Direct Sum¶

Definition¶

Let $V_1,V_2,\ldots,V_m$ be subspaces of $V$ , and let $W=V_1+V_2+\cdots+V_m$ $V_{m}$ . If every $\boldsymbol{w}\in W$ can be written in exactly one way as

\boldsymbol{w}=\boldsymbol{v}_1+\boldsymbol{v}_2+\cdots+\boldsymbol{v}_m,

with $\boldsymbol{v}_i\in V_i$ $V_{i}$ , then $W$ is the direct sum of the subspaces, denoted

W=V_1\oplus V_2\oplus\cdots\oplus V_m.

Interpretation¶

A direct sum is a sum of subspaces in which every vector of the sum has only one decomposition: knowing $\boldsymbol{w}$ determines each piece $\boldsymbol{v}_i\in V_i$ $V_{i}$ . The subspaces do not overlap in a way that would allow a vector to be split in two different ways.

Two lines V1 and V2 through the origin in the plane, with a vector w written uniquely as the sum of v1 on V1 and v2 on V2.

In the figure, $V_1$ and $V_2$ are two different lines through the origin of $\mathbb{R}^2$ . The parallelogram rule writes $\boldsymbol{w}$ as $\boldsymbol{v}_1+\boldsymbol{v}_2$ with $\boldsymbol{v}_1$ on $V_1$ and $\boldsymbol{v}_2$ on $V_2$ , and because the two lines meet only at $\boldsymbol{0}$ , there is no other way to do it. Every vector of the plane can be split like this, so $\mathbb{R}^2=V_1\oplus V_2$ $V_{2}$ .

For two subspaces, uniqueness is equivalent to $V_1\cap V_2=\{\boldsymbol{0}\}$ ${0}$ . If a nonzero $\boldsymbol{u}$ lay in both, then $\boldsymbol{0}+\boldsymbol{0}$ and $\boldsymbol{u}+(-\boldsymbol{u})$ would be two different decompositions of $\boldsymbol{0}$ . Conversely, if $\boldsymbol{v}_1+\boldsymbol{v}_2=\boldsymbol{v}_1'+\boldsymbol{v}_2'$ $v_{2}^{'}$ , then $\boldsymbol{v}_1-\boldsymbol{v}_1'=\boldsymbol{v}_2'-\boldsymbol{v}_2$ lies in both subspaces, so it is $\boldsymbol{0}$ and the two decompositions agree. The idea is the same as in linear independence: uniqueness everywhere follows from uniqueness for $\boldsymbol{0}$ .

For three or more subspaces, checking the intersections in pairs is not enough. In $\mathbb{R}^2$ , the lines spanned by $(1,0)$ , $(0,1)$ and $(1,1)$ meet pairwise only in $\boldsymbol{0}$ , yet $(1,1)$ is both $(1,0)+(0,1)+\boldsymbol{0}$ and $\boldsymbol{0}+\boldsymbol{0}+(1,1)$ $(1, 1)$ , so their sum is not direct.

Key Results¶

For two subspaces, $V_1+V_2$ is direct exactly when $V_1\cap V_2=\{\boldsymbol{0}\}$ ${0}$ .

Examples¶

The $x$ -axis and $y$ -axis form a direct sum equal to $\mathbb{R}^2$ .
Let $V_1=\{(a,0)\colon a\in\mathbb{R}\}$ and $V_2=\{(0,b)\colon b\in\mathbb{R}\}$ $R}$ . The vector $(3,-2)$ has the decomposition $(3,0)+(0,-2)$ $(0, - 2)$ . Any decomposition $(a,0)+(0,b)=(3,-2)$ forces $a=3$ and $b=-2$ $- 2$ , so it is unique.

Prerequisites¶

Linear Dependence and Independence: uniqueness of the decomposition is the absence of a nontrivial relation between components.
Dimension: for finite-dimensional subspaces, $\dim(V_1\oplus V_2)=\dim V_1+\dim V_2$ $dim V_{2}$ .

Sources¶

Math II: Linear Algebra · Week 2

Keywords: internal direct sum, external direct sum Symbols: element of, intersection, real numbers

Last updated: 20 October 2026 03:02