§4.4 Similarity and Diagonalization

An illustrative example Similar matrices Diagonalization

Tom Lewis

Fall Semester 2020 An illustrative example Similar matrices Diagonalization Outline

An illustrative example

Similar matrices

Diagonalization An illustrative example Similar matrices Diagonalization

An illustrative example An illustrative example Similar matrices Diagonalization

Problem Let T be the transformation in the plane that reﬂects a vector through the line passing through the origin with direction [3, 4]. Find the matrix of the transformation T. An illustrative example Similar matrices Diagonalization

A “natural” basis for T

T(v) u=[3,4]

v=[4,-3] An illustrative example Similar matrices Diagonalization

In summary Let

−0.28 0.96 3 4 1 0 A = , P = , D = 0.96 0.28 4 −3 0 −1

Then

A = PDP −1 or P −1AP = D

A and D represent the same transformation, T, but with respect to diﬀerent bases. An illustrative example Similar matrices Diagonalization

Similar matrices An illustrative example Similar matrices Diagonalization

The similarity condition above can also be expressed as

A = PBP −1.

A and B are the same transformation represented through diﬀerent bases. An illustrative example Similar matrices Diagonalization

Example Let 10 12 1 0 A = and B = −6 −7 0 2 Observe that

10 12 4 −3 1 0 −2 −3 = . −6 −7 −3 2 0 2 −3 −4

This shows that A ∼ B. An illustrative example Similar matrices Diagonalization

Properties (a), (b), and (c) are the deﬁning properties of an equivalence relation: matrix similarity is an example of an equivalence relation. An illustrative example Similar matrices Diagonalization An illustrative example Similar matrices Diagonalization

Example The matrices

1 2 1 −1 A = and B = 0 1 1 2

are not similar:

det A = 1 and det B = 3 An illustrative example Similar matrices Diagonalization

Diagonalization An illustrative example Similar matrices Diagonalization

Equivalently, A = PDP −1. An illustrative example Similar matrices Diagonalization

Problem Let D be an n × n diagonal matrix and let P be an invertible n × n matrix with columns p1,..., pn . Thus   λ1 ··· 0  . ..  D =  . . 0  and P = p1 ··· pn . 0 ··· λn

Let A = PDP −1.

For each i, 1 6 i 6 n, show that

Api = λi pi .

Thus A has n linearly independent eigenvectors. An illustrative example Similar matrices Diagonalization

Example Suppose 3 4 −4 0 3 −4 A = . 2 3 0 7 −2 3 Then A has eigenvalues −4 and 7 and corresponding eigenvectors 3 4 and . 2 3 An illustrative example Similar matrices Diagonalization

1. If A = PDP −1, then A has n linearly independent eigenvectors. 2. If A has n linearly independent eigenvectors, then A = PDP −1 for some invertible matrix P and some diagonal matrix D. An illustrative example Similar matrices Diagonalization

Problem −13 24 Show that the matrix A = is diagonalizable. −8 15 An illustrative example Similar matrices Diagonalization

Solution Observe that

−13 − λ 24 2 pA(λ) = = λ − 2λ − 3 = (λ + 1)(λ − 3) −8 15 − λ

The spectrum of A is {−1, 3}.

Let p1 and p2 be the eigenvectors corresponding to λ = −1 and λ = 3.

The eigenvectors p1 and p2 are linearly independent since they correspond to distinct eigenvalues; thus, A is diagonalizable. An illustrative example Similar matrices Diagonalization An illustrative example Similar matrices Diagonalization

Problem Show that the matrix

 5 6 0  A = −3 −4 0  . −3 −3 −1

is diagonalizable. An illustrative example Similar matrices Diagonalization

Solution The characteristic polynomial of A is

2 pA(λ) = −(λ + 1) (λ − 2).

Thus A has only two distinct eigenvalues: −1 and 2.

−1 0 −2 E−1 = span  1  , 0 and E2 = span  1       0 1   1 

 λ = −1 λ = 2   alg. mult. 2 1 geo. mult. 2 1 An illustrative example Similar matrices Diagonalization

The ﬁnal form Here is the diagonal representation of the matrix A:

 5 6 0  −1 0 −2 −1 0 0  1 2 0 −3 −4 0  =  1 0 1   0 −1 0  1 1 1 −3 −3 −1 0 1 1 0 0 2 −1 −1 0 An illustrative example Similar matrices Diagonalization An illustrative example Similar matrices Diagonalization

Problem 3 1 Let A = . Show that A is not diagonalizable. 0 3 An illustrative example Similar matrices Diagonalization

Solution The characteristic polynomial of A is

2 pA(λ) = (λ − 3) .

Thus A has only one eigenvalue: 3.

1 E = span 3 0

λ = 3 alg. mult. 2 geo. mult. 1 The matrix A is not diagonalizable. An illustrative example Similar matrices Diagonalization

Problem 4 −6 The matrix A = has has eigenvalues λ = 2 and λ = 1 1 −1 3 2 with corresponding eigenvectors and . 1 1 1. Find a diagonalization of A. 2. Express A5. 3. Find the limit of 2−n An as n → .

∞