Question

Suppose vectors b₁,…,bₚ​​ span a subspace W, and let {a₁​​,…,aq} be any set in W containing more than p vectors. Let B=[b₁​​]…,bₚ​​​​] and A=a₁​​∣…∣aq​​].

a. Explain why for each vector aj​​, there exists a vector cj​ in Rp such that aj​=Bcj​​. b. Let C=[c1​​​.... ​cq​​​]. Explain why there is a nonzero vector u such that Cu=0.
c. Use B and C to show that Au=0. Explain why this shows that the columns of A are linearly dependent.

Answer 1

Vectors aᵢ in W can be expressed as Bcᵢ with B being the basis of W. More vectors than the dimension of W implies the existence of a nonzero vector u such that Cu=0, indicating linear dependence. Consequently, Au=0, which shows the columns of A are also linearly dependent.

Step-by-step explanation:

When dealing with the span of vectors b₁, …, bₖ that create a subspace W, and there is a set {a₁ … aₖ} within W that has more than p vectors, we encounter a few important concepts. For each vector aᵢ, it can be represented as aᵢ = Bcᵢ, where B is the matrix formed by the vectors b₁, …, bₖ, and cᵢ is a coefficient vector in Rⁿ. This representation affirms that every vector in the set can be expressed as a linear combination of the vectors in the basis B.

With set {c₁ … cₖ} forming matrix C, and since there are more vectors in A than the dimension of the subspace, by the pigeonhole principle, there must be a nonzero vector u such that Cu = 0. This result reflects the linear dependence among the column vectors of C.

Understanding how multiplication works, we also see that if Cu = 0, it follows that Au = B(Cu), which also equals 0. Hence, the columns of matrix A are linearly dependent.