Exist. of a Lin. Map σ on X⊗Y to Z that Mat. a Bilin. Map on X×Y to Z

# The Existence of a Linear Map σ on X⊗Y to Z that Matches a Bilinear Map on X×Y to Z

 Theorem 1: Let $X$, $Y$, and $Z$ be normed spaces. If $T : X \times Y \to Z$ is a bilinear map then there exists a unique linear map $\tilde{T} : X \otimes Y \to Z$ such that for all $x \in X$ and for all $y \in Y$ we have that $\tilde{T} (x \otimes y) = T(x, y)$.

If we let $\tau : X \times Y \to X \otimes Y$ be defined for all $(x, y) \in X \times Y$ by $\tau(x, y) = x \otimes y$, then Theorem 1 states that for any bilinear map $T : X \times Y \to Z$ there exists a unique map $\tilde{T} : X \otimes Y \to Z$ such that $T = \tilde{T} \circ \tau$.

• Proof: Suppose that $\displaystyle{\sum_{r=1}^{k} x_r \otimes y_r = 0}$. We want to then show that $\displaystyle{\sum_{r=1}^{k} T(x_r, y_r) = 0}$.
• Let $\{ a_i \}$ be a basis for $\mathrm{span} (x_1, x_2, ..., x_k)$ and let $\{ b_j \}$ be a basis for $\mathrm{span} (y_1, y_2, ..., y_k)$. Then each $x_r$, $1 \leq r \leq k$ can be written uniquely as a linear combination of $\{ a_i \}$, and similarly, each $y_r$, $1 \leq r \leq k$ can be written uniquely as a linear combination of $\{ b_j \}$. So for each $1 \leq r \leq k$ write:
(1)
\begin{align} \quad x_r = \sum_{i} \alpha_{ir} a_i \: , \quad \alpha_{ir} \in \mathbf{F} \end{align}
(2)
\begin{align} \quad y_r = \sum_{j} \beta_{jr} b_j \: , \quad \beta_{jr} \in \mathbf{F} \end{align}
• Then:
(3)
\begin{align} \quad 0 &= \sum_{r=1}^{k} x_r \otimes y_r \\ &= \sum_{r=1}^{k} \left [ \left [ \sum_{i} \alpha_{ir} a_i\right ] \otimes \left [ \sum_{j} \beta_{jr} b_j \right ] \right ] \\ &= \sum_{r=1}^{k} \sum_{i} \sum_{j} \alpha_{ir} \beta_{jr} a_i \otimes b_j \\ &= \sum_{i} \sum_{j} \sum_{r=1}^{k} \alpha_{ir} \beta_{jr} a_i \otimes b_j \end{align}
• Since $\{ a_i \} \subset X$ is linearly independent in $X$ and $\{ b_j \} \subset Y$ is linearly independent in $Y$, we know by one of the propositions on the Linear Independence Properties of Tensor Products of Normed Linear Spaces page we have $\{ a_i \otimes b_j \}_{i, j}$ is linearly independent in $X \otimes Y$. Thus the above equation implies that for all $i$ and all $j$:
(4)
\begin{align} \quad \sum_{r=1}^{k} \alpha_{ir} \beta_{jr} = 0 \end{align}
• Therefore we have that:
(5)
\begin{align} \quad \sum_{r=1}^{k} T(x_r, y_r) &= \sum_{r=1}^{k} T \left ( \sum_{i} \alpha_{ir} a_i, \sum_{j} \beta_{jr} b_j \right ) \\ &= \sum_{r=1}^{k} \sum_{i} \sum_{j} \alpha_{ir} \beta_{jr} T(a_i, b_j) \\ &= \sum_{i} \sum_{j} \underbrace{\sum_{r=1}^{k} \alpha_{ir} \beta_{jr}}_{=0 \: \forall i, \: \forall j} T(a_i, b_j) \\ &= 0 \end{align}
• So we define $\tilde{T} : X \otimes Y \to Z$ for all $u = \sum_{r=1}^{k} x_k \otimes y_k \in X \otimes Y$ by:
(6)
\begin{align} \quad \tilde{T} (u) = \left ( \sum_{r=1}^{k} x_r \otimes y_r \right ) := \sum_{r=1}^{k} T(x_r, y_r) \end{align}
