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

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    Yup the matrix should be correct now. – 2017-02-15
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    Im currently at the step where im trying to find the inverse $E^{-1}$ – 2017-02-15
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    I used this software tool and i got the answer that i wrote above http://wims.unice.fr/~wims/wims.cgi?session=0YD058B781.12&+lang=en&+module=tool%2Flinear%2Fmatrix.en&+cmd=resume&+preset=A – 2017-02-15
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    Look at the results here - which look correct to me - click the link: [*WA*](http://www.wolframalpha.com/input/?i=jordan+normal+form+%7B%7B0,1,0,0%7D,%7B-a%5E2+-+b,+0,+b,+0%7D,%7B0,0,0,1%7D,%7Bb+,0,+-a%5E2+-+b,+0%7D%7D) – 2017-02-15
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    $\lambda$ are the eigenvalues – 2017-02-15
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    Is $\lambda$ supposed to also be a matrix? Maybe the matrix of eigenvalues of the original matrix? – 2017-02-15
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    Just a note that the "inverse tool" at http://lem.ma/aV should be able to handle the inverse of matrix! – 2017-02-15
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    There is a slight modification from the usual form the matrix should be of this form $J=\begin{bmatrix}e^{\lambda_{1} t} & 0 & 0 & 0\\ 0 & e^{\lambda_{2} t} & 0 & 0\\ 0 & 0 & e^{\lambda_{3} t} & 0\\0 & 0 & 0 &e^{\lambda_{4} t} \end{bmatrix}$ while the one given by link have the form $J=\begin{bmatrix} {\lambda_{1} } & 0 & 0 & 0\\ 0 & {\lambda_{2} } & 0 & 0\\ 0 & 0 & {\lambda_{3} } & 0\\0 & 0 & 0 & {\lambda_{4} } \end{bmatrix}$ – 2017-02-15
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    How do u matrixexp? Thanks for ur help – 2017-02-15

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