4 Linear algebra

4.5 Sums and intersections

Proposition 4.5.1.

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

  1. 1.

    XYV.

  2. 2.

    X+Y={𝐱+𝐲:𝐱X,𝐲Y}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.
    • 𝟎VXY as X and Y are subspaces so contain 𝟎V.

    • Let 𝐱,𝐲XY. X is a subspace, so closed under addition, so 𝐱+𝐲X. For the same reason 𝐱+𝐲Y. Therefore 𝐱+𝐲XY.

    • Let λ be a scalar and 𝐱XY. X is a subspace, so closed under scalar multiplication, so λ𝐱X. For the same reason λ𝐱Y. Therefore λ𝐱XY.

  2. 2.
    • 𝟎V is in X and Y as they are subspaces, so 𝟎V+𝟎V=𝟎V is in X+Y.

    • Any two elements of X+Y have the form 𝐱1+𝐲1 and 𝐱2+𝐲2, where 𝐱iX and 𝐲iY.

      (𝐱1+𝐲1)+(𝐱2+𝐲2)=(𝐱1+𝐱2)+(𝐲1+𝐲2)

      by associativity and commutativity. But 𝐱1+𝐱2X as X is a subspace and 𝐲1+𝐲2Y as Y is a subspace, so this is in X+Y which is therefore closed under addition.

    • Let λ be a scalar.

      λ(𝐱1+𝐲1)=λ𝐱1+λ𝐲1

      λ𝐱1X as X is a subspace so closed under scalar multiplication, λ𝐲1Y for the same reason, so their sum is in X+Y which is therefore closed under scalar multiplication. ∎