Computing a Matrix Exponential

Consider the matrix

$$e^{At}$$

here $t$ is a scalar parameter, and $A$ is an $n\times n$ matrix.

Lemma: For $A$ and $t$ as above, the eigenvalues of $At$ are $t$ times the eigenvalues of A.

Proof: let $\mu$ be the eigenvalue of $A$ then we show that $\lambda=\mu t$ is an eigenvalue of $At$:
Since $\mu$ is an eigenvalue of $A$ then $\det(A-\mu I)=0$. Hence $\det(At-\mu I)= \det (t(A-\mu I))=t^n\det (A-\mu 5)=t^n(0)=0$ $\Box$

Lemma:

$$e^{At}=It +At+\frac{1}{2!}At^2 +\cdots =\sum^{\infty}_{n=0}\frac{1}{n!}A^nt^n\quad \leftarrow$$$$\mbox { not generally useful}$$

for any $A$. $\Box$ Theorem: With $A$ as above. Then

$$e^{At}=\alpha_{n-1}A^{n-1}t^{n-1}+\alpha_{n-2}A^{n-2}t^{n-2}+\cdots \alpha_2 A^2t^2+\alpha_1 At+\alpha_0 I$$

$$\alpha_i\quad i= 0, 1, \cdots, n-1$$$$\mbox { are functions of }t \mbox { which must be determined for each } A.$$

Theorem: A as above. Then

$$r(\lambda)\equiv \alpha_{n-1}\lambda^{n-1}+\alpha_{n-2}\lambda^{n-2}+\cdots \alpha_2\lambda^2+\alpha_1\lambda +\alpha_0$$

then if $\lambda_i$ is an eigenvalue of $At$ then $e^{\lambda i}=r(\lambda i)$

furthermore if $\lambda_i$ is eigenvalue of multiplicity $k$, $k> 2$ then the following equations are true:

$$e^{\lambda i}=\frac{d}{d\lambda}r(\lambda)\vert _{\lambda=\lambda... ..._i}=\cdots =\frac{d^{k-1}}{d\lambda^{k-1}}r(\lambda)\vert _{\lambda=\lambda_i}$$

Example Suppose $A$ is $4 \time 4$ matrix, with eigenvalues $5$ and $2$, with multiplicity $k=3$ and $k=1$, respectively. Then $\lambda = 5t$ and $\lambda=2t$ are eigenvalues of $At$.

Here $n=4$, thus

$$r(\lambda) = \alpha_3\lambda^3+\alpha^2\lambda^2+\alpha_1\lambda+\alpha_0$$

$$r'(\lambda) = 3\alpha_3\lambda^2+2\alpha_2\lambda+\alpha_1$$

$$r''(\lambda) = 6\alpha_3\lambda+2\alpha_2$$

Since $\lambda = 5t$ is eigenvalue of multiplicity $3\Rightarrow e^{5t}=r(5t) =r'(5t)$ and $e^{5t}=r''(5t)$. Thus

$$e^{5t}= \alpha_3(5t)^3+\alpha_2(5t)^2+\alpha_1(5t)+\alpha_0$$ (187)

$$e^{5t}=3\alpha_3(5t)^2+2\alpha_2(5t)+\alpha_1$$ (188)

$$e^{5t}= 6 \alpha_3 (5t)+ 2 \alpha_2$$ (189)

$$\mbox {also }\lambda=2t\quad \Rightarrow e^{2t}=\alpha_3(2t)^3+\alpha_2(2t)^2+\alpha_1(2t)=\alpha_0$$ (190)

Thus Equations (188)-(191) are 4 equations in 4 unknowns $\alpha_0$, $\alpha_1$, $\alpha_2$, $\alpha_3$ therefore $e^{At} =\alpha_3A^3t^3+\alpha_2A^2t^2+\alpha_1At+\alpha_0 I$ can be calculated. $\Box$

Matrix Polynomials and the Cayley-Hamilton Theorem

Let $A$ be an $n\times n$ matrix with constant entries denote $\lambda_i$ and $u_i$ be the associated eigenvalues and right eigenvectors, so that

$$Au_i=\lambda_iu_i\qquad i=1,2\ldots n$$ (191)

here $\lambda_i\in {\cal C}$ is the $i^{th}$ eigenvalue $u_i$ is the $i^{th}$ eigenvector with components $(u^1_i, u^2_i, u^3_i, \cdots u^n_i)^T$.

