# Linear Maps Defined by Bases

Recall from the Linear Maps page that if $V$ and $W$ are vector spaces then a function $T : V \to W$ is said to be a linear map or linear transformation from $V$ to $W$ if for all vectors $u, v \in V$ we have that $T(u + v) = T(u) + T(v)$ (the additivity property) and for all vectors $u \in U$ and scalars $a \in \mathbb{F}$ we have that $T(au) = aT(u)$ (the homogeneity property).

Now suppose that $V$ is a finite-dimensional vector space and say $\mathrm{dim} (V) = n$. Let $W$ be a vector space. Then there exists a basis $\{ v_1, v_2, …, v_n\}$ of $V$. Furthermore, let $\{w_1, w_2, …, w_n \}$ be any set of vectors in $W$. In the following theorem we will see that a unique linear map $T : V \to W$ defined by $T(v_j) = w_j$ for each $j = 1, 2, …, n$ exists – that is, there exists a linear map $T$ which maps specific basis vectors in $V$ to specific vectors in $W$ for each basis of $V$.

Theorem 1: Let $V$ be a finite-dimensional vector space and let $W$ be a vector space. Let $\mathrm{dim} (V) = n$, and let $\{ v_1, v_2, …, v_n \}$ be a basis of $V$ and let $\{ w_1, w_2, …, w_n \}$ be any set of vectors in $W$. Then there exists a unique linear map $T : V \to W$ defined by $T(v_j) = w_j$ for each $j = 1, 2 , …, n$. |

**Proof:**We will first prove the existence of such a linear map. Let $\{ v_1, v_2, …, v_n \}$ be a basis of $V$. Then for any vector $v \in V$ there exists scalars $a_1, a_2, …, a_n \in \mathbb{F}$ such that:

(1)

- Define the linear map $T$ by:

(2)

- Now note that $T(v_1) = w_1$, $T(v_2) = w_2$, …, $T(v_n) = w_n$. More precisely, $T(v_j) = w_j$ for each $j = 1, 2, …, n$.

- We now need to show that the additivity and homogeneity properties hold for the function $T$. We will first show that the additivity property holds. Let $u, v \in V$. Then for some set of scalars $b_1, b_2, …, b_n \in \mathbb{F}$ we have that $u = b_1v_1 + b_2v_2 + … + b_nv_n$ and we already have from earlier that $v = a_1v_1 + a_2v_2 + … + a_nv_n$ and so:

(3)

- Thus the additivity property holds. Now we will show that the homogeneity property holds. Let $c \in \mathbb{F}$ and let $v \in V$ we such that $v = a_1v_1 + a_2v_2 + … + a_nv_n$ (like from earlier). Then we have that:

(4)

- Therefore the homogeneity property holds and indeed $T$ is a linear map.

- We now need to show that this linear map is unique. Notice that the homogeneity and additivity of $T$ implies that $T$ is uniquely determined by the vectors in $\mathrm{span} (v_1, v_2, …, v_n) = V$, since for every vector $v \in V$, we can write $v = a_1v_1 + a_2v_2 + … + a_nv_n$ and:

(5)