# Eigenvalues and Laplace

1. Mar 9, 2008

### soulflyfgm

hi, can some one give me any hints how to solve this problem? thank you

i tried to type it here but it dint come up so i uploaded http://tinypic.com/view.php?pic=2hgtqoz&s=3" with the problem.

Thank you so much

Recall that for an nxn matrix A with distinct eigenvalues $$\lambda$$ $$_{F}[/ktex], k=1,2,...,n e^{At} = \sum^{n}_{k=1} Z_{k}e^{\lambda_{k}t} By taking the Laplace Transform of both sides (or otherwise) show that \sum^{n}_{k=1}Z_{k}= I_{n} Where I_{n} is the nxn identity matrix Last edited by a moderator: Apr 23, 2017 2. Mar 9, 2008 ### trambolin Although stated in a very horrible way, what is asked here boils down to show that, if A is diagonalizable, one can compute $e^{At}$ by only computing the diagonal entries after diagonalizing. Keep in mind that $\mathcal{L}(e^{At}) = (sI - A)^{-1}$. So once you have the diagonal form employ the inverse laplace transformation. I am sure that you can do it, so think about it for a while... 3. Mar 9, 2008 ### mrandersdk okay maybe i read it totally wrong but if you know that [tex] e^{At} = \sum^{n}_{k=1} Z_{k}e^{\lambda_{k}t}$$

for all t then use it fot t=0 and get

$$I = e^{0} = e^{A0} = \sum^{n}_{k=1} Z_{k}e^{\lambda_{k}0} = \sum^{n}_{k=1} Z_{k}$$

and you are done?

4. Mar 9, 2008

### trambolin

The sum works with a general form of the following
$$\left[ {\begin{array}{*{20}c} 1 & 0 \\ 0 & 0 \\ \end{array}} \right] + \left[ {\begin{array}{*{20}c} 0 & 0 \\ 0 & 1 \\ \end{array}} \right] = I$$

So it is directly related with the diagonal form... Thus you can also prove for non-zero t

5. Mar 10, 2008

### soulflyfgm

ok i have proven this so far
(SI-A)L(e^(At)) = Identity...but i do not how to prove that this is equal to $$\sum^{n}_{k=1}Z_{k}$$

Any hints? thank you so much

6. Mar 10, 2008

### trambolin

OK, but that one tells you nothing, how about this? Let A is a diagonal matrix, then for 2x2 case,

$$e^{At}= e^{\left[ {\begin{array}{*{20}c} \alpha & 0 \\ 0 & \beta \\ \end{array}} \right]t} = \left[ {\begin{array}{*{20}c} e^{\alpha t} & 0 \\ 0 & e^{\beta t} \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} 1 & 0 \\ 0 & 0 \\ \end{array}} \right]e^{\alpha t} + \left[ {\begin{array}{*{20}c} 0 & 0 \\ 0 & 1 \\ \end{array}} \right]e^{\beta t}$$
And, then from the previous post,
$$\left[ {\begin{array}{*{20}c} 1 & 0 \\ 0 & 0 \\ \end{array}} \right] + \left[ {\begin{array}{*{20}c} 0 & 0 \\ 0 & 1 \\ \end{array}} \right] = I$$

Can you see the pattern now?

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