Can someone clarify for me the side note found in Axler's Linear Algebra Done Right, that is: Direct sums of subspaces are analogous to disjoint union of subsets.

I am not sure what exactly is a disjoint union of subsets (having checked online definitions) and thus its relation to direct sums.


  • $\begingroup$ Also, shouldn't the (external) direct sum of vector spaces have as basis the union of bases for those spaces, and this union will have been disjoint? $\endgroup$ Jul 27, 2018 at 9:01

4 Answers 4


A set $S$ is the ${disjoint}$ sum of sets $A$ and $B$ if the following is true.

  • $S = A \cup B$ and
  • $A$ and $B$ have no elements in common, i.e. $A \cap B = \emptyset$.

Now let $V$ be a vector-space over a field $\mathbb{F}$. We want to say that $V$ is the 'disjoint' sum of two subspaces $U, W$ of $V$. We cannot say that $U$ and $W$ are disjoint as sets because any subspace of $V$ contains $0$. What we can do is to minimise their intersection: other than the 0-vector, $U$ and $W$ should not have any common members. Moreover, to get all of $V$, we must require that every vector $v$ in $V$ can somehow be constructed from elements of $U$ and $W$. This gives rise to the following definition of direct sum:

  • $S = U + W$ and
  • $U$ and $W$ have no non-zero elements in common, i.e. $U \cap W = \{0\}$.

Clearly this definition of direct sum in a vector space is almost the same as that of disjoint sum of sets. The only thing that remains in need of explanation is the sum of subspaces $U + W$. But that's simple:

$$U + W = \{u + w\ |\ u \in U, w \in W\}$$

In other words, we are lifting vector addition from vectors to sets of vectors.

As an example, the vector space $\mathbb{R}^2$, i.e. the euclidean plane, is the direct sum of the x-axis and the y-axis. In other words $\mathbb{R}^2$ is the direct sum of the subspace $\{ (x, 0)\ |\ x \in \mathbb{R}\}$ and $\{ (0, y)\ |\ y \in \mathbb{R}\}$.

It turns out that the formal similarity between these two constructions is not accidental. Both are instances of a more general construction called co-product or categorical sum which is defined for categories.


This answer is unlikely to be useful to the OP, but contains things which should probably be said here in case this question is referred to later by users with a different background.

The disjoint union of sets and the direct sum of vector spaces are both examples of a coproduct in a category. Given two objects $X_1$ and $X_2$ in a category $\mathcal{C}$, the coproduct $X=X_1\coprod X_2$ is an object with maps $i_j\colon X_j\to X$ such that for any object $Y$ of $\mathcal{C}$ and any maps $f_j\colon X_j\to Y$, there is a unique map $f\colon X\to Y$ such that $f_j=f\circ i_j$ (composing right to left).

In the category of sets, the maps $i_j$ are the inclusions of $X_j$ into $X$, and in the category of vector spaces, the map $i_1$ (resp. $i_2$) is the embedding of $X_1$ (resp. $X_2$) into $X$ as $X_1\times\{0\}$ (resp. $\{0\}\times X_2$).


The idea is that for vector spaces $W$ and $V$, $W\oplus V$ joins the two vector spaces without blending their contents.

$W\oplus V$ contains a copy of $W$ ($W\oplus \{0\}$) and a copy of $V$ ($\{0\}\oplus V$) and the intersection of the two copies is as small as possible ($\{0\}\oplus \{0\}$).

The intersection of two subspaces can't get any smaller, so this is the closest thing to their intersection being empty as we can get.

So, $W$ and $V$ coexist in $W\oplus V$, but they "don't intersect" in the sense that their intersection is trivial. That's the resemblance to the union of disjoint sets.


For finite subsets of some set $U$, one has $$ \#(A_1\cup A_2\cup\cdots\cup A_n)\leq\#A_1+\#A_2+\cdots+\#A_n, $$ with equality if and only if the sets $A_1,A_2,\ldots,A_n$ are all disjoint,which is a condition one can test by considering the $A_i$ pairwise (i.e., one may test that $\#(A_i\cap A_j)=0$ whenever $i\neq j$). For finite dimensional subspaces of some vector space $V$ one has$$ \dim(V_1+V_2+\cdots+V_n)\leq\dim V_1+\dim V_2+\cdots+\dim V_n, $$ with equality if and only if the subspaces $V_1,V_2,\ldots,V_n$ form a direct sum $V_1\oplus V_2\oplus\cdots\oplus V_n$ (this is just notation to indicate the sum is direct); this is a condition one cannot test by considering the $V_i$ pairwise (i.e., it does not suffice to test that $\dim(V_i\cap V_j)=0$ whenever $i\neq j$).

So direct sums have something similar to disjoint unions, but not everything.


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