# 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 { α _{1}, α_{2}, …, α_{n} } 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 {β_{1}, β_{2}, …, β_{m}} be a linearly independent set of vectors in V. None of β_{i} is a zero vector. Since { α_{1}, α_{2}, …, α_{n} } is a basis of V and β_{i} is non-zero vector in V.

β_{1} = c_{1} α_{1} + c_{2} α_{2} + … + c_{n} α_{n} , c_{i} ∈ F, not all of which are zero. Let c_{j} ≠ 0, then by the Replacement Theorem, { α_{1}, α_{2}, …, α_{i-1}, β_{1}, α_{i+1}, …, α_{n} } is a basis of V.

β_{2} = d_{1} α_{1} + d_{2} α_{2} + … + d_{i-1} α_{i-1} + d_{i} β_{1} + d_{i+1} α_{i+1} + … + d_{n} α_{n}, d_{i} ∈ F, not all zero.

It is clear that at least one of d_{1}, d_{2},…, d_{i-1}, d_{i+1}, …, d_{n} is non-zero. For, if is not true, then β_{2} = d_{i} β_{1} which shows that β_{1} and β_{2} are linearly dependent, which is a contradiction to the initial assumption that β_{1}, β_{2}, …, β_{m} are linearly independent.

Let d_{j} ≠ 0, j≠i. Then, by the Replacement Theorem, { α_{1}, α_{2}, …, α_{i-1}, β_{1}, α_{i+1}, …, α_{n} } 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:

β_{1}, β_{2}, …, β_{m} all came to the new basis containing α’s. In this case we have m < n.

β_{1}, β_{2}, …, β_{m} 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 β_{1}, β_{2}, …, β_{n} will replace α_{1}, α_{2}, …, α_{n} and form a new basis.

Then the remaining β’s viz β_{n+1}, β_{n+2}, …, β_{m} will be linear combination of β_{1}, β_{2}, …, β_{n} which means that { β_{1}, β_{2}, …, β_{m} } 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 { α_{1}, α_{2}, …, α_{n} } and {β_{1}, β_{2}, …, β_{m}} be the two bases of a finite dimensional vector space V.

Since { α_{1}, α_{2}, …, α_{n} } is a basis of V and {β_{1}, β_{2}, …, β_{m}} is a linearly independent set of vector, we have, by the above theorem m ≤ n …….(1)

Again {β_{1}, β_{2}, …, β_{m}} is a basis of V and { α_{1}, α_{2}, …, α_{n} } 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 = { α_{1}, α_{2}, …, α_{m} } 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 { α_{1}, α_{2}, …, α_{m}, β} is linearly independent.

Let us suppose that c_{1}, c_{2}, …, c_{m}, b ∈ F such that

c_{1} α_{1} + c_{2} α_{2} + … + c_{m} α_{m} + 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^{-1}c_{1} α_{1} – b^{-1}c_{2} α_{2} – … – b^{-1}c_{m} α_{m}

This shows that β is a linear combination of α_{1}, α_{2}, …, α_{m} and hence β ∈ L(S), which is not true, as β ∈ V – L(S).

Hence b = 0

Then from (1), c_{1} α_{1} + c_{2} α_{2} + … + c_{m} α_{m} = θ

But α_{1}, α_{2}, …, α_{m} are linearly independent vectors, hence c_{1 }= c_{1 }= … = c_{1} = 0.

This shows that the set S_{1} = { α_{1}, α_{2}, …, α_{m}, β} is a linearly independent set.

Now, if L(S_{1}) = V, then S_{1} is a basis of V. if L(S_{1}) ≠ V, then we suppose that γ ∈ V – L(S_{1}). Then proceeding exactly in the same way as above, we construct a set S_{2} = { α_{1}, α_{2}, …, α_{m}, β, γ} of linearly independent vectors.

We check if L(S_{2}) = V, then S_{2} 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 R ^{3} where W = {(x, y, z) ∈ R^{3} | 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 c_{1}, c_{2} exists in F such that c_{1} α + c_{2} β = θ

⇒ c_{1} (1, 0, -1) + c_{2} (0, 1, -1) = (0, 0, 0)

⇒ c_{1} = 0, c_{2} = 0, – c_{1 }– c_{2} = 0

Therefore, c_{1} = 0, c_{2} = 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 R^{3}.

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

Since R^{3} = 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 V _{4} of dimension 4 over a field F. Extend it to a basis for V_{4}.**

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

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

= {c_{1} + c_{2}, c_{1} + 2c_{2}, c_{1} + c_{2}, c_{1} + 2c_{2} | c_{1}, c_{2} ∈ F}

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

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

Let L(S_{1}) = { d_{1} (1, 1, 1, 1) +d_{2} (1, 2, 1, 2) +d_{3} (0, 3, 2, 3) | d_{1}, d_{2} , d_{3} ∈ F}

= { d_{1}+ d_{2}, d_{1 }+ 2d_{2} +3d_{3}, d_{1 }+ d_{2} +2d_{3}, d_{1 }+ 2d_{2} +3d_{3} | d_{1}, d_{2} , d_{3} ∈ F}

It follows from the structure of L(S_{1}) that the vector (2, 6, 4, 5) is not in L(S_{1}).

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