Linear Indep. Prop. of Tensor Products of Normed Linear Spaces

# Linear Independence Properties of Tensor Products of Normed Linear Spaces

Recall from The Algebraic Tensor Product of Two Normed Linear Spaces page that if $X$ and $Y$ are normed linear spaces and if $x \in X$, $y \in Y$ then we defined $x \otimes y : X^* \times Y^* \to \mathbf{F}$ for all $f \in X^*$ and all $g \in Y^*$ by:

(1)
\begin{align} \quad (x \otimes y)(f, g) = f(x)g(y) \end{align}

We noted that $x \otimes y$ is a bilinear map on $X^* \times Y^*$ to $\mathbf{F}$ for each $x \in X$ and each $y \in Y$ and we said that the linear span of $\{ x \otimes y : x \in X, y \in Y \}$ in $\mathrm{BL}(X^*, Y^*, \mathbf{F})$ is called the algebraic tensor product of $X$ and $Y$ and is denoted $X \otimes Y$.

We will now prove some important results of tensor products regarding linear independence.

 Proposition 1: Let $X$ and $Y$ be normed spaces. If $u \in X \otimes Y$ then there exists linearly independent sets $\{ x_1, x_2, ..., x_n \}$ and $\{ y_1, y_2, ..., y_n \}$ for which $\displaystyle{u = \sum_{i=1}^{n} x_i \otimes y_i}$.
• Proof: Let $u \in X \otimes Y$. By definition, $X \otimes Y$ is the linear span of the set $\{ x \otimes y : x \in X, y \in Y \}$, and so $u$ can be expressed as a finite linear combination of elements from $X \otimes Y$.
• Let $\{ x_1, x_2, ..., x_n \} \subset X$ and $\{ y_1, y_2, ..., y_n \} \subset Y$ such that $n$ is chosen as small as possible (which is guaranteed by the well-ordering principle) for which $\displaystyle{u = \sum_{i=1}^{n} x_i \otimes y_i}$. We aim to show that $\{ x_1, x_2, ..., x_n \}$ and $\{ y_1, y_2, ..., y_n \}$ are linearly independent sets.
• If $n = 1$ then certainly the singleton sets $\{ x_1 \}$ and $\{ y_1 \}$ are linearly independent in $X$ and $Y$ respectively.
• So let $n \geq 2$ and suppose that $\{ y_1, y_2, ..., y_n \}$ is not linearly independent. Without loss of generality, assume that $y_n$ can be written as a linear combination of $\{ y_1, y_2, ..., y_{n-1} \}$. Then there exists $a_1, a_2, ..., a_{n-1} \in \mathbf{F}$ for which:
(2)
\begin{align} \quad y_n = \sum_{i=1}^{n-1} a_iy_i \end{align}
• Then observe that for all $f \in X^*$ and $g \in Y^*$ we have that:
(3)
\begin{align} \quad \left [ \sum_{i=1}^{n-1} (x_i + a_ix_n) \otimes y_i \right ](f, g) &= \left [(x_1 + a_1x_n) \otimes y_1 + (x_2 + a_2x_n) \otimes y_2 + ... + (x_{n-1} + a_{n-1}x_n) \otimes y_{n-1} \right ] (f, g) \\ &= [(x_1 + a_1x_n) \otimes y_1](f, g) + [(x_2 + a_2x_n) \otimes y_2](f, g) + ... + [(x_{n-1} + a_{n-1}x_n) \otimes y_{n-1}] (f, g) \\ &= f(x_1 + a_1x_n)g(y_1) + f(x_2 + a_2x_n)g(y_2) + ... + f(x_{n-1} + a_{n-1}x_n)g(y_{n-1}) \\ &= f(x_1)g(y_1) + a_1f(x_n)(y_1) + f(x_2)g(y_2) + a_2f(x_n)g(y_2) + ... + f(x_{n-1})g(y_{n-1}) + a_{n-1}f(x_n)g(y_{n-1}) \\ &= \sum_{i=1}^{n-1} f(x_i)g(y_i) + \sum_{i=1}^{n-1} f(x_n) g(a_i y_i) \\ &= \sum_{i=1}^{n-1} (x_i \otimes y_i)(f, g) +\left (x_n \otimes \sum_{i=1}^{n-1} a_i y_i \right ) (f, g) \\ &= \sum_{i=1}^{n-1} (x_i \otimes y_i)(f, g) + (x_n \otimes y_n)(f, g) \\ &= \sum_{i=1}^{n} (x_i \otimes y_i) (f, g) \\ &= u(f, g) \end{align}
• Therefore $u = \sum_{i=1}^{n-1} (x_i + a_ix_n) \otimes y_i$. But this contradicts the minimality of $n$. Therefore the assumption that $\{ y_1, y_2, ..., y_n \}$ is linearly dependent is false. Thus $\{ y_1, y_2, ..., y_n \}$ is linearly independent.
• A similar argument shows that $\{ x_1, x_2, ..., x_n \}$ is linearly independent, for if not, say $x_n = \sum_{i=1}^{n-1} a_ix_i$ then we also have that:
(4)
\begin{align} \quad u = \sum_{i=1}^{n-1} x_i \otimes (y_i + a_iy_n) \end{align}
• Again, contradicting the minimality of $n$. Thus we conclude that that if $u \in X \otimes Y$ then there exists linearly independent sets $\{ x_1, x_2, ..., x_n \} \subset X$ and $\{ y_1, y_2, ..., y_n \} \subset Y$ for which $\displaystyle{u = \sum_{i=1}^{n} x_i \otimes y_i}$.
 Proposition 2: Let $X$ and $Y$ be normed spaces and let $\{ x_1, x_2, ..., x_n \} \subset X$, $\{ y_1, y_2, ..., y_n \} \subset Y$. If $\displaystyle{\sum_{i=1}^{n} x_i \otimes y_i =0}$ and $\{ x_1, x_2, ..., x_n \}$ is linearly independent in $X$ then $y_1, y_2, ..., y_n = 0$.

