Direct Sum Proof I am reading Axler's Linear Algebra Done Right Book and I am on the part about the direct sum. He gives the following proposition:

When he assumes that $a$ and $b$ hold to prove that the proof gives the completeness part of iff, he has to prove that these make the vectors add such that the sum of vectors can only be represented uniquely. Here is his proof:
 
I understand that this means the sum is represented uniquely with two but what about with three. I feel that if it works for two it would have to work for three but I don't know why. For example $(1, 0, 0) - (1, 0, 0) - (0, 0, 0) = 0$ but $(1, 0, 0)$ does not equal $(0, 0, 0)$ or in another case $(1, 1, 0) - (1, 0, 0) - (0, 1, 0) = 0$ but none of these equal each other and they still sum to zero so that $0=(a-b-c)$; this could be extended to be like the proofs that $0 = (a_1-b_1-c_1) + (a_2-b_2-c_2) + ... + (a_n-b_n-c_n)$. What am I missing or not understanding?
Thanks,
Jackson
Also I read some of the related questions such as Proving uniqueness but in the comments it assumed that the set was only of one or two elements.
 A: Sometimes I think the art of teaching is 90% mind reading. The fact is, I'm not well versed in these subtle and powerful techniques to read your mind over the internet, so unfortunately, I can't tell you with any certainty exactly what you've gotten wrong, or what you're missing.
You seem to have slightly misread, or maybe misunderstood, or perhaps you've let your intuition take you further than the text. It would appear that you're familiar with the $n = 2$ case. Let's run through that quickly. Fix any $w \in V$ and suppose $w = u_1 + u_2 = v_1 + v_2$, where $u_i, v_i \in U_i$ for $i = 1, 2$. Then $(u_1 - v_1) + (u_2 - v_2) = w - w = 0$. Since $u_i - v_i \in U_i$ for $i = 1, 2$, it follows, from property $(b)$, that $u_i - v_i = 0$, i.e. $u_i = v_i$, hence the representation of $w$ as a sum of vectors from $U_1$ and $U_2$ is unique. Therefore, the sum is direct.
So, what I think you might have done is incorrectly generalised this in your head. The argument on paper notwithstanding, you've noticed, in the crucial part of the argument, the two subtractions of two vectors and generalised it to two subtractions of $n$ vectors. The argument actually involves $n$ subtractions of two vectors.
At this point, I'd probably read the argument again. It's basically the same as the $2$ case. There's not a lot more to it. You get a sum of $(u_1 - v_1) + (u_2 - v_2) + \ldots + (u_n - v_n)$, rather than stopping at $2$. Because of property (b) (which deals with such sums of vectors, each from a different $U_i$, that equal $0$), each $u_i - v_i = 0$, so $u_i = v_i$ and the sum is unique.
Hope that helps!
