Question

please help guys
A= [0 0 -2;1 2 1;1 0 3]

find a matrix P diagonazlizes A.
what is A^30

Answer 1

If \(D\) is the diagonalization of \(A\), then we can write:\[A=PDP^{-1}\]where:\[P=\left[\begin{matrix}e_1 & e_2 & e_3\end{matrix}\right]\]and \(e_i\) represents the i'th eigenvector of \(A\).
You first need to find the eigenvalues of A using:\[\left|A-\lambda I\right|=0\]solving this should give you the three eigenvalues of \(A\), \(\lambda_1, \lambda_2, \lambda_3\).
Then use each of the eigenvalues to solve:\[(A-\lambda I)x=0\]this should give you the three eigenvectors \(e_1, e_2, e_3\). Hence you have found \(P\).
Ti find \(A^{30}\) you can use the property:\[A^n=(PDP^{-1})^n=PDP^{-1}*PDP^{-1}*PDP^{-1}*...\]and note that on either side of the internal multiplications you have \(P^{-1}*P=I\) and therefore these collapse to give:\[A^n=PD^nP^{-1}\]and calculating \(D^n\) for the diagonal matrix \(D\) is trivial.