Let $X$ and $Y$ be subspaces of a vector space $V$.
$X\cap Y\u2a7dV$.
$X+Y=\{\mathbf{x}+\mathbf{y}:\mathbf{x}\in X,\mathbf{y}\in Y\}\u2a7dV$.
To show something is a subspace we have to check the three properties: containing the zero vector, closure under addition, and closure under scalar multiplication.
${\mathrm{\U0001d7ce}}_{V}\in X\cap Y$ as $X$ and $Y$ are subspaces so contain ${\mathrm{\U0001d7ce}}_{V}$.
Let $\mathbf{x},\mathbf{y}\in X\cap Y$. $X$ is a subspace, so closed under addition, so $\mathbf{x}+\mathbf{y}\in X$. For the same reason $\mathbf{x}+\mathbf{y}\in Y$. Therefore $\mathbf{x}+\mathbf{y}\in X\cap Y$.
Let $\lambda $ be a scalar and $\mathbf{x}\in X\cap Y$. $X$ is a subspace, so closed under scalar multiplication, so $\lambda \mathbf{x}\in X$. For the same reason $\lambda \mathbf{x}\in Y$. Therefore $\lambda \mathbf{x}\in X\cap Y$.
${\mathrm{\U0001d7ce}}_{V}$ is in $X$ and $Y$ as they are subspaces, so ${\mathrm{\U0001d7ce}}_{V}+{\mathrm{\U0001d7ce}}_{V}={\mathrm{\U0001d7ce}}_{V}$ is in $X+Y$.
Any two elements of $X+Y$ have the form ${\mathbf{x}}_{1}+{\mathbf{y}}_{1}$ and ${\mathbf{x}}_{2}+{\mathbf{y}}_{2}$, where ${\mathbf{x}}_{i}\in X$ and ${\mathbf{y}}_{i}\in Y$.
$$({\mathbf{x}}_{1}+{\mathbf{y}}_{1})+({\mathbf{x}}_{2}+{\mathbf{y}}_{2})=({\mathbf{x}}_{1}+{\mathbf{x}}_{2})+({\mathbf{y}}_{1}+{\mathbf{y}}_{2})$$ |
by associativity and commutativity. But ${\mathbf{x}}_{1}+{\mathbf{x}}_{2}\in X$ as $X$ is a subspace and ${\mathbf{y}}_{1}+{\mathbf{y}}_{2}\in Y$ as $Y$ is a subspace, so this is in $X+Y$ which is therefore closed under addition.
Let $\lambda $ be a scalar.
$$\lambda ({\mathbf{x}}_{1}+{\mathbf{y}}_{1})=\lambda {\mathbf{x}}_{1}+\lambda {\mathbf{y}}_{1}$$ |
$\lambda {\mathbf{x}}_{1}\in X$ as $X$ is a subspace so closed under scalar multiplication, $\lambda {\mathbf{y}}_{1}\in Y$ for the same reason, so their sum is in $X+Y$ which is therefore closed under scalar multiplication. ∎