cayley-Hamilton law

1)
there is $A=\left(\begin{array}{ccc}4 & -2 & 4\\-2 & 1 & -2\\4 & -2 & 4\end{array}\right)$
A.we define $f:R^{3}xR^{3}->R$ by $f(x,y)=(Ax,y)$ for every $y=(y_{1},y_{2},y_{3})^{t}$ and $x=(x_{1},x_{2},x_{3})^{t}$ in $R^{3}$
does f define inner product in $R^{3}$ ?
how i tried:
in the book by definition inner product is following these rules:
(u,v)=(v,u)
(u+v,w)=(u,w)+(v,w)
(ku,v)=k(u,v)
$(v,v)\geq0$
i dont know how to apply these rules here
?
B.find a simetric matrices B so $B^{2}=A$
(clue: use cayley-Hamilton law)
?
how i tried:
hamilton law say that the representative matrice turns to zero the minimal polinomial M(A)=0
any guidence
?

Re: cayley-Hamilton law

a) are you sure about the definition of f? an inner product should go from $\mathbb{R}^3 \times \mathbb{R}^3 \rightarrow \mathbb{R}$ .

perhaps you mean $f(x,y) = <(Ax)^T,y>$ ? without some further clarification, your question doesn't make much sense.

Re: cayley-Hamilton law

yes you are correct i had typing mistake its
$f:R^{3}xR^{3}->R$