# Can you calculate \qquad \qquad e^{ ( ( ln(2), 1, 1, 1 ), ( 0, ln(2), 1, 1), ( 0, 0, ln(2), 1 ), ( 0, 0, 0, ln(2) ) ) } \qquad  ?

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#### Explanation

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#### Explanation:

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Feb 14, 2018

$\left(\begin{matrix}2 & 2 & 3 & \frac{13}{3} \\ 0 & 2 & 2 & 3 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 2\end{matrix}\right)$

#### Explanation:

Lets start by writing
$\left(\begin{matrix}\ln \left(2\right) & 1 & 1 & 1 \\ 0 & \ln \left(2\right) & 1 & 1 \\ 0 & 0 & \ln \left(2\right) & 1 \\ 0 & 0 & 0 & \ln \left(2\right)\end{matrix}\right)$

in the form $\ln \left(2\right) {I}_{4} + A$, where ${I}_{4}$ is the $4 \setminus \times 4$ identity matrix and $A$ is the upper triangular matrix

$A = \left(\begin{matrix}0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{matrix}\right)$

Since ${I}_{4}$ commutes with all $4 \times 4$ matrices in general, and $A$ in particular, the exponential matrix is simply

$\exp \left(\ln \left(2\right) {I}_{4} + A\right) = \exp \left(\ln \left(2\right) {I}_{4}\right) \exp \left(A\right) = {e}^{\ln \left(2\right)} {I}_{4} \exp \left(A\right) = 2 \exp \left(A\right)$

So the calculation essentially reduces to the determination of ${e}^{A}$. Note that the standard approach using diagonalization does not work here, since $A$ is not diagonalizable. However, the characteristic equation for $A$ is easily seen to be
${\lambda}^{4} = 0$
and thus the Cayley-Hamilton theorem says that
${A}^{4} = 0$
so that the infinite series for ${e}^{A}$ truncates :

e^A = I+A+1/{2!} A^2 + 1/{3!}A^3

It is easy to calculate

${A}^{2} = \left(\begin{matrix}0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{matrix}\right) \mathmr{and} {A}^{3} = \left(\begin{matrix}0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{matrix}\right)$

Thus
${e}^{A} = \left(\begin{matrix}1 & 1 & \frac{3}{2} & \frac{13}{6} \\ 0 & 1 & 1 & \frac{3}{2} \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1\end{matrix}\right)$

Our solution is twice this matrix!

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