0 and, due to (44.3),
also f1(a2)
0. Hence not all characteristic roots of j 1(A)f1(A)
equal zero, that is j 1(a2)f1(a2) 0 (cf. Theorem
23). In an analogous
manner, it is proved that also F2 W and
F1 I.44. Decomposition into components with a single characteristic root.
For a complete system of similarity invariants, we will proceed as follows: In every similarity class, we will select a matrix of an especially simple form, a so-called normal form from which the invariants of the class can be read off.v Since matrices of a similarity class are concerned with different modes of presentation and the same linear mapping, it is natural to introduce this concept.
Let A be an (n,n)-matrix, which we will interpret as the matrix of a linear mapping F of an n-dimensional vector space V, and j (x) the characteristic polynomial of A. Since by the Cayley-Hamilton relation j (A) = O, you find
j (F) = W. (44.1)
We will assume that A has at least two different characteristic roots. You can then decompose j (x) into a product
j (x) = j 1(x)j 2(x) (44.2)
of two polynomials j 1(x) and j 2(x) of positive degree, which do not have zeroes in common. According to Theorem 1, there exist then two polynomials f1(x) and f2(x) for which
j 1(x)f1(x) + j 2(x)f2(x) = 1. (44.3)
Now let
j 1(F)f1(F) = F1, j 2(F)f2(F) = F2.
Then, by (44.3),
F1 + F2 = I. (44.4)
Equations (44.1) and (44.2) yield
F1F2 = F2F1 = j 1(F)f1(F)j 2(F)f2(F) = j (F)f1(F)f2(F) = W. (44.5)
Moreover, F1 as well as F2 differ from I and W, because, if F1 = W and therefore, by (44.4), at the same
time F2 = I, then you
should have j1(A)f1(A)=0.
In particular, also all characteristic roots of j 1(A)f1(A)
must vanish. Now let a2
be a characteristic root of A and also a zero of j 2(x). Then j 1(a2) 0 and, due to (44.3),
also f1(a2)
0. Hence not all characteristic roots of j 1(A)f1(A)
equal zero, that is j 1(a2)f1(a2) 0 (cf. Theorem
23). In an analogous
manner, it is proved that also F2 W and
F1 I.
We consider now the two partial spaces V1 = F1V and V1 = F1V. These two partial spaces do not contain common vectors apart from the null-vector, because, if x is in V1 as well as in V2, then x = F1y = F2z. It follows from (44.5) that F1x = F1F2z = W z = 0 and F2x = F2F1y = W y = 0, whence x=Ix = (F1 + F2)x = F1x + F2x = 0. By (44.4), for every vector of V,
x=Ix = (F1 + F2)x = F1x + F2x = x1 + x2,
where x1 is in V1 and x2 in V2, whence V = V1 + V2. Moreover, V1 as well as V2, are admissible with respect to the mapping F, for, if x1 is any vector of V1, that is, x1 = F1x, then
F1x = x = FF1x = F1Fx = F1y
with y = Fx, so that F1x is again a vector of V1. In an analogous manner, you show that also V2 is admissible.
If you now establish a base for V, which is composed of the two bases of V1 and V2, respectively, then, according to what has been discussed in Section 43, there corresponds to the mapping F a matrix, which decomposes completely into two diagonal boxes:
In the case that one of these diagonal boxes or both of them have different characteristic roots, you can continue the direct decomposition. Eventually you obtain
If the matrix A has different characteristic roots a1, a2 ··· , ar, then there exists a matrix similar to A of the form
where Ai is a matrix of the type (ni,ni) with the single characteristic root ai.