Jordan Canonical Form of complicated Matrix

Question

$$B:= \left[ \begin{matrix} -1 & 9 &0 &0 &0 \\ 0 & -1 & 0 & 0 & 0 \\ 0&3&-1&0&0 \\ 0 & 1 & 1 & 1 & 0 \\ 0&-2&-2&0&1 \\ \end{matrix}\right]$$ How do I find the Jordan Canonical Form of $B$? Also, I was wondering if there was any way to compute a formula for $B^n$ far all integers $n\geq5$? Or if thats not possible here?

Answer 1

Find the eigenvalues first. There are two different eigenvalues: $\lambda=-1$ with algebraic multiplicity 3, and $ \lambda=1$ with algebraic multiplicity 2.

When $\lambda=-1$, two eigenvectors can be found (aka geometric multiplicity=2) using $A-\lambda I=0$. Notice the algebraic multiplicity (3) exceeds geometric multiplicty (2) by 1 for this eigenvalue.

When $\lambda=1$, two eigenvectors can be found (aka geometric multiplicity=2) using $A-\lambda I=0$.

Thus the Jordan Canonical Form of B is:
$$\begin{bmatrix}-1&1&0&0&0\\0&-1&0&0&0\\0&0&-1&0&0\\0&0&0&1&0\\0&0&0&0&1\end{bmatrix}$$ Make sure to double check the calculations.