Proof For `H` + `K`, we have

`u`, `v` ∈ `H` + `K`

⇒ `u` = `u`_{H} + `u`_{K},
`u`_{H} ∈ `H`,
`u`_{K} ∈ `K`;
`v` = `v`_{H} + `v`_{K},
`v`_{H} ∈ `H`,
`v`_{K} ∈ `K`

⇒ **u** + **v** =
`u`_{H} + `u`_{K} + `v`_{H} + `v`_{K} =
(`u`_{H} + `v`_{H}) + (`u`_{K} + `v`_{K}) ∈ `H` + `K`.

The verification of `u` ∈ `H` + `K` ⇒ `c`**u** ∈ `H` + `K` is similar.

For `H`∩`K`, we have

`u`, `v` ∈ `H`∩`K`

⇒ `u` ∈ `H`, `v` ∈ `H`, `u` ∈ `K`, `v` ∈ `K`

⇒ **u** + **v** ∈ `H` (since `H` is a subspace), **u** + **v** ∈ `K` (since `K` is a subspace)

⇒ **u** + **v** ∈ `H`∩`K`.

The verification of `u` ∈ `H`∩`K` ⇒ `c`**u** ∈ `H`∩`K` is similar.

We note that the sum and the intersection are the adaptations of the union and the intersection of sets to vector spaces. While the intersection remains the same, the sum is different form the union, because the union of two subspaces is not a subspace.

Example In this example, we saw that `H` = {`f` ∈ `F`(**R**): `f`(2) = 0} is a subspace of `F`(**R**). Similarly, `K` = {`f` ∈ `F`(**R**): `f`(1) = 0} is also a subspace of `F`(`R`). The intersection `H`∩`K` consists of all functions `f` satisfying `f`(2) = `f`(1) = 0. In particular, such functions form a subspace. By the similar reason, subsets such as {`f` ∈ `F`(**R**): `f`(3) = `f`(100) = `f`(-1) = `f`(-0.5) = 0} are also subspaces of `F`(**R**).

We claim that `H` + `K` = `F`(**R**). This means that any function `f` can be written as `f` = `g` + `h`, such that `g`(2) = 0 and `h`(1) = 0. This can be done by choosing `g`(`t`) = `f`(2)(`t` - 1) and `h`(`t`) = `f`(`t`) - `f`(2)(`t` - 1), for example.

Example In this example, we saw the solutions `H` = {(`x`, `y`, `z`): `x` + `y` + `z` = 0} of a homogeneous equation is a subspace of **R**^{3}. Similarly, `K` = {(`x`, `y`, `z`): 2`x` - `y` + 3`z` = 0} is also a subspace. Their intersection `H`∩`K` consists of all solutions of the 2 by 3 homogeneous system `x` + `y` + `z` = 0, 2`x` - `y` + 3`z` = 0. By a simple computation, the solutions of the system are found to be `y`(-4, 1, 3), with `y` arbitrary. Thus the subspace `H`∩`K` actually consists of all scalar mutiples of (-4, 1, 3).

We claim the sum `H` + `K` = **R**^{3}. This means that any (`x`, `y`, `z`) can be written as (`x`, `y`, `z`) = (`x`_{1}, `y`_{1}, `z`_{1}) + (`x`_{2}, `y`_{2}, `z`_{2}), satisfying `x`_{1} + `y`_{1} + `z`_{1} = 0 and 2`x`_{2} - `y`_{2} + 3`z`_{2} = 0. We try to find such a combination by taking `z`_{2} = 0. Then `y`_{2} = 2`x`_{2}, the combination becomes (`x`, `y`, `z`) = (`x` - `x`_{2}, `y` - 2`x`_{2}, `z`) + (`x`_{2}, 2`x`_{2}, 0), and the problem is reduced to (`x` - `x`_{2}) + (`y` - 2`x`_{2}) + `z` = 0. Thus we find `x`_{2} = (`x` + `y` + `z`)/3 and finally get the combination

(`x`, `y`, `z`) = ( (2`x` - `y` - `z`)/3, (-2`x` + `y` - 2`z`)/3, `z` ) + ( (`x` + `y` + `z`)/3, (2`x` + 2`y` + 2`z`)/3, 0 )

we are looking for.