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Given: $$ A \in M_{n\times n}(\mathbb R) \; , \; A - A^2 = I $$

Then we have to prove that $A$ does not have real eigenvalues. How do we prove such a thing?

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closed as off-topic by quid Jul 6 at 15:05

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    $\begingroup$ No work or motivation shown - vote to close. $\endgroup$ – user641 May 1 '13 at 17:24
  • $\begingroup$ Why so many upvotes $\endgroup$ – Lost1 Jun 25 '14 at 15:09
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Hint: Apply your equation to an eigenvector.

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$A-A^2=I\to A^2-A=-I\to A^2-A+I=0$

Since A satisfies its own characteristic equation (Cayley-Hamilton Theorem), replace A with $\lambda$ and see what you get....

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  • $\begingroup$ Why did you replace $I$ with $1$? $\endgroup$ – TheNotMe May 1 '13 at 16:39
  • $\begingroup$ sorry...shouldn't have $\endgroup$ – Eleven-Eleven May 1 '13 at 16:41
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    $\begingroup$ When you have done linear algebra for long enough, many people get tired of $I$ and just write $1$ instead. It is not that much different from writing the zero vector, or operator, as $0$. $\endgroup$ – Harald Hanche-Olsen May 1 '13 at 16:41
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    $\begingroup$ You identify the I with 1 when doing these types of problems, especially Cayley Hamilton since you are replacing real or complex numbers with Matrices. $\endgroup$ – Eleven-Eleven May 1 '13 at 16:43
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    $\begingroup$ A satisfies its own characteristic equation, but not every polynomial equation which A satisfies is necessarily its characteristic equation. Therefore, I think you're missing a step in this solution. $\endgroup$ – Kallus May 1 '13 at 16:49
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Hint: Is there a real number $\lambda$ with $\lambda-\lambda^2=1$?

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  • $\begingroup$ Obviously not. But I don't know, I can't relate the hint. Why would $A-A^2 = I$ mean that there might be $\lambda - \lambda^2 = 1$ ? $\endgroup$ – TheNotMe May 1 '13 at 16:38
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    $\begingroup$ Combine my hint what that of @Julian. $\endgroup$ – Harald Hanche-Olsen May 1 '13 at 16:39
  • $\begingroup$ Apparently I am weak on the theorm side of diagonalization. I know that if $A$ is diagonalization then $Av = \lambda v$. So $\lambda v - \lambda^2v = Iv = v$ and we divide by $v$ But.. is this correct? Because the question did not tell us if $A$ is diagonalizational or not $\endgroup$ – TheNotMe May 1 '13 at 16:42
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    $\begingroup$ No diagonalization needed! You are on the wrong track. Start with @Julian's hint, then come back to mine. $\endgroup$ – Harald Hanche-Olsen May 1 '13 at 16:44
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    $\begingroup$ Under @JulianKuelshammer's assumption, $A^2v=A(Av)=A(\lambda v)=\lambda Av=\lambda\lambda v=\lambda^2v$. Do you get the point now? $\endgroup$ – Harald Hanche-Olsen May 1 '13 at 16:54
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Assume $A-A^2=1$. Let $Q(X):=X^2-X+1$. Then $Q(A) = 0$.

We also denote the characteristic polynomial of $A$ by $\chi_A$. Furthermore, we denote the minimal polynomial of $A$ by $\mu_A$. This is the smallest polynomial, w.r.t. degree, that satisfies $\mu_A(A)=0$. This polynomial divides all other polynomials $P$ that fulfil $P(A)=0$.

Let $\lambda$ be an arbitrary root of $\chi_A$. Then $\lambda$ is an eigenvalue of $A$ to some eigenvector $v\neq 0$. For any polynomial $P$, $P(\lambda)$ is then an eigenvalue of $P(A)$. In particular, $\mu_A(\lambda)$ is an eigenvalue of $\mu_A(A)$, i.e. $\mu_A(A)\cdot v = \mu_A(\lambda)\cdot v$. However, $\mu_A(A)=0$ by definition of $\mu_A$, thus $\mu_A(\lambda)\cdot v = 0\cdot v = 0$. Since $v \neq 0$, we conclude $\mu_A(\lambda)=0$.

Therefore, any root of $\chi_A$ is also a root of $\mu_A$. Furthermore, since $Q(A)=0$, $\mu_A$ divides $Q$, therefore any root of $\mu_A$ must be a root of $Q$. However, $Q$ has no root in $\mathbb{R}$.

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$A$ satisfies $x^2-x+1=0$ whose discriminant is negative.

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By using index notation, $A-A^2=I$ can be written as $A_{ij}-A_{ik}A_{kj}=\delta_{ij}$. By definition: $A_{ij}n_i=\lambda n_j$. So that, $A_{ij}n_i-A_{ik}A_{kj}n_i=\delta_{ij}n_i$, hence $\lambda n_j -\lambda n_k A_{kj}=n_j$, whence $\lambda n_j -\lambda^2 n_j=n_j$, or $(\lambda^2-\lambda+1)n_j=0$, $n_j\neq 0$ an eigenvector. $\lambda^2-\lambda+1=0$ has not real roots.

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