+ a_{1}λ + a_{0} where n is a positive integer, define
′ n− 1 n−2
p (λ) ≡ nanλ + (n− 1)an−1λ +⋅⋅⋅+ a1
In other words, you use the usual rules of differentiation in calculus to write down this formal derivative. It
has absolutely no physical significance in this context because the coefficients are just elements of some
field, possibly ℤ_{p}. It is a purely algebraic manipulation. A term like ka where k ∈ ℕ and a ∈ F
means to add a to itself k times. There are no limits or anything else. However, this has certain
properties. In particular, the “derivative” of a sum equals the sum of the derivatives. This is
fairly clear from the above definition. You just need to always be considering polynomials. Also
A short computation shows that this is indeed the case. Then by induction one can conclude
that
(∏p ) ∑p ∏
pi(λ) = p′j(λ) pi(λ)
i=1 j=1 i⁄=j
In particular, if
p∏ ki
p(λ) = (λ − μi)
i=1
then
∑p ∏
p′(λ) = kj(λ− μj)kj−1 (λ − μi)ki
j=1 i⁄=j
I want to emphasize that this is an arbitrary field of scalars, but if one is only interested in the real or
complex numbers, then all of this follows from standard calculus theorems.
Proposition 6.4.1Suppose the minimum polynomial p
(λ)
of an n×n matrix A completely factorsinto linear factors. Then A is diagonalizableif and only if p
(λ)
,p^{′}
(λ)
are relatively prime.
Proof:Suppose p
(λ)
,p^{′}
(λ)
are relatively prime. Say
∏n
p(λ) = (λ − μi)ki , μi are distinct
i=1
From the above discussion,
∑p ∏
p′(λ) = kj(λ− μj)kj−1 (λ − μi)ki
j=1 i⁄=j
and p^{′}
(λ )
,p
(λ)
are relatively prime if and only if each k_{i} = 1. Then by Corollary 6.3.8 this is true if and
only if A is diagonalizable. ■
Example 6.4.2Find whether the matrix
( )
1 − 1 2
A = |( 0 1 2 |)
1 − 1 1
is diagonalizable. Assume the field of scalars is ℂ because in this field, the minimum polynomial will factorthanks to the fundamental theorem of algebra.
Then we need to have for a linear combination involving a,b,c,d as scalars
a+ b + 3c + 5d = 0
2c +6d = 0
b+ 2c+ 3d = 0
− b − 3c− 6d = 0
Then letting d = 1, this gives only one solution,
a = 1,b = 3,c = − 3
and so the candidate for the minimum polynomial is
λ3 − 3λ2 + 3λ + 1
In fact, this does work as is seen by substituting A for λ. So is this polynomial and its derivative relatively
prime?
λ3 − 3λ2 + 3λ + 1 = 1 (λ − 1)(3λ2 − 6λ +3) + 2
3
and clearly
( )
3λ2 − 6λ + 3
and 2 are relatively prime. Hence this matrix is diagonalizable. Of course,
finding its diagonalization is another matter. For this algorithm for determining whether two polynomials
are relatively prime, see Problem 34 on Page 95.
Of course this was an easy example thanks to Problem 12 on Page 386. because there are three distinct
eigenvalues, one real and two complex which must be complex conjugates. This problem says that
eigenvectors corresponding to distinct eigenvalues are an independent set.
Consider the following example in which the eigenvalues are not distinct, consisting of a,a.
Then we need to have for a linear combination involving scalars x,y,z
x+ (a +1)y + (a2 + 2a)z = 0
( y+ 2a)z = 0
x+ (a − 1)y + a2 − 2a z = 0
Then some routine row operations yield x = a^{2}z, y = −2az and z is arbitrary. For the minimum
polynomial, we take z = 1 because this is a monic polynomial. Thus the minimum polynomial
is
a2 − 2aλ+ λ2 = (λ − a)2
and clearly this and its derivative are not relatively prime. Thus this matrix is not diagonalizable for any
choice of a.