5.2 Computing the exponential: the Jordan form
By (iv), it is enough to compute \(e^{Jt}\) for a matrix \(J\) similar to \(A\), and the natural choice is the Jordan form. Every complex square matrix is similar to a block diagonal matrix of Jordan blocks \(J_m(\lambda ) = \lambda I + N\), where \(N\) has ones on the superdiagonal and zeros elsewhere; \(N\) is nilpotent, \(N^m = 0\).
- (i)
- For an \(m \times m\) Jordan block, \[ e^{J_m(\lambda )t} = e^{\lambda t}\Bigl (I + tN + \frac {t^2}{2!}N^2 + \dots + \frac {t^{m-1}}{(m - 1)!}N^{m-1}\Bigr ), \] an upper triangular matrix whose \((i, i + k)\) entry is \(e^{\lambda t}t^k/k!\).
- (ii)
- For the real \(2 \times 2\) block \(C = \begin {bmatrix} a & -b\\ b & a\end {bmatrix}\), which has eigenvalues \(a \pm ib\), \[ e^{Ct} = e^{at}\begin {bmatrix} \cos bt & -\sin bt\\ \sin bt & \cos bt\end {bmatrix} . \]
- (iii)
- Consequently every component of every solution of \(x' = Ax\) is a linear combination of the functions \(t^ke^{at}\cos bt\) and \(t^ke^{at}\sin bt\), where \(a \pm ib\) runs over the eigenvalues of \(A\) and \(k\) is less than the size of the largest Jordan block for that eigenvalue.
Proof. (i) \(\lambda I\) and \(N\) commute, so \(e^{(\lambda I + N)t} = e^{\lambda t}e^{Nt}\) by Theorem 5.1.2(iii), and the series for \(e^{Nt}\) stops at \(N^{m-1}\). Since \(N^k\) has ones on the \(k\)th superdiagonal, the entries are as stated. (ii) \(C = aI + bR\) with \(R = \begin {bmatrix} 0 & -1\\ 1 & 0\end {bmatrix}\), \(R^2 = -I\), and \(aI\) commutes with \(R\), so \(e^{Ct} = e^{at}e^{bRt}\) with \(e^{bRt} = \sum _k(bt)^kR^k/k! = \cos bt\,I + \sin bt\,R\), splitting the series into even and odd terms. (iii) Every solution is \(e^{At}x_0 = Pe^{Jt}P^{-1}x_0\), and the entries of \(e^{Jt}\) are of the stated forms by (i); a complex conjugate pair of eigenvalues contributes \(e^{(a \pm ib)t} = e^{at}(\cos bt \pm i\sin bt)\), and real solutions are real combinations of these. □
Powers of \(t\) appear only when \(A\) is not diagonalisable. Then the eigenvectors do not span the space, and the missing directions are filled by generalised eigenvectors, solutions of \((A - \lambda I)^kv = 0\), along which solutions grow or decay like \(te^{\lambda t}\), \(t^2e^{\lambda t}\), and so on.
Example 5.2.2. \(A = \begin {bmatrix} 1 & 1\\ -1 & 3\end {bmatrix}\) has characteristic polynomial \(\lambda ^2 - 4\lambda + 4\), a double eigenvalue \(2\), and \(A - 2I = \begin {bmatrix} -1 & 1\\ -1 & 1\end {bmatrix}\) has rank one, so \(A\) is not diagonalisable. Since \((A - 2I)^2 = 0\), \(e^{At} = e^{2t}e^{(A - 2I)t} = e^{2t}\bigl (I + t(A - 2I)\bigr ) = e^{2t}\begin {bmatrix} 1 - t & t\\ -t & 1 + t\end {bmatrix}\), with no need for the Jordan basis.
The trick of the example works in general: if \(A\) has a single eigenvalue \(\lambda \) of algebraic multiplicity \(n\), then \((A - \lambda I)^n = 0\) by the Cayley–Hamilton theorem, and \(e^{At} = e^{\lambda t}\sum _{k<n}t^k(A - \lambda I)^k/k!\).
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