Question

(a) Let λ be a real number. Compute A − λI.

(b) Find the eigenvalues of A, that is, find the values of λ for which the matrix A − λI is not invertible. (Hint: There should be exactly 2. Label the larger one λ1 and the smaller λ2.)

(c) Compute the matrices A − λ1I and A − λ2I.

(d) Find the eigenspace associated with λ1, that is the set of all solutions v = v1 v2 to (A − λ1I)v = 0.

(e) Find the eigenspace associated with λ2 similarly.

Repeat the process to find the eigenvalues and corresponding eigenspaces for

A = [ 0 2 0

2 0 0

1 1 4 ]

(Note that this matrix has three eigenvalues, not 2.)

Answer 1

(a). We have A – λI₃ =

-λ	2	0
2	-λ	0
1	1	4-λ

(b). The characteristic equation of A is det(A – λI₃)= 0 or, λ³-4λ²-4λ+16 = 0 or, (λ-4)( λ+2)( λ-2)= 0. Thus, the eigenvalues of A are λ₁ =4, λ₂=-2 and λ₃ = 2.

(c). The matrix A – λ₁I₃ is

-4	2	0
2	-4	0
1	1	0

The matrix A – λ₂I₃ is

2	2	0
2	2	0
1	1	6

The matrix A – λ₃I₃ is

-2	2	0
2	-2	0
1	1	2

(d). The RREF of the matrix A – λ₁I₃ is

1	0	0
0	1	0
0	0	0

If v = (x,y,z)^T, then the equation (A – λ₁I₃)v = 0 is equivalent to x = 0 and y = 0 so that v = (0,0,z)^T= z(0,0,1)^T. Hence the eigenvector of A associated with the eigenvalue ʎ₁= 4 is v₁ = (0,0,1)^T.

The RREF of the matrix A – λ₂I₃ is

1	1	0
0	0	1
0	0	0

If v = (x,y,z)^T, then the equation (A – λ₂I₃)v = 0 is equivalent to x+y = 0 or, x = -y and z = 0 so that v = (-y,y,0)^T = y(-1,1,0)^T. Hence the eigenvector of A associated with the eigenvalue ʎ₂= -2 is v₂ = (-1,1,0)^T.

The RREF of the matrix A – λ₃I₃ is

1	0	1
0	1	1
0	0	0

If v = (x,y,z)^T, then the equation (A – λ₃I₃)v = 0 is equivalent tox+z = 0 or, x = -z and y+z = 0 or, y = -z so that v = (-z,-z,z)^T = z(-1,-1,1)^T. Hence the eigenvector of A associated with the eigenvalue ʎ₃= 2 is v₃ = (-1,-1,1)^T.

(e ). E_ʎ1 = span{(0,0,1)^T }, E_ʎ2 = span{(-1,1,0)^T } and E_ʎ3 = span{(-1,-1,1)^T },