And also assuming that the matrix is not diagonalazible.

For instance.let the matrix be :

$$A=\begin{bmatrix} 2 & 1 \\ 0 & 2 \\ \end{bmatrix}$$

I'm going to take the exponential of it which is $e^A$, not trying to solve a differential equation here so just taking $e^A$ not $e^{At}$.

We all know the formula for the exponential. But the thing is I completely forgot how that formula converges. Anyway since A is not diagonal we can't take the exponential of the entries, what w

  • $\begingroup$ Besides just using the power series, which converges everywhere? $\endgroup$
    – Bib-lost
    Aug 3, 2016 at 10:30

1 Answer 1


Write $A=2I+N$ where $$N=\begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix}$$

Then $\exp(A)=\exp(2I)\exp(N)$ since $I$ and $N$ commute.

$\exp(N)$ is not too difficult to compute as $N^2=0$...


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