X⊗Y and Y⊗X are Isomorphic Linear Spaces

Definition: Let $X$ and $Y$ be normed linear spaces. If $u = \sum_{i=1}^{n} x_i \otimes y_i \in X \otimes Y$ then the Transpose of $u$ is the tensor $u^t = \sum_{i=1}^{n} y_i \otimes x_i \in Y \otimes X$.

The following proposition tells us that if $X$ and $Y$ are normed linear spaces then $X \otimes Y$ is isomorphic to $Y \otimes X$.

Proposition 1: Let $X$ and $Y$ be normed linear spaces. Then $X \otimes Y$ and $Y \otimes X$ are isomorphic linear spaces.

Recall that two linear spaces are isomorphic if there exists a bijective linear operator between them.

• Proof: Define a map $T : X \otimes Y \to Y \otimes X$ for all $u \in X \otimes Y$ by $T(u) = u^t$.

• 1. Showing that $T$ is a linear operator on $X \otimes Y$: Let $u, w \in X \otimes Y$ with $u = \sum_{i=1}^{n} x_i \otimes y_i$ and with $w = \sum_{j=1}^{m} x_j' \otimes y_j'$, and let $\alpha \in \mathbb{F}$. Then:

(1)

\begin{align} \quad T(u + w) = T \left ( \sum_{i=1}^{n} x_i \otimes y_i + \sum_{j=1}^{m} x_j' \otimes y_j' \right ) = \sum_{i=1}^{n} y_i \otimes x_i + \sum_{j=1}^{m} y_j' \otimes x_j' = u^t + w^t = T(u) + T(w) \end{align}

(2)

\begin{align} \quad T(\alpha u) = T \left ( \alpha \sum_{i=1}^{n} x_i \otimes y_i \right ) = T \left ( \sum_{i=1}^{n} (\alpha x_i) \otimes y_i \right ) = \sum_{i=1}^{n} y_i \otimes (\alpha x_i) = \alpha \sum_{i=1}^{n} y_i \otimes x_i = \alpha u^t = \alpha T(u) \end{align}

• So indeed, $T$ is a linear operator from $X \otimes Y$ to $Y \otimes X$.

• 2. Showing that $T$ is bijective: Let $u = \sum_{i=1}^{n} x_i \otimes y_i \in X \otimes Y$. Suppose that $T(u) = 0$. Then $u^t = 0$, i.e.:

(3)

\begin{align} \quad \sum_{i=1}^{n} y_i \otimes x_i = 0 \end{align}

• So for all $g \in Y^*$ and for all $f \in X^*$ we have that:

(4)

\begin{align} \quad 0 = \sum_{i=1}^{n} (y_i \otimes x_i)(g, f) = \sum_{i=1}^{n} g(y_i)f(x_i) = \sum_{i=1}^{n} f(x_i)g(y_i) = \sum_{i=1}^{n} (x_i \otimes y_i)(f, g) = u(f, g) \end{align}

• Thus $u = 0$. So $\ker(T) = \{ 0 \}$ and thus, $T$ is injective. Clearly $T$ is also surjective, since given $v = \sum_{i=1}^{n} y_i \otimes x_i \in Y \otimes X$ we have that $v^t \in X \otimes Y$ is such that $T(v^t) = (v^t)^t = v$. Thus $T$ is bijective and we conclude that $T$ is an isomorphism of $X \otimes Y$ to $Y \otimes X$.