47. Similarity to diagonal matrices

We will deal here only with one of the many applications of Jordan's normal form - with the question when a matrix A is similar to a diagonal matrix. Obviously, this is just the case when all elementary components in the Jordan normal form of A had the degree 1. However, this statement does not help much, because the real construction of the normal form is very laborious. A more convenient criterion is

Theorem 30

A is similar to a diagonal matrix if and only if there exists a polynomial f(x) without multiple roots for which f(A) = O.

Proof: Denote by a₁, ··· , a_r the different characteristic roots of A. If P^-1AP=D is a diagonal matrix, the numbers a₁, ··· , a_r appear, possibly several times, in the principal diagonal.

Letting

y (x) = (x - a₁) ··· (x - a_r),

y (D) becomes a diagonal matrix with the diagonal elements y(a₁), ··· , y(a_r) at certain multiplicities. Since

y(a₁) = ··· = y(a_r) = 0,

you have y (D) = O, whence

y (A) = y (PDP^-1) = Py (D)P^-1 = O,

and since y (x) does not have multiple roots, one part of the Rule has been proved.

Conversely, let f(x) be a polynomial without multiple roots, for which f(A) = O. Since therefore all the characteristic roots of f(A) vanish, you must have f(a₁)=···=f(a_r)=0, whence f(x) contains every linear factor x - a_i, and indeed in the first power. Therefore you can write

f(x) = (x - a₁)f₁(x),

where f₁(a₁) 0. If C = P^-1AP is the Jordan normal form of A, then f(C)=P^-1f(A)P=O. According to the decomposition (46.5),

f(C) = f(D₁) + f(D₂) + · · · + f(D_r).

In particular, f(C) = O yields f(D₁) = O. However,

f(D₁) = (D₁ - a₁E)f₁(D₁). (47.1)

Since D₁ has only the roots a₁, the single characteristic root f(a₁) of f₁(D₁) differs from zero, so that f₁(D₁) is a regular matrix. Therefore, by (47.1), it follows from f(D₁) = O that

D₁ - a₁E = O.

However, this shows that all elementary components of D₁ must be of degree 1. Correspondingly, you prove that also the elementary components, belonging to the other characteristic roots, must have the degree 1. This proves Theorem 30.

Another criterion yields

Theorem 31

A matrix A is similar to a diagonal matrix if and only if for every characteristic root a_i of A there follows from

(A - a_iE)² = o (47.2)

that

(A - a_iE)x = o. (47.3)

Proof: Since there follows certainly (47.2) from (47.3), the proof with consideration of Theorem 14 shows that A is similar to a diagonal matrix if and only if

Rank (A - a_iE)² = Rank (A - a_iE).

However, with P^-1AP = C,

Rank (A - a_iE)² = Rank P^-1(A - a_iE)²P = Rank (C - a_iE)²

and

Rank (A - a_iE) = Rank P^-1(A - a_iE)P = Rank (C - a_iE).

Now introduce for C the Jordan normal form of A. Then, if C is a diagonal matrix, you have obviously in the principal diagonal of C- a_iE and (A - a_iE)² the same number of non-zero elements, whence

Rank (C - a_iE)² = Rank (C - a_iE).

Conversely, if this equation holds, (46.7) and (46.8) show that necessarily k = 1 in the given notation. However, this shows that all elementary components are of degree 1, whence C is a diagonal matrix.

Exercises

31. Every square matrix A can certainly be interchanged with all polynomials in A. Show that: If A is

a) a diagonal matrix with all different diagonal elements or
b) a single elementary component

then polynomials in A are the only matrices which can be interchanged with A.

32. Every square matrix is similar to its transposed matrix.

33. Find the Jordan normal form of the (n,n)-matrices:

a) When every element = 1.
b) When all elements in and below the main diagonal equal 1, those above it equal 0.
c) When it is Hermitian?

Answers

last