Scalar multiplication

Equations (8.20) are not sufficiently specific for practical purposes, so it is important to consider special cases leading to simpler relations. When the pore orientations are isotropically distributed, the second order tensors k, 3 and y are isotropic and are therefore scalar multiples of the unit tensor. Thus equation (8.20) simplifies to... 

Chapter 9 dealt with the basic operations of addition of two matrices with the same dimensions, of scalar multiplication of a matrix with a constant, and of arithmetic multiplication element-by-element of two matrices with the same... 

For simplicity we speak of a mechanism or a reaction, rather than a mechanism vector or reaction vector. The distinction lies in the fact that a reaction r (or mechanism) is essentially the same whether its rate of advancement is p or a, whereas pr and or are different vectors (for p a). Therefore, a reaction could properly be defined as a one-dimensional vector space which contains all the scalar multiples of a single reaction vector, but the mathematical development is simpler if a reaction is defined as a vector. This leaves open the question of when two reactions, or two mechanisms, are essentially different from a chemical viewpoint, which will be taken up... 

In a similar way scalar multiplications of Eq. (1) by E and Eq. (2) by B/p0 yields, after subtraction of the resulting equations, the local energy balance equation... 

Exercise 1.19 Define an arrow in R to be an ordered pair (p, pf), where Pi and p2 are each a triple of real numbers. (Think of pi as the initial point and p as the endpoint.) Define a relation on the set of arrows by (/ b pf) (qt q .) if and only if P2 — Pi = qi qi- Show that this is an equivalence relation. Now think of each arrow as a point in R. Does the usual addition in R survive the equivalence relation If so, is the resulting addition on equivalence classes of arrows the same as the addition of 3-vectors you learned in linear algebra What about scalar multiplication in R . Find an injective and surjective linear function from R / to R. (Hint it will help to introduce some notation for (r, r2, r, r4, r, r(,) e R in the equivalence class corresponding to (si, 52, 53) e R. )... 

Show that satisfies the criteria of Definition 1.3. Show that f g if and only if f -g is a constant function. Show that addition, scalar multiplication and differentiation are well defined on equivalence classes. Show that evaluation is not well defined given a point c e [a, b, find two functions in S that are equivalent but take different values at c. On the other hand, differences of evaluations are well defined show that f(fi — f a) is well defined on equivalence classes. 

Does addition of functions survive the equivalence Does scalar multiplication (by complex numbers) survive the equivalence Does multiplication of two functions survive the equivalence ... 

In this chapter we introduce complex linear algebra, that is, linear algebra where complex numbers are the scalars for scalar multiplication. This may feel like review, even to readers whose experience is limited to real linear algebra. Indeed, most of the theorems of linear algebra remain true if we replace R by C because the axioms for a real vector space involve only addition and multiplication of real numbers, the definition and basic theorems can be easily adapted to any set of scalars where addition and multiplication are defined and reasonably well behaved, and the complex numbers certainly fit the bill. However, the examples are different. Furthermore, there are theorems (such as Proposition 2.11) in complex linear algebra whose analogues over the reals are false. We will recount but not belabor old theorems, concentrating on new ideas and examples. The reader may find proofs in any number of... 

For example, the real line R is not a complex vector space under the usual multipUcation of real numbers by complex numbers. It is possible for the product of a complex number and a real number to be outside the set of real numbers for instance, (z)(3) = 3i R. So the real line R is not closed under complex scalar multiplication. 

The trivial complex vector space has one element, the zero vector 0. Addition is defined by 0 -I- 0 = 0 for any complex number c, define the scalar multiple of 0 by c to be 0. Then all the criteria of Definition 2.1 are trivially true. For example, to check distributivity, note that for any c e C we have... 

If a subset W of a vector space V satisfies the definition of a vector space, with addition and scalar multiplication defined by the same operation as in V, then W is called a vector subspace or, more succinctly, a subspace of V. For example, the trivial subspace 0 is a subspace of any vector space. 

Because the sum of two continuous functions is continuous, and any scalar multiple of a continuous function is continuous, C[—1, 1] is indeed a vector space. 

The notion of a linear transformation is crucial. A function from a (complex) vector space to a (complex) vector space is a (complex) linear transformation if it preserves addition and (complex) scalar multiplication. Here is a more explicit definition. 

Proof. If A is diagonal, then an easy computation shows that AD — DA = 0. To prove the other implication, suppose that AD — DA = 0, Let e denote the fth standard basis vector of C . Then 0 = AD — DAid = DuAci — DAa. So Aci is an eigenvector of D with eigenvalue D. Because Da Djj unless i = j, it follows that Aa must be a scalar multiple of a for each L Hence A must be diagonal. ... 

Exercise 2.1 Consider the set of homogeneous polynomials in two variables with real coefficients. There is a natural addition of polynomials and a natural scalar multiplication of a polynomial by a complex number. Show that the set of homogeneous polynomials with these two operations is not a complex vector space. 

Exercise 2.3 Show that C (with the usual addition and multiplication) is itself a complex vector space of dimension 1. Then show thatC with the usual addition but with scalar multiplication by real numbers only is a real vector space of dim ension 2. 

Exercise 2.6 Let V be an arbitraty complex vector space of dimension n. Show that by restricting scalar multiplication to the reals one obtains a real vector space of dimension 2n. 

The functions in A form a complex vector space under the usual addition and scalar multiplication of functions. 

This fact will be at the heart of the proof of our main result in Section 6.5. Proof. First, we show that V satisfies the hypotheses of the Stone-Weierstrass theorem. We know that V is a complex vector space under the usual addition and scalar multiplication of functions adding two polynomials or multiplying a polynomial by a constant yields a polynomial. The product of two polynomials is a polynomial. To see that V is closed under complex conjugation, note that for any x e [—1, 1] and any constant complex numbers flo, , a sN[Pg.102]

Proposition 6,3 Suppose (G, V, p) is a finite-dimensional irreducible representation. Then every linear operator T . V V that commutes with p is a scalar multiple of the identity. In other words. ifT. V Visa homomorphism of representations, then T is a scalar multiple of the identity. 

Proof. Suppose that (G, V, p) is irreducible and the linear transformation T V —> V commutes with p. We must show that T is a scalar multiple of the identity. Because V is finite dimensional there must be at least one eigenvalue Z of T (by Proposition 2.11). By Proposition 5.2. the eigenspace corresponding to A must be an invariant space for p. This space is not trivial, so because p is irreducible it must be all of V. In other words, T =. 1. So T is a scalar multiple of the identity. ... 

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