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

Definition (The sum of subspaces).

Recall that the sum of subspaces $U$ and $V$ is
\[U+V=\{\mathbf{x}+\mathbf{y} \mid \mathbf{x}\in U, \mathbf{y}\in V\}.\] The sum $U+V$ is a subspace.
(See the post “The sum of subspaces is a subspace of a vector space” for a proof.)

Proof.

Let $n=\dim(U)$ and $m=\dim(V)$.
Let
\[B_1=\{\mathbf{u}_1, \dots, \mathbf{u}_n\}\] be a basis of the vector space $U$ and let
\[B_2=\{\mathbf{v}_1, \dots, \mathbf{v}_m\}\] be a basis of the vector space $V$.

An arbitrary element of the vector space $U+W$ is of the form $\mathbf{x}+\mathbf{y}$, where $\mathbf{x}\in U$ and $\mathbf{y} \in V$.

Since $B_1$ is a basis of $U$, we can write
\[\mathbf{x}=r_1\mathbf{u}_1+\cdots +r_n \mathbf{u}_n\] for some scalars $r_1, \dots, r_n\in K$.
Also, since $B_2$ is a basis of $V$, we can write
\[\mathbf{y}=s_1\mathbf{v}_1+\cdots +s_m \mathbf{v}_m\] for some scalars $s_1, \dots, s_m\in K$.

Thus, we have
\begin{align*}
\mathbf{x}+\mathbf{y}&=r_1\mathbf{u}_1+\cdots +r_n \mathbf{u}_n+s_1\mathbf{v}_1+\cdots +s_m \mathbf{v}_m,
\end{align*}
and hence $\mathbf{x}+\mathbf{y}$ is in the span $S:=\Span(\mathbf{u}_1, \dots, \mathbf{u}_n, \mathbf{v}_1, \dots, \mathbf{v}_m)$.

Thus we have $U+W \subset S$ and it yields that \begin{align*}
\dim(U+W) \leq \dim(S)\leq n+m=\dim(U)+\dim(V).
\end{align*}
This completes the proof.

Related Question.

See the post ↴
The rank of the sum of two matrices
for a proof of this problem.

The post Dimension of the Sum of Two Subspaces first appeared on Problems in Mathematics.

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.

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

The post The Sum of Subspaces is a Subspace of a Vector Space first appeared on Problems in Mathematics.