The associative law of multiplication enables us to write the product of three (or more) factors without any parentheses; in particular we may consider the product of any finite number, say, \(m\) , of factors all equal to \(A\) . This product depends only on \(A\) and on \(m\) (and not, as we just remarked, on any bracketing of the factors); we shall denote it by \(A^{m}\) . The justification for this notation is that, although in general transformation multiplication is not commutative, for the powers of one transformation we do have the usual laws of exponents, \(A^{n} A^{m}=A^{n+m}\) and \((A^{n})^{m}=A^{n m}\) . We observe that \(A^{1}=A\) ; it is customary also to write, by definition, \(A^{0}=1\) . With these definitions the calculus of powers of a single transformation is almost exactly the same as in ordinary arithmetic. We may, in particular, define polynomials in a linear transformation. Thus if \(p\) is any polynomial with scalar coefficients in a variable \(t\) , say \(p(t)=\alpha_{0}+\alpha_{1} t+\cdots+\alpha_{n} t^{n}\) , we may form the linear transformation

\[p(A)=\alpha_{0} 1+\alpha_{1} A+\cdots+\alpha_{n} A^{n}.\] 

The rules for the algebraic manipulation of such polynomials are easy. Thus \(p(t) q(t)=r(t)\) implies \(p(A) q(A)=r(A)\) (so that, in particular, any \(p(A)\) and \(q(A)\) are commutative); if \(p(t)=\alpha\) (identically), we shall usually write \(p(A)=\alpha\) (instead of \(p(A)=\alpha \cdot 1\) ); this is in harmony with the use of the symbols \(0\) and \(1\) for linear transformations.

If \(p\) is a polynomial in two variables and if \(A\) and \(B\) are linear transformations, it is not usually possible to give any sensible interpretation to \(p(A, B)\) . The trouble, of course, is that \(A\) and \(B\) may not commute, and even a simple monomial, such as \(s^{2} t\) , will cause confusion. If \(p(s, t)=s^{2} t\) , what should we mean by \(p(A, B)\) ? Should it be \(A^{2} B\) , or \(A B A\) , or \(B A^{2}\) ? It is important to recognize that there is a difficulty here; fortunately for us it is not necessary to try to get around it. We shall work with polynomials in several variables only in connection with commutative transformations, and then everything is simple. We observe that if \(A B=B A\) , then \(A^{n} B^{m}=B^{m} A^{n}\) , and therefore \(p(A, B)\) has an unambiguous meaning for every polynomial \(p\) . The formal properties of the correspondence between (commutative) transformations and polynomials are just as valid for several variables as for one; we omit the details.

For an example of the possible behavior of the powers of a transformation we look at the differentiation transformation \(D\) on \(\mathcal{P}\) (or, just as well, on \(\mathcal{P}_n\) , for some \(n\) ). It is easy to see that for every positive integer \(k\) , and for every polynomial \(x\) in \(\mathcal{P}\) , we have \((D^{k} x)(t)=\frac{d^{k} x}{d t^{k}}\) . We observe that whatever else \(D\) does, it lowers the degree of the polynomial on which it acts by exactly one unit (assuming, of course, that the degree is \(\geq 1\) ). Let \(x\) be a polynomial of degree \(n-1\) , say; what is \(D^{n} x\) ? Or put it another way: what is the product of the two (commutative) transformations \(D^{k}\) and \(D^{n-k}\) (where \(k\) is any integer between \(0\) and \(n\) ), considered on the space \(\mathcal{P}_n\) ? We mention this example to bring out the disconcerting fact implied by the answer to the last question; the product of two transformations may vanish even though neither one of them is zero. A non-zero transformation whose product with some non-zero transformation is zero is called a divisor of zero.

EXERCISES