Sums of Subspaces

Suppose $V_1,\dots ,V_m$ are subspaces of $V$. The sum of $V_1,\dots ,V_m$, denoted by $V_1+\cdots +V_m$, is the set of all possible sums of elements from $V_1,\dots ,V_m$. More precisely:

$$V_1+\cdots +V_m=\{v_1+\cdots +v_m:v_1\in V_1,\dots ,v_m\in V_m\}$$

Sums of subspaces in the theory of vector spaces are analogous to unions of subsets in set theory. Given two subspaces of a vector space, the smallest subspace containing them is their sum. Analogously, given two subsets of a set, the smallest subset containing them is their union.

Basic Examples

Let’s look at some examples of sums of subspaces.

Example: Sum of axis subspaces in ${\mathbb{R}}^2$.

Suppose $U,V$ are subspaces of ${\mathbb{R}}^2$, $U$ is the subspace of all vectors on the $x$-axis, $U=\{(x,0):x\in \mathbb{R}\}$ and $V$ is the subspace of all vectors on the $y$-axis, $W=\{(0,y):y\in \mathbb{R}\}$. This would mean $U$ contains vectors like $(1,0),(-2,0),(3,0)$, and $W$ contains vectors like $(0,1),(0,-2),(0,3)$.

Then, the sum $U+W$ is the set of all possible sum combinations of a vector from $U$ and a vector from $W$. In other words, for any $(x,0)\in U$ and $(0,y)\in W$, where $x,y$ can be any real numbers, their sum is

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

This means that every vector in $U+W$ has the form $(x,y)$, where $x,y\in \mathbb{R}$. Thus, we actually have

$$U+W={\mathbb{R}}^2$$

Example: Sum of subspaces for ${\mathbb{F}}^3$

Suppose $U$ is the set of all elements of ${\mathbb{F}}^3$ whose second and third coordinates equal $0$, and $W$ is the set of all elements of ${\mathbb{F}}^3$ whose first and third coordinates equal $0$:

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

Then

$$U+W=\{(x,y,0)\in {\mathbb{F}}^3:x,y\in \mathbb{F}\}$$

Example: Sum of subspaces of ${\mathbb{F}}^4$

Suppose

$$\begin{array}{r}U=\{(x,x,y,y)\in {\mathbb{F}}^4:x,y\in \mathbb{F}\}\\ W=\{(x,x,x,y)\in {\mathbb{F}}^4:x,y\in \mathbb{F}\}\end{array}$$

We could say that:

• $U$ is the set of elements of ${\mathbb{F}}^4$ whose first two coordinates equal each other and whose third and fourth coordinates equal each other.
• $W$ is the set of elements of ${\mathbb{F}}^4$ whose first three coordinates equal each other.

To find a description of $U+W$, consider a typical element $(a,a,b,b)$ of $U$ and a typical element $(c,c,c,d)$ of $W$, where $a,b,c,d\in \mathbb{F}$. We have

$$(a,a,b,b)+(c,c,c,d)=(a+c,a+c,b+c,b+d)$$

which shows that every element of $U+W$ has its first two coordinates equal to each other. Thus

$$U+W\subseteq \{(x,x,y,z)\in {\mathbb{F}}^4:x,y,z\in \mathbb{F}\}$$

where the $\subseteq$ is the symbol for “is a subset of”.

To prove the inclusion in the other direction, suppose $x,y,z\in \mathbb{F}$. Then

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

where the first vector on the right is in $U$ and the second vector on the right is in $W$. Thus, $(x,x,y,z)\in U+W$, showing that the inclusion holds in the other direction:

$$\{(x,x,y,z)\in {\mathbb{F}}^4:x,y,z\in \mathbb{F}\}\subseteq U+W$$

Hence, we have:

$$U+W=\{(x,x,y,z)\in {\mathbb{F}}^4:x,y,z\in \mathbb{F}\}$$

which shows that $U+W$ is the set of elements of ${\mathbb{F}}^4$ whose first two coordinates equal each other.

Sum of subspaces is the smallest containing subspace

The next result states that the sum of subspaces is a subspace, and is in fact the smallest subspace containing all the summands (which means that every subspace containing all the summands also contains the sum).

Sum of subspaces if the smallest containing subspace

Suppose $V_1,\dots ,V_m$ are subspaces of $V$. Then $V_1+\cdots +V_m$ is the smallest subspace of $V$ containing $V_1,\dots ,V_m$.

Proof.

First, we want to show that $V_1+\cdots +V_m$ contains the additive identity $0$ and is closed under addition and scalar multiplication. This implies that $V_1+\cdots +V_m$ is a subspace of $V$.

Additive Identity/Zero Element: A subspace must contain the additive identity $0$, and since each $V_i$ for $i=1,\dots ,m$ is a subspace, they each contain the zero vector. Therefore, the sum $V_1+\cdots +V_m$ also contains the zero vector because you can write $0=0+\cdots +0$ as a sum of elements from the subspaces.

Closed under addition: If we take two elements from $V_1+\cdots +V_m$, say $v_1+\cdots +v_m$ and $w_1+\cdots +w_m$, where $v_i\in V_i$ and $w_i\in V_i$ for each $i$, then their sum is:

$$(v_1+\cdots +v_m)+(w_1+\cdots +w_m)=(v_1+w_1)+\cdots +(v_m+w_m)$$

Since each $V_i$ is closed under addition, $v_i+w_i\in V_i$ for each $i$. Thus, the sum is still in $V_1+\cdots +V_m$, which proves that $V_1+\cdots +V_m$ is closed under addition.

Closed under scalar multiplication: If you take an element $(v_1+\cdots +v_m)\in (V_1+\cdots +V_m)$ and a scalar $\alpha \in \mathbb{F}$, then:

$$\alpha (v_1+\cdots +v_m)=\alpha v_1+\cdots +\alpha v_m$$

Since each $V_i$ is closed under scalar multiplication, $a{v}_i\in V_i$ for each $i$. Therefore, $\alpha (v_1+\cdots +v_m)\in (V_1+\cdots +V_m)$, showing closure under scalar multiplication.

Thus, we have shown that $V_1+\cdots +V_m$ is a subspace of $V$.

Second, we want to argue that the subspaces $V_1,\dots ,V_m$ are all contained in $V_1+\cdots +V_m$. To see this, we can note that an element of $V_i$ can be written as $v_i+0+\cdots +0$ where all other components are zero. Hence, all $V_1,\dots ,V_m$ are contained within $V_1+\cdots +V_m$.

Every subspace of $V$ containing $V_1,\dots ,V_m$ contains $V_1+\cdots +V_m$ (because subspaces must contain all finite sums of their elements). Thus $V_1+\dots V_m$ is the smallest subspace of $V$ containing $V_1,\dots ,V_m$.

Lastly, we want to show that $V_1+\cdots +V_m$ is the smallest subspace that contains all $V_1,\dots ,V_m$. Suppose there is a subspace $W\subseteq V$ that contains all $V_1,\dots ,V_m$. Since $W$ contains all the individual subspaces, it must also contain all infinite sums of elements from those subspaces:

$$v_1+v_2+\cdots +v_m\in W$$

But this is precisely what $V_1+\cdots +V_m$ represents: the set of all infinite sums of elements from the subspaces $V_1,\dots ,V_m$. Thus, any subspace that contains $V_1,\dots ,V_m$ must also contain $V_1+\cdots +V_m$. This makes $V_1+\cdots +V_m$ the smallest subspace that contains all of $V_1,\dots ,V_m$.