Proof of subspace

Let α = (A 1, A2, A3), β = (B 1, B2, B3) ∈ u and any λ, μ∈R, and prove that λα+μβ∈U is the subspace of R3. The specific steps are as follows:

Let α = (A 1, A2, A3), β = (B 1, B2, B3) ∈ U and any λ, μ ∈ r.

Then there is

λα+μβ=(λa 1+μb 1,λa2+μb2,λa3+μb3)

Because a2=a 1+a3, b2=b 1+b3.

So λ a2+μ b2 = λ (a1+a3)+μ (b1+B3) = (λ a1+μ b1)+(λ a3+μ B3).

So λ α+μ β ∈ U. So U is the subspace of R 3.

Extended data

Application of subspace

Let v be a nonempty set and p be a domain. If:

1, an operation is defined in V, which is called addition, that is, any two elements α and β in V correspond to the only determined element α+β in V according to certain rules, which is called the sum of α and β.

2. An operation is defined between the elements of P and V, which is called scalar multiplication (also called quantity multiplication), that is, any element α in V and any element K in P correspond to a uniquely determined element kα in V according to certain laws, which is called the product of K and α.

3. Addition and scalar multiplication satisfy the following conditions:

(1)α+β=β+α, for any α, β ∈ v.

(2)α+(β+γ)=(α+β)+γ, for any α, β, γ∈V 。

(3) There is an element 0∈V, which has α+0=α for all α∈V, and element 0 is called the zero element of V. 。

(4) For any α∈V, there is a negative element β∈V that makes α+β=0, and β is called α and recorded as-α.

(5) For the unit element 1 in P, there is 1α=α(α∈V).

(6) For any k, l∈P, α∈V, there is (kl)α=k(lα).

(7) For any k, l∈P, α∈V, there is (k+l)α=kα+lα.

(8) For any k∈P, α, β∈V, there is k(α+β)=kα+kβ,

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