0
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ It sounds like you're just considering the case where $U_i$ is one-dimensional and $\oplus U_i=V$. $\endgroup$ Dec 28, 2015 at 21:33
  • $\begingroup$ So you mean the resulting $ m $ vectors may not spans $ V $ and hence may not be a basis for $ V? $ $\endgroup$
    – user298251
    Dec 28, 2015 at 21:49
  • $\begingroup$ 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. $\endgroup$ Dec 28, 2015 at 21:55

2 Answers 2

2
$\begingroup$

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 $V$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy