3. (a) [9 marks] Let vi,V2,... , Vk be a collection of k vectors in Rn, (i) Give
ID: 3138550 • Letter: 3
Question
3. (a) [9 marks] Let vi,V2,... , Vk be a collection of k vectors in Rn, (i) Give the definition of a linear combination of the vectors v1, V2,..., Vk, and define spanfvi,..., vk), the span of the vectors vi, V2,..., Vk. (ii) Describe a procedure for checking whether a given vector w E RT is contained in spanvi,..., vkl (b) [10 marks] Let 4 W- and x 4 Find a basis for spansu, v, w, x), and hence decide whether the set of vectors [u, v, w,x) is linearly independent (c) [6 marks] For each of the statements (i)-(iii), state whether it is true or false (i) Every subspace of Rn has only one basis (ii) If b?, b2 and bj are linearly independent vectors in R3, then b1, b2, b3) is a basis for R (ii) If an nxn matrix A is invertible, then the columns of A are linearly dependentExplanation / Answer
3.(a).(i).If, a1,a2,…,ak are scalars then a1v1 +a2v2+…+ak vk is called a linear combination of the vectors v1,v2,…,vk. Span { v1,v2,…,vk } is the set of all linear combinations of the vectors v1,v2,…,vk.
(ii). Let A be the n x k+1 matrix with v1,v2,…,vk,w as columns. We use row operations to reduce A to its RREF. If the (k+1)th column of the RREF of A can be expressed as a linear combination of its first k columns, then w is contained in span { v1,v2,…,vk }.
(b). Let A = [u,v,w,x] =
2
1
3
0
-1
-6
4
-11
0
2
-2
4
-2
-5
1
-8
The RREF of A is
1
0
2
-1
0
1
-1
2
0
0
0
0
0
0
0
0
Now, it is apparent that only u,v are linearly independent and w,x are linear combinations of u,v. Hence{u,v} is a basis for span{u,v,w,x}. Further, the set {u,v,w,x} is linearly dependent.
( c).(i). False. If dim(V) = k, then any linearly independent set of k vectors in V ( a subspace of Rn) is a basis for V.
(ii). True. dim ( R3 ) = 3 so that any linearly independent set of 3 vectors in R3 is a basis for R3.
(iii). False. If A is invertible, then the columns of A are linearly independent.
2
1
3
0
-1
-6
4
-11
0
2
-2
4
-2
-5
1
-8
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