Notation for Matrix Exponentials

Question

Just wondering if this is acceptable notation: For $$ A= \begin{bmatrix} 1 & 4\\ 2 & 3 \end{bmatrix}, $$ $$ e^{At}=\sum_{n=0}^{\infty}\frac{(At)^n}{n!}=I+At+\frac{(At)^2}{2!}+\frac{(At)^3}{3!}+\cdots $$ $$ I+ \begin{bmatrix} 1 & 4\\ 2 & 3 \end{bmatrix} t+ \frac{1}{3} \sum_{n=2}^{\infty} \begin{bmatrix} 1 & -2\\ 1 & 1 \end{bmatrix} \begin{bmatrix} 5^n & 0\\ 0 & (-1)^n \end{bmatrix} \begin{bmatrix} 1 & 2\\ -1 & 1 \end{bmatrix} t^n $$ Honestly looks very, very informal to me but I don't know any other way to represent a matrix exponential. Any suggestions?

Answer 1

Moo’s comment to your question provides some excellent links to materials on matrix exponentials. From them you can learn, among other things, that if $A$ is diagonalizable into $B\Lambda B^{-1}$, then $e^{tA}=Be^{t\Lambda}B^{-1}$, and that $e^{t\Lambda}=\operatorname{diag}(e^{\lambda_1t},\dots,e^{\lambda_nt})$ where the $\lambda_k$ are the eigenvalues of $A$ (repeated according to their multiplicities). Those notes go through some practical ways to compute the exponential of a non-diagonalizable matrix without having to compute a full Jordan decomposition, but upon a quick scan I didn’t see any mention of a way to compute the exponential of a diagonalizable matrix without computing an eigenbasis for it.

If a matrix $A$ is diagonalizable, it can be decomposed as $\lambda_1P_1+\cdots+\lambda_nP_n$, where the $\lambda_k$ are the distinct eigenvalues of $A$ and the $P_k$ are projections onto the corresponding eigenspaces such that $P_iP_j=0$ when $i\ne j$. Using this property and the fact that for any projection $P^2=P$, we can see that $e^{tA}=e^{\lambda_1t}P_1+\cdots+e^{\lambda_nt}P_n$. Thus, if you know that a matrix is diagonalizable, for instance when its eigenvalues are distinct, you can use this decomposition to compute its exponential.

This is particularly easy in the $2\times2$ case. Define $$P_1={A-\lambda_2I\over\lambda_1-\lambda_2} \\ P_2={A-\lambda_1I\over\lambda_2-\lambda_1}.$$ You can verify that $A=\lambda_1P_1+\lambda_2P_2$. For your matrix, the eigenvalues are $5$ and $-1$, so we have $$P_1=\frac16\begin{bmatrix}2&4\\2&4\end{bmatrix} \\ P_2=-\frac16\begin{bmatrix}-4&4\\2&-2\end{bmatrix}$$ and $$e^{tA}=\frac13e^{5t}\begin{bmatrix}1&2\\1&2\end{bmatrix}+\frac13e^{-t}\begin{bmatrix}2&-2\\-1&1\end{bmatrix}=\frac13\begin{bmatrix}e^{5t}+2e^{-t}&2e^{5t}-2e^{-t}\\e^{5t}-e^{-t}&2e^{5t}+e^{-t}\end{bmatrix}.$$ This method of constructing the eigenspace projections has a fairly straightforward generalization to higher dimensions.