# Direct sum and basis for a vector space

I was reading Axler "Linear Algebra Done Right" and he defines the concept of direct sum as following:

Suppose $U_{1}, U_{2}, \dots , U_{m}$ are subspaces of a vector space $V.$ The sum $U_{1} + U_{2} + \dots + U_{m}$ is called the direct sum if each element of $U_{1} + U_{2} + \dots + U_{m}$ can be written in only one way as a sum $u_{1} + u_{2} + \dots + u_{m}$ where each $u_{i}$ is in $U_{i}.$ So if you pick $m$ vectors $u_{i}$ each in $U_{i},$ then those vectors form a basis for $V,$ is it true to conclude that way?

• It sounds like you're just considering the case where $U_i$ is one-dimensional and $\oplus U_i=V$. Dec 28, 2015 at 21:33
• So you mean the resulting $m$ vectors may not spans $V$ and hence may not be a basis for $V?$
– user298251
Dec 28, 2015 at 21:49
• To take a dramatic example: say you pick the zero vector every time (it belongs to each $U_i$ after all). Then you get only the set $\{0\}$, which neither spans any nontrivial vector space nor is ever linearly independent. Dec 28, 2015 at 21:55

## 2 Answers

The answer to your question is no. For instance, if any of the $u_i = 0$, then $(u_i)$ isn't linearly independent and so it cannot form a basis for $V$.

However, if $V = \bigoplus_i U_i$ and you pick a basis for each $U_i$, then the union of these bases is going to be a basis for $V$.

If for each $1 \leq i \leq m$ you choose a basis $\mathcal{B}_i = (u^i_1, \dotsc, u^i_{n_i})$ of $U_i$ then you get that $$\mathcal{B} = (u^1_1, \dotsc, u^1_{n_1}, u^2_1, \dotsc, u^2_{n_2}, \dotsc, u^m_1, \dotsc, u^m_{n_m})$$ is a basis of $V$.

It is a generating set because each $v \in V$ can be written as $v = \sum_{i=1}^m u_i$ with $u_i \in U_i$ and then each $u_i$ can be written as a linear combination $u_i = \sum_{j=1}^{n_i} \lambda^i_j u^i_j$, resulting in $$v = \sum_{i=1}^m u_i = \sum_{i=1}^m \sum_{j=1}^{n_i} \lambda^i_j u^i_j.$$

If on the other hand $$0 = \sum_{i=1}^m \underbrace{\sum_{j=1}^{n_i} \lambda^i_j u^i_j}_{\in U_i}$$ then by the uniqueness it follows that $\sum_{j=1}^{n_i} \lambda^i_j u^i_j = 0$ for each $i$, so by the linear independence of $\mathcal{B}_i$ it follows that $\lambda^i_j = 0$ for all $i,j$. So $\mathcal{B}$ is also linear independent.