# Rigorous Explanation of Internal Direct Sum Criterion

In "Sheldon Axler's Linear Algebra Done Right, 3rd edition", on page 21 "internal direct sum", or direct sum as the author uses, is defined as such:

Let $$U_1, U_1,...,U_m$$ be subspaces of $$V$$ vector space over field $$F$$, if any $$v\in V$$ can be written uniquely as a sum of $$u_1+u_2+...+u_m$$ where each $$u_i\in U_i$$, then the sum ($$U_1+U_2+...+U_m$$) is internal direct sum. [1]

On page 23 following this definition, a condition to check whether or not a sum of subspaces is internal direct is stated as follows:

A direct sum is internal direct sum if and only if the only way to write $$0$$ vector as the sum of $$u_1+u_2+...+u_m$$ where each $$u_i\in U_i$$ is equal to $$0$$. [2]

Question (1): How do we infer [2] from [1]?

Question (2): Why does checking whether or not $$0$$ vector can be uniquely written as the sum of $$u_1+u_2+...+u_m$$ where each $$u_i=0$$ suffice to say that the sum is internal direct sum? In other words, why is it not possible to be able to write $$0$$ vector uniquely while other vectors $$v\neq 0$$ are not writable uniquely, or vice versa?

I checked out many books about Linear Algebra as to a rigorous explanation for that, but I could not find anything satisfactory. Being these said, what I look for is some eloquent explanation to all these questions I stated above.

• Your question 1 and question 2 are actually the same question. Just so you know. May 23 '19 at 10:50

The core idea is that if you have two different ways to decompose the same vector $$u_1 + \cdots + u_m = v_1 + \cdots + v_m$$ (where $$u_i, v_i\in U_i$$) then the zero vector can be decomposed as $$0 = (u_1 - v_1) + (u_2 -v_2) + \cdots + (u_m - v_m)$$ and since the two original decompositions were different, not all these terms can cancel, and we have found a second way to decompose the $$0$$ vector.
So if there is some vector which can be decomposed in two different ways, then the $$0$$ vector can be decomposed in two different ways. Contrapositively, if the $$0$$ vector can only be decomposed as $$0 = u_1 +\cdots + u_m$$ with $$u_1 = \cdots = u_m = 0$$, then any other vector also only has one deomposition.