Recall from the Corollaries to the Hahn-Banach Theorem page that if $X$ is a normed space and $x, y \in X$ are such that $f(x) = f(y)$ for all $f \in X^*$ then $x = y$. In the case when $y = 0$ we have that if $f(x) = 0$ for all $f \in X^*$ then $x = 0$. We use this important corollary in the proof below.

• Proof; Let $f \in X^*$ and let $g \in Y^*$. Since $\displaystyle{\sum_{i=1}^{n} x_i \otimes y_i = 0}$ we have that:
(5)
\begin{align} \quad 0(f, g) &= \left ( \sum_{i=1}^{n} x_i \otimes y_i \right )(f, g) \\ &= ( x_1 \otimes y_1 + x_2 \otimes y_2 + ... + x_n \otimes y_n)(f, g) \\ &= f(x_1)g(y_1) + f(x_2)g(y_2) + ... + f(x_n)g(y_n) \\ &= f \left ( g(y_1)x_1 + g(y_2)x_2 + ... + g(y_n)x_n \right ) \\ &= f \left ( \sum_{i=1}^{n} g(y_i)x_i \right ) \end{align}
• But observe that $0(f, g) = f(0)g(0) = 0$ for all $f \in X^*$ and for all $g \in Y^*$. Thus for all $f \in X^*$ we have that:
(6)
\begin{align} \quad f \left ( \sum_{i=1}^{n} g(y_i)x_i \right ) = 0 \end{align}
• This implies that:
(7)
\begin{align} \quad \sum_{i=1}^{n} g(y_i)x_i = 0 \end{align}
• But since $\{ x_1, x_2, ..., x_n \}$ is a linearly independent set, the above equation implies that $g(y_i) = 0$ for each $1 \leq i \leq n$. Since this holds for all $g \in Y^*$ we must have that $y_i = 0$ for each $1 \leq i \leq n$. $\blacksquare$
 Proposition 3: Let $X$ and $Y$ be normed spaces. If $\{ x_1, x_2, ..., x_m \} \subset X$ and $\{ y_1, y_2, ..., y_n \} \subset Y$ are linearly independent in $X$ and $Y$ respectively, then $\{ x_i \otimes y_j : 1 \leq i \leq m, 1 \leq j \leq n \}$ is linearly independent in the algebraic tensor product $X \otimes Y$.
• Proof: Let $a_{ij} \in \mathbf{F}$, $1 \leq i \leq m$, $1 \leq j \leq n$ and suppose that:
(8)
\begin{align} \quad 0 &= \sum_{i=1}^{m} \sum_{j=1}^{n} a_{ij} x_i \otimes y_j \\ &= \sum_{i=1}^{m} x_i \otimes \sum_{j=1}^{n} a_{ij} y_j \end{align}
• Since $\{ x_1, x_2, ..., x_m \}$ is linearly independent in $X$, with the above equation and proposition 2 we must have that for each $1 \leq i \leq m$ that:
(9)
\begin{align} \quad \sum_{j=1}^{n} a_{ij} y_j = 0 \end{align}
• But since $\{ y_1, y_2, ..., y_n \}$ is linearly independent, we have that the above equation implies that $a_{ij} = 0$ for each $1 \leq j \leq n$ as well. Thus we conclude that $\{ x_i \otimes y_j : 1 \leq i \leq m, 1 \leq j \leq n \}$ is linearly independent in $X \otimes Y$. $\blacksquare$