Space is direct sum of subspaces - propostion conditions giving me problems In Sheldon Axler's "Linear Algebra Done Right" - 2$^{\textrm{nd}}$ Edition, on the section for Direct Sums, the following proposition is stated.



Following this is the proof of this 'if and only if' proposition - first in the forward direction ($V=U_1 \oplus \cdots \oplus U_n \,\implies \,$ (a) & (b) ). This is the part I am having trouble understanding.



Here is my problem: To prove (b), the author seemed to simply use (b), thus proving it using itself, which is a tautology! How can the author say "Then each $u_j$ must be 0, and then justify that with the "uniqueness part of the definition", which is exactly what the author is trying to prove? Have I misunderstood something?
 A: By the definition of the direct sum, there is a unique way to write $0$ as the sum $u_1+\dots +u_n$, where $u_i \in U_i$. 
But at the same time, we know that $0 \in U_i$ $\forall i$, and $0+\dots+0=0$. 
If there was also a set of nonzero (or better not all zero) vectors $u_i$, $i=1, \dots, n$, such that $u_1+\dots +u_n = 0$ it would violate uniqueness.
This proves the "only if" part for (b), which indeed follows easily from the definition. 
Recall, however, that the definition of the direct sum requires that all vectors in $V$ are written uniquely as a sum $u_1+\dots +u_n$, $u_i \in U_i$. 
The proposition essentially states that it suffices to only check that this property holds for $0$. 
If it holds for $0$ then it will hold for all points in $V$. In other words, the value of the proposition comes in the "if" part; not the "only if" that easily followed from the definition.
A: The author defines the direct sum here:



Note that he says that in a direct sum (by definition), "each vector in $V$ can be uniquely represented in the form above".  It is from this that we are able to prove (b).
