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I just solved another problem but my result is different from the solution and I suspect this could be a mistake in the solution.

Can anyone confirm that this is a mistake in the solution?

The matrix $J$ is given as follows:

$$ J = \begin{pmatrix} -3 & 1 & 0\\ 0 & -3 & 0 \\ 0 & 0 & -1 \end{pmatrix} $$

and the matrix exponential is given as follows:

$$\exp(t J) = \begin{pmatrix} e^{-3t} & t e^{-3t} & 0\\ 0 & e^{-3t} & 0 \\ 0 & 0 & e^{-t} \end{pmatrix} $$

But it should be:

$$ \left( \begin{matrix} e^{-3t} & e^t & 1\\ 1 & e^{-3t} & 1\\ 1 & 1 & e^{-t} \end{matrix} \right)$$

Right? Because $e^{t 0} = 1$ and $e^{1t} = e^t$.

Rodrigo de Azevedo
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learner
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1 Answers1

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The matrix exponential is defined by $$\exp{tJ} = \sum_{k=0}^\infty \frac{(tJ)^k}{k!}$$ Then there is a theorem that if $J= D+N$ and $DN=ND$ we can split the exponential as follows: $\exp({tJ}) = \exp({tD})\exp({tN})$. Another theorem (about the so-called Jordan decomposition) states that any matrix $A$ can be uniquely divided as $A=D+N$ with $DN=ND$ where $D$ is diagonalizable and $N$ is nilpotent. This decomposition is obvious in that case because $J$ already has Jordan normal form. We are doing this here with $$D= \begin{pmatrix} -3&0&0\\0&-3&0\\0&0&-1\end{pmatrix}$$ $$N=\begin{pmatrix} 0&1&0\\0&0&0\\0&0&0\end{pmatrix}$$ It is easy to see that these two matrices commute (by simply calculating). For a diagonal matrix it is true that $$\exp(tD) = \begin{pmatrix} e^{-3t}&0&0\\0&e^{-3t}&0\\0&0&e^{-1t}\end{pmatrix}$$ because the powers of $D$ that are added in the infinite sum will just give the sum $$\sum_{k=0}^\infty \frac{(t d_{ii})^k}{k!}$$ in the $i$-th column and $i$-th row - That only works for diagonal matrices because otherwise the powers $D^k$ will not be that easy to describe.

For $N$ we use the definition: $$\exp(tN)=I + tN + t^2 \frac{N^2}{2} + \dots$$ We see in our case that $N^2=0$ so all higher powers are also $=0$. That means $$\exp(tN)=\begin{pmatrix} 1&t&0\\0&1&0\\0&0&1\end{pmatrix}$$

From that we get $$\exp(tJ) = \exp(tD)\exp(tN) = \begin{pmatrix} e^{-3t}&te^{-3t}&0\\0&e^{-3t}&0\\0&0&e^{-t}\end{pmatrix}$$ just as the solution given to you states.

Lukas
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