Examples of Bilinear Mappings

# Examples of Bilinear Mappings

Recall from the Bilinear Mappings and Multilinear Mappings page that if $X$, $Y$, and $Z$ are linear spaces over $\mathbf{F}$ then a bilinear map from $X$ and $Y$ to $Z$ is a function $T : X \times Y \to Z$ such that for every fixed $y \in Y$ the map $x \to T(x, y)$ is linear and for every fixed $x \in X$ the map $y \to T(x, y)$ is linear.

Furthermore, if $X$, $Y$, and $Z$ are normed linear spaces then a bilinear map $T : X \times Y \to Z$ is said to be bounded if there exists an $M > 0$ such that $\| T(x, y) \|_Z \leq M \| x \|_X \| y \|_Y$ for all $x \in X$ and for all $y \in Y$.

We will now look at some examples of bilinear maps.

## Example 1

Let $X$ be a normed linear space. Define $T : X^* \times X \to \mathbf{F}$ for all $f \in X^*$ and all $x \in X$ by:

(1)
\begin{align} \quad T(f, x) = f(x) \end{align}

Note that for all fixed $f \in X^*$ and all $x, y \in X$, $\alpha \in \mathbf{F}$ we have that $T(f, x+y) = f(x + y) = f(x) + f(y) = T(f, x) + T(f, y)$ and $T(f, \alpha x) = f(\alpha x) = \alpha f(x) = \alpha T(f, x)$. Similarly, for all fixed $x \in X$ and all $f, g \in X^*$, $\alpha \in \mathbf{F}$ we have that $T(f + g, x) = (f + g)(x) = f(x) + g(x) = T(f, x) + T(g, x)$ and $T(\alpha f, x) = (\alpha f)(x) = \alpha f(x) = \alpha T(f, x)$. So $T$ is a bilinear map.

## Example 2

Let $X = \mathbb{F}^n$ (where $\mathbb{F} = \mathbb{R}$ or $\mathbb{C}$). Define $T : X \times X \to \mathbf{F}$ for all $\vec{x} = (x_1, x_2, ..., x_n), \vec{y} = (y_1, y_2, ..., y_n) \in X$ by:

(2)
\begin{align} \quad T(\vec{x}, \vec{y}) = x_1y_1 + x_2y_2 + ... + x_ny_n \end{align}

Then $T$ is simply the dot product on $\mathbb{R}^n$ and it is easy to verify that the dot product on $\mathbb{R}^n$ is a bilinear map.

## Example 3

Let $X$, $Y$, $Z$, and $W$ be normed linear spaces and let $T : X \times Y \to Z$ be a bilinear map and let $S : W \to Y$ be a linear map. Define $A : X \times W \to Z$ for all $x \in X$ and all $w \in W$ by:

(3)
\begin{align} \quad A(x, w) = T(x, S(w)) \end{align}

Note that for all fixed $x \in X$ and all $w_1, w_2 \in W$, $\alpha \in \mathbf{F}$ we have that:

(4)
\begin{align} \quad A(x, w_1 + w_2) = T(x, S(w_1 + w_2)) = T(x, S(w_1) + S(w_2)) = T(x, S(w_1)) + T(x, S(w_2)) = A(x, w_1) + A(x, _2) \end{align}
(5)
\begin{align} \quad A(x, \alpha w_1) = T(x, S(\alpha w_1)) = T(x, \alpha S(w_1)) = \alpha T(x, S(w_1)) = \alpha A(x, w_1) \end{align}

So for each fixed $x \in X$ the map $w \to A(x, w)$ is linear. Similarly, it can be shown that for each fixed $w \in W$ the map $x \to A(x, w)$ is linear. Therefore $A$ is a bilinear map.