1. ## Eigenvectors

The matrix $\displaystyle A=\left(\begin{array}{ccc}3&-1&0\\-4&-6&-6\\5&11&10\end{array}\right)$

has eigenvectors $\displaystyle \left(\begin{array}{c}1\\-1\\1\end{array}\right)$, $\displaystyle \left(\begin{array}{c}1\\2\\-3\end{array}\right)$ and $\displaystyle \left(\begin{array}{c}1\\1\\-2\end{array}\right)$ with corresponding eigenvalues 4, 1 and 2 respectively.
The matrix B has eigenvalues 2, 3, 1 with corresponding eigenvectors
$\displaystyle \left(\begin{array}{c}1\\-1\\1\end{array}\right)$, $\displaystyle \left(\begin{array}{c}1\\2\\-3\end{array}\right)$ and $\displaystyle \left(\begin{array}{c}1\\1\\-2\end{array}\right)$
Find a matrix P and a diagonal matrix D such that
$\displaystyle (A+B)^4=PDP^{-1}$

Not required to evaluate $\displaystyle P^{-1}$

A start and some hints would be very nice.

2. Originally Posted by I-Think
The matrix $\displaystyle A=\left(\begin{array}{ccc}3&-1&0\\-4&-6&-6\\5&11&10\end{array}\right)$

has eigenvectors $\displaystyle \left(\begin{array}{c}1\\-1\\1\end{array}\right)$, $\displaystyle \left(\begin{array}{c}1\\2\\-3\end{array}\right)$ and $\displaystyle \left(\begin{array}{c}1\\1\\-2\end{array}\right)$ with corresponding eigenvalues 4, 1 and 2 respectively.
The matrix B has eigenvalues 2, 3, 1 with corresponding eigenvectors
$\displaystyle \left(\begin{array}{c}1\\-1\\1\end{array}\right)$, $\displaystyle \left(\begin{array}{c}1\\2\\-3\end{array}\right)$ and $\displaystyle \left(\begin{array}{c}1\\1\\-2\end{array}\right)$
Find a matrix P and a diagonal matrix D such that
$\displaystyle (A+B)^4=PDP^{-1}$

Not required to evaluate $\displaystyle P^{-1}$

A start and some hints would be very nice.
Notice that the eigenvectors for B are the same as those of A. What do you know about the matrix whose columns are those three eigenvectors?

3. And of course, if v is an eigenvector of A with eigenvalue $\displaystyle \lambda_A$ and an eigenvector of B with eigenvalue $\displaystyle \lambda_B$ then $\displaystyle (A+ B)v= Av+ Bv= \lambda_A v+ \lambda_B v= (\lambda_A+ \lambda_B) v$ so v is an eigenvector for A+ B with eigenvalue $\displaystyle \lambda_A+ \lambda_B$. That gives you the diagonal matrix D.

4. Thanks Opalq, I actually realize that they had the same eigenvectors.
And HallsofIvy, that was actually the first part of the question.

Just one last question, for the matirx P, does it matter what order the eigenvectors are placed ?

5. Originally Posted by I-Think
Just one last question, for the matrix P, does it matter what order the eigenvectors are placed ?
No, but the eigenvalues on the diagonal of D must occur in the same order as their corresponding eigenvectors in the columns of P.