Direct Sums of Subspaces

Suppose $V_1,\dots ,V_m$ are subspaces of $V$. Every element of the sums of subspaces $V_1+\cdots +V_m$ can be written in the form

$$v_1+\cdots +v_m$$

where each element $v_k\in V_k$.

The sum $V_1+\cdots +V_m$ is called the direct sum if each element on $V_1+\cdots +V_m$ can be written in only one way as a sum $v_1+\cdots +v_m$, where each $v_k\in V_k$.

If $V_1+\cdots +V_m$ is a direct sum, then $V_1\oplus \cdots \oplus V_m$ denotes $V_1+\cdots +V_m$, with the $\oplus$ notation serving as an indication that this is a direct sum.

Sums of subspaces are analogous to unions of subsets. Similarly, direct sums of subspaces are analogous to disjoint unions of subsets. No two subspaces of a vector space can be disjoint, because both contain $0$. So disjointness is replaced, at least in the case of two subspaces, with the requirement that the intersection equal $\{0\}$.

Basic Examples

Example: Direct sum of two subspaces

Suppose $U$ is the subspace of ${\mathbb{F}}^3$ of vectors whose last coordinate equals $0$, and $W$ is the subspace of ${\mathbb{F}}^3$ of vectors whose first two coordinates equal $0$:

$$\begin{array}{rl}U& =\{(x,y,0)\in {\mathbb{F}}^3:x,y\in \mathbb{F}\}\\ W& =\{(0,0,z)\in {\mathbb{F}}^3:z\in \mathbb{F}\}\end{array}$$

Then, ${\mathbb{F}}^3=U\oplus W$.

Example: A non-direct sum

Suppose

$$\begin{array}{rl}{V}_1& =\{(x,y,0)\in {\mathbb{F}}^3:x,y\in \mathbb{F}\},\\ V_2& =\{(0,0,z)\in {\mathbb{F}}^3:z\in \mathbb{F}\},\\ V_3& =\{(0,y,y)\in {\mathbb{F}}^3:y\in \mathbb{F}\}\end{array}$$

Then ${\mathbb{F}}^3=V_1+V_2+V_3$ because every vector $(x,y,z)\in {\mathbb{F}}^3$ can be written as

$$(x,y,z)=(x,y,0)+(0,0,z)+(0,0,0)$$

where the first vector on the right side is in $V_1$, the second vector is in $V_2$, and the third vector is in $V_3$.

However, ${\mathbb{F}}^3$ is not a direct sum of $V_1,V_2,V_3$, because the vector $(0,0,0)$ can be written in more than one way as a sum $v_1+v_2+v_3$, with each $v_k\in V_k$. Specifically, we have:

$$(0,0,0)=(0,1,0)+(0,0,1)+(0,-1,-1)$$

and

$$(0,0,0)=(0,0,0)+(0,0,0)+(0,0,0)$$

where the first vector is in $V_1$, the second vector is in $V_2$, and the third vector is in $V_3$.

Thus, the sum $V_1+V_2+V_3$ is not a direct sum.

Condition for Direct Sums

The definition of direct sum requires every vector in the sum to have a unique representation as an appropriate sum. The next result shows that when deciding whether a sum of subspaces is a direct sum, we only need to consider whether $0$ can be uniquely written as an appropriate sum.

Condition for Direct Sums

Suppose $V_1,\dots ,V_m$ are subspaces of $V$. Then, $V_1+\cdots +V_m$ is a direct sum if and only if the only way to write $0$ as a sum $v_1+\cdots +v_m$, where each $v_k\in V_k$, is by taking each $v_k$ equal to $0$.

Proof.

First, we show necessity: assuming that $V_1+\cdots +V_m$ is a direct sum, the only unique way to write the zero vector is if each $v_k=0$.

Suppose $V_1+\cdots +V_m$ is a direct sum. All of the subspaces $V_1,\dots ,V_m$ are subspaces of $V$, which means all of these subspaces share the same additive identity $0$. Clearly, adding all of these additive identities $0_1+\cdots +0_m$, where each $0_k\in V_k$, is equal to $0$. Then, the definition of direct sum implies that this is the only way to write $0$, since this $0$ must unique.

Thus, the definition of direct sum implies that the only way to write $0$ as a sum $v_1+\cdots +v_m$, where each $v_k\in V_k$, is by taking each $v_k$ equal to $0$.

Second, we show sufficiency: assuming that the only way to express the zero vectors is by having all $v_k=0$, this implies that $V_1+\cdots +V_m$ is a direct sum.

Suppose that the only way to write $0$ as a sum $v_1+\cdots +v_k$, where $v_k\in V_k$, is by taking each $v_k$ equal to $0$.

To show that $V_1+\cdots +V_m$ is a direct sum, let some $v\in V_1+\cdots +V_k$. We can write

$$v=v_1+\cdots +v_m$$

for some $v_1\in V_1,\dots ,v_m\in V_m$. To show that this representation is unique, suppose we also have

$$v=u_1+\cdots +u_m,$$

where $u_1\in V_1,\dots ,u_m\in V_m$. Subtracting these two equations, we have:

$$0=(v_1-u_1)+\cdots +(v_m-u_m).$$

Because $v_1-u_1\in V_1,\dots ,v_m-u_m\in V_m$, the equation above implies that each $v_k-u_k$ equals $0$. This shows that any two representations of the same vector must be equal, so every vector actually only has one unique representation.

Direct Sum of Two Subspaces

The next result gives a simple condition for testing whether a sum of two subspaces is a direct sum.

Direct sum of two subspaces

Suppose $U$ and $W$ are subspaces of $V$. Then

$$U+W\text{ is a direct sum}⟺U\cap V=\{0\}$$

The symbol $⟺$ means “if and only if ”; this symbol could also be read to mean “is equivalent to”.

Proof.

First, we show that if we assume $U+W$ is a direct sum, we must have $U\cap W=\{0\}$.

Suppose $U+W$ is a direct sum. Suppose some element $v$ is in both $U$ and $V$, or $v\in U\cap W$. Then, we can write the zero vector as $0=v+(-v)$, where $v\in U$ and $-v\in W$. Since $U+W$ is a direct sum, the only way for us express $0$ as a sum of vectors from $U$ and $W$ is if both vectors are $0$. Thus, we have $v=0$. Thus, $U\cap W=\{0\}$, completing the proof in one direction.

Now we want to prove the other direction; assuming that $U\cap W=\{0\}$, we want to show that $U+W$ is a direct sum.

Suppose $u\in U,w\in W$, and

$$0=u+w.$$

To complete the proof, we only need to show that $u=w=0$, which we showed above for the condition for direct sums. The equation above implies that $u=-w$. Since we are dealing with subspaces that are closed under scalar multiplication, $u=-w\in W$ means that $u\in W$ as well, which in turn means $u\in U\cap W$. Since our beginning assumption was $U\in W=\{0\}$, we must have $u=0$, which by the equation above implies that $w=0$, completing the proof.

The result above deals only with the case of two subspaces. When asking about a possible direct sum with more than two subspaces, it is not enough to test that each pair of the subspaces intersect only at $\{0\}$. To see this, consider the the non-direct sum example above. In that non-example of a direct sum, we have $V_1\cap V_2=V_1\cap V_3=V_2\cap V_3=\{0\}$.