Sum of Subspaces is the Smallest Containing Subspace

The next result states that the sums 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 $V_1,\dots ,V_m$.