Exponential of a matrix

Question

Can someone show me a step by step solution to find \begin{equation} e ^{At} \end{equation} where A = $\begin{bmatrix}2 & -2\\3 & -2\end{bmatrix}$

Answer 1

One common method is to diagonalize the matrix as an intermediate step. That is, suppose we can write $A = PDP^{-1}$, where $D$ is diagonal and $P$ is (of course) invertible. We then have $$ e^{PDP^{-1}} = Pe^DP^{-1} $$ Exponentiating diagonal matrices is easy: if $d_i$ are the diagonal entries of $D$, then $$ e^D = \pmatrix{e^{d_1}&&\\&\ddots&\\&&e^{d_n}} $$ In this case, $A$ has eigenvalues $\pm i \sqrt 2$, and so must be diagonalizable.

Answer 2

First we need to diagonalize it. I get $$\det\begin{bmatrix}2-\lambda&-2\\3&-2-\lambda\end{bmatrix}=\lambda^2-4+6=\lambda^2+2=0$$ so $$\lambda=\pm\sqrt2$$ EDIT where by $\sqrt2$ I mean $i\sqrt2$ and the rest is pretty much all wrong :(

The corresponding eigenvectors are $$E_{\sqrt2}=\ker\begin{bmatrix}2-\sqrt2&-2\\3&-2-\sqrt2\end{bmatrix}=\text{span}\begin{pmatrix}2\\2-\sqrt2\end{pmatrix}$$ and $$E_{-\sqrt2}=\ker\begin{bmatrix}2+\sqrt2&-2\\3&-2+\sqrt2\end{bmatrix}=\text{span}\begin{pmatrix}2\\2+\sqrt2\end{pmatrix}$$ Therefore we have $A=S\Lambda S^{-1}$ where $$\Lambda=\begin{bmatrix}\sqrt2&0\\0&-\sqrt2\end{bmatrix}$$ $$S=\begin{bmatrix}2&2\\2-\sqrt2&2+\sqrt2\end{bmatrix}$$ I'm assuming (and really hoping...) that by $e^{At}$ you mean $$e^{At}=\left(1+A+{1\over 2!}A^2+...\right) t$$ We have $$1+A+{1\over 2!}A^2+...=1+S\Lambda S^{-1}+{1\over 2!}S\Lambda^2 S^{-1}=S(1+\Lambda+{1\over 2!}\Lambda^2+...)S^{-1}=S\begin{pmatrix}e^{t\sqrt2}&0\\0&e^{-t\sqrt2}\end{pmatrix}S^{-1}$$ EDIT: I have added some $t$'s which is apparently a scalar to the picture... Can you take it from here?

Answer 3

I'd first diagonalize the matrix, i.e., write it as $A = M D M^{-1}$ with $D$ diagonal, as that way you have $A^n = M D^n M^{-1}$. Say $D$'s diagonal is $d_1, d_2, \ldots, d_s$, write this as $D = \mathop{diag}(d_1, \ldots, d_s)$. Then: $$ e^{A t} = \sum_{n \ge 0} \frac{A^n t^n}{n!} = M \cdot \sum_{n \ge 0} \frac{D^n t^n}{n!} \cdot M^{-1} = M \cdot \mathop{diag}(e^{d_1 t}, e^{d_2 t}, \ldots, e^{d_s t}) \cdot M^{-1} $$ Obviously you'd have to check for convergence, etc.