45. Decomposition into elementary components*

* Cf. Pickert Linear Algebra (Encyclopedia of Mathematical Sciences I1,6) Leipzig 1953

During a further transformation, one can restrict considerations, according to Theorem 27, to matrices with a single characteristic root. Let M be a matrix of type (m,m) with the single characteristic root b. Assume to start with that b=0. Then the characteristic equation of M is x^m = 0, whence there exists the Cayley-Hamilton relation M^m = O. There corresponds to the similarity class of M a linear mapping Y of an m-dimensional vector space W.

A lower power of M than the m-th can already equal the null-matrix, say

M^p = O, but M^p-1 O (p m).

Correspondingly,

Y ^p = W, but Y ^p-1 W.

If here p = 1, that it M = O, no further transformation is necessary. Now let in the sequel p > 1.

Let for k = 0, 1, ··· , p be that partial space of W which during the mapping is mapped onto o, be denoted by W_k. Now, due to Y ^p = W , you have that W^p=W and, due to Y ⁰ = I , that W₀ consists only of the null-vector. In the sequence of the vector spaces

W_p (= W), W_p-1 , ··· ,W₁, W₀ (= o), (45.1)

each is a genuine partial space of the preceding one. To start with, W_k is contained in W_k+1 (k + 1 p), because every vector x of W_k has the property Y ^kx = o, whence, obviously,

Y ^k+1(Y ^p-k-1 x) = Y ^px = Y o = W x = o,

whence x is contained in W_k+1. If you should have W_k+1 = W_k, you would have for every vector y of W, which satisfies the equation Y ^k+1y = o also
Y ^ky = o. For every vector x of W, y = Y ^p-k-1 x is now a vector which with Y ^k+1 is mapped on to o, because

Y ^k(Y ^p-k-1 x) = Y ^p x = W x = o.

Hence in the case W_k+1 = W_k also

Y ^k(Y ^p-k-1x) =Y ^p-1x = o.

If y is a vector in W_k+1, then Yy lies W_k, because while y lies in W_k+1, Y^k+1y=o; since you can write Y^k+1y = Y^k(Yy), Yy is mapped by Y^k= onto o, that is, it lies in W_k.

A system x₁, ··· ,x_s of vectors of W is said to be linearly independent over a partial space T, if there is not contained in T a linear combination l₁x₁, ··· ,l_sx_s apart from the trivial one with l₁ = ··· = l_s = 0. If T is the partial space, which only contains the null-vector, the linear independence over T means that x₁,···,x_s are linearly independent. A base of W over T is a system x₁, ··· , x_r comprising as many vectors as possible which are linearly independent over T. If you add to a base of T a base of W over T, you obtain a base of W, because, if t₁, ··· , t_t is a base on T, then the vectors

x₁, ··· , x_r , t₁, ··· , t_t (45.2)

are linearly independent, since in a relation

l₁x₁ + ··· + l_rx_r + m₁t₁ + ··· + m_tt_t = o

must, first of all, all l_t vanish, because otherwise there would exist in T a non-trivial linear combination of x₁, ··· , x_r; moreover, due to the linear independence of t₁, ··· , t_t, also all m_t must vanish . Next, let x be any vector of W. Then the vectors x, x₁, ··· , x_r form a system of r + 1 vectors, which must be linearly dependent over T. As a consequence, T contains a linear combination

t = lx + l₁x₁ + ··· + l_rx_r .

Since x₁, ··· , x_r are linearly independent over T, l 0, whence

x = -(l₁/l)x₁ - ··· - (l_r/l)x_r - (1/l)t.

However, since t is a linear combination of t₁, ··· , t_t, you can represent x as a linear combination of the vectors (45.2).

Auxiliary Theorem: If x₁, ··· , x_s is a system of vectors of W_k+1, which are linearly independent over W_k , the vectors Yx₁, ··· ,Yx_s are linearly independent over W_k-1.

Proof: If Yx₁, ··· ,Yx_s over W_k-1 were linearly dependent, there would not exist a non-trivial linear combination

x = g₁Yx₁+ ··· +g_sYx_s over W_k-1

which is contained in W_k-1. But then you would have

Y ^k-1x = Y ^k(g₁x₁+ ··· +g_sx_s) = o,

that is, g₁x₁+ ··· +g_sx_s is contained in W_k, which contradicts the assumption, whence the Theorem has been proved.

Construct now the following special base for W. Let be an arbitrary base of W over W_p-1. Then, according to the auxiliary Theoren\m, are vectors of W_p-1, which are linearly independent over W_p-2. If in case of need you extend it by to a base of W_p-2 over W_p-3, there arises the base of W_p-2 over W_p-3 with the form

Continuing in this manner until we arrive at a base of W₁ over W₀, that is, since W₀ = o, at

Combine all such relative bases to a base E of W in the sequence

If you subject these vectors to the mapping Y, you obtain with preservation of the sequence

Hence YE = CE with an (m,m)-matrix C, the form of which you can describe as follows:

We will understand by an elementary component C_k(0) of degree k > 1, belonging to the characteristic root 0, a (k,k)-matrix of the form

in which there occur units (1) in the diagonal above the principal diagonal and otherwise zeroes. An elementary component, belonging to 0, of degree 1 is the (1,1)-matrix 0. The matrix decomposes entirely into

* If m_k = m_k+1, there do not occur elementary components of degree p - k.

Hitherto, it has been assumed that the matrix M, from which you started, had the characteristic root 0. However, if b is the characteristic root of M, then, by Theorem 23, M₀=M-bE has the characteristic root 0, and the preceding reasoning can be applied to M₀. Since

P^-1MP = P^-1M₀P + bE,

decomposes P^-1MP = completely into elementary components, which belong to b

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