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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.

Thanks.

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  • $\begingroup$ 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

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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.

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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. $\{0\}\times X_2$).

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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.

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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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