Find a basis for the eigenspace corresponding to the smaller

Find a basis for the eigenspace corresponding to the smaller eigenvalue. You may use any method but must provide an exact answer. No decimal approximations. Use your basis from (b) to generate one eigenvector. Show your work on how you did this. Your answer can\'t simply be a basis vector.

Solution

The characteristic polynomial is: det(I A) = ¯ ¯ ¯ ¯ ¯ ¯ 1 7 4 1 1 4 1 3 6 ¯ ¯ ¯ ¯ ¯ ¯ = ( 1) ¯ ¯ ¯ ¯ 1 4 3 6 ¯ ¯ ¯ ¯ 7 ¯ ¯ ¯ ¯ 1 4 1 6 ¯ ¯ ¯ ¯ + 4 ¯ ¯ ¯ ¯ 1 1 1 3 ¯ ¯ ¯ ¯ = ( 1)( 2 7 + 18) 7( 2) + 4( 4) = ( 3 8 2 + 25 18) + (7 + 14) + (4 16) = 3 8 2 + 22 20 = ( 2)( 2 6 + 10) = ( 2)( (3 + i))( (3 i)) The eigenvalues are = 2, = 3 + i, and = 3 i To find a basis for the eigenspace corresponding to eigenvalue = 2, we compute the nullspace of the matrix I A: 1 7 4 1 1 4 1 3 4 1 0 4 0 1 0 0 0 0 Thus, the equations for the nullspace are: x1 = 4x3 x2 = 0 x3 = x3 Thus, a basis for the eigenspace corresponding to eigenvalue = 2 is 4 0 1 . (Any multiple of this vector is also a basis for the eigenspace.) To find a basis for the eigenspace corresponding to eigenvalue = 3 + i, we compute the nullspace of the matrix I A: 2 + i 7 4 1 2 + i 4 1 3 3 + i 1 2 + i 4 2 + i 7 4 1 3 3 + i (2 + i) row 1 + row 1 1 2 + i 4 0 4 4i 4 4i 0 1 + i 1 + i × 1 4 1 2 + i 4 0 1 i 1 i 0 1 + i 1 + i + row 2 5 1 2 + i 4 0 1 i 1 i 0 0 0 ÷(1 i) 1 2 + i 4 0 1 i 0 0 0 (2 + i) row 2 1 0 3 + 2i 0 1 i 0 0 0 Thus, the equations for the nullspace are: x1 = (3 2i)x3 x2 = ix3 x3 = x3 Thus, a basis for the eigenspace corresponding to eigenvalue = 3 + i is 3 2i i 1 . (Any multiple of this vector is also a basis for the eigenspace.) A basis for the eigenspace corresponding to eigenvalue = 3i is the conjugate of the above basis. Thus, a basis for the eigenspace corresponding to eigenvalue = 3 i is 3 + 2i i 1 . (Any multiple of this vector is also a basis for the eigenspace.)