# Matrix exponential using Jordan form

I'm trying to calculate the matrix exponential $$e^{At}$$ for $$A=\frac{1}{2}\begin{bmatrix}-1&1&-1\\2&-2&0\\1&-1&-1\end{bmatrix}$$

I found the eigenvalues $$\lambda_1=\lambda_2=-1, \lambda_3=0$$ which give the eigenvectors $$v_{\lambda_{1,2}}=\begin{bmatrix}0\\1\\1\end{bmatrix},v_{\lambda_3}=\begin{bmatrix}1\\1\\0\end{bmatrix}$$ I now would like to find the matrices $$S$$ and $$J$$ so that $$A=SJS^{-1}$$ Constructing these matrices requires another column vector. I now try to find the generalized eigenvector for eigenvalue $$-1$$ as follows. $$(A-(-1)I)\mathbf{v}=\begin{bmatrix}1/2&1/2&-1/2\\1&0&0\\1/2&-1/2&1/2\end{bmatrix}\mathbf{v}=\begin{bmatrix}0\\1\\1\end{bmatrix}$$ But this has no solution, so I am stuck.

• You may want to look at the examples in en.m.wikipedia.org/wiki/Generalized_eigenvector ; then you have to compute the generalized eigenvector for the eigenvalue $0$, since it has an algebraic multiplicity of 2 – Maximilian Janisch Sep 7 at 19:55
• Do you have to use the Jordan normal form to solve this? There’s a way to do it using Cayley-Hamilton that doesn’t require computing any eigenvectors at all. – amd Sep 7 at 20:10
• You might also want to recompute those eigenvalues. The ones you have say that $A$ is a rank-one matrix, which it certainly isn’t. Also, $Av_{\lambda_3}\ne-2v_{\lambda_3}$, so you probably forgot to include the factor of $1/2$. – amd Sep 7 at 20:12
• The characteristic polynomial is $-\lambda^3 - 2 \lambda^2 - \lambda=0$ which produces the eigenvalues $\lambda_1=\lambda_2=-1$ and $\lambda_3=0$. – Axion004 Sep 8 at 1:37
• Since the eigenvalue $-1$ has algebraic multiplicity 2, I have to use the generalized eigenvector to complete the set of vectors, but it doesn't seem to work. I'll edit the question. – gilianzz Sep 8 at 9:42

Wolfram alpha gives $$\begin{eqnarray} J&=& \begin{bmatrix} -1&1&0\\ 0&-1&0\\ 0&0&0 \end{bmatrix},\ \mbox{and}\\ S &=& \begin{bmatrix} 0&1&1\\ 1&-1&1\\ 1&0&0 \end{bmatrix}\ . \end{eqnarray}$$ Since $$J^n = \left(-1\right)^n\begin{bmatrix} 1&-n&0\\ 0&1&0\\ 0&0&0 \end{bmatrix},$$ then $$e^{tJ} = \sum_{n=0}^\infty\frac{t^n J^n}{n!}=\begin{bmatrix} \sum_{n=0}^\infty \frac{(-t)^n}{n!}&-\sum_{n=0}^\infty n\frac{(-t)^n}{n!}&0\\ 0&\sum_{n=0}^\infty \frac{(-t)^n}{n!}&0\\ 0&0&0 \end{bmatrix}\\ =\ \begin{bmatrix} e^{-t}&te^{-t}&0\\ 0&e^{-t}&0\\ 0&0&0 \end{bmatrix}\ .$$ So $$\begin{eqnarray} e^{tA} &=& S\, e^{tJ} S^{-1}\\ &=&\frac{1}{2} \begin{bmatrix} e^{-t}&-e^{-t}&e^{-t}\\ (t-1)\,e^{-t}&(1-t)\,e^{-t}&(1+t)\,e^{-t}\\ te^{-t}&-te^{-t}&(2+t)\,e^{-t} \end{bmatrix}\ . \end{eqnarray}$$