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Compute $e^M$ where $M=\begin{bmatrix}8 & -1\\4 & 4\end{bmatrix}$

Because M is not diagonalizable i try to use Jordan decomposition so i find the Jordan matrix to be $J=\begin{bmatrix}6 & 1\\0 & 6\end{bmatrix}$ but i cannot find the other matrix $S$. I found the eigenvector to be $V_1=\begin{bmatrix}1\\2 \end{bmatrix}$ then i try to find the generalized vectors for the eigenvalue $\lambda = 6$ but i'm doing something wrong on the way. Any help would be appreciated. Thanks

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The generalized eigenvector you want is any solution to the equation $$ (M - 6I)x = v_1 $$ We can solve this equation by row-reduction: $$ \pmatrix{2&-1&1\\4&-2&2} \leadsto \pmatrix{2&-1&1\\0&0&0} $$ So, we can take for instance the generalized eigenvector $\pmatrix{0\\-1}$.

More generally, any vector of the form $t\pmatrix{0\\-1} + (1-t)\pmatrix{1/2 \\ 0}$ for $t \in \Bbb R$ will do.

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  • $\begingroup$ But if i take $x=0$ then $y$ shouldn't be equal to $-1$? So the vector would be $\begin{bmatrix}0\\-1\end{bmatrix}$? $\endgroup$ – Mattematics Aug 26 '15 at 15:52
  • $\begingroup$ You're right; my mistake $\endgroup$ – Omnomnomnom Aug 26 '15 at 15:54
  • $\begingroup$ Thank you very much for answering :) $\endgroup$ – Mattematics Aug 26 '15 at 15:56
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As a remark, if the Jordan form $J$ is known to be a $2\times2$ Jordan block with eigenvalue $\lambda$, you don't really need to calculate $S$ in order to find $e^M$. More specifically, suppose $M=SJS^{-1}$ where $$ J=\pmatrix{\lambda&1\\ 0&\lambda}=\lambda I+N. $$ Then \begin{align} e^J &= I + (\lambda I+N) + \frac{(\lambda I+N)^2}{2!} + \frac{(\lambda I+N)^3}{3!} + \ldots\\ &= e^{\lambda}I + \left(1 + \frac{2(\lambda)}{2!} + \frac{3(\lambda^2)}{3!} + \ldots\right)N\quad(\text{because } N^2=N^3=\ldots=0)\\ &= e^{\lambda} (I+N)\\ &= e^{\lambda} (J+(1-\lambda)I) \end{align} and it follows that $e^M = e^{\lambda}(M+(1-\lambda)I)$.

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