# Linear Algebra – Subspace of a Vector Space

Linear Algebra / Tuesday, August 13th, 2019

# Subspace of Vector Space

If V is a vector space over a field F and W ⊆ V, then W is a subspace of vector space V if under the operations of V, W itself forms vector space over F.

It is clear that {θ} and V are both subspaces of V. These are trivial subspaces. Any subspace of a vector space V other than {θ} and V itself is called a proper subspace of V.

 Theorem

A non-empty subset of W of vector space V over a field F is a subspace of V if and only if
α ∈ W, β ∈ W ⇒ α + β ∈ W
α ∈ W, c ∈ F ⇒ cα ∈ W

Proof:

Let us suppose that the conditions (1) and (2) hold in W. Let α, β ∈ W. Since F is a field, 1 ∈ F and so -1 ∈ F.

By condition (2), we have (-1).β ∈ W or, – β ∈ W

Hence by condition (1), α + (-β) ∈ W or, α – β ∈ W whenever α, β ∈ W

This shows that (W, +) is a subgroup of the additive group (V, +). But since V is abelian, W is abelian. Hence using condition (2) and the heredity property, we can conclude that W is a vector space over the field F. Hence W is a subspace of V over F. Thus the sufficiency of the conditions is established.

The necessity of the conditions (1) and (2) follows from the definition of a vector space.

Note: The above theorem may be stated in the alternative form giving only one condition as follows:

A non-empty set W of a vector space V over a field F is a subspace of V if and only if a.α + b.β ∈ W for all α, β ∈ W and all a, b∈ F.

 Example 01

Let S be the subset of R3 defined by S = {(x, y, z) ∈ R3 | y = z =0}. Then S is a non-empty subset of R3, since (0, 0, 0) ∈ S.
Let α = (a, 0, 0) and β = (b, 0, 0) ∈ S, where a, b ∈ R
Then α + β = (a, 0, 0) + (b, 0, 0) = (a + b, 0, 0) ∈ S
Thus α ∈ S, β ∈ S ⇒ α + β ∈ S
Let c ∈ R, then c.α = c. (a, 0, 0) =(ca, 0, 0) ∈ S
Hence, by definition, S is a subspace of R3.

 Example 02

Let V = R3 and W ⊆ V such that W = {(a, b, c) ∈ Q3 | a, b, c ∈ Q} i.e., W consists of those vectors whose components are rational numbers.

Clearly, α = (4, 2, 3) ∈ W and √2 ∈ R.
But √2.α =  √2 . (4, 2, 3) = (4√2, 2√2, 3√2) ∉ W
Hence the condition (2) for being a subspace is violated. So W as defined above is not a subspace of V.

 Example 03

Let V be the vector space of all functions from the real field R into R, and let W = {f : f(-x) = – f(x), x ∈ R}, i.e., W consists of the odd real valued functions defined over R.

The function f which assigns 0 to every x ∈ R, is the zero function, denoted by O. Thus O(x) = 0 ∈ R.

Then clearly, O(x) = 0 = – 0 = – O(x)

This implies that the zero function O ∈ W.

Suppose g(x) and h(x) ∈ W, then g(-x) = -g(x), h(-x) = -h(x). Then for any a, b ∈ R, we have

(ag + bh)(-x) = ag(-x) + bh(-x) = -ag(x) – bh(x) = -(ag(x) + bh(x)) = -(ag + bh)(x)

Hence, g, h ∈ W ⇒ ag +bh ∈ W; a, b ∈ R

So W is a subspace of V over the field of real numbers R.

 Theorem

Intersection of two subspaces of a vector space V over a field F is a subspace of V over F.

Proof:

Let W1 and W2 be two subspaces of a vector space V over a field F. Let W = W1 ∩ W2

Suppose the α, β ∈ W, then α, β ∈ W1 and α, β ∈ W2. But W1 and W2 are subspaces of V. Hence, α, β ∈ W1, and W2 and for any c ∈ F, cα ∈ W1, and W2.

This implies cα ∈ W.

So for α, β ∈ W, we have α + β ∈ W1 ∩ W2 = W and cα ∈ W1 ∩ W2 = W, c ∈ F

Hence W is a subspace of V.

Note: The union of two subspaces of a vector space is not, in general, a subspace of V.

Let W1 and W2 be the two subspaces of the vector space R2 such that

W1 = {(x, y) ∈ R2 | y = 0}

And W2 = {(x, y) ∈ R2 | x = 0}

Let α = (1, 0) and β = (0, 1)

Clearly, α ∈ W1 and β ∈ W2

Now, α + β = (1, 0) + (0, 1) = (1, 1)

But α + β = (1, 1) ∉ W1 and α + β ∉ W2

Hence, α + β ∉ W1 ∪ W2 , although α ∈ W1 ∪ W2 and β ∈ W1 ∪ W2.

This shows that W1 ∪ W2 is not a subspace of V.

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