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?
Best Answer
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 $V$.