# Let $\alpha$ and $\beta$ be two distinct eigenvalues of $A$ then $A^3 = \frac{\alpha^3-\beta^3}{\alpha-\beta}A-\alpha\beta(\alpha+\beta)I$?

Let $\alpha$ and $\beta$ be two distinct eigenvalues of a $2\times2$ matrix $A$. Then which of the following statements must be true.

1 - $A^n$ is not a scalar multiple of identity matrix for any positive positive integer $n$.

2 - $A^3 = \dfrac{\alpha^3-\beta^3}{\alpha-\beta}A-\alpha\beta(\alpha+\beta)I$

For the statement 1 I picked up a diagonal matrix with diagonal entries 1 and -1 whose square comes out to be identity matrix. Thus statement may be false. But for the second statement i am not able to figure out a way to start. This probably may be easy but I am not able to get this. Please post a small hint so that I may proceed further.

• Note that $\alpha$ and $\beta$ are two of the roots of the cubic $x^3-(\alpha^2+\alpha\beta+\beta^2)x+\alpha\beta(\alpha+\beta)$... – J. M. is a poor mathematician May 12 '12 at 0:30

By the Cayley–Hamilton theorem, we have $A^2-(\alpha+\beta)A+\alpha\beta I=0$ and so $A^3=(\alpha+\beta)A^2-\alpha\beta A=(\alpha+\beta)((\alpha+\beta)A-\alpha\beta I)-\alpha\beta A$, which is option #2 after simplification.
More generally, $A$ satisfies $p(A) = 0$ if and only if the polynomial $p(x)$ is divisible by the minimal polynomial of $A$, which (because the eigenvalues are distinct) is the characteristic polynomial $q(x) = (x-\alpha)(x-\beta)$. And $p(x)$ is divisible by $q(x)$ if and only if $p(\alpha) = p(\beta) = 0$. In particular, this will be true for $p(x) = x^n - c x - d$ (where $n$ is an integer $\ge 2$) if and only if $\alpha^n - c \alpha - d = \beta^n - c \beta - d = 0$; solving for $c$ and $d$ we get $c = \dfrac{\alpha^n - \beta^n}{\alpha - \beta}$ and $d = \alpha \beta \dfrac{\beta^{n-1} - \alpha^{n-1}}{\alpha - \beta}$.
Notice, using that $( \alpha-\beta )( \alpha^2 + \alpha \beta+ \beta^2 )=\alpha^3-\beta^3$, that the right hand side is equal to $$T= \alpha^2(a-\beta I) + \beta^2(A-\alpha I) + \alpha\beta A$$ Now, pick $v\in \ker(A-\alpha I)$ then $Tv=\alpha^3 v$. Similarly, if $w\in \ker(A-\beta I)$ then $Tw= \beta^3 w$. This tells us that $T=A^3$ on $\ker(A-\alpha I) \oplus \ker(A-\beta I) = k^2$ (where $k$ is your field) since we have two distinct eigenvalues. So $T=A^3$