How to express $A^{-1}$ in the form of $\alpha I+\beta A$, where $\alpha, \beta \in \mathbb{R}$.

Question

Let $A\in \mathbb{M}_3(\mathbb{R})$ be a symmetric matrix whose eigen-values are $1,1$ and $3$. Express $A^{-1}$ in the form $\alpha I +\beta A$, where $\alpha, \beta \in \mathbb{R}$.

Answer 1

Let $\alpha,\beta\in \Bbb R$ such that $A^{-1}=\alpha I+\beta A$. Then $$A(\alpha I+\beta A)=\alpha A+ \beta A^2=I$$

This is equivalent to say that $\beta A^2 + \alpha A-I=0$, that is $A$ is a solution of the polynomial $$p(x)=\beta x^2+\alpha x-1=0.$$ This is also equivalent to say that the eigenvalues $1,3$ are solutions of $p(x)$. Let $q(x)=(x-1)(x-3)=x^2-4x+3$, then $$p(x)=-\frac13q(x)=-\frac13 x^2+\frac43 x -1.$$ Hence $\alpha=\frac43$ and $\beta=-\frac13$.

Answer 2

The minimal polynomial of $A$ is $p(x)=(x-1)(x-3)=x^2-4x+3$. By Cayley-Hamilton:

$$A^2-4A+3I=0.$$

This gives

$$A-4I+3A^{-1}=0.$$