Image of Vector Subspace under Linear Transformation is Vector Subspace

Theorem

Let $K$ be a field.

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

Let $U$ be a vector subspace of $X$.

Let $T : X \to Y$ be a linear transformation.

Then $T \sqbrk U$ is a vector subspace of $Y$.

Proof

Since $U$ is a non-empty set, we can apply the One-Step Vector Subspace Test to $T \sqbrk U$.

Let $\lambda \in K$ and $u, v \in T \sqbrk U$.

Then there exists $x, y \in U$ such that:

$u = T x$

and:

$v = T y$

Then:

\(\ds \lambda u + v\)

\(=\)

\(\ds \lambda T x + T y\)

\(\ds \)

\(=\)

\(\ds \map T {\lambda x + y}\)

Definition of Linear Transformation

Since $U$ is a vector subspace, we have $\lambda x + y \in U$.

So $\map T {\lambda x + y} \in T \sqbrk U$.

That is, $\lambda u + v \in T \sqbrk U$.

So $T \sqbrk U$ is a vector subspace of $Y$.

$\blacksquare$