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.

  • $\begingroup$ Your question 1 and question 2 are actually the same question. Just so you know. $\endgroup$
    – Arthur
    May 23, 2019 at 10:50

1 Answer 1


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.


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