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In lessons, we have been taught to find a general solution finding a matrix to the power n by diagonalising it but for an assignment have been asked to find a general formula for: $$\begin{bmatrix}2 & 1\\0 & 2\end{bmatrix} ^ n$$ By expending it by hand for the first few terms, I found this to be: $$\begin{bmatrix}2^n & u_n\\0 & 2^n\end{bmatrix}$$ where $u_n = 2u_{n-1} + 2^{n-1}$

How would I convert this to a general formula and is there a better way to solve this kind of question as several similar ones were set.

KikenKat
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2 Answers2

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It is useful to write $$ A = \begin{bmatrix}2 & 1\\0 & 2\end{bmatrix} = \begin{bmatrix}2 & 0\\0 & 2\end{bmatrix} + \begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}. $$ Then note that the scalar matrix $$ S = \begin{bmatrix}2 & 0\\0 & 2\end{bmatrix} $$ is in the center of the matrix ring, that is, $S B = B S$ for each matrix $B$, and that the matrix $$ N = \begin{bmatrix}0 & 1\\0 & 0\end{bmatrix} $$ is nilpotent. More precisely, $N^{2} = 0$.

Now you can compute with the usual binomial $$ A^{n} = (S + N)^{n} = S^{n} + n S^{n-1} N. $$

Andreas Caranti
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Hint: In your example, the factorization $$ \left[\begin{array}{@{}cc@{}} 2 & 1 \\ 0 & 2 \\ \end{array}\right] = \left[\begin{array}{@{}cc@{}} 2 & 0 \\ 0 & 2 \\ \end{array}\right] \left[\begin{array}{@{}cc@{}} 1 & \frac{1}{2} \\ 0 & 1 \\ \end{array}\right] $$ (with commuting factors) is helpful.

Andrew D. Hwang
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