DIAGONALIZATION. In order to see the implications of this definition, let us consider the following example Example 1. Consider the matrix

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1 DIAGONALIZATION Definition We say that a matrix A of size n n is diagonalizable if there is a basis of R n consisting of eigenvectors of A ie if there are n linearly independent vectors v v n such that for some numbers λ λ n Av λ v Av n λ n v n In order to see the implications of this definition let us consider the following example Example Consider the matrix A and let us compute the eigenvalues of A by looking at the characteristic polynomial 3 λ 5 pλ det λ 3λ λ 2 4 λ 2 + λ 2 λ λ + 2 so that the eigenvalues of A are λ and λ 2 Let us now find the eigenspaces of A -Eigenspace: The Eigenspace corresponding to λ is determined by the null space of 2 5 A I 2 5 { } 5/2 and NulA I is the subspace span 2-Eigenspace: We now consider the matrix 5 5 A + 2I 2 2 { } and NulA + 2I is the subspace span Let us now consider the vectors v and v 2 given 5/2 v v 2

2 2 DIAGONALIZATION ie the vectors v and v 2 form bases of the eigenspaces corresponding to λ and λ 2 respectively in particular Av v Av 2 2v 2 Clearly the vectors v and v 2 are linearly independent and therefore they form a basis for R 2 It follows that A is diagnonalizable because the set C {v v 2 } forms a basis of R 2 Let us now use B to denote the standard basis of R 2 ie B {e e 2 } with We have bases and from e e 2 B {e e 2 } C {v v 2 } { Av v Av 2 2v 2 we take coordinates with respect to the basis C so that and therefore [Av ] C [v ] C e [Av 2 ] C 2[v 2 ] C 2e 2 [Av ] C [Av 2 ] C e 2e 2 2 On the other hand using the change-of-basis matrix P we have and in addition we observe that and we have shown that [Av ] C P Av [Av 2 ] C P Av 2 [v ] C e P v e v P e [v 2 ] C e 2 P v 2 e 2 v 2 P e 2 [Av ] C [v ] C e P A P e e [Av 2 ] C 2[v 2 ] C 2e 2 P A P e 2 2e 2

3 In matrix notation we have P A P e or and DIAGONALIZATION 3 P A P e 2 P A P 2 A P 2 e 2e 2 2 P P D P Let us now briefly recall how to compute the matrix P when B is the standard basis: in this case if we let P be the matrix with columns v v 2 ie if P v v 2 then P P so that at the end we obtain A P DP We have illustrated the proof of the following theorem Theorem Suppose that A is diagonalizable ie suppose that there is a basis C {v v n } of R n consisting of Eigenvectors of A ie there are numbers λ λ n such that Av λv Av n λ n v n Then there is an invertible matrix P such that A P DP where D is diagonal In fact we have λ D λ n and the matrix P can be taken to be P v v n ie a matrix whose columns are the eigenvectors of A in the same order as the elements λ λ n in the diagonal Definition 2 Given a linear transformation T : R n R n and a basis C {v v n } of R n the matrix of T in the basis C is the matrix A C [T v ] C [T v n ] C Note that in this case if A is the matrix of T in the standard basis B {e e n } of R n then A C P A P P AP A P A C P

4 4 DIAGONALIZATION where again the columns of P are the vectors v v n This says that A is diagonalizable if and only there exists a basis of R n such that in that basis the matrix of the linear transformation T x Ax is diagonal Let us now explore a simple computational application of diagonalizable matrices Suppose that as before that we have the matrix 3 5 A 2 4 with eigenvalues λ and λ 2 and eigenspaces spanned by the vectors 5/2 v v 2 respectively Then A P DP where and D 2 P v v 2 5/2 then in order to compute powers of A we observe that and note that in general A 2 P DP P DP P D 2 P A 3 P D 2 P P DP P D 3 P A n P D n P D n n 2 n which gives us a very convenient representation of the powers of A Let us consider further examples: Example 2 Consider the matrix A the characteristic polynomial of A is pλ deta λ I λ 2λ 5 2 so the eigenvalues of A are λ 2 and λ 5 Let us see what happens when λ 2 Observe that A 2 I

5 and the null space of A 2 I is span DIAGONALIZATION 5 A 5 I For λ 5 we have and observe that the rank of A 5 I is 2 and the null space of A 2 I is which implies that A is not diagonalizable since R 3 does not have a basis consisting of eigenvectors of A Note in fact that so that the 5-Eigenspace is span /2 Example 3 If now A / the characteristic polynomial of A is again λ 2λ 5 2 and for λ 2 we obtain A 2I 3 span 2 and also A 5 I 3 span /3 Note that in this case the matrix P is given by /3 /3 P /3 The previous two examples illustrate the following general result Theorem 2 Let A be an n n matrix If the characteristic polynomial of A has the form then pλ aλ λ m λ λ k m k

6 6 DIAGONALIZATION For each λ j the dimension of the Eigenspace corresponding to λ j is less than or equal to m j 2 A is diagonalizable if and only if for each j k the dimension of the Eigenspace corresponding to λ j is precisely m j 3 In particular if pλ factors as a product of n linear factors A is diagonalizable Remark: We will deal with the case of complex eigenvalues in the future