Find the eigenvalues and eigenvectors of the following matrix and express the

matrix in the form of $P=Ee^{\lambda t}E^{-1}$ where $E$ are the eigenvectors and $\lambda$ are the eigenvalues

\begin{bmatrix}0 & 1 & 0 & 0\\ -a^2-b & 0 & b & 0\\ 0 & 0 & 0 & 1\\b & 0 & -a^2-b & 0\end{bmatrix} Find the matrix $P$

What i tried. First i tried finding the eigenvalues by taking the characteristic polynomials and then solving for for the characteristic polynomials to get the four eigenvalues. Then for each eigenvalues i tired finding the corresponding vector basis. Combining the vector basis gives $E$ the eigenvector. Then taking the inverse gives $E^{-1}$ following which we do a matrix multiplication to get the matrix $P$. While i know the steps behind solving this problem my diffculty lies in the inherent tediouness in solving each step to get the correct answer Is there a simpler way to solve this problem without having to go through all the difficult steps or at least is there a mathmatical software that could help me solve this problem? Could anyone hep me with finding the matrix $P$. Thanks

I worked out the four eigenvectors

$$\lambda_{1}=-ia$$ $$\lambda_{2}=ia$$

$$\lambda_{3}=-\sqrt{-a^2-2b}$$ $$\lambda_{4}=\sqrt{-a^2-2b}$$

Eigenvector $E$ is $$\begin{bmatrix}1 & 1 & 1 & 1\\ -ia & ia & -\sqrt{-a^2-2b}& \sqrt{-a^2-2b}\\ 1 & 1 & -1 & -1\\-ia & ib & \sqrt{-a^2-2b}& -\sqrt{-a^2-2b}\end{bmatrix}$$

  • $\begingroup$ Yup the matrix should be correct now. $\endgroup$ – ys wong Feb 15 '17 at 18:23
  • $\begingroup$ Im currently at the step where im trying to find the inverse $E^{-1}$ $\endgroup$ – ys wong Feb 15 '17 at 18:25
  • $\begingroup$ I used this software tool and i got the answer that i wrote above wims.unice.fr/~wims/… $\endgroup$ – ys wong Feb 15 '17 at 18:29
  • $\begingroup$ Look at the results here - which look correct to me - click the link: WA $\endgroup$ – Moo Feb 15 '17 at 18:31
  • $\begingroup$ $\lambda$ are the eigenvalues $\endgroup$ – ys wong Feb 15 '17 at 18:33

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