Show that the matrix $A=\begin{bmatrix}
1 & 0 \\
2 & 1 \\
\end{bmatrix}$ can be decomposed as a sum of a unit and nilpotent matrix.
Hence evaluate the matrix $A^{2007}$.
I read about nilpotent matrices the other day. It said "A square matrix such that is the zero matrix for some positive integer matrix power" is nilpotent.
But how to use this in this problem is beyond me.
Best Answer
$A=\begin{bmatrix} 1 & 0 \\ 2 & 1 \\ \end{bmatrix} = I + N = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 2 & 0 \end{bmatrix}$
Observe that the second matrix is nilpotent for $k = 2$.
Now apply binomial theorem:
$A^{2007} = (I + N) ^ {2007} = \sum_{i = 0}^{2007}(\binom{2007}{i} * I^{i} * N^{2007 - i})$ (Holds since the matrices commute)
and note that all the terms except for the following 2 become 0 (because $N$ is nilpotent):
$A ^{2007} = (I + N)^{2007} = \binom{2007}{2007} $ * $I^{2007}$ * $N^0$ + $\binom{2007}{1} * I^{2006} * N $
Which gives
$ A ^ {2007} = \begin{bmatrix} 1 & 0 \\ 4014 & 1 \end{bmatrix}$
Edit:
As pointed out in the comments by Prof. Marc, things work out here since the matrices commute $(A * I = I * A \ \forall A$ where $I$ is the identity matrix) . Otherwise, the binomial theorem does not hold.