# 4.5 Sums and intersections

###### Proposition 4.5.1.

Let $X$ and $Y$ be subspaces of a vector space $V$.

1. 1.

$X\cap Y\leqslant V$.

2. 2.

$X+Y=\{\mathbf{x}+\mathbf{y}:\mathbf{x}\in X,\mathbf{y}\in Y\}\leqslant V$.

###### Proof.

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.

1. 1.
• $\mathbf{0}_{V}\in X\cap Y$ as $X$ and $Y$ are subspaces so contain $\mathbf{0}_{V}$.

• 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 $\lambda\mathbf{x}\in X\cap Y$.

2. 2.
• $\mathbf{0}_{V}$ is in $X$ and $Y$ as they are subspaces, so $\mathbf{0}_{V}+\mathbf{0}_{V}=\mathbf{0}_{V}$ 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. ∎