# Linear Algebra - eigenvalues

I have this problem:

Let $A$ be any $n \times n$ matrix, defined over the real numbers, such that $A-A^2=I$. Then prove that $A$ does not have any real eigenvalues.

What I did:

$$A-A^2=I$$ $$A-A^2-I=0$$ $$A(A-I)-I=0$$

Now I need to show that $A(A-I) \neq I$. Assume that $A=2I$. (If $A=I$ then $0 \neq I$.) Then $$A(A-I)=I$$ $$2I(2I-I)=I$$ $$2I(I)=I$$ $$2I \neq I$$

This equation doesn't have a solution, but I don't think I proved correctly.

Any help will be appreciated, thanks.

• $A$ is a $n\times n$ matrix, ok, but in what ring, field is it defined? – Bman72 Oct 7 '14 at 12:37

If $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $(A-A^2)v = \lambda v - \lambda^2 v = (\lambda - \lambda^2)v$; hence if $A-A^2 = I$, then you'd be solving $(\lambda - \lambda^2)v = v$ implying that $\lambda-\lambda^2 = 1$, which has no real solutions (but certainly does have complex solutions).
$$A-A^2=I\iff A^2-A+I=0\implies A$$
is a zero of the polynomial $\;p(x)=x^2-x+1\;$. This polynomial has no roots in $\;\Bbb R\;$ and thus the claim is true if $\;A\;$ is defined as a real matrix, but this polynomial has two different complex non-real roots and if $\;A\;$ is defined over $\;\Bbb C\;$ then it does have two (different) eigenvalues, at least.
Sure you mean that $A$ does not have a real eigenvalue. Indeed the polynomial $P(x)=x^2-x+1$ annihilates $A$ so any eigenvalue of $A$ is a root of $P$. We verify easily that $P$ has two complex non real roots since $\Delta =1-4<0$.