### Eigenvalue and Eigenvector

##### 3. Diagonalization

The eigenvalues and eigenvectors have been defined in terms of linear transformations and vectors. In this section, we give matrix interpretation.

Let A be an n by n matrix. Let v1, v2, ..., vk be vectors and let λ1, λ2, ..., λk be numbers. Denote

 P = [v1 v2 ... vk], D = [ λ1 0 . . 0 ]. 0 λ2 . . 0 : : : 0 0 . . λk

Then the equalities

Av1 = λ1v1, Av2 = λ2v2, ..., Avk = λkvk,

are equivalent to

AP = [Av1 Av2 ... Avk] = [λ1v1 λ2v2 ... λkvk] = [v1 v2 ... vk]D = PD.

Moreover, v1, v2, ..., vk form a basis if and only if P is an invertible matrix, and the equality AP = PD becomes A = PDP-1.

We just proved that the two properties in the following definition are equivalent.

A matrix is diagonalizable if it has any of the following equivalent properties

• A has a basis of eigenvectors.
• A is similar to a diagonal matrix.

Moreover, any expression A = PDP-1 is a diagonalization of A.

The first property can also be used to define the diagonalizability for linear transformations T: VV. Then the second property has the following interpretation. Let A = [T(B)]B be the matrix of T with respect to a basis B = {b1, b2, ..., bn} of V. Let E = {v1, v2, ..., vn} be a basis of eigenvectors for T. The fact T(vi) = λivi simply means that the matrix D = [T(E)]E of T with respect to the basis E is diagonal. By this discussion, we conclude

A = [T(B)]B = [E]B [T(E)]E [E]B-1 = PDP-1.

Example In an earlier example, we found the matrix

 A = [ 13 -4 ] -4 7

has a basis v1 = (1, 2), v2 = (-2, 1) of eigenvectors, with eigenvalues λ1 = 5, λ2 = 15. Thus A = PDP-1, with

 P = [v1 v2] = [ 1 2 ], D = [ λ1 0 ] = [ 5 0 ]. -2 1 0 λ2 0 15

In another earlier example, we found the matrix

 [ 1 3 -3 ] -3 7 -3 -6 6 -2

has a basis (1, 1, 0), (1, 0, -1), (1, 1, 2) of eigenvectors, with corresponding eigenvalues 4, 4, 2. This gives us

 [ 1 3 -3 ] = [ 1 1 1 ] [ 4 0 0 ] [ 1 1 1 ]-1. -3 7 -3 1 0 1 0 4 0 1 0 1 -6 6 -2 0 -1 2 0 0 2 0 -1 2

Note that by rearranging the order of the eigenvectors in the basis, with corresponding change of order for eigenvalues, we have other diagonalizations of the same matrix

 [ 1 3 -3 ] = [ 1 1 1 ] [ 2 0 0 ] [ 1 1 1 ]-1, -3 7 -3 1 0 1 0 4 0 1 0 1 -6 6 -2 2 -1 0 0 0 4 2 -1 0
 [ 1 3 -3 ] = [ 1 1 1 ] [ 4 0 0 ] [ 1 1 1 ]-1. -3 7 -3 1 1 0 0 2 0 1 1 0 -6 6 -2 0 2 -1 0 0 4 0 2 -1

Example Consider the matrix

 A = [ 2 3 -3 ]. 0 2 -3 0 0 1

Being upper triangular, it is easy to see that the characteristic polynomial is (2 - λ)2(1 - λ) and there are two eigenvalues λ1 = 2, λ2 = 1.

For λ1 = 2, we find v1 = (1, 0, 0) form a basis of nul(A - 2I). For λ2 = 1, we find v2 = (-6, 3, 1) form a basis of nul(A - I). Since we only get two eigenvectors, A does not have a basis of eigenvectors and is not diagonalizable.

In terms of matrices, we have

AP = [ 2 3 -3 ] [ 1 -6 ] = [ 1 -6 ]
 [ 2 0 ] = PD. 0 1
0 2 -3 0 3 0 3
0 0 1 0 1 0 1

However, the matrix

 P = [ 1 -6 ] 0 3 0 1

is not invertible, and we cannot multiply P-1 on the right of AP = PD to get diagonalization.