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.

Let $A$ and $B$ be $m\times n$ matrices.
Prove that
\[\rk(A+B) \leq \rk(A)+\rk(B).\]

Prove a Given Subset is a Subspace and Find a Basis and Dimension
Let
\[A=\begin{bmatrix}
4 & 1\\
3& 2
\end{bmatrix}\]
and consider the following subset $V$ of the 2-dimensional vector space $\R^2$.
\[V=\{\mathbf{x}\in \R^2 \mid A\mathbf{x}=5\mathbf{x}\}.\]
(a) Prove that the subset $V$ is a subspace of $\R^2$.
(b) Find a basis for […]

The Subset Consisting of the Zero Vector is a Subspace and its Dimension is Zero
Let $V$ be a subset of the vector space $\R^n$ consisting only of the zero vector of $\R^n$. Namely $V=\{\mathbf{0}\}$.
Then prove that $V$ is a subspace of $\R^n$.
Proof.
To prove that $V=\{\mathbf{0}\}$ is a subspace of $\R^n$, we check the following subspace […]

Linear Transformation and a Basis of the Vector Space $\R^3$
Let $T$ be a linear transformation from the vector space $\R^3$ to $\R^3$.
Suppose that $k=3$ is the smallest positive integer such that $T^k=\mathbf{0}$ (the zero linear transformation) and suppose that we have $\mathbf{x}\in \R^3$ such that $T^2\mathbf{x}\neq \mathbf{0}$.
Show […]

Show the Subset of the Vector Space of Polynomials is a Subspace and Find its Basis
Let $P_3$ be the vector space over $\R$ of all degree three or less polynomial with real number coefficient.
Let $W$ be the following subset of $P_3$.
\[W=\{p(x) \in P_3 \mid p'(-1)=0 \text{ and } p^{\prime\prime}(1)=0\}.\]
Here $p'(x)$ is the first derivative of $p(x)$ and […]

True or False. The Intersection of Bases is a Basis of the Intersection of Subspaces
Determine whether the following is true or false. If it is true, then give a proof. If it is false, then give a counterexample.
Let $W_1$ and $W_2$ be subspaces of the vector space $\R^n$.
If $B_1$ and $B_2$ are bases for $W_1$ and $W_2$, respectively, then $B_1\cap B_2$ is a […]

Dual Vector Space and Dual Basis, Some Equality
Let $V$ be a finite dimensional vector space over a field $k$ and let $V^*=\Hom(V, k)$ be the dual vector space of $V$.
Let $\{v_i\}_{i=1}^n$ be a basis of $V$ and let $\{v^i\}_{i=1}^n$ be the dual basis of $V^*$. Then prove that
\[x=\sum_{i=1}^nv^i(x)v_i\]
for any vector $x\in […]

[…] Here we used the fact that for two subspaces $U$ and $V$ of a vector space, we have [dim(U+V)leq dim(U)+dim(V).] For a proof of this fact, see the post “Dimension of the sum of two subspaces“. […]

## 2 Responses

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

[…] Here we used the fact that for two subspaces $U$ and $V$ of a vector space, we have [dim(U+V)leq dim(U)+dim(V).] For a proof of this fact, see the post “Dimension of the sum of two subspaces“. […]