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let $ A\in M_{nxn}\left(\mathbb{R}\right) $ with the minimal polynomial $ m_{A}\left(x\right)=x^{2}+1 $ and let $f\in \mathbb{R}[x] $ be polynomial such $f(A) $ is not a scalar matrix. prove that $ f(A) $ does not have real values. Hint: check for the case $\deg(f)=1 $ first.

Here's what I tried:

If $\deg(f)=1 $ then we can assume $ f\left(x\right)=x-\lambda $ I want to find the characteristic polynomial of $f(A) $ So i have to calculate the determinant of $ f\left(A\right)=A-\lambda I $

$ \det\left(xI-\left(A-\lambda I\right)\right)=\det\left(\left(\lambda+x\right)I-A\right) $

I dont know how to continue.

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  • $\begingroup$ They are called “polynomial” and “polynomials”, not “polynom” and “polynoms”. $\endgroup$ – Arturo Magidin May 13 at 19:25
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    $\begingroup$ Ok thanks. Any thoughts about what I wrote? $\endgroup$ – FreeZe May 13 at 20:22
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There is no need to calculate the characteristic polynomial of $A-\lambda I$.

We know that $A$ itself does not have any eigenvalues, because if $\lambda$ is an eigenvalue of $A$ then $t-\lambda$ must divide the minimal polynomial. Since the minimal polynomial is $x^2+1$, there are no (real) eigenvalues.

Now we can simply use the following fact:

Theorem. Let $M$ be an $n\times n$ matrix. If $\mu$ is an eigenvalue of $M$, and $\sigma, \rho$ are any scalars, then $\sigma\mu+\rho$ is an eigenvalue of $\sigma M+\rho I$.

Proof. Let $\mathbf{x}$ be an eigenvector of $M$ corresponding to $\mu$. Then $M\mathbf{x}=\mu\mathbf{x}$. Therefore, $$(\sigma M+\rho I)\mathbf{x} = \sigma M\mathbf{x} + \rho\mathbf{x} = \sigma\mu\mathbf{x}+\rho\mathbf{x} = (\sigma\mu+\rho)\mathbf{x}.$$ Thus, $\mathbf{x}$ is an eigenvector fo $\sigma M+\rho I$ with eigenvalue $\sigma\mu+\rho$. $\Box$

So, if $f(x) = x-\lambda$, and $f(A) = A-\lambda I$ had a real eigenvalue $\mu$, then $A=f(A)+\lambda I$ would have $\mu+\lambda$ as an eigenvalue, which is impossible because we already know that $A$ does not have real eigenvalues.

Note, however, that the problem does not assert that $f$ is monic.

But if $f(x) = ax+b$, then $A = \frac{1}{a}(f(A)-bI) = \frac{1}{a}f(A) - \frac{b}{a}I$, so you can still apply the theorem above.

If $f$ has degree greater than $1$, then consider writing $f(x) = q(x)(x^2+1) + r(x)$, with $r=0$ or $\deg(r)\lt 2$.

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  • $\begingroup$ For the case $ /deg{f} > 1 $ I proved by induction that $ A^n =A $ if $n=1mod4 $ $A^n=-Id $ if $ n=2mod4$ $A^n=-A$ if $n=3mod4$ and $ A^n = Id $ if $n=0mod4 $ but obviously your solution is simpler. Thanks $\endgroup$ – FreeZe May 13 at 23:39
  • $\begingroup$ @Waizman: use a \pmod{b} to produce $a\pmod{b}$’ and a\bmod b to get $a\bmod b$. $\endgroup$ – Arturo Magidin May 13 at 23:44

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