Recall from §5.2.5 that a matrix A representing a map A is diagonal with respect to a basis if and only if the basis consists of eigenvectors.
Thisisaspecificindividual’scopyofthenotes.Itisnottobecopiedand/orredistributed. Definition. A linear map A : Fn 7→ Fn is said to be diagonalizable if Fn has a basis consisting of
eigenvectors of A.
Further, we have shown that a map A with n distinct eigenvalues has a basis of eigenvectors. It follows that if a matrix A has n distinct eigenvalues, then it is diagonalizable by means of a similarity transformation using the transformation matrix that changes to a basis of eigenvectors.
Definition. More generally we say that a n × n matrix A [over F] is diagonalizable if A is similar with a diagonal matrix, i.e. if there exists an invertible transformation matrix P [with entries in F] such that P−1AP is diagonal.
5.5.1 When is a matrix diagonalizable?
While the requirement that a matrix A has n distinct eigenvalues is a sufficient condition for the matrix to be diagonalizable, it is not a necessary condition. The requirement is that Fn has a basis consisting of eigenvectors; this is possible even if the eigenvalues are not distinct as the following example shows.
Example. In example (ii) on page 84 we saw that if a map A is represented with respect to a given basis by a matrix still construct linearly independent eigenvectors
x1= matrix for transforming components from the original basis to the eigenvector basis is given by
P = From the expression for an inverse (4.9)
P−1=1
Hence from (5.21) the map A is represented with respect to the eigenvector basis by
P−1AP = 1
We need to be able to count the number of linearly independent eigenvectors.
Proposition. Suppose λ1, . . . , λrare distinct eigenvalues of a linear map A : Fn7→ Fn, and let Bi denote a basis of the eigenspace Eλi. Then
B= B1∪ B2∪ . . . ∪ Br
is a linearly independent set.
Thisisaspecificindividual’scopyofthenotes.Itisnottobecopiedand/orredistributed. Proof. Argue by contradiction. Suppose that the set B is linearly dependent, i.e. suppose that there
exist cij ∈ F, not all of which are zero, such that
to (5.27) to conclude that
0 = Y
i.e. if no eigenvalue has a non-zero defect, then B is a basis [of eigenvectors], and a matrix A representing A is diagonalizable. However, a matrix with an eigenvalue that has a non-zero defect does not have sufficient linearly independent eigenvectors to be diagonalizable, cf. example (iii) on page 84.
Procedure (or ‘recipe’). We now can construct a procedure to find if a matrix A is diagonalizable, and to diagonalize it where possible:
(i) Calculate the characteristic polynomial pA= det(A − λI).
(ii) Find all distinct roots, λ1, . . . , λr of pA.
(iii) For each λi find a basis Bi of the eigenspace Eλi.
(iv) If no eigenvalue has a non-zero defect, i.e. if (5.28a) is true, then A is diagonalizable by a transformation matrix with columns that are the eigenvectors.
(v) However, if there is an eigenvalue with a non-zero defect, i.e. if X
i
dim Eλi ≡X
i
mλi< n , (5.28b)
then A is not diagonalizable.
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Thisisaspecificindividual’scopyofthenotes.Itisnottobecopiedand/orredistributed. 5.5.2 Canonical form of 2 × 2 complex matrices
We claim that any 2 × 2 complex matrix A is similar to one of (i) : λ1 0
Proof. If A has distinct eigenvalues, λ1 6= λ2 then, as shown above, A is diagonalizable by a similarity transformation to form (i). Otherwise, λ1= λ2 = λ, and two cases arise according as dim Eλ= 2 and dim Eλ= 1. If dim Eλ= 2 then Eλ= C2. Let B = {u, v} be a basis of two linearly independent vectors. Since Au = λu and Av = λv, A transforms to a diagonal matrix in this basis, i.e. there exists a transformation matrix P such that
P−1AP =λ 0
0 λ
= λI , (5.30)
i.e. form (ii). However, we can say more than this since it follows from (5.30) that A = λI (see also the exercise on page 91 following (5.24)).
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If dim Eλ = 1, take non-zero v ∈ Eλ, then {v} is a basis for Eλ. Extend this basis for Eλ to a basis B = {v, w} for C2 by choosing w ∈ C2\Eλ. If Aw = αv + βw it follows that there exists a transformation matrix P, that transforms the original basis to B, such that
A = Pe −1AP =λ α 0 β
.
However, if eA, and thence A, is to have an eigenvalue λ of algebraic multiplicity of 2, then A =e λ α
we conclude that u is an eigenvector of eA (and w is a generalised eigenvector of eA). Moreover, since Aw = u + λw ,e
if we consider the alternative basis eB= {u, w} it follows that there is a transformation matrix, say Q, such that
(PQ)−1A(PQ) = Q−1AQ =e λ 1
0 λ
. We conclude that A is similar to a matrix of form (iii).
Remarks.
Thisisaspecificindividual’scopyofthenotes.Itisnottobecopiedand/orredistributed. (b) It is possible to show that canonical forms, or Jordan normal forms, exist for any dimension n.
Specifically, for any complex n × n matrix A there exists a similar matrix eA = P−1AP such that Aeii = λi, Aei i+1= {0, 1}, Aeij= 0 otherwise, (5.31) i.e. the eigenvalues are on the diagonal of eA, the super-diagonal consists of zeros or ones, and all other elements of eA are zero.29
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Unlectured example. For instance, suppose that as in example (iii) on page 84,
A =
Recall that A has a single eigenvalue λ = −2 with an algebraic multiplicity of 3 with a defect of 1.
Consider a vector w that is linearly independent of the eigenvectors (5.11c) (or equivalently (5.11d)), e.g. where it is straightforward to check that u is an eigenvector. Now form
P = (u w v) =
where v is an eigenvector that is linearly independent of u. We can then show that
P−1= and thence that A has a Jordan normal form
P−1AP =
5.5.3 Solution of second-order, constant coefficient, linear ordinary differential equations Consider the solution of
¨
x + b ˙x + cx = 0 , (5.32a)
where b and c are constants. If we let ˙x = y then this second-order equation can be expressed as two first-order equations, namely
29This is not the whole story, since eA is in fact block diagonal, but you will have to wait for Linear Algebra to find out about that.
Thisisaspecificindividual’scopyofthenotes.Itisnottobecopiedand/orredistributed. This is a special case of the general second-order system
˙x = Ax , (5.33a)
where
x =x y
and A =A11 A12
A21 A22
. (5.33b)
Let P be a transformation matrix, and let z = P−1x, then
˙z = Bz , where B = P−1AP . (5.34)
By appropriate choice of P, it is possible to transform A to one of three canonical forms in (5.29). We consider each of the three possibilities in turn.
(i) In this case
˙z =λ1 0 0 λ2
z , (5.35a)
i.e. if
z =z1
z2
, then z˙1= λ1z1 and z˙2= λ2z2. (5.35b) This has solution
z =α1eλ1t α2eλ2t
, (5.35c)
where α1 and α2 are constants.
(ii) This is case (i) with λ1= λ2= λ, and with solution z =α1
α2
eλt. (5.36)
(iii) In this case
˙z =λ 1
0 λ
z , (5.37a)
i.e.
˙
z1= λz1+ z2 and z˙2= λz2, (5.37b) which has solution
z =α1+ α2t α2
eλt. (5.37c)
We conclude that the general solution of (5.33a) is given by x = Pz where z is the appropriate one of (5.35c), (5.36) or (5.37c).
Remark. If λ in case (iii) is pure imaginary, then this is a case of resonance in which the amplitude of an oscillation grows algebraically in time.