Find basis for the subspace of R4 spanned by 2 2 3 1 2 2 1 2

Find basis for the subspace of R^4 spanned by {(2, 2, -3, -1), (2, 2, -1, 2), (-3, -1, -3, 2), (-1, -3, 2, -3)}. Ans: {(l, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)} Find a basis for and the dimension of the solution space of Ax = 0. A = [1 2 5 2 8 14 -9 -38 -65] Ans: {(-1, 5, 1)}, dim = 1 Find a basis for the solution space of the following homogeneous system of linear equations.

Solution

17. Let A be the matrix with the given vectors as columns. We will reduce A to i=ts RREF as under:

Multiply the 1st row by ½; Add -2 times the 1st row to the 2nd row

Add 3 times the 1st row to the 3rd row; Add 1 times the 1st row to the 4th row

Interchange the 2nd row and the 3rd row; Multiply the 2nd row by ½

Add -3 times the 2nd row to the 4th row; Multiply the 3rd row by ½

Add -47/4 times the 3rd row to the 4th row; Multiply the 4th row by 2/15

Add 1 times the 4th row to the 3rd row; Add -1/4 times the 4th row to the 2nd row

Add 1/2 times the 4th row to the 1st row; Add 15/4 times the 3rd row to the 2nd row

Add 3/2 times the 3rd row to the 1st row; Add -1 times the 2nd row to the 1st row

Then the RREF of A is I₄. This implies that the given vectors are linearly independent and form a basis for R⁴. Also, another basis for R⁴ is {(1,0,0,0)^T, (0,1,0,0)^T, (0,0,1,0)^T, (0,0,0,1)^T}.

18. We will reduce A to its RREF as under:

Add -2 times the 1st row to the 2nd row

Add -5 times the 1st row to the 3rd row

Multiply the 2nd row by ¼

Add -4 times the 2nd row to the 3rd row

Add -2 times the 2nd row to the 1st row

Then the RREF of A is

1

0

1

0

1

-5

0

0

0

Thus, if X =(x,y,z)^T, then the equation AX = 0 is equivalent to x+z = 0 or, x = -z, and y =5z = 0 or, y = 5z so that X = (-z,5z,z)^T= z(-1,5,1)^T. Therefore, a basis for the required solution space of AX = 0 is {(-1,5,1)^T}.Its dimension is 1.

1	0	1
0	1	-5
0	0	0