# Eigenvalues and Cayley-Hamilton theorem

Let $A$ be an invertible matrix such that $A^{-1}=I-A$. Show that $A$ doesn't have real eigenvalues.

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Since $$A^{-1}=I-A$$ then we have $$A^2-A+I=0$$ hence the polynomial $P(x)=x^2-x+1$ annihilates the matrix $A$ and then the eigenvalues of $A$ belong to the set of roots of $P$ which are complex non real since the discriminant of $P$ is $\Delta=-3<0$.

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Let $\lambda$ be any eigen value. Then $\lambda\neq 0$ since $A$ is invertible. Now, using the following facts:

1. $\frac1\lambda$ is eigenvalue for $A^{-1}$.

2. $1-\lambda$ is eigen value of $I-A$.

I'm sure in this point you can finish by yourself, can't you?

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Yea I did. thankyou – user137205 Mar 22 '14 at 12:49

If $A$ had an eigenvector $v$ for $\lambda$ (which eigenvalue cannot be $0$ since $A^{-1}$ exists), applying both sides of the euqation to $v$ would give $$\lambda^{-1}v=v-\lambda v$$ and since $v\neq 0$ for an eigenvector, this requires $$\lambda^{-1}=1-\lambda$$ to hold. But that equation has no solutions for $\lambda$ in the real numbers, as you can easily check.

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