Characteristic polynomial of $A^2$, given the characteristic polynomial of $A$ 
Let $A \in M_{3\times 3}(\mathbb{R})$. Its characteristic polynomial is $P_A(t) = t^3+t^2+t-3$. Find the coefficient of the characteristic polynomial of  $A^2$.

I tried to solve it by finding the factors of A, but it looks like it doesn't have factors in the real numbers $\mathbb{R}$ only above $\mathbb{C}$.
 A: $P_A(t)P_A(-t)$ is a multiple of $P_A$ so it will be zero when $t=A$.
It is also a polynomial, degree six, with only even-degree terms;
so it is a cubic in $t^2$.  This cubic is the polynomial $P_{A^2}$
A: Solution 1:
The matrix $A$ has three distinct eigenvalues since $P_A'(t)=3t^2+2t+1=2t^2+(t+1)^2\gt 0$ for every $t\in\mathbb R$. And, each eigenvalue of $A^2$ is a square of some eigenvalue of $A$ and so, similarly as $A$, the matrix $A^2$ has three distinct eigenvalue. So, we should find a polynomial $q(t)$ such that each root of $q$, is a square of some root of $P_A$. Thus, in the characteristic polynomial of $A$ we should change $t$ to $\sqrt t$. Therefore, $$t\sqrt t+t+\sqrt t-3=\sqrt t(1+t)+t-3=0$$ And from here: $$\sqrt t(1+t)=3-t$$ And by squaring both side:
$$P_{A^2}(t)=t(1+t)^2-(3-t)^2=t^3+t^2+7t-9$$
Solution 2:
Let $X\in M_3(\mathbb R)$ and $\lambda_i(X)$ for $i=1,2,3$ denote the eigenvalues of $X$. The characteristic polynomial of $X$ is as follows:  $$P_X(t)=t^3-(\sum_i\lambda_i(X))t^2+(\sum_{i< j}\lambda_i(X)\lambda_j(X))t-\prod_i\lambda_i(X).$$ Because of the characteristic polynomial of $A$ is: $$P_A(t)=t^3+t^2+t-3$$ we get that: $$\sum_i\lambda_i(A)=-1\,,\sum_{i<j}\lambda_i(A)\lambda_j(A)=1\,,\prod_i\lambda_i(X)=3.$$ 
Now by 
$$\begin{align}
\sum_i\lambda_i(A)^2=(\sum_i\lambda_i(A))^2-2\sum_{i<j}\lambda_i(A)\lambda_j(A)\\
\end{align}
$$
$$\begin{align}
\sum_{i<j}\lambda_i(A)^2\lambda_j(A)^2&=(\sum_{i<j}\lambda_i(A)\lambda_j(A))^2-2(\prod_i\lambda_i(A))(\sum_i\lambda_i(A))
\\
\end{align}
$$
$$\begin{align}
\prod_i\lambda_i(A)^2=(\prod_i\lambda_i(A))^2
\\
\end{align}
$$
we get 
$$
\begin{align}
\sum_i\lambda_i(A)^2=-1\\
 \sum_{i<j}\lambda_i(A)^2\lambda_j(A)^2=7\\ \prod_i\lambda_i(A)^2=9
\end{align}
$$
So, $$P_{A^2}(t)=t^3+t^2+7t-9.$$
