📂Linear Algebra

The Basis of the Domain Generates the Image of the Linear Transformation

Theorem¹

Let’s suppose we have a given linear transformation $T : V \to W$. Assume $V$ is finite-dimensional, and let $S = \left\{ \mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{n} \right\}$ be a basis of $V$. Then, the image of any $\mathbf{v} \in V$ can be represented as follows.

$$T(\mathbf{v}) = c_{1}T(\mathbf{v}_{1}) + c_{2}T(\mathbf{v}_{2}) + \cdots c_{n}T(\mathbf{v}_{n})$$

Here, $c_{i}$ are coefficients that satisfy $\mathbf{v} = \sum c_{i}\mathbf{v}_{i}$. In other words, $\left\{ T(\mathbf{v}_{i}) \right\}$ generates the range of $T$.

$$R(T) = \span( T(S) ) = \span \left( \left\{ T(\mathbf{v}_{1}), T(\mathbf{v}_{2}), \dots, T(\mathbf{v}_{n}) \right\} \right)$$

Explanation

This means knowing how the bases transform under the linear transformation $T$ allows us to know the image of all $\mathbf{v} \in V$.

Proof

By the Uniqueness of Basis Representation, for all $\mathbf{v} \in V$, the following linear combination uniquely exists.

$$\mathbf{v} = c_{1}\mathbf{v}_{1} + c_{2}\mathbf{v}_{2} + \cdots + c_{n}\mathbf{v}_{n}$$

Then, due to the linearity of $T$, the following holds true.

$$\begin{align*} T ( \mathbf{v} ) &= T( c_{1}\mathbf{v}_{1} + c_{2}\mathbf{v}_{2} + \cdots c_{n}\mathbf{v}_{n} ) \\ &= T( c_{1}\mathbf{v}_{1} ) + T( c_{2}\mathbf{v}_{2} ) + \cdots + T( c_{n}\mathbf{v}_{n} ) \\ &= c_{1}T(\mathbf{v}_{1} ) + c_{2}T( \mathbf{v}_{2} ) + \cdots + c_{n}T( \mathbf{v}_{n} ) \end{align*}$$

■