## Loser66 one year ago If $$A=\left[\begin{matrix}1&0\\0&2\end{matrix}\right]$$, then $$e^{tA}=\left[\begin{matrix}e^t&0\\0&e^{2t}\end{matrix}\right]$$ Now, if $$B=\left[\begin{matrix}0&1\\0&0\end{matrix}\right]$$, then $$e^{tB}=\left[\begin{matrix}1&t\\0&1\end{matrix}\right]$$ I don't get why t there. Please, help

1. Loser66

@oldrin.bataku

2. jtvatsim

Are you specifically referring to the $e^{tB}$ matrix?

3. Loser66

Yes, sir

4. jtvatsim

OK, lemme check. I believe exponentiation of matrices is defined by its infinite series, but let me see if there is a more elegant approach.

5. jtvatsim

OK, I got it. I'll let @oldrin.bataku reply first and see if they have any additional insights.

6. jtvatsim

We first need that $e^x = 1 + x + x^2/2! + x^3/3! + \cdots$Are you familiar and/or comfortable with that?

7. Loser66

Yes, I do

8. jtvatsim

OK, great. This definition can be extended to matrices so that $e^{tB} = I + tB + (tB)^2/2! + (tB)^3/3! + \cdots$ In general, $e^A = \sum_{n=0}^\infty \frac{A^n}{n!}$

9. jtvatsim

Expanded, we have $e^{tB} = I + tB + t^2B^2/2! + t^3B^3/3! + \cdots$

10. jtvatsim

OK, so far?

11. Loser66

Yes

12. jtvatsim

Alright, awesome! Now, the annoying thing is the powers of matrices. We really hope that we have a "nice" matrix, otherwise this can get messy. Fortunately, B isn't that bad.

13. jtvatsim

$B = \left[\begin{matrix}0 & 1 \\ 0 & 0\end{matrix}\right], \ \ B^2 = \left[\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right]$

14. jtvatsim

The zero matrix times any matrix is still the zero matrix, so all powers of B above 1 will be zeroed out.

15. Loser66

I know it

16. jtvatsim

Hence, our problem reduces to this: $e^{tB} = I + tB + 0 + 0 + \cdots = I + tB$ and the result follows immediately.

17. Loser66

yyyyyyyyyyyyyes!

18. jtvatsim

$I + tB = \left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right] + \left[\begin{matrix}0 & t \\ 0 & 0\end{matrix}\right] = \left[\begin{matrix}1 & t \\ 0 & 1\end{matrix}\right]$

19. jtvatsim

Yay!! Indeed, any questions?

20. Loser66