# The Sum of Subspaces is a Subspace of a Vector Space

## Problem 430

Let $V$ be a vector space over a field $K$.
If $W_1$ and $W_2$ are subspaces of $V$, then prove that the subset
$W_1+W_2:=\{\mathbf{x}+\mathbf{y} \mid \mathbf{x}\in W_1, \mathbf{y}\in W_2\}$ is a subspace of the vector space $V$.

## Proof.

We prove the following subspace criteria:

1. The zero vector $\mathbf{0}$ of $V$ is in $W_1+W_2$.
2. For any $\mathbf{u}, \mathbf{v}\in W_1+W_2$, we have $\mathbf{u}+\mathbf{v}\in W_1+W_2$.
3. For any $\mathbf{v}\in W_1+W_2$ and $r\in K$, we have $r\mathbf{v}\in W_1+W_2$.

Since $W_1$ and $W_2$ are subspaces of $V$, the zero vector $\mathbf{0}$ of $V$ is in both $W_1$ and $W_2$.
Thus we have
$\mathbf{0}=\mathbf{0}+\mathbf{0}\in W_1+W_2.$ So condition 1 is met.

Next, let $\mathbf{u}, \mathbf{v}\in W_1+W_2$.
Since $\mathbf{u}\in W_1+W_2$, we can write
$\mathbf{u}=\mathbf{x}+\mathbf{y}$ for some $\mathbf{x}\in W_1$ and $\mathbf{y}\in W_2$.
Similarly, we write
$\mathbf{v}=\mathbf{x}’+\mathbf{y}’$ for some $\mathbf{x}’\in W_1$ and $\mathbf{y}’\in W_2$.

Then we have
\begin{align*}
\mathbf{u}+\mathbf{v}&=(\mathbf{x}+\mathbf{y})+(\mathbf{x}’+\mathbf{y}’)\\
&=(\mathbf{x}+\mathbf{x}’)+(\mathbf{y}+\mathbf{y}’).
\end{align*}
Since $\mathbf{x}$ and $\mathbf{x}’$ are both in the vector space $W_1$, their sum $\mathbf{x}+\mathbf{x}’$ is also in $W_1$.
Similarly we have $\mathbf{y}+\mathbf{y}’\in W_2$ since $\mathbf{y}, \mathbf{y}’\in W_2$.

Thus from the expression above, we see that
$\mathbf{u}+\mathbf{v}\in W_1+W_2,$ hence condition 2 is met.

Finally, let $\mathbf{v}\in W_1+W_2$ and $r\in K$.
Then there exist $\mathbf{x}\in W_1$ and $\mathbf{y}\in W_2$ such that
$\mathbf{v}=\mathbf{x}+\mathbf{y}.$ Since $W_1$ is a subspace, it is closed under scalar multiplication. Hence we have $r\mathbf{x}\in W_1$.
Similarly, we have $r\mathbf{y}\in W_2$.

It follows from this observation that
\begin{align*}
r\mathbf{v}&=r(\mathbf{x}+\mathbf{y})\\
&=r\mathbf{x}+r\mathbf{y}\in W_1+W_2,
\end{align*}
and thus condition 3 is met.

Therefore, by the subspace criteria $W_1+W_2$ is a subspace of $V$.

## Related Question.

Let $U$ and $V$ be finite dimensional subspaces in a vector space over a scalar field $K$.
Then prove that
$\dim(U+V) \leq \dim(U)+\dim(V).$

For a proof, see the post “Dimension of the sum of two subspaces“.

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### 1 Response

1. 06/05/2017

[…] mid mathbf{x}in U, mathbf{y}in V}.] The sum $U+V$ is a subspace. (See the post “The sum of subspaces is a subspace of a vector space” for a […]

##### Idempotent Matrices are Diagonalizable

Let $A$ be an $n\times n$ idempotent matrix, that is, $A^2=A$. Then prove that $A$ is diagonalizable.

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