10.5: The Matrix Exponential via Eigenvalues and Eigenvectors
In this module we exploit the fact that the matrix exponential of a diagonal matrix is the diagonal matrix of element exponentials. In order to exploit it we need to recall that all matrices are almost diagonalizable. Let us begin with the clean case: if \(A\) is n-by-n and has \(n\) distinct eigenvalues, \(\lambda_{j}\), and therefore \(n\) linear eigenvectors, \(s_{j}\), then we note that
\[\forall j, j \in \{1, \cdots, n\} : (As_{j} = \lambda_{j}s_{j}) \nonumber\]
maybe written as
\[A = S \Lambda S^{-1} \nonumber\]
where \(S = \begin{pmatrix} {s_{1}}&{s_{2}}&{\cdots}&{s_{n}} \end{pmatrix}\) is the full matrix of eigenvectors and \(\Lambda = diag (\lambda_{1}, \lambda_{2}, \cdots, \lambda_{n}\) is the diagonal matrix of eigenvalues. One cool reason for writing \(A\) as in Equation is that
\[A^2 = S \Lambda S^{-1} S \Lambda S^{-1} = S \Lambda^{2} S^{-1} \nonumber\]
and, more generally
\[A^k = S \Lambda^{k} S^{-1} \nonumber\]
If we now plug this into the definition in The Matrix Exponential as a Sum of Powers, we find
\[e^{At} = Se^{\Lambda t} S^{-1} \nonumber\]
where \(e^{\Lambda t}\) is simply
\[diag (e^{\lambda_{1}t}, e^{\lambda_{2}t}, \cdots, e^{\lambda_{1}t}) \nonumber\]
Let us exercise this on our standard suite of examples.
If
\[A = \begin{pmatrix} {1}&{0}\\ {0}&{2} \end{pmatrix} \nonumber\]
then
\[S = I \Lambda = A \nonumber\]
and so \(e^{At} = e^{\Lambda t}\)
As a second example let us suppose
\[A = \begin{pmatrix} {0}&{1}\\ {-1}&{0} \end{pmatrix} \nonumber\]
and compute, in matlab,
>> [S, Lam] = eig(A) S = 0.7071 0.7071 0 + 0.7071i 0 - 0.7071i Lam = 0 + 1.0000i 0 0 0 - 1.0000i >> Si = inv(S) Si = 0.7071 0 - 0.7071i 0.7071 0 + 0.7071i >> simple(S*diag(exp(diag(Lam)*t))*Si) ans = [ cos(t), sin(t)] [-sin(t), cos(t)]
If
\[A = \begin{pmatrix} {0}&{1}\\ {0}&{0} \end{pmatrix} \nonumber\]
then matlab delivers
>> [S, Lam] = eig(A) S = 1.0000 -1.0000 0 0.0000 Lam = 0 0 0 0
So zero is a double eigenvalue with but one eigenvector. Hence SS is not invertible and we can not invoke. The generalization is often called the Jordan Canonical Form or the Spectral Representation. The latter reads
\[A = \sum_{j=1}^{h} \lambda_{j}P_{j}+D_{j} \nonumber\]
where the \(\lambda_{j}\) are the distinct eigenvalues of \(A\) while, in terms of the resolvent \(R(z) = (zI-A)^{-1}\)
\[P_{j} = \frac{1}{2 \pi i} \int R(z) dz \nonumber\]
is the associated eigen-projection and
\[D_{j} = \frac{1}{2 \pi i} \int R(z)(z-\lambda_{j}) dz \nonumber\]
is the associated eigen-nilpotent. In each case, \(C_{j}\) is a small circle enclosing only \(\lambda_{j}\)
Conversely we express the resolvent
\[R(z) = \sum_{j=1}^{h} \frac{1}{z-\lambda_{j}}P_{j}+\sum_{k=1}^{m_{j}-1}\frac{1}{(z-\lambda_{j})^{k+1}}D^{k}_{j} \nonumber\]
where
\[m_{j} = \dim (\mathscr{R}(P_{j})) \nonumber\]
with this preparation we recall Cauchy's integral formula for a smooth function f
\[f(a) = \frac{1}{2\pi i} \int \frac{f(z)}{z-a} dz \nonumber\]
where \(C(a)\) is a curve enclosing the point \(a\)
\[f(A) = \frac{-1}{2\pi i} \int f(z)R(z) dz \nonumber\]
where \(C(r)\) encloses ALL of the eigenvalues of \(A\). For \(f(z) = e^{zt}\) we find
\[e^{At} = \sum_{j=1}^{h} e^{\lambda_{j}t} (P_{j}+\sum_{k=1}^{m_{j}-1} \frac{t^k}{k!}D^{k}_{j}) \nonumber\]
with regard to our example we find, \(h=1, \lambda_{1}=0, P_{1}=I, m_{1}=2, D_{1}=A\) so
\[e^{At} = I+tA \nonumber\]
Let us consider a slightly bigger example, if
\[A = \begin{pmatrix} {1}&{1}&{0}\\ {0}&{1}&{0}\\ {0}&{0}&{2} \end{pmatrix} \nonumber\]
then
>> R = inv(s*eye(3)-A) R = [ 1/(s-1), 1/(s-1)^2, 0] [ 0, 1/(s-1), 0] [ 0, 0, 1/(s-2)]
and so \(\lambda_{1} = 1\) and \(\lambda_{2} = 2\) while
\[P_{1} = \begin{pmatrix} {1}&{0}&{0}\\ {0}&{1}&{0}\\ {0}&{0}&{0} \end{pmatrix} \nonumber\]
and so \(m_{1} = 2\)
\[D_{1} = \begin{pmatrix} {0}&{1}&{0}\\ {0}&{0}&{0}\\ {0}&{0}&{0} \end{pmatrix} \nonumber\]
and
\[P_{2} = \begin{pmatrix} {0}&{0}&{0}\\ {0}&{0}&{0}\\ {0}&{0}&{1} \end{pmatrix} \nonumber\]
and \(m_{2}=1\) and \(D_{2}=0\)
\[e^{At} = e^{t}(P_{1}+tD_{1})+e^{2t}P_{2} \nonumber\]
\[\begin{pmatrix} {e^t}&{te^t}&{0}\\ {0}&{e^t}&{0}\\ {0}&{0}&{e^{2t}} \end{pmatrix} \nonumber\]