# minimal polynomial, characteristic polynomial proof

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.

• They are called “polynomial” and “polynomials”, not “polynom” and “polynoms”. – Arturo Magidin May 13 at 19:25
• Ok thanks. Any thoughts about what I wrote? – FreeZe May 13 at 20:22

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 $$\deg(r)\lt 2$$.

• 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 – FreeZe May 13 at 23:39
• @Waizman: use a \pmod{b} to produce $a\pmod{b}$’ and a\bmod b to get $a\bmod b$. – Arturo Magidin May 13 at 23:44