Question 7.3.3: For a scalar t, determine the matrix exponential e^At, where......
For a scalar t, determine the matrix exponential e^{At}, where
A = \begin{pmatrix}−α &β\\ α &−β\end{pmatrix} with α + β ≠ 0.
Solution 1: The characteristic equation for A isλ^2 + (α + β)λ = 0, so the eigenvalues of A are λ_1 = 0 and λ_2 = −(α + β). Note that A is diagonalizable because no eigenvalue is repeated—recall (7.2.6).
If no eigenvalue of A is repeated, then A is diagonalizable. (7.2.6)
Using the function f(z) = e^{zt}, the spectral representation (7.3.6) says that
= f(λ_1)G_1 + f(λ_2)G_2 + · · · + f(λ_k)G_k, (7.3.6)
e^{At} = f(A) = f(λ_1)G_1 + f(λ_2)G_2 = e^{λ_{1}t}G_1 + e^{λ_{2}t}G_2.The spectral projectors G_1 and G_2 are determined from (7.3.11) to be
G_i = \prod\limits^{k}_{\substack{j=1\\j≠i}} (A − λ_jI) / \prod\limits^{k}_{\substack{j=1\\j≠i}} (λ_i − λ_j) for i = 1, 2, . . . , k. (7.3.11)
G_1 = \frac{A − λ_2I}{−λ_2} = \frac{1}{α + β} \begin{pmatrix}β& β\\ α &α \end{pmatrix} and G_2 = \frac{A}{λ_2} = \frac{1}{α + β} \begin{pmatrix}α &−β\\ −α &β\end{pmatrix},so
e^{At} = G_1 + e^{−(α+β)t} G_2 = \frac{1}{α + β} \left[\begin{pmatrix}β &β\\ α&α\end{pmatrix} + e^{−(α+β)t} \begin{pmatrix}α &−β\\ −α &β\end{pmatrix}\right].Solution 2: Compute eigenpairs (λ_1, x_1) and (λ_2, x_2), construct P = [x_1 | x_2], and compute
e^{At} = P \begin{pmatrix}f(λ_1)& 0\\ 0 &f(λ_2)\end{pmatrix}P^{-1} = P \begin{pmatrix}e^{λ_{1}t}& 0\\ 0 &e^{λ_{2}t}\end{pmatrix}P^{-1}.The computational details are called for in Exercise 7.3.2.