Matrix to the nth exponent is it's own derivative. Coincidence or not? [duplicate]

$$\begin{bmatrix} X & 0\\ 1 & X\\ \end{bmatrix} = \begin{bmatrix} X & 0\\ 1 & X\\ \end{bmatrix}$$ $$\begin{bmatrix} X & 0\\ 1 & X\\ \end{bmatrix}^2 = \begin{bmatrix} X^2 & 0\\ 2X & X^2\\ \end{bmatrix}$$ $$\begin{bmatrix} X & 0\\ 1 & X\\ \end{bmatrix}^3 = \begin{bmatrix} X^3 & 0\\ 2X^2+X^2 & X^3\\ \end{bmatrix}$$ $$\begin{bmatrix} X & 0\\ 1 & X\\ \end{bmatrix}^n = \begin{bmatrix} X^n & 0\\ nX^{n-1} & X^n\\ \end{bmatrix}^n$$ Here you can see that the this matrix on the n-th exponent is it's own derivative. Is there any explanation for this ?

$$n = 1,2,3 \rightarrow\text{induction holds}$$

$$n \rightarrow n+1$$ $$\begin{bmatrix} X & 0\\ 1 & X\\ \end{bmatrix}^{n+1} = \begin{bmatrix} X & 0\\ 1 & X\\ \end{bmatrix}^n \begin{bmatrix} X & 0\\ 1 & X\\ \end{bmatrix}=\begin{bmatrix} X^n & 0\\ nX^{n-1} & X^n\\\end{bmatrix}\begin{bmatrix} X & 0\\ 1 & X\\\end{bmatrix} = \begin{bmatrix} X^{n+1} & 0\\ (n+1)X^n & X^{n+1}\\\end{bmatrix}$$

• Notice your matrix looks like $XI + \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$, for $I$ the identity matrix. Then if we write $\epsilon$ for the second matrix of that sum (which I'm too lazy to tex again), we see $\epsilon^2 = 0$. So $(XI + \epsilon)^2 = X^I + 2X\epsilon + \epsilon^2$. Now perhaps this notation is more suggestive. This $\epsilon$ matrix is acting as an infinitessimal, and so we're doing an informal sort of calculus – HallaSurvivor Dec 1 '20 at 8:24
• Yours is a duplicate of a lovely question I saw a while ago where I first learned this fact. Give me a second to find it. It goes into much greater depth! ^_^ – HallaSurvivor Dec 1 '20 at 8:25
• – HallaSurvivor Dec 1 '20 at 8:26
• It works with $\epsilon$ upper triangular instead of lower triangular (as in your question), but you can see the exact same idea works – HallaSurvivor Dec 1 '20 at 8:27
• But the short answer is "no, this is not a coincidence!" ^_^ – HallaSurvivor Dec 1 '20 at 8:28