43. Linear mapping

It is important for the sequel to obtain a less formal, conceptive view of similarity.

Let V be an n-dimensional vector space. A linear mapping of V onto itself is a function F, which allots uniquely to every vector x of V a unique image vector Fx, also contained in V, and possesses the properties:

F (x + y) = F x + F y, F (lx) = l·F x ( l is a number). (43.1)

The totality of all such image vectors will be denoted by FV. Obviously, FV is a vector space, contained in V; it can happen that FV is a genuine partial space of V.

Special linear mappings are the zero-mapping W, which allots to every vector the null-vector, and the identical mapping which maps every vector onto itself.

If e₁ ,··· , e_n is a base of V and

x = x₁e₁+ ··· + x_ne_n,

then, by (43.1),

Fx = x₁F e₁+ ··· + x_nF e_n.

Hence the image of every vector is already determined when the images of a base are known. Let the images of the base vectors be

Combining the base vectors e₁, ··· , e_n in a column E, you can rewrite (43.2)

FE = AE with A = (a_ik). * (43.3)

*The fact the the column E consists of vectors rather than numbers is of no consequence for the formal manipulations.

Through the establishment of a fixed base, there corresponds to every linear mapping a well-defined (n,n)-matrix, and conversely such a matrix determines always exactly one linear mapping. To the null-mapping W, there corresponds the null-matrix, to the identical mapping I, the unit matrix.

We examine now how the matrix, allotted to the mapping F, changes as one transits from the base e₁, ··· , e_n to a new base f₁, ··· , f_n of V. You can express the vectors e₁, ··· , e_n in terms of the new base by

Combining the f₁, ··· , f_n in a column F, you can write

E = PF with P = (p_ik), (43.4)

where, by Theorem 9, |P| 0.

Substitution of (43.4) into (43.3) yields

F (PF) = APF. (43.5)

By (43.1), F (PF) = P(FF), whence (43.5) yields

PF = P^-1APF

and there corresponds to a transition to another base the replacement of A by a similar matrix. Conversely, similar matrices belong to the same linear mapping when suitable bases are employed. This result yields

Theorem 28

Classes of similarity correspond uniquely invertibly to linear mapping of V onto itself.

In this context. also the complete decomposition of a matrix A into diagonal minors can be interpreted conceptually. Let

have square partial matrices of the type (n₁,n₁) and (n₂,n₂), respectively, where n₁ + n₂ = n. The mapping F then maps everyone of the n₁ base vectors into a linear combination of just these vectors. corresponding result applies to the n₂ remaining base vectors , the images of which express themselves alone in terms of these n₂ vectors. The entire n-dimensional vector space is decomposed into two partial spaces V₁ and V₂ with dimensions, covered by the and , respectively. Obviously, the two partial spaces share only the null-vector. You say that V decomposes into the direct sum of V₁ and V₂ and write V=V₁+V₂. Since each of the two partial spaces V₁ and V₂ is mapped into itself, they are referred to as admissible partial spaces with respect to the mapping F. As mapping of V₁, the mapping F corresponds in itself to the matrix A₁, as mapping of V₂ to the matrix A₂.

Conversely, if V is the direct sum of two admissible partial spaces V₁ and V₂ of dimensions n₁ and n₂, respectively, and you employ a base of V, which has been composed out of one base of V₁ and one of V₂, then the matrix belonging to the mapping F decomposes completely into two diagonal boxes with n₁ and n₂ rows, respectively.

Analogous arguments for the decomposition into more than two partial spaces and diagonal boxes, respectively, are obvious.

The linear mapping F can be iterated. You define

F ^kx = F (F^k-1 x) (k = 0, 1, ··· ; F ⁰ = I).

(43.3) yields

F(FE) = F(AE) = A(FE) = A(AE) = A²E,

and, in general,

F^kE = A^kE (k = 0, 1, ··· ; A⁰ = E).

You an also define polynomials in a linear mapping F. If at a definite base of V there corresponds to the mapping F the matrix A and

f(x) = c₀ + c₁x + ··· + c_mx^m

is an arbitrary polynomial, then f(F) is that linear mapping, to which is allotted at the selected base the polynomial

f(A) = c₀E + c₁A + ··· + c_mA^m

as matrix. This definition of f(F) does not depend on the base of the vector space, for, if you introduce by (43.4) a new base, you have to replace A by P^-1AP and f(A) by f(P^-1AP). On the other hand, there corresponds to f(F) at the new base the matrix P^-1f(A)P, However, you have

P^-1f(A)P = f(P^-1AP).

If f(x) and g(x) are two polynomials and

s(x) = f(x) + g(x), p(x) = f(x) g(x),

then s(A) = f(A) + g(A) and p(A) = f(A) g(A) and therefore

s(F) = f(F) + g(F), p(F) = f(F) g(F).

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