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$$B:= \left[ \begin{matrix} -1 & 9 &0 &0 &0 \\ 0 & -1 & 0 & 0 & 0 \\ 0&3&-1&0&0 \\ 0 & 1 & 1 & 1 & 0 \\ 0&-2&-2&0&1 \\ \end{matrix}\right]$$ 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?

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  • $\begingroup$ For computing the Jordan Canonical Form, there is a standard, algorithmic approach. Once you've got this form though, you can express the matrix as $B = SJS^{-1}$ where J denotes the form. Exponentiating is easy, as the inside copies of $S$ and $S^{-1}$ will cancel out. $\endgroup$ – Alfred Yerger May 7 '16 at 21:26
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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.

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  • $\begingroup$ this is great. How'd you find the determinate of the 5x5 matrix with the diagonals having the eigenvalues though? I'm stuck at that part $\endgroup$ – user316861 May 8 '16 at 23:06
  • $\begingroup$ @user316861 Do you mean the determinant of your initial matrix $B$? $\endgroup$ – Irregular User May 10 '16 at 19:03

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