How can I raise this matrix to a variable power?

Question

I want to evaluate this:

$$ \begin{bmatrix} 1 & 1 & 0\\ 0 & 1 & -1\\ 0 & 0 & 1 \end{bmatrix}^n $$

How would I approach this?

This isn't a homework problem. I was just thinking about a linear algebra approach to the problem of summing all numbers from $n$ to $1$ and found that

$$ M^n\begin{bmatrix} 0\\ n\\\ 1 \end{bmatrix}_0 = \sum_{i = 1}^n i $$

(where $M$ is the matrix above and $_0$ is the first element of the product vector). Is it possible to calculate the matrix exponentiation without already knowing how to calculate the sum?

Answer 1

Here's a general theory for applying functions to square matrices. Any square matrix $M$ has a Jordan normal form: a decomposition $M = SJS^{-1}$ where $S$ is invertible and $J$ is block-diagonal $J = J_1 \oplus J_2 \oplus \cdots\,$ with the Jordan blocks $J_i$ of the form

$$J_i = \begin{bmatrix} \lambda_i & 1\\ & \lambda_i & \ddots\\ && \ddots & 1\\ &&& \lambda_i \end{bmatrix}$$

for some block size and some scalar $\lambda_i$. Then to apply a (sufficiently well-behaved) function to the matrix, we have $f(M) = S \, f(J) \, S^{-1}$.

In your scenario, we have the function $f : x \mapsto x^n$. To compute the Jordan normal form of $M$, it's already very close to the desired form so we can just determine it by inspection:

$$M = \begin{bmatrix} 1 & 1 & 0\\ 0 & 1 & -1\\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -1 \end{bmatrix} = SJS^{-1}.$$

Then from the result in the page above, we get

$$f(J) = f\left(\begin{bmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{bmatrix}\right) = \begin{bmatrix} f(1) & f'(1) & \frac{f''(1)}{2!}\\ 0 & f(1) & f'(1)\\ 0 & 0 & f(1) \end{bmatrix} = \begin{bmatrix} 1 & n & \frac{n(n-1)}{2}\\ 0 & 1 & n\\ 0 & 0 & 1 \end{bmatrix}.$$

Taking the similarity transformation back, we find that

$$M^n = S f(J) S^{-1} = \begin{bmatrix} 1 & n & -\frac{n(n-1)}{2}\\ 0 & 1 & -n\\ 0 & 0 & 1 \end{bmatrix}.$$

As an aside, when the matrix is diagonalizable (which yours is not), the formula simplifies because we can take $M = S D S^{-1}$ where $D = \mathrm{diag}(d_1, d_2, \ldots)$ is diagonal, hence applying $f$ to $D$ is simply $f(D) = \mathrm{diag}(f(d_1), f(d_2), \ldots)$ (i.e., we don't need to compute off-diagonal elements and take derivatives).