function $T(-,v_2)$ is linear.
    
    So we have produced an operation $\sharp : \cB\cL(V_1,V_2; W) \to \cL(V_1, \cL(V_2,W))$ sending $T \mapsto T^\sharp$. We next need to show that $\sharp$ is linear. In other words, we need to show that if $T,T' \in \cB\cL(V_1,V_2; W)$ and $\lambda \in \bF$, then 
    $$ (\lambda T + T')^\sharp \overset? = \lambda T^\sharp + (T')^\sharp $$
    is an equality in $\cL(V_1, \cL(V_2,W))$. But the left-hand side is the function $(\lambda T+T')(v_1,-) : V_2 \to W$ and the right-hand side is the function $\lambda T(v_1,-) + T'(v_1,-) : V_2 \to W$. In other words, we need to show that for any $v_2 \in V_2$, we have
    $$ (\lambda T+T')(v_1,v_2) \overset?= \lambda T(v_1,v_2) + T'(v_1,v_2)$$
    but this is just how addition and scalar multiplication of functions is defined.
    
    Summarizing so far, we have produced a linear transformation $\sharp : \cB\cL(V_1,V_2; W) \to \cL(V_1, \cL(V_2,W))$. We now must show that it is an isomorphism. We will do this by exhibiting its inverse function.
    
    Suppose that $S \in \cL(V_1,\cL(V_2,W))$. In other words, for any $v_1$, $S(v_1) \in \cL(V_2,W)$ is a linear map. Define $S^\flat : V_1 \times V_2 \to W$ by $S^\flat(v_1,v_2) = S(v_1)(v_2)$. We claim that $S^\flat$ is bilinear. First, fixing $v_1$, we definitely have $S^\flat(v_1,-) = S(v_1) : V_2 \to W$ is a linear map, so that does half of the bilinearity. Next, fix $v_2$; then we claim that $S^\flat(-, v_2) : V_1 \to W$ is linear. Unpacked, we are claiming that, for any $v_1,v_1'\in V_1$ and any $\lambda \in \bF$, then
    $$ S^\flat(\lambda v_1 + v_1', v_2) \overset?= \lambda S^\flat(v_1, v_2) + S^\flat(v_1', v_2)$$
    is an equality in $W$. In other words, we are claiming that
    $$ S^\flat(\lambda v_1 + v_1', -) \overset?= \lambda S^\flat(v_1, -) + S^\flat(v_1', -)$$
    is an equality of functions $V_2 \to W$. But the left-hand side is precisely $S(\lambda v_1+v_1') \in \cL(V_2,W)$ and the right-hand side is precisely $\lambda S(v_1) + S(v_1') \in \cL(V_2,W)$, and these are equal by linearity of $S$.
    
    Summarizing so far, we have a linear transformation $\cB\cL(V_1,V_2; W) \to \cL(V_1, \cL(V_2,W))$ sending $T \mapsto T^\sharp$ defined by $T^\sharp(v_1)(v_2) = T(v_1,v_2)$, and we have a function $\cL(V_1,\cL(V_2,W)) \to \cB\cL(V_1,V_2; W)$ sending $S \mapsto S^\flat$ defined by $S^\flat(v_1,v_2) = S(v_1)(v_2)$. Then we obviously have $(T^\sharp)^\flat = T$ and $(S^\flat)^\sharp = S$, and so these two transformations are inverses. Since one was linear, both are, and they give the desired isomorphism.
    
    \item {\bf Suppose that $V_1$ and $V_2$ are finite dimensional. Assume that there exists a vector space $V$ such that, for any vector space $W$, the vector spaces $\cB\cL(V_1,V_2; W)$ and $\cL(V,W)$ are isomorphic. What is the dimension of $V$?
    
    \textbf{Remark:} In fact, such a $V$ does exist. It is called the \emph{tensor product} of $V_1$ with $V_2$, and denoted $V_1 \otimes V_2$.}
    
    If there is supposed to be an isomorphism  $\cB\cL(V_1,V_2; W) \cong \cL(V,W)$ for any $W$, then in particular there should be an isomorphism for $W = \bF$. But $\dim \cL(V,\bF) = \dim V$, whereas, using (c), we learn that
    $$ \dim \cB\cL(V_1,V_2; \bF) = \dim \cL(V_1, \cL(V_2,\bF)) = \dim V_1 \times \dim \cL(V_2,\bF) = \dim V_1 \times \dim V_2.$$
    So we must have \fbox{$\dim (V_1 \otimes V_2) = \dim V_1 \times \dim V_2$}.
  \end{enumerate}
\end{enumerate}



\end{document}  