# Finding the n-th power of a matrix using Jordan's normal form

There is a matrix $B$ given $$B = \begin{bmatrix} 5&2&-6\\-5&-2&-6\\4&1&-6\end{bmatrix}$$ I am to find $B^n$ using Jordan's normal form. However I have no idea how to deal with a non - nilpotent matrix.
I managed to find eigenvalues: $\lambda_1 = 3, \lambda_2 = -2i\sqrt{3}, \lambda_3 = 2i\sqrt{3}$

• The three eigenvalue are different, so the matrix is diagonalizable. You don't have any nilpotent matrix here. – Emilio Novati Jun 20 '17 at 8:53
• Corrected and updated my answer below. Please give it a check, I think it will be helpful for you ! Also, try to prove the product for $n$-times. – Rebellos Jun 20 '17 at 10:17

Since you have found the eigenvalues, find then the eigenvectors $v_1$, $v_2$, $v_3$. After that, since $λ_1 \neq λ_2 \neq λ_3$, your matrix is diagonalizable and you do not need a Jordan form, since you do not have any nilpotent matrix here.

Any way, finding the $n$-th power by diagonalization :

$$B = VΛV^{-1}$$

where :

$$Λ = Diag(λ_1,λ_2,λ_3)$$

$$V = \big[v_1|v_2|v_3\big]$$

$$V^{-1} = \big[v_1|v_2|v_3\big]^{-1}$$

So, you have a product of n-times :

$$B^n = (VΛV^{-1})(VΛV^{-1})\dots(VΛV^{-1})$$

If you perform the calculations, you get to :

$$B^n = VΛ^nV^{-1}$$

In order to understand how you get this, try a smaller power, such as $2$ :

$$B^2 = (VΛV^{-1})(VΛV^{-1}) = VΛ(V^{-1}V)ΛV^{-1} = VΛ^2V^{-1}$$

• Are you sure about that last form? – robjohn Jun 20 '17 at 10:00
• @robjohn Corrected it, I'm day-walking probably. – Rebellos Jun 20 '17 at 10:13
• $V$ and $V^{-1}$ should be swapped in the last two equations. – Bernard Jun 20 '17 at 10:28