I am trying to show that if a square matrix $A$ is block upper triangular,
$A = \begin{bmatrix} B & C \\ 0 & I \end{bmatrix}$
that $A^k$ takes the following form, provided $B - I$ is invertible:
$A^k = \begin{bmatrix} B^k & (B^k-I)(B-I)^{-1}C \\ 0 & I \end{bmatrix}$
I've started computing $A^k$ for the first couple values:
$A^2 = \begin{bmatrix} B & C \\ 0 & I \end{bmatrix} \begin{bmatrix} B & C \\ 0 & I \end{bmatrix} = \begin{bmatrix} B^2 + 0 & BC + C \\ 0 + 0 & I \end{bmatrix} = \begin{bmatrix} B^2 & (B + I)C \\ 0 & I \end{bmatrix}$
$A^3 = \begin{bmatrix} B^2 & (B + I)C \\ 0 & I \end{bmatrix} \begin{bmatrix} B & C \\ 0 & I \end{bmatrix} = \begin{bmatrix} B^3 & B^2C + (B+I)CI \\ 0 & I \end{bmatrix} = \begin{bmatrix} B^3 & (B^2 + B + I)C \\ 0 & I \end{bmatrix}$
And I can see that the upper right entry is:
$\displaystyle \sum_{i=0}^{k-1} B^iC $
But why is this equal to $(B^k-I)(B-I)^{-1}C$?
And how do I demontrate this expression for $A^k$, for arbitrary $k$?
Consider $$\left(\sum_{i=0}^{k-1} B^{i}\right)(B-I).$$ Multiplying out, this is equal to $$\sum_{i=0}^{k-1} B^{i+1} - \sum_{i=0}^{k-1} B^{i}.$$ Note that $$\sum_{i=0}^{k-1} B^{i+1} = B^{k} + \sum_{i=0}^{k-2} B^{i+1} = B^{k} + \sum_{i=1}^{k-1} B^{i},$$ while $$\sum_{i=0}^{k-1} B^{i} = I + \sum_{i=1}^{k-1} B^{i}.$$ So, we have $$\sum_{i=0}^{k-1} B^{i+1} - \sum_{i=0}^{k-1} B^{i} = B^{k} - I,$$ hence $$\left(\sum_{i=0}^{k-1} B^{i}\right)(B-I) = B^{k} - I.$$ Therefore, if $B-I$ is invertible, we can multiply both sides on the right by $(B-I)^{-1}$ and conclude that $$\sum_{i=0}^{k-1} B^{i} = (B^{k}-I)(B-I)^{-1}.$$
To demonstrate that this is true for an arbitrary $k$, try induction. In other words, suppose we have $$A^{m} = \begin{bmatrix} B^{m} & (B^{m} - I)(B-I)^{-1}C \\ 0 & I \end{bmatrix},$$ and use this to try and prove that $$A^{m+1} = \begin{bmatrix} B^{m+1} & (B^{m+1} - I)(B-I)^{-1}C \\ 0 & I \end{bmatrix}.$$
Then, since we know the claim is already true for $A^{1}$, it follows that it is true for $A^{m}$.