• Then $\tilde{T} (x \otimes y) = T(x, y)$ for all $x \in X$ and $y \in Y$. Now there are a few things to check. First we will check that $\tilde{T}$ is well-defined, that is, the values of $\tilde{T}$ are independent of how we choose to write $u$ as a linear combination in $X \otimes Y$.
• Showing that $\tilde{T}$ is well-defined: Let $u \in X \otimes Y$ and suppose that $u = \sum_{r=1}^{k} x_r \otimes y_r$ and $u = \sum_{r=1}^{k'} x_r' \otimes y_r'$. Then:
(7)
\begin{align} \quad 0 = u' - u = \sum_{r=1}^{k'} x_r' \otimes y_r' - \sum_{r=1}^{k} x_r \otimes y_r \end{align}
• Then as we have proven above:
(8)
\begin{align} \quad 0 = \sum_{r=1}^{k'} T(x_r', y_r') - \sum_{r=1}^{k} T(x_r, y_r) \end{align}
• So $\tilde{T} (u)$ is well-defined regardless of the way we choose to write $u$ as a linear combination of elements of the form $x_k \otimes y_k$.
• Showing that $\tilde{T}$ is linear: Let $u, v \in X \otimes Y$ with $u = \sum_{r=1}^{k} x_r \otimes y_r$ and let $v = \sum_{r=1}^{k'} x_r' \otimes y_r'$, and let $\alpha \in \mathbf{F}$. Then:
(9)
\begin{align} \quad \quad \tilde{T} (u + v) = \tilde{T} \left ( \sum_{r=1}^{k} x_r \otimes y_r + \sum_{r=1}^{k'} x_r' \otimes y_r' \right ) = \sum_{r=1}^{k} T(x_r, y_r) + \sum_{r=1}^{k'} T(x_r', y_r') = \tilde{T}(u) + \tilde{T}(v) \end{align}
(10)
\begin{align} \quad \quad \tilde{T} (\alpha u) = \tilde{T} \left ( \alpha \sum_{r=1}^{k} x_r \otimes y_r \right ) = \tilde{T} \left ( \sum_{r=1}^{k} \alpha x_r \otimes y_r \right ) = \alpha \sum_{r=1}^{k} T(x_r, y_r) = \alpha \tilde{T} (u) \end{align}
• Therefore $\tilde{T}$ is linear.
• Showing that $\tilde{T}$ is unique: Suppose there exists another linear map $\tilde{T}^* : X \otimes Y \to Z$ such that $\tilde{T}^*(x \otimes y) = T(x, y)$ for all $x \in X$ and for all $y \in Y$. Let $u = \sum_{r=1}^{k} x_k \otimes y_k \in X \otimes Y$. Then by the linearity of $\tilde{T}^*$ and $\tilde{T}$ we have that:
(11)
\begin{align} \quad \tilde{T}^*(u) &= \tilde{T}^* \left ( \sum_{r=1}^{k} x_k \otimes y_k \right ) \\ &= \sum_{r=1}^{k} \tilde{T}^* (x_k \otimes y_k) \\ &= \sum_{r=1}^{k} T(x_k, y_k) \\ &= \sum_{r=1}^{k} \tilde{T}(x_k \otimes y_k) \\ &= \tilde{T} \left ( \sum_{r=1}^{k} x_k \otimes y_k \right ) \\ &= \tilde{T} (u) \end{align}
• So $\tilde{T} = \tilde{T}^*$ on all of $X \otimes Y$, so $\tilde{T}$ is unique. $\blacksquare$

Theorem 1 is very significant. It tells us that if $X$, $Y$, and $Z$ are normed linear spaces then every bilinear map $T : X \times Y \to Z$ has a corresponding linear map $\tilde{T} : X \otimes Y \to Z$ such that $T(x, y) = \tilde{T} (x \otimes y)$. In other words, $X \otimes Y$ somehow turns bilinear maps into linear maps. This is very useful because results regarding bilinear maps can often be reduced to problems regarding linear maps and linear maps are well understood.