Dimension Of Vector Space


Linear Algebra / Tuesday, October 30th, 2018
(Last Updated On: November 8, 2018)

Definition of Dimension of a Vector Space

If a vector space V over a field F has a basis consisting of a finite number of elements, then the space is said to be finite dimensional; the number of elements in a basis is called the dimension of vector space and is written as dim V. If dim V = n, V is said to be n-dimensional. If V is not finite dimensional, it is called infinite dimensional. The null vector space has no basis. It is also said to be dimension zero.

 Theorem

If { α1, α2, …, αn } be a basis of a finite dimensional vector space V over a field F, then any set of linearly independent vectors of V contains at most n vectors.

Proof:

Let {β1, β2, …, βm} be a linearly independent set of vectors in V. None of βi is a zero vector. Since { α1, α2, …, αn } is a basis of V and βi is non-zero vector in V.

β1 = c1 α1 + c2 α2 + … + cn αn , ci ∈ F, not all of which are zero. Let cj ≠ 0, then by the Replacement Theorem, { α1, α2, …, αi-1, β1, αi+1, …, αn } is a basis of V.

β2 = d1 α1 + d2 α2 + … + di-1 αi-1 + di β1 + di+1 αi+1 + … + dn αn, di ∈ F, not all zero.

It is clear that at least one of d1, d2,…, di-1, di+1, …, dn is non-zero. For, if is not true, then β2 = di β1 which shows that β1 and β2 are linearly dependent, which is a contradiction to the initial assumption that β1, β2, …, βm are linearly independent.

Let dj ≠ 0, j≠i. Then, by the Replacement Theorem, { α1, α2, …, αi-1, β1, αi+1, …, αn } is a new basis of V.

Proceeding in this way we observe that at each step one α is replaced by one β, and a new set of basis of V is obtained.

The following possibilities occur:

β1, β2, …, βm all came to the new basis containing α’s. In this case we have m < n.

β1, β2, …, βm exhaust all α’s and form a new basis. In this case m = n.

The case m > n is not possible. For, it m > n, then β1, β2, …, βn will replace α1, α2, …, αn and form a new basis.

Then the remaining β’s viz βn+1, βn+2, …, βm will be linear combination of β1, β2, …, βn which means that { β1, β2, …, βm } is a linearly dependent set of vectors, a contradiction.

Hence m ≤ n.

 Example 01

Any two bases of a finite dimensional vector space V have the same number of vectors.

Proof:

Let { α1, α2, …, αn } and {β1, β2, …, βm} be the two bases of a finite dimensional vector space V.

Since { α1, α2, …, αn } is a basis of V and {β1, β2, …, βm} is a linearly independent set of vector, we have, by the above theorem m ≤ n …….(1)

Again {β1, β2, …, βm} is a basis of V and { α1, α2, …, αn } is a linearly independent set of vector, we have, by the above theorem n ≤ m …….(2)

Combining (1) and (2), n = m, and the theorem is proved.

 Extension Theorem

A linearly independent set of vectors in a finite dimensional vector space V over a field F is either a basis of V, or can be extended to a basis of V.

Proof:

Let S = { α1, α2, …, αm } be a linearly independent set of vectors in V. L(S) being the smallest subspace containing S, hence L(S) ⊆ V.

If L(S) = V, then S is a basis of V and there is nothing to prove.

So we suppose that L(S) ⊂ V.

Let β ∈ V – L(S)

We shall prove that the set of vector { α1, α2, …, αm, β} is linearly independent.

Let us suppose that c1, c2, …, cm, b ∈ F such that

c1 α1 + c2 α2 + … + cm αm + bβ = θ ……..(1)

It follows that b = 0. For if b ≠ 0, then b-1 exists in F, and multiplying equation (1) by b-1, we get

β = – b-1c1 α1 – b-1c2 α2 – … – b-1cm αm

This shows that β is a linear combination of α1, α2, …, αm and hence β ∈ L(S), which is not true, as β ∈ V – L(S).

Hence b = 0

Then from (1), c1 α1 + c2 α2 + … + cm αm = θ

But α1, α2, …, αm are linearly independent vectors, hence c1 = c1 = … = c1 = 0.

This shows that the set S1 = { α1, α2, …, αm, β} is a linearly independent set.

Now, if L(S1) = V, then S1 is a basis of V. if L(S1) ≠ V, then we suppose that γ ∈ V – L(S1). Then proceeding exactly in the same way as above, we construct a set S2 = { α1, α2, …, αm, β, γ} of linearly independent vectors.

We check if L(S2) = V, then S2 is a basis of V ; if not, we proceed as above. And after a finite number of steps we can find a basis of V containing S.

 Example 01

Find the basis and the dimension of the subspace W of R3 where W = {(x, y, z) ∈ R3 | x + y + z = 0}

Let ξ = (a, b, c) be an arbitrary vector of W.

Then from the structure of W, we have a + b + c = 0 or, c = – a – b

Hence, ξ = (a, b, – a – b) = a(1, 0, -1) + b(0, 1, -1)

Let α = (1, 0, -1) and β = (0, 1, -1)

Then ξ is a linear combination of vectors α and β.

Therefore, ξ ∈ L{α, β}

We shall show that α and β are linearly independent. Let us suppose that the scalars c1, c2 exists in F such that c1 α + c2 β = θ

⇒ c1 (1, 0, -1) + c2 (0, 1, -1) = (0, 0, 0)

⇒ c1 = 0, c2 = 0, – c1 – c2 = 0

Therefore, c1 = 0, c2 = 0

Hence α, β are linearly independent.

Therefore, {α, β} is a basis of W and dim W =2.

 Theorem

Let U and W be two subspaces of a finite dimensional vector space V over a field F. Then dim (U + W) = dim U + dim W – dim (U ∩ W)

 Example 02

Suppose U and W are the xy plane and yz plane respectively, in R3.

U = {(a, b, 0)} and W = {(0, b, c)}

Since R3 = U + W, dim (U + W) = 3

We have, dim (U + W) = dim U + dim W – dim (U ∩ W)

⇒ 3 = 2 + 2 – dim (U ∩ W)

Therefore, dim (U ∩ W) = 1

 Example 03

Let {(1, 1, 1, 1), (1, 2, 1, 2)} be a linearly independent subset of a vector space V4 of dimension 4 over a field F. Extend it to a basis for V4.

Let S = {(1, 1, 1, 1), (1, 2, 1, 2)}

Then L(S) = {c1 (1, 1, 1, 1) + c2 (1, 2, 1, 2) | c1, c2 ∈ F}

= {c1 + c2, c1 + 2c2, c1 + c2, c1 + 2c2 | c1, c2 ∈ F}

We observe that L(S), the first and the 3rd coordinates are equal. It follows that the vector (0, 3, 2, 3) is not in L(S).

Thus we have an enlarged independent set S1 = {(1, 1, 1, 1), (1, 2, 1, 2), (0, 3, 2, 3)}

Let L(S1) = { d1 (1, 1, 1, 1) +d2 (1, 2, 1, 2) +d3 (0, 3, 2, 3)  | d1, d2 , d3 ∈ F}

= { d1+ d2, d1 + 2d2 +3d3, d1 + d2 +2d3, d1 + 2d2 +3d3 | d1, d2 , d3 ∈ F}

It follows from the structure of L(S1) that the vector (2, 6, 4, 5) is not in L(S1).

Hence the vectors (1, 1, 1, 1), (1, 2, 1, 2), (0, 3, 2, 3), (2, 6, 4, 5) are linearly independent and since V4 is of dimension 4, they form a basis of V4.

 

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