Question 5.E.13.16: Given a square matrix X, the matrix exponential e^X is defin......
Given a square matrix X, the matrix exponential e^X is defined as
e^X = I + X + \frac{X^2}{2!} + \frac{X^3}{3!} + · · · = \sum\limits_{n=0}^{∞} \frac{X^n}{n!}.
It can be shown that this series converges for all X, and it is legitimate to differentiate and integrate it term by term to produce the statements de^{At}/dt = Ae^{At} = e^{At} A and ∫ e^{At} A dt = e^{At}.
(a) Use the fact that lim_{t→∞} e^{−A^{T}At} = 0 for all A ∈ ℜ^{m×n} to show A^† = ∫_0^∞ e^{−A^{T}At} A^{T} dt.
(b) If lim_{t→∞} e^{−A^{k+1}t} = 0, show A^D = ∫_0^∞ e^{−A^{k+1}t}A^{k} dt, where k = index(A).^{60}
(c) For nonsingular matrices, show that if lim_{t→∞} e^{−At} = 0, then A^{−1} = ∫_0^∞ e^{−At} dt.
^{60} A more useful integral representation for A^D is given in Exercise 7.9.22 (p. 615).
(a) Use the fact that A^T = A^T P_{R(A)} = A^T AA^† (see Exercise 5.13.4) to write
\int_{0}^{∞} e^{−A^{T}At} A^T dt = \int_{0}^{∞} e^{−A^{T}At} A^T AA^† dt = \left(\int_{0}^{∞} e^{−A^{T}At} A^T A dt\right) A^†
= \left[− e^{−A^{T}At}\right]_{0}^{∞} A^† = \left[0 − (−I)\right] A^† = A^†.
(b) Recall from Example 5.10.5 that A^k = A^{k+1} A^D = A^k AA^D, and write
\int_{0}^{∞} e^{−A^{k+1}t} A^k dt = \int_{0}^{∞} e^{−A^{k+1}t} A^k AA^D dt = \left(\int_{0}^{∞} e^{−A^{k+1}t} A^{k+1} A dt\right) A^D
= \left[− e^{−A^{k+1}t}\right]_{0}^{∞} A^D = \left[0 − (−I)\right] A^D = A^D.
(c) This is just a special case of the formula in part (b) with k = 0. However, it is easy to derive the formula directly by writing
\int_{0}^{∞} e^{−At} dt = \int_{0}^{∞} e^{−At} AA^{−1} dt = \left(\int_{0}^{∞} e^{−At} A dt\right) A^{-1}
= \left[e^{−At}\right]_{0}^{∞} A^{−1} = \left[0 − (−I)\right] A^{-1} = A^{-1}.