# Prove: If $g(A)$ is not scalar ($g(A) \neq \lambda I$) $\rightarrow$ $g(A)$ has no real eigenvalues for a given matrix and minimum polynomial

Given $A \in M_{n x n} (\mathbb R)$ such that $m_A(x) = x^2 + 1$ (the minimum polynomial), and let $g \in \mathbb R[x]$. Prove: If $g(A)$ is not scalar ($g(A) \neq \lambda I$) $\rightarrow$ $g(A)$ has no real eigenvalues

I can't really find the connection. So if $m_A(X)$ is like that, it is obvious that it has no -real- eigenvalues, since the minimum polynomial includes all the roots of the characteristic polynomial. Or am I getting something wrong?

• You just pointed out that $A$ has no real eigenvalues but you didn't prove it holds for $g(A)$. Commented Jun 1, 2013 at 10:22
• I think I should figure out what does applying a polynomial to a matrix do to its' minimum polynomial. Is that relevant? Commented Jun 1, 2013 at 10:28

Hint: Let $g(x) = (x^2+1)p(x) + qx + r$ for some real polynomial $p$ and some real numbers $q$ and $r$. Then $g(A) = qA + rI$. If $g(A)$ is not a scalar matrix, then $q\neq0$.

• Does $g(x)$ have to be in this form? If so, why? If not, why did you take this form? Commented Jun 1, 2013 at 10:54
• @TheNotMe Do you know Euclidean division of polynomial? Commented Jun 1, 2013 at 11:05
• I do, I just don't understand what the connection here. Why does $g(x)$ look like this? Commented Jun 1, 2013 at 11:19
• @TheNotMe Divide $g(x)$ by $x^2+1$. Call the quotient $p(x)$. By Euclidean division, the degree of the remainder is a smaller than the degree of the divisor $x^2+1$, hence it must be of the form $qx+r$. Commented Jun 1, 2013 at 11:46
• Haha, that I understand. I am more confused about why $g(x)$ has to have $(x^2 + 1)p(x)$ as an element. Commented Jun 1, 2013 at 11:51

Fact $(1)$(Can easily be proved by induction using $A^2=-I$):$A^{2k-1}=\epsilon _1A$ and $A^{2k}=\epsilon_2 I$, $\forall k\in \mathbb{N},\epsilon_1,\epsilon_2\in\{1,-1\}$

Let $g(x)=\sum_{i=0}^{n}a_ix^i$($a_i\in R$)

Using Fact $(1)$ we have

$g(A)=aA+bI$($a,b\in R$ and clearly $a\ne 0$ for then $g(A)=bI$ which is impossible)

So if $g(A)$ has real eigen values it will imply ,

$(aA+bI)v=\lambda v$

$\Rightarrow A=\frac{(\lambda-b)}{a}v$ .This is impossible as $A$ doesnt have any real eigen value.

• Weird. $m_A(A) = 0 = A^2 + I \rightarrow A^2 = -I$ and not $A^2 = -A$. Commented Jun 1, 2013 at 13:12
• Now I think its ok.@TheNotMe Commented Jun 1, 2013 at 14:42