Image of Generating Set of Vector Space under Linear Transformation is Generating Set of Image

Theorem
Let $K$ be a field.

Let $X$ be a vector space over $K$.

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

Let $G$ be a generating set for $X$.

Then $T \sqbrk G$ be a generating set for $T \sqbrk X$.

Proof
Let $y \in T \sqbrk X$.

Then there exists $x \in X$ such that $y = T x$.

Since $G$ generates $X$, there exists $n \in \N$, $x_1, \ldots, x_n \in G$ and $\alpha_1, \ldots, \alpha_n \in K$ such that:


 * $\ds x = \sum_{i \mathop = 1}^n \alpha_i x_i$

Then from the linearity of $T$ we have:


 * $\ds y = T x = \sum_{i \mathop = 1}^n \alpha_i T x_i$

with $T x_i \in T \sqbrk G$ for each $1 \le i \le n$.

So:


 * $T \sqbrk X \subseteq \map \span {T \sqbrk G}$

and so:


 * $T \sqbrk X = \map \span {T \sqbrk G}$

So $T \sqbrk G$ is a generating set for $T \sqbrk X$.