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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.

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  • $\begingroup$ Did you mean "express $A^{-1}$ in the form $\alpha I + \beta A$"? $\endgroup$ – Ben Grossmann Dec 4 '18 at 19:05
  • $\begingroup$ Yes @Omnomnomnom. $\endgroup$ – little o Dec 4 '18 at 19:06
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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 $A^{-1} = \frac 13(-A + 4I)$.

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  • $\begingroup$ Very nice idea @Omnomnomnom. Thank you very much for your help. $\endgroup$ – little o Dec 4 '18 at 19:15

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