Premultiply (192) by $A$

$$A^2u_i=\lambda_iAu_i=\lambda_i(\lambda_iu_i)=\lambda^2_i u_i$$

In fact, premultiplying (192) by $A^{m-1}$ shows that $A^m$ has eigenvalues $\lambda^{m}_{i}$ and eigenvectors $u_i$    i.e.

$$A^m u_i=\lambda^m_i u_i$$

Note: One can use a similar argument to show that $A^T$ has the same eigenvalues as those of $A$:

$$\det(\lambda I=A^T)=\det(\lambda I-A^{T})^T=\det(\lambda I-A)$$

$\therefore$ the characteristic equation for $A$ and $A^T$ are the same.

Let $p(\lambda)=\lambda^r+p_1\lambda^{r-1}+p_2\lambda^{r-2}\cdots p_{r-1}\lambda+pr$

the $au$ arbitrary polynomial of degree $r$. Hence, for the matrix $A$ of size $n\times n$

$$p(A)=A^r+p_1A^{r-1}\cdots p_{r}-1A+prI$$

where $I$ is the identity matrix of size $n\times n$. If $u_i$ are eigenvectors of $A$ then

$$p(A)u_i=A^r+u_1+p_1A^{r-1}u_i+\cdots p_{r}u_i=p(\lambda_i)u$$

showing that the e'values and the e'vectors of $p(A)$ ARE $p(\lambda_i)$ and $u_i$ for $i-1,2\cdots n$.

Cayley-Hamilton Theorem: Every matrix satisfies its own characteristic equation, i.e.

$$k(A)=A^n+k_1 A^{n-1}+k_2A^{n-2}\cdots k_n I=0$$

ex) $A = \left[\begin{array}{ll} 1 & 3\\ 2 & 2 \end{array}\right]$ $\Rightarrow$ characteristic polynomial is $\lambda^2-3\lambda-4=0$

So $k(A)=A^2-3A-4I\equiv 0$

$\left[\begin{array}{ll} 7 & 9\\ 6 & 10\end{array}\right] - 3\left[\begin{arra... ...& 1\end{array}\right] =\left[\begin{array}{ll} 0 & 0\\ 0 & 0\end{array}\right]$

Now, consider $e^{tA}$     ($A$ is constant entry $n\times n$ matrix): it satisfies

$$\frac{d }{d t}e^{tA}=Ae^{tA}$$

$$\mbox { and } \frac{d^k}{dt^k}e^{tA}=A^ke^{tA}\quad k\ge 0$$

In fact for every polynomial $p$

$$p\left(\frac{d}{dt}\right)e^{tA}=p(A)e^{tA}$$

the solution of $$p\left(\frac{d}{dt}\right)z=0$$     is      $z=\sum^n_{j=1}c_jz_j(t)$

where $\left\{c_j\right\}^n_1$ are constant coefficients. Similarly

$$e^{tA}=\sum^n_{j=1}C_jz_j(t)$$ (192)

$\left\{C_j\right\}^n_1$ are constant matrices, derived by taking derivatives of (193) with respect to $t$ and evaluating them at $t=0$. The $k^{th}$ derivatives of (193) is

$$A^k=\sum^n_{j=1}C_jy_j^{(k)}(0).$$ (193)

If the independent solutions $\left\{y_j\right\}^n_{j=}$ are chosen to satisfy $y_j^{k-1}(0)=\delta_{jk}$

$\Rightarrow$ from (194)     $e^{tA}=\sum^n_{j=1}A^{j-1}yj(t).$

Moreover, if $p$ has simple roots and the $\left\{yj\right\}^n_{j=1}$ are chosen to be

$$y_j^{(t)}=e^{\lambda_jt}, \quad \lambda_j$$$$\mbox { the roots of } p,$$

then (194) becomes the set of $n$ equations

$$A^k=\sum^n_{j}=C_j\lambda^k_j\quad k=0,1\ldots n-1$$

which can be solved for $$\left\{C_j\right\}^n_{j=1}$$ which are spectal projections corresponding to the e'values $$\left\{\lambda_j\right\}^n_{j=1}$$.

$\Box$