Please help me with this one:
$\displaystyle y'=\begin{pmatrix}
2 & 2&0 \\
2& 2 &0 \\
0& 0 & 4
\end{pmatrix} y$
Thanks.
Find the eigenvalue of the matrix. (Hint: one of the eigenvalues is 4. What are the eigenvalues of $\displaystyle \begin{pmatrix}2 & 2 \\ 2 & 2 \end{pmatrix}$?)
Find the corresponding eigenvectors. Fortunately, for this problem there are three independent eigenvectors so the matrix is "diagonalizable" (every symmetric matrix is diagonalizable). That is, if P is the matrix having the eigenvectors as columns, $\displaystyle P^{-1}AP= D$ where A is the given matrix and P is the diagonal matrix having the eigenvalues on the main diagonal.
Multiply the entire equation by P to get Py'= (Py)'= PAy. Let $\displaystyle X=Py$ so that $\displaystyle y= P^{-1}X$. Then the equation becomes $\displaystyle X'= PAP^{-1}Py= DX$. X'= DX is easy to solve and then y= P^{-1}X.