# Linear Algebra : Span and Subspace

(The material is from Basis and Dimension. Please explain each step of your solution for # 8, 9. Thank you.)

Proposition 3.7:

Let V in Rn be a k-dimensional subspace. Then any k vectors that span V must be linearly independent and any k linearly independent vectors in V must span V.

Â© BrainMass Inc. brainmass.com October 4, 2022, 4:18 pm ad1c9bdddfhttps://brainmass.com/math/linear-algebra/linear-algebra-span-subspace-175057

## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please see the attached file for the complete solution.

Thanks for using BrainMass.

(The material is from Basis and Dimension. Please explain each step of your solution for # 8, 9. Thank you.)

Proposition 3.7:

Let V in Rn be a k-dimensional subspace. Then any k vectors that span V must be linearly independent and any k linearly independent vectors in V must span V.

Question 8.

Proof. We prove it by contradiction. Suppose that l<k, and . Since is a basis of V, every vector of (in V) can be expressed as a linear combination of . This is impossible, as is a basis of V and so are linear independent. So, .

Question 9.

Proof.

(1) We prove that any k vectors must form a basis of V, so they are independent.

Proof. Let , and we know that V is k-dimensional. If are not linear independent, then we choose the maximum number of vectors of , say where l<k. Then will form a basis of V, contradicting that the dimension of V is k.

(2) We prove that any linear independent k vectors in V must span V.

Proof. Since are linear independent, if , then there is at least one vector, say w, in V such that can not be expressed as a linear combination of . In other words, these k+1 vectors , w are linear independent. This is a contradiction, as the subspace V is k-dimensional (which means that there are no such k+1 linear independent vectors)

Â© BrainMass Inc. brainmass.com October 4, 2022, 4:18 pm ad1c9bdddf>https://brainmass.com/math/linear-algebra/linear-algebra-span-subspace-175057