# Linear Independent Vectors

Definition 1: Vectors X1, …, Xk of the same size and shape are independent if for any scalar values b1, … bk, if b1 X1 +⋯+ bXk = 0, then b1 = … = bk  = 0.

Vectors X1, …, Xk are dependent if they are not independent, i.e. there are scalars b1, … bk, at least one of which is non-zero, such that b1 X1 +⋯+ bk Xk = 0.

Observation: If X1, …, Xk  are independent, then Xj ≠ 0 for all j.

Property 1X1, …, Xk  are dependent if and only if at least one of the vectors can be expressed as a linear combination of the others.

Proof: Suppose X1, …, Xk  are dependent. Then there are scalars b1, … bk, at least one of which is non-zero such that b1 X1 +⋯+ bk Xk = 0. Say bi ≠ 0. Then

Now suppose that Xi = $\sum_{j \neq i} b_j X_j$. Then b1 X1 +⋯+ bk Xk = 0, where b= -1, and so X1, …, Xk  are dependent.

Definition 2: The span of independent vectors X1, …, Xk  consists of all the vectors which are a linear combination of these vectors. If W is any set of vectors, then the vectors X1, …, Xk are said to be a basis of W if they are independent and their span equals W. We call a set of vectors W closed if W is the span of some set of vectors. Since we will only consider finite sets of vectors, henceforth we will assume that any closet set is finite.

Property 2: If X1, …, Xk is a basis of W, then every element in W has a unique representation as a linear combination of X1, …, Xk.

Proof: That one such representation exists follows from the definition of span. We now show uniqueness. Suppose there are two such representations, as follows:

$\sum_{j = 1}^k b_j X_j$  = $\sum_{j = 1}^k c_j X_j$

Then $\sum_{j = 1}^k (b_j - c_j) X_j$ = 0. But since X1, …, Xk are independent, bj – cj = 0 for all j, i.e. bj = cj for all j, and so the two representations are equal.

Property 3: If B is a set of independent vectors such that the span of B is a subset of a closed (finite) set of vectors W, then B can be expanded to be a (finite) basis for W.

Proof: Let W = {X1, …, Xn}. We now build a finite set of vectors  recursively as follows. Start with B0 = B. For each k from 0 to n–1, define Bk+1 = Bk ∪ {Xk} if Xk is not already in the span of Bk and Bk+1 = Bk otherwise. Clearly each Bk is an independent set of vectors and the span of Bn is W.

Corollary 1: Every closed set of vectors W has a (finite) basis.

Proof: The corollary is Property 3 with B = Ø.

Property 4: If Y1,…,Yn is a basis for W and X1, …, Xm is a set of independent vectors in W, then m ≤ n.

Proof: We now show by induction that for each k, 1 ≤ km, there are vectors Z1,…,Zk such that {Z1,…,Zk} ⊆ {Y1,…,Yn} and Z1, …, ZkXk+1, …, Xm  are independent.

We now assume the assertion is true for all values less than k and show it is true for k, where km, i.e. we need to prove that for at least one jZ1,…,Z, YjXk+1,…, Xm are independent. If there is no such j then by Property 1, each Yj can be expressed as a linear combination of Z1,…,ZXk+1,…, Xm. But since Xk is in and Y1,…,Yn, is a basis for W, we can express Xk as a linear combination of the Y1,…,Yn  and therefore as a linear combination of Z1,…,ZXk+1,…, Xm. Thus by Property 1, Z1,…,Z, XkXk+1,…, Xare not independent, a violation of the induction hypothesis.

We conclude there is some j for which Z1, …, ZYjXk+1,…, Xm are independent. We set Zequal to any such Yjwhich yields the desired result, namely that Z1, …, ZXk+1,…, Xm are independent.

Where k = m, we conclude that {Z1, …, Zm} ⊆ {Y1, …, Yn} and that Z1, …, Zm are independent. Since Z1, …, Zare independent, they are all distinct, but since {Z1, …, Zm} ⊆ {Y1, …, Yn}, it follows that ≤ n.

Corollary 2: Any two bases for a finite set of vectors have the same number of elements.

Proof: Suppose X1, …, Xm  and Y1, …, Yn are bases for W. By Property 4, m ≤ n and also nm, and so m = n.

Corollary 3: Let Y1, …, Ym be a basis for W and let X1, …, Xm be a set of independent vectors in W. Then X1, …, Xm is also basis for W.

Proof: Suppose X1, …, Xm is not a basis for W. By Property 3, it can be expanded to become a basis for W. This basis has n elements for some n > m, but this contradicts Property 4 since Y1, …, Ym is a basis for W. Thus X1, …, Xm must also be a basis for W.

Corollary 4: Any set of n linearly independent n × 1 column vectors is a basis for the set of n × 1 column vectors. Similarly, any set of n linearly independent 1 × n row vectors is a basis for the set of 1 × n row vectors.

Proof: Let Cj be the jth column of the identity matrix In. It is easy to see that for any n, C1, …, Cn forms a basis for the set of all n × 1 column vectors. The result for column vectors now follows by Corollary 3. The proof for row vectors is similar.

Definition 3: The dimension of a closed set of vectors W is the size of any basis of W.

Observation: This definition makes sense since, as we have seen from the above, any closed set of vectors has a basis, and any two bases have the same number of elements. Further, note that any element in W can be expressed uniquely as a linear combination of the elements in any basis.

### 2 Responses to Linear Independent Vectors

1. Michael says:

I just discovered your site. It is very instructive!

Quick question: in the summation as part of the first proof on this page, should the b(j) term be -b(j)?

• Charles says:

Michael,
Yes, you are right. Thanks for catching this error. I have now updated the referenced webpage.
Charles