# Express $A^{-1}$ in the form $\alpha I + \beta A$.

Let $$A \in \Bbb {M_3} (\Bbb R)$$ whose eigen values are $$1,1,3$$. Then express $$A^{-1}$$ in the form $$\alpha I + \beta A$$, $$\alpha,\beta \in \Bbb R$$.

What I found is that $$A^{-1} = \frac {1} {3} (A^2 -5A+7I)$$. How do I express it in the desired form? Please help me in this regard.

Thank you very much.

• Did you mean "express $A^{-1}$ in the form $\alpha I + \beta A$"? – Ben Grossmann Dec 4 '18 at 19:05
• Yes @Omnomnomnom. – little o Dec 4 '18 at 19:06

Your formula (computed using the characteristic polynomial) is correct. If $$A$$ fails to be diagonalizable, then $$A$$ is non-derogatory and, as my post here explains, no further reduction will be possible. In particular, we can conclude that since $$\{I,A,A^2\}$$ is a linearly independent set, the set $$\{A^{-1},I,A\}$$ will also be linearly independent.
However, if $$A$$ is diagonalizable, then its minimal polynomial will be $$(x-1)(x-3) = x^2 - 4x + 3$$, which is to say that $$A$$ will satisfy $$A^2 - 4A + 3I = 0$$ which we can rearrange to find that $$A^{-1} = \frac 13(-A + 4I)$$.