📂Linear Algebra

Linear Transformations Between Finite-Dimensional Vector Spaces

Theorem¹

Let $V, W$ be a vector space. Let $\left\{ \mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{n} \right\}$ and $\left\{ \mathbf{w}_{1}, \mathbf{w}_{2}, \dots, \mathbf{w}_{n} \right\}$ be bases of $V, W$, respectively. Then there exists a unique linear transformation $T : V \to W$ that satisfies $T(\mathbf{v}_{i}) = \mathbf{w}_{i}$.

Corollary²

Let $V, W$ be a vector space. Let $S = \left\{ \mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{n} \right\}$ be a basis of $V$. If $U, T : V \to W$ is a linear transformation and $U(\mathbf{v}_{i}) = T(\mathbf{v}_{i})$, then it follows that $U = T$.

Generalization³

Let $V, W$ be a vector field, and let $\beta$ be a basis of $V$. Then for some function $f : \beta \to W$, there exists a unique linear transformation that satisfies the following.

$$T : V \to W \quad \text{ by } \quad T(x) = f(x) \quad \forall x \in \beta$$

Proof

Let $\mathbf{x} \in V$. Since $\left\{ \mathbf{v}_{i} \right\}$ is a basis,

$$\mathbf{x} = \sum a_{i} \mathbf{v}_{i}$$

there exist unique constants $a_{i}$ that satisfy the equation. Now, define $T$ as follows.

$$T : V \to W \quad \text{ by } \quad T(\mathbf{x}) = \sum a_{i}\mathbf{w}_{i}$$

Then $T(\mathbf{v}_{i}) = \mathbf{w}_{i}$ is satisfied.

• Linearity

If $\mathbf{x}, \mathbf{y} \in V$, then

$$\mathbf{x} = \sum a_{i} \mathbf{v}_{i}, \quad \mathbf{y} = \sum b_{i}\mathbf{v}_{i}$$

$$c \mathbf{x} + \mathbf{y} = \sum (ca_{i} + b_{i})\mathbf{v}_{i}$$

Therefore,

$$T(c \mathbf{x} + \mathbf{y}) = \sum (ca_{i} + b_{i})\mathbf{w}_{i} = c\sum a_{i}\mathbf{w}_{i} + \sum b_{i}\mathbf{w}_{i} = cT(\mathbf{x}) + T(\mathbf{y})$$

• Uniqueness

Let the linear transformation $U : V \to W$ satisfy $U(\mathbf{v}_{i}) = \mathbf{w}_{i}$. Then, for $\mathbf{x} = \sum a_{i} \mathbf{v}_{i} \in V$,

$$U(\mathbf{x}) = \sum a_{i} U(\mathbf{v}_{i}) = \sum a_{i} \mathbf{w}_{i} = T(\mathbf{x})$$

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