We consider next the so-called Neumann series \(\sum_{n=0}^{\infty} A^{n}\) , where \(A\) is a linear transformation with norm \(< 1\) on a finite-dimensional vector space. If we write \[S_{p}=\sum_{n=0}^{p} A^{n},\] then \begin{align} (1-A) S_{p}=S_{p}-A S_{p}=1-A^{p+1} \tag{1}. \end{align} To prove that \(S_{p}\) has a limit as \(p \to \infty\) , we consider (for any two indices \(p\) and \(q\) with \(p>q\) ) \[\|S_{p}-S_{q}\| \leq \sum_{n=q+1}^{p}\|A^{n}\| \leq \sum_{n=q+1}^{p}\|A\|^{n}.\] Since \(\|A\|<1\) , the last written quantity approaches zero as \(p, q \to \infty\) ; it follows that \(S_{p}\) has a limit \(S\) as \(p \to \infty\) . To evaluate the limit we observe that \(1-A\) is invertible. (Proof: \((1-A) x=0\) implies that \(A x=x\) , and, if \(x \neq 0\) , this implies that \(\|A x\|=\|x\|>\|A\| \cdot\|x\|\) , a contradiction.) Hence we may write (1) in the form \begin{align} S_{p}=(1-A^{p+1})(1-A)^{-1}=(1-A)^{-1}(1-A^{p+1}); \tag{2} \end{align} since \(A^{p+1} \to 0\) as \(p \to \infty\) , it follows that \(S=(1-A)^{-1}\) .

As another example of an infinite series of transformations we consider the exponential series. For an arbitrary linear transformation \(A\) (not necessarily with \(\|A\|<1\) ) we write \[S_{p}=\sum_{n=0}^{p} \frac{1}{n!} A^{n}.\] Since we have \[\|S_{p}-S_{q}\| \leq \sum_{n=q+1}^{p} \frac{1}{n!}\|A\|^{n},\] and since the right side of this inequality, being a part of the power series for \(\exp \|A\|=e^{\|A\|}\) , converges to \(0\) as \(p, q \to \infty\) , we see that there is a linear transformation \(S\) such that \(S_{p} \to S\) . We write \(S=\exp A\) ; we shall merely mention some of the elementary properties of this function of \(A\) .

Consideration of the triangular forms of \(A\) and of \(S_{p}\) shows that the proper values of \(\exp A\) , together with their algebraic multiplicities, are equal to the exponentials of the proper values of \(A\) . (This argument, as well as some of the ones that follow, applies directly to the complex case only; the real case has to be deduced via complexification.) From the consideration of the triangular form it follows also that the determinant of \(\exp A\) , that is, \(\prod_{i=1}^{N} \exp \lambda_{i}\) , where \(\lambda_{1}, \ldots, \lambda_{N}\) are the (not necessarily distinct) proper values of \(A\) , is the same as \(\exp (\lambda_{1}+\cdots+\lambda_{N})=\exp(\operatorname{tr} A)\) . Since \(\exp \zeta \neq 0\) , this shows, incidentally, that \(\exp A\) is always invertible.

Considered as a function of linear transformations the exponential retains many of the simple properties of the ordinary numerical exponential function. Let us, for example, take any two commutative linear transformations \(A\) and \(B\) . Since \(\exp (A+B)-\exp A \exp B\) is the limit (as \(p \to \infty\) ) of the expression \begin{align} &\sum_{n=0}^{p} \frac{1}{n!}(A+B)^{n}-\sum_{m=0}^{p} \frac{1}{m!} A^{m} \cdot \sum_{k=0}^{p} \frac{1}{k!} B^{k} \\ = &\sum_{n=0}^{p} \frac{1}{n!} \sum_{j=0}^{n}\binom{n}{j} A^{j} B^{n-j}-\sum_{m=0}^{p} \sum_{k=0}^{p} \frac{1}{m!k!} A^{m} B^{k}, \end{align} we will have proved the multiplication rule for exponentials when we have proved that this expression converges to zero. (Here \(\binom{n}{j}\) stands for the combinatorial coefficient \(\frac{n!}{j!(n-j)!}\) .) An easy verification yields the fact that for \(k+m \leq p\) the product \(A^{m} B^{k}\) occurs in both terms of the last written expression with coefficients that differ in sign only. The terms that do not cancel out are all in the subtrahend and are together equal to \[\sum_{m} \sum_{k} \frac{1}{m!k!} A^{m} B^{k},\] the summation being extended over those values of \(m\) and \(k\) that are \(\leq p\) and for which \(m+k>p\) . Since \(m+k>p\) implies that at least one of the two integers \(m\) and \(k\) is greater than the integer part of \(p/2\) (in symbols \(\lfloor p/2 \rfloor\) ), the norm of this remainder is dominated by \begin{align} & \sum_{m=0}^{\infty} \sum_{k=\lfloor p/2 \rfloor}^{\infty} \frac{1}{m!k!}\|A\|^{m}\|B\|^{k} +\sum_{k=0}^{\infty} \sum_{m=\lfloor p/2 \rfloor}^{\infty} \frac{1}{m!k!}\|A\|^{m}\|B\|^{k} \\ = &\Big(\sum_{m=0}^{\infty} \frac{1}{m!}\|A\|^{m}\Big) \Big(\sum_{k=\lfloor p/2 \rfloor}^{\infty} \frac{1}{k!}\|B\|^{k}\Big) + \Big(\sum_{k=0}^{\infty} \frac{1}{k!}\|B\|^{k}\Big) \Big(\sum_{m=\lfloor p/2 \rfloor}^{\infty} \frac{1}{m!}\|A\|^{m}\Big) \\ = &(\exp \|A\|) \alpha_{p}+(\exp \|B\|) \beta_{p}, \end{align} where \(\alpha_{p} \to 0\) and \(\beta_{p} \to 0\) as \(p \to \infty\) .

Similar methods serve to treat \(f(A)\) , where \(f\) is any function representable by a power series, \[f(\zeta)=\sum_{n=0}^{\infty} \alpha_{n} \zeta^{n},\] and where \(\|A\|\) is (strictly) smaller than the radius of convergence of the series. We leave it to the reader to verify that the functional calculus we are here hinting at is consistent with the functional calculus for normal transformations. Thus, for example, \(\exp A\) as defined above is the same linear transformation as is defined by our previous notion of \(\exp A\) in case \(A\) is normal.

EXERCISES