Linear Maps Defined by Bases

# 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)
\begin{align} \quad v = a_1v_1 + a_2v_2 + ... + a_nv_n \end{align}
• Define the linear map $T$ by:
(2)
\begin{align} \quad T(v) = T(a_1v_1 + a_2v_2 + ... + a_nv_n) = a_1w_1 + a_2w_2 + ... + a_nw_n \end{align}
• 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)
\begin{align} \quad T(u + v) = T((b_1v_1 + b_2v_2 + ... + b_nv_n) + (a_1v_1 + a_2v_2 + ... + a_nv_n)) = T([b_1 + a_1]v_1 + [b_2 + a_2]v_2 + ... + [b_n + a_n]v_n) \\ \quad = [b_1 + a_1]w_1 + [b_2 + a_2]w_2 + ... + [b_n + a_n]w_n = (b_1w_1 + b_2w_2 + ... + b_nw_n) + (a_1w_1 + a_2w_2 + ... + a_nw_n) \\ \quad = T(u) + T(v) \end{align}
• 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)
\begin{align} \quad T(cv) = T(c(a_1v_1 + a_2v_2 + ... + a_nv_n)) = T(ca_1v_1 + ca_2v_2 + ... + ca_nv_n) \\ \quad = (ca_1)w_1 + (ca_2)w_2 + ... + (ca_n)w_n = c(a_1w_1 + a_2w_2 + ... + a_nw_n) = cT(v) \end{align}
• 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)
\begin{align} \quad T(v) = T(a_1v_1 + a_2v_2 + ... + a_nv_n) = a_1T(v_1) + a_2T(v_2) + ... + a_nT(v_n) = a_1w_1 + a_2w_2 + ... + a_nw_n \quad \blacksquare \end{align}