Question

We have W = {w^1,...,w^n} as a spanning set for a subspace S in R^j, and V = {v^1,...,v^m} a linearly independent subset of S. Let A be the matrix whose columns are the v^m, B the matrix whose columns are the w^n. Please show that there is an N by M matrix C such that A = BC.

Answer 1

Since $v^1, \ldots, v^m \in S$ and $S$ is the spanning set of vectors $w^1, \ldots, w^n$ so each of vectors $v^k$ is a non-trivial linear combination of vectors $w^1, \ldots, w^n$. Thus

$$v ^ {k} = w ^ {1} c _ {1 k} + w ^ {2} c _ {2 k} + \dots + w ^ {n} c _ {n k}, \qquad k = \overline {{1 , m}}$$

Rewriting this equality we get:

$$v ^ {k} = (w ^ {1}, w ^ {2}, \ldots , w ^ {n}) \left( \begin{array}{c} c _ {1 k} \\ c _ {2 k} \\ \dots \\ c _ {n k} \end{array} \right)$$

Writing the statements for all $k = \overline{1,m}$ together we get:

$$(v ^ {1}, v ^ {2}, \ldots , v ^ {m}) = (w ^ {1}, w ^ {2}, \ldots , w ^ {n}) \left( \begin{array}{c c c} c _ {1 1} & \dots & c _ {1 m} \\ \vdots & \ddots & \vdots \\ c _ {n 1} & \dots & c _ {n m} \end{array} \right)$$

Using the notations given and denoting

$$C = \left( \begin{array}{c c c} c _ {1 1} & \dots & c _ {1 m} \\ \vdots & \ddots & \vdots \\ c _ {n 1} & \dots & c _ {n m} \end{array} \right)$$

we get:

$$A = B C$$