If \(V\) is a finite dimensional vector space and \[V^m \coloneqq V \oplus V \oplus \cdots \oplus V \quad \text{($m$ summands)},\] then for each \(m\geq 1\), \(V^m\) is finite dimensional and \(\dim V^m = m(\dim V)\).
Solution
Let \(e_1,\ldots,e_n\) be a basis of \(V\), \(n=\dim V\), and for \(1\le j\le m\) let \(\iota_j\colon V\to V^m\) be the inclusion of the \(j\)th summand.
1. The \(mn\) vectors \(\iota_j(e_i)\), \(1\le i\le n\), \(1\le j\le m\), form a basis of \(V^m\).
An element \((v_1,\ldots,v_m)\in V^m\) equals \(\sum_j\iota_j(v_j)\), and writing \(v_j=\sum_i c_{ij}e_i\) gives \((v_1,\ldots,v_m)=\sum_{i,j}c_{ij}\iota_j(e_i)\), so the vectors span. If \(\sum_{i,j}c_{ij}\iota_j(e_i)=0\), then the \(j\)th coordinate \(\sum_i c_{ij}e_i\) vanishes for each \(j\), so every \(c_{ij}=0\) by independence of \(e_1,\ldots,e_n\).
2. Q.E.D.
By step 1, \(V^m\) has a finite basis of \(mn\) elements, so it is finite dimensional and \(\dim V^m=m\dim V\).