Vector space/Product/Dimension/Exercise/Solution

The product space has dimension $n + m$ . To proof this let $v_{1}, \dots, v_{n}$ be a basis of $V$ and let $w_{1}, \dots, w_{m}$ be a basis of $W$ . We claim that the elements

(v_{j}, 0), j \in {1, \dots, n}, and (0, w_{i}), i \in {1, \dots, m},

form a basis of $V \times W$ .

Let $(v, w) \in V \times W$ . Then there are representations

v = \sum_{j = 1}^{n} a_{j} v_{j} and w = \sum_{i = 1}^{m} b_{i} w_{i} .

Therefore

\begin{matrix} (v, w) & = (\sum_{j = 1}^{n} a_{j} v_{j}, \sum_{i = 1}^{m} b_{i} w_{i}) \\ = (\sum_{j = 1}^{n} a_{j} v_{j}, 0) + (0, \sum_{i = 1}^{m} b_{i} w_{i}) \\ = \sum_{j = 1}^{n} a_{j} (v_{j}, 0) + \sum_{i = 1}^{m} b_{i} (0, w_{i}), \end{matrix}

which means that we have a generating system.

To show linear independence, suppose that

\sum_{j = 1}^{n} a_{j} (v_{j}, 0) + \sum_{i = 1}^{m} b_{i} (0, w_{i}) = 0 = (0, 0) .

The same computation shows

(\sum_{j = 1}^{n} a_{j} v_{j}, \sum_{i = 1}^{m} b_{i} w_{i}) = (0, 0)

and this means

\sum_{j = 1}^{n} a_{j} v_{j} = 0 and \sum_{i = 1}^{m} b_{i} w_{i} = 0.

Since each family forms a basis, we conclude $a_{j} = 0$ for all $j$ and $b_{i} = 0$ for all $i$ .

To the exercise