1-4 5 2 3 4. [5 points] a) Let A = 0 1 -2 6 05 2 1-2 2| , a 3 × 5 matrix. Let W
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Question
1-4 5 2 3 4. [5 points] a) Let A = 0 1 -2 6 05 2 1-2 2| , a 3 × 5 matrix. Let W be the solution set of the equation Ax = 0. Then W is a subspace of Rs (you don't need to verify this). Find, with justification, a set S of vectors in W which simultaneously 1) spans W and 2) is linearly independent Remark: S is not unique, but it will necessarily contain exactly 3 vectors. b) Let U be the span of the columns of A, which is a subspace of R3 (you don't need to verify this). Find, with justification, a subset T of the columns of A so that Span(T) U, and with T being linearly independent. Remark: T is not unique, but it will necessarily contain exactly 2 vectors. Remark the sizes of the sets S and T above sum to 5, which is the number of columns of A. This is a special case of the "rank-nullity theorem," which we will cover at the end of chapter 4.Explanation / Answer
W,the solution set of the equation AX = 0 is N(A), the null space of A, and U is Col(A), the column space of A. To determine W and U, we will reduce A to its RREf as under:
Add -1 times the 1st row to the 3rd row
Multiply the 2nd row by ½
Add -2 times the 2nd row to the 3rd row
Add 4 times the 2nd row to the 1st row
Then the RREF of A is
1
0
7
-2
7
0
1
½
-1
1
0
0
0
0
0
Now, if X = (x,y,z,v,w)T, then the equation AX = 0 is equivalent to x+7z-2v+7w = 0 or, x = -7z+2v-7w…(1) and y+z/2-v+w = 0 or, y = -z/2 +v-w…(2). Then X = (-7z+2v-7w, -z/2 +v-w,z,v,w)T = z/2(-14,-1,1,0,0)T +v(2,1,0,1,0)T+w(-7,-1,0,0,1)T. Hence W = span(S) , where S = {(-14,-1,1,0,0)T,(2,1,0,1,0)T,(-7,-1,0,0,1)T}. The vectors in the spanning set of W are linearly independent as the RREF of the matrix with these vectors as columns is
1
0
0
0
1
0
0
0
1
0
0
0
0
0
0
(b) From the RREF of A, it is apparent that only its first two columns are linearly independent and that its remaining columns are linear combinations of the first two columns. Then U = span (T), where T = { (1,0,1)T,(-4,2,-2)T}.
1
0
7
-2
7
0
1
½
-1
1
0
0
0
0
0
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