Show that, for all $n > 0$, $A^n = {a^n\over a − b} (A − bI) + {b^n\over b − a} (A − aI)$.

Question

Let $A ∈ M_{2×2}(\mathbb{C})$ be a matrix having distinct eigenvalues $a \neq b$. Show that, for all $n > 0$, $A^n = {a^n\over a − b} (A − bI) + {b^n\over b − a} (A − aI)$.

I'm trying to prove this with induction. I have done the base case for $n=1$ which is a little messy, but it works. Now I have assumed the result is true for some $k$ and I considered $A^{k+1}=A^kA=[{a^k\over a − b} (A − bI) + {b^k\over b - a} (A − aI)]A$

i'm having some trouble showing this, any solutions/hints are greatly appreciated.

Answer 1

Hint Since the eigenvalues are distinct, the matrix is diagonalizable, and so we may choose a basis $({\bf e}_a, {\bf e}_b)$ of eigenvectors of $A$, so that $A {\bf e}_a = a A$ and $B {\bf e}_a = a B$. In particular, we can write any vector ${\bf x} \in \Bbb C^2$ as $${\bf x} = u {\bf e}_a + v {\bf e}_b$$ for some $u, v \in \Bbb C$, and we can apply the expression for $A^{n + 1}$ to $\bf x$ so written.

Answer 2

There exists an invertible matrix $T$ such that $TAT^{-1}=\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}$ . If you show that $$T({a^n\over a − b} (A − bI) + {b^n\over b − a} (A − aI))T^{-1}=\begin{bmatrix} a^n & 0 \\ 0 & b^n \end{bmatrix}=TA^nT^{-1}$$ you win.