Is there any intuitive explanation to $A^2=\operatorname{tr}(A)A-\det(A)I_{2\times 2}$ for $A \in \mathbb{R}^{2\times 2}$? 
$A^2=\operatorname{tr}(A)A-\det(A)I_{2\times 2}$ for $A \in \mathbb{R}^{2\times 2}$


This equation is easy to prove by denoting $$A = \begin{bmatrix}
 a & b\\
 c & d
\end{bmatrix}$$ but I am wondering if there is anything behind this equation, or it's just a coincidence.
 A: The characteristic polynomial of a $2 \times 2$ matrix $A$ is $$f(t) = t^2 - \operatorname{tr}(A)t + \operatorname{det}(A).$$
So your identity is a special case of the Cayley-Hamilton theorem.

Some more thoughts: The set of ($2 \times 2$)-matrices form a finite-dimensional vector space, so the matrices $I, A, A^2, A^3, A^4, A^5, \ldots$ cannot be linearly independent.
This means that you can find some $k \geq 1$ such that $A^k$ is expressed as a linear combination of $I$, $A$, $\ldots$, $A^{k-1}$.
Here $k = 1$ doesn't work in general, since not all $(2 \times 2)$-matrices are scalar matrices. And you must have $k \leq 3$, since the space of $(2 \times 2)$-matrices has dimension $4$.
It turns out $k = 2$ works, i.e. you can find a polynomial identity of the form $$A^2 = c A + d I$$ with $c,d \in \mathbb{R}$, which at this point is not too surprising.
You could solve for $c$ and $d$ to find that in general $c = tr(A)$ and $d = det(A)$ works. There are many ways to "find" the Cayley-Hamilton identity in this special case.
