Let \(V\) be an \(n\)-dimensional vector space with basis \(\mathcal B=\{b_1,\ldots,b_n\}\). Construct a basis of \(V^{\oplus m}\) and compute its dimension.
Solution
For \(1\le i\le m\) and \(1\le k\le n\), let \[ e_{i,k}=(0,\ldots,0,b_k,0,\ldots,0)\in V^{\oplus m}, \] where \(b_k\) occurs in the \(i\)th summand. Set \[ \mathcal B^{(m)}=\{e_{i,k}:1\le i\le m,\ 1\le k\le n\}. \]
To see that it spans, let \(v=(v_1,\ldots,v_m)\). Write \[ v_i=\sum_{k=1}^n a_{i,k}b_k. \] Then \[ v=\sum_{i=1}^m\sum_{k=1}^n a_{i,k}e_{i,k}. \]
For linear independence, suppose \[ \sum_{i,k}a_{i,k}e_{i,k}=0. \] Looking in the \(i\)th summand gives \[ \sum_{k=1}^n a_{i,k}b_k=0. \] Since \(\mathcal B\) is a basis, every \(a_{i,k}=0\).
Hence \(\mathcal B^{(m)}\) is a basis and \[ \dim(V^{\oplus m})=mn=m\dim V. \]