Activity 7.4.1.
Supose that \(A\) is diagonalizable. How to compute \(e^{At}\text{?}\)
Section 7.4 Application: Find \(e^{At}\)
It is very well known that
\begin{equation*}
e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\ldots
\end{equation*}
One can define
\begin{equation*}
e^{A}=I+A+\frac{A^{2}}{2!}+\frac{A^{3}}{3!}+\ldots
\end{equation*}
It is a fact that
\begin{equation*}
I+A+\frac{A^{2}}{2!}+\frac{A^{3}}{3!}+\ldots
\end{equation*}
is convergent. In this class, we are able to compute \(e^{A}\) if \(A\) is diagonalizable.
Example: Compute \(e^{A}\) for \(A= \begin{pmatrix}
2 \amp -12 \\
1 \amp -5
\end{pmatrix}\)
Solution:
-
Find the characteristic polynomial \(\det(A-\lambda I_{2})=\lambda^{2}+3\lambda+2\text{.}\) The eigenvalues of \(A\) is \(-1,-2\text{.}\)
-
Find the corresponding eigenvectors. Thus we find a diagonal matrix \(D\) and an invertible matrix \(P\) such that\begin{equation*} P^{-1}AP=D\qquad \text{ or }\qquad A=PDP^{-1}. \end{equation*}
-
Therefore,\begin{align*} e^{A} \amp= e^{PDP^{-1}}\\ \amp= I+PDP^{-1}+\frac{(PDP^{-1})(PDP^{-1})}{2!}+\ldots \\ \amp= PP^{-1}+PDP^{-1}+\frac{PD^{2}P^{-1}}{2!}+\ldots\\ \amp= P(I+D+\frac{D^{2}}{2!}+\ldots)P^{-1}\\ \amp= P\left( \begin{pmatrix} 1 \amp 0\\ 0 \amp 1 \end{pmatrix} +\begin{pmatrix} -1 \amp 0\\ 0 \amp -2 \end{pmatrix}+\frac{1}{2!}\begin{pmatrix} (-1)^{2} \amp 0\\ 0 \amp (-2)^{2} \end{pmatrix}+\ldots\right)P^{-1}\\ \amp = P\begin{pmatrix} 1+(-1)+\frac{(-1)^{2}}{2!}+\ldots \amp 0\\ 0 \amp 1+(-2)+\frac{(-2)^{2}}{2!}+\ldots \end{pmatrix}P^{-1}\\ \amp = P\begin{pmatrix} e^{-1} \amp 0\\ 0 \amp e^{-2} \end{pmatrix}P^{-1}\\ \amp=\begin{pmatrix} \frac{4}{e}-\frac{3}{e^{2}} \amp \frac{12}{e^{2}}-\frac{12}{e}\\ \frac{1}{e}-\frac{1}{e^{2}} \amp \frac{4}{e^{2}}-\frac{3}{e} \end{pmatrix} \end{align*}
Solution.
Since \(A\) is diagonalizable, there exists an invertible matrix \(P\) and a diagonal matrix \(D=\begin{pmatrix}
\lambda_1\amp \amp \\
\amp \ddots \amp \\
\amp \amp \lambda_n
\end{pmatrix}\) such that \(P^{-1}AP=D.\) It is easy to see that
\begin{equation*}
P^{-1}(At)P=P^{-1}(A(tI_n))P=P^{-1}AP(tI_n)=D(tI_n)=tD.
\end{equation*}
That is,
\begin{equation*}
e^{At}=P\begin{pmatrix}
e^{\lambda_1} \amp \amp\\
\amp \ddots\amp\\
\amp \amp e^{\lambda_n}\\
\end{pmatrix}P^{-1}
\end{equation*}
Exercise: compute \(e^{At}\text{,}\) where
\begin{equation*}
A = \begin{pmatrix}
1 \amp 2 \amp -2 \\
-2 \amp 5 \amp -2 \\
-6 \amp 6 \amp -3
\end{pmatrix}
\end{equation*}
