4 Linear algebra 4.4 Subspaces 4.6 Linear independence

4.5 Sums and intersections

Proposition 4.5.1.

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

1.

$X\cap Y\leqslant V$ .
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.
- •
  
  $\mathbf{0}_{V}\in X\cap Y$ 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.
- •
  
  $\mathbf{0}_{V}$ 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. ∎