Question 7.E.3.2: For the matrix A in Example 7.3.3, verify with direct comput......

From Example 7.3.3, the eigenvalues are λ_1 = 0  and  λ_2 = −(α + β), and associated eigenvectors are computed in the usual way to be

x_1 = \begin{pmatrix}β/α\\1\end{pmatrix}   and   x_2 = \begin{pmatrix}-1\\1\end{pmatrix},

so

P = \begin{pmatrix}β/α &−1\\ 1&1\end{pmatrix}   and   P^{-1} = \frac{1}{1 + β/α} \begin{pmatrix}1&1\\ −1 &β/α\end{pmatrix}.

Thus

P \begin{pmatrix}e^{λ_{1}t}& 0\\ 0 &e^{λ_{2}t}\end{pmatrix} P^{-1} = \frac{α}{α + β} \begin{pmatrix}β/α &−1\\1&1\end{pmatrix} \begin{pmatrix}1&0\\ 0 &e^{−(α+β)t}\end{pmatrix} \begin{pmatrix}1&1\\ −1 &β/α\end{pmatrix}

= \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]

= e^{λ_{1}t}G_1 + e^{λ_{2}t}G_2.