math111_logo Properties

2. About several matrices

In this part, we consider the eigenvalues and eigenvectors of several matrices.

Let v be an eigenvector of A and B with eigenvalues λ and μ. Then v is an eigenvector of aA + bB with eigenvalue + , and an eigenvector AB with eigenvalues λμ.

Proof From Av = λv, Bv = μv, we have

(aA + bB)v = aAv + bBv = v + v = ( + )v,
(AB)v = A(Bv) = A(μv) = λμv.

Example Consider the matrices

A = [ 13 -4 ], B = [ 6 2 ].
-4 7 2 9

From an earlier example, we know v1 = (1, 2), v2 = (-2, 1) are eigenvectors of A with eigenvalues 5 and 15. It is also easy to verify that they are eigenvectors of B with eigenvalues 10 and 5. We may then conclude that v1 = (1, 2), v2 = (-2, 1) are eigenvectors of the following matrices with the following eigenvalues:

A + B = [ 19 -2 ], λ1 = 5 + 10 = 15, λ2 = 15 + 5 = 20.
-2 16
A + 2B = [ 25 0 ], λ1 = 5 + 20 = 15, λ2 = 15 + 10 = 25.
0 25
AB = [ 70 -10 ], λ1 = 5×10 = 50, λ2 = 15×5 = 75.
-10 55
AB - B2 = [ 30 -40 ], λ1 = 50 - 100 = -50, λ2 = 75 - 25 = 50.
-40 -30

Let AB = BA. Let v be an eigenvector of A with eigenvalue λ and Bv0. Then Bv is again an eigenvector of A with eigenvalue λ.

Proof A(Bv) = (AB)v = (BA)v = B(Av) = B(λv) = λBv.

Example Let A be a 3 by 3 matrix commuting with

B = [ 0 2 0 ].
1 -1 1
3 -5 3

Suppose v1 = (1, 1, 0), v2 = (1, 1, 2) are eigenvectors of A with eigenvalues 4 and -2. Then Bv1 = (2, 0, -2) and Bv2 = (2, 2, 4) are also eigenvectors of A with eigenvalues 4 and -2. Note that v1, v2, Bv1 form a basis of eigenvectors of A with eigenvalues 4, -2, 4. Therefore for

P = [v1 v2 Bv1] = [ 1 1 2 ], D = [ 4 0 0 ],
1 1 0 0 -2 0
0 2 -2 0 0 4

we have

A = PDP-1 = [ 1 3 -3 ].
-3 7 -3
-6 6 -2

The matrix appeared in this example. You may also directly verify that AB = BA.

Finally, we remark that the properties also holds for linear transformations T, S: VV. For example, if v is an engenvector for both T and S, then v is an engenvector for aT + bS and TS.


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