Linear maps defined on basis vectors

An arbitrary map

\[ f : V \to W \]

encodes a lot of information: one needs to specify \(f(v)\) for every \(v \in V\). For linear maps, this simplifies drastically:

Proposition 4.40 (Related exercises: Exercise 6.13, Exercise 4.44, Exercise 4.18)

Let \(v_1, \dots, v_n\) be a basis of a vector space \(V\). Let \(W\) be another vector space and \(w_1, \dots, w_n\) arbitrary vectors (they need not be linearly independent, or span \(W\) etc.) Then there is a unique linear map \(f: V \to W\) such that

\[ f(v_i) = w_i. \]

(4.41)

Proof. Recall Proposition 3.64: given a basis \(v_1, \dots, v_n\) of a vector space, any vector \(v \in V\) can be uniquely expressed as a linear combination

\[ v = b_1 v_1 + \dots + b_n b_n = \sum_{i=1}^n b_i v_i, \]

i.e., we can express \(v\) in such a form and the real numbers \(b_i\) are uniquely determined by \(v\). Moreover, we can think of these numbers \(b_1, \dots, b_n\) as the coordinates of \(v\) (with respect to our coordinate system given by the basis). Namely, given another vector \(v' = \sum_{i=1}^n b'_i v_i\) and some \(a \in {\bf R}\), we have

\[ \begin{align*} v+v' & = \sum_{i=1}^n (b_i + b'_i) v_i \\ av & = \sum_{i=1}^n (ab_i) v_i. \end{align*} \]

Now, given \(v \in V\), we define

\[ f(v) := \sum_{i=1}^n b_i w_i. \]

(4.42)

In particular, for \(v = v_i\), this satisfies \(f(v_i) = w_i\). The map \(f\) is linear; this follows from the preceding discussion.

Conversely, if a linear map \(f\) satisfies \(f(v_i) = w_i\), for \(v\) as above, it necessarily satisfies

\[ f(v) = f (\sum_{i=1}^n b_i v_i) = \sum_{i=1}^n b_i f(v_i) = \sum_{i=1}^n b_i w_i. \]

So, the map defined in is the only linear map satisfying . ◻

Example 4.43 (Related exercises: Exercise 4.5)

We consider \(V = {\bf R}^3\), with the basis

\[ v_1 = e_1 = (1,0,0), v_2=e_2 = (0,1,0), v_3 = (0,1,-1). \]

(Note that \(e_1, e_2\) are part of the standard basis of \({\bf R}^3\).) According to Proposition 4.40, there is a unique linear map \(f: {\bf R}^3 \to {\bf R}^3\) such that

\[ f(v_1) = (2,-1,0), f(v_2) = (1,-1,1), f(v_3) = (0,2,2). \]

We determine \(f(e_3)\), where \(e_3 = (0,0,1)\) is the third standard basis vector. We have

\[ e_3 = v_2-v_3. \]

Thus

\[ f(e_3) = f(v_2-v_3) = f(v_2)-f(v_3) = (1,-1,1)-(0,2,2) = (1,-3,-1). \]

Thus, with respect to the standard basis \(e_1, e_2, e_3\) (which is distinct from the one above!), the matrix of \(f\) is given by

\[ A = \left ( \begin{array}{ccc} 2 & 1 & 1 \\ -1 & -1 & -3 \\ 0 & 1 & -1 \end{array} \right ). \]

That is, \(f\) agrees with the map

\[ f: {\bf R}^3 \to {\bf R}^3, v \mapsto Av. \]