Linear Transformation defined from Graph

Theorem

Let $K$ be a field.

Let $X$ and $Y$ be vector spaces over $K$.

Let $X \times Y$ be the direct product of $X$ and $Y$.

Let $U$ be a vector subspace of $X \times Y$ such that:

whenever $\tuple { {\mathbf 0}_X, y} \in U$, we have $y = {\mathbf 0}_Y$

Define:

$\map D T = \set {x \in X : \tuple {x, y} \in U \text { for some } y \in Y}$

Then there exists a linear transformation $T : \map D T \to Y$ with graph $\map \GG T = U$.

Proof

We first show that $\map D T$ is a vector subspace of $X$.

We have $\tuple { {\mathbf 0}_X, {\mathbf 0}_Y} \in U$, so ${\mathbf 0}_X \in \map D T$.

Hence $\map D T \ne \O$.

Let $x, y \in \map D T$ and $\alpha \in \C$.

Then there exists $u, v \in Y$ such that $\tuple {x, u} \in U$ and $\tuple {y, v} \in U$.

Since $U$ is a vector subspace we have:

$\tuple {x, u} + \alpha \tuple {y, v} = \tuple {x + \alpha y, u + \alpha v} \in U$

So $x + \alpha y \in \map D T$.

Hence from the One-Step Vector Subspace Test, we have that $\map D T$ is a vector subspace of $X$.

Suppose that $\tuple {x, y_1}, \tuple {x, y_2} \in U$.

Then $\tuple { {\mathbf 0}_X, y_1 - y_2} = \tuple {x, y_1} - \tuple {x, y_2} \in U$ since $U$ is a vector subspace of $X \times Y$.

So $y_1 - y_2 = {\mathbf 0}_Y$ and hence $y_1 = y_2$.

Hence we can define $T x$ as the unique $y$ such that $\tuple {x, y} \in U$ for $x \in \map D T$.

Clearly we will have $\map G T = U$.

We just need to show that $T$ is linear.

Let $x, y \in \map D T$ and $\alpha \in \C$.

Then we have $\tuple {x, T x} + \alpha \tuple {y, T y} = \tuple {x + \alpha y, T x + \alpha T y} \in U$.

Since $\map T {x + \alpha y}$ is unique with $\tuple {x + \alpha y, \map T {x + \alpha y} } \in U$, we have:

$\map T {x + \alpha y} = T x + \alpha T y$

Hence $T$ is linear.

$\blacksquare$