Diagonalisation and characteristic polynomial Let $A\in M_3(\Bbb R)$ which is not a diagonal matrix. Let $P$ be a polynomial
(in one variable), with real coefficients and of degree 3 such that $P(A) = 0$.
Pick out the true statements:
a. $P = cP_A$ where $c \in\Bbb R$ and $P_A$ is the characteristic polynomial of $A$;
b. if $P$ has a complex root (i.e. a root with non-zero imaginary part), then
$P = cP_A$, with $c$ and $P_A$ as above;
c. if $P$ has a complex root, then $A$ is diagonalizable over $\Bbb C$.
I guess b and c are true although I am unable to disprove a. Because A could satisfy $(X-A)$ but this is not a polynomial. Now if $P$ has a complex root, then its conjugate too is its root as the polynomial has  real coefficients. As we all know An odd degree polynomial has always a real root, what we have is three distinct roots. Since $A$ satisfies the polynomial the only difference between the characteristic polynomial and $P$ has to be a constant.
 A: Part a
Counterexample: consider
$$
A = \pmatrix{1&1&1\\1&1&1\\1&1&1}
$$
We note that $P_A(x) = x^2(x-3)$ (you should verify that this is the case).  However, $A^2 - 3A = 0$, which is to say that $Q(x) = x^2 - 3x = x(x-3)$ is a polynomial for which $P(A) = 0$. It follows that $P(x) = (x-1)Q(x) = x^3 - 4x^2 + 3x$ is a polynomial of degree $3$ for which $P(A)=0$ that is not a scalar multiple of $P_A$.
Note that any non-diagonal matrix of the form
$$
P\;
\pmatrix{
\lambda_1&0&0\\
0&\lambda_2&0\\
0&0&\lambda_2
}\;P^{-1}
$$
With $\lambda_1 \neq \lambda_2$ will also be a suitable counterexample.
Part b
We note that $A$ satisfies a polynomial of the form 
$$
P(X) = c(x-z)(x-\overline{z})(x-r)
$$
For $z \in \mathbb{C}$ and $r \in \mathbb{R}$. Thus, the minimal polynomial $q_A(x)$ of $A$ must divide $P$.  The only factor of $P$ that could be the minimal polynomial of a real, non-diagonal $3 \times 3$ matrix is 
$$
q_A(x) = (x-z)(x-\overline{z})(x-r) 
$$
which divides and thus must be equal to $p_A(x)$.  Thus, $P(x) = cp_A(x)$
Part c
$A$ has $3$ distinct eigenvalues, therefore it has $3$ linearly independent eigenvectors. It follows that $A$ is diagonalizable.
A: Part a
Counterexample: take 
$$
A = \begin{bmatrix}
0 & 1 & 0\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{bmatrix}
$$
and $P(x) = x^{2} (x + 1)$; here $P_{A} = x^{3}$.
Part b and c
These are true! 
Since $P$ is a real polynomial, it must have a real root, and two conjugate complex roots. (I am using complex in the sense of OP, that is, non-real.) So $P$ has distinct roots. Now the minimal polynomial of $A$ divides $P$, so it has also distinct roots, which implies $A$ is diagonalizable over the complex numbers.
This shows c. To show b, use the argument given in the other answer.
A: The hypotheses of this question are rather strange, but this is what they give. The polynomial $P$ is a multiple of the minimal polynomial$~\mu$ of$~A$. Now $\mu$ could be a monic polynomial of degree $1\leq d\leq 3$ with real coefficients, but since $A$ is given to not be a multiple of the identity (which would in particular be diagonal), one must have $d\in\{2,3\}$. Since $A$ has size $3$, its (real) characteristic polynomial$~\chi$ has odd degree and therefore a real root, and this real eigenvalue must also be a root of$~\mu$; when $d=2$ this means that the quadradic polynomial $\mu$ must be split over the real numbers: $\mu=(X-\alpha)(X-\beta)$. When $d=3$ we get no such condition, but now one must have $\mu=\chi$ by the Cayley-Hamilton theorem.
From the above we see when $d=3$ that $P$ must be a scalar multiple of $\chi=\mu$, but there is no reason why it should be the case if $d=2$, as the factor of degree$~1$ by which $\mu$ is multiplied to get $P$ is independent of what $\chi$ is; therefore a. is false. (For a very easy example one could take $A\neq0$ with $A^2=0$; then $\mu=X^2$, $\chi=X^3$, but one may choose $P=aX^3+bX^2$ with nonzero $\def\R{\Bbb R}a,b\in\R$; more generally one could arrange $\mu$ to be any reducible quadratic polynomial, one of whose roots$~\alpha$ will be a multiple root of$~\chi$, and take $P=Q\mu$ where $Q\in\R[X]$ is of degree$~1$ and does not have $\alpha$ as root).
If $P$ has a non-real complex root$~z$ then one cannot have $d=2$, as we have seen that $\mu$ then has real roots, and $z$ cannot be root of the remaining factor$~Q$ of$~P$, since $Q$ has real coefficients (as $P$ was required to) and is of degree$~1$. Then $d=3$, and hence $P$ is a scalar multiple of $\chi$; point b. is true.
Finally, still assuming $P$ has a non-real root$~z$, this means that (being real) it has both $z$ and its complex conjugate $\overline z$ as complex roots, and also one real root. Then being annihilated by $P$ which has three distinct simple roots in$~\Bbb C$, the matrix $A$ will be be diagonalisable over$~\Bbb C$ (though it is not diagonalisable over$~\R$, since $\mu$ having root$~z$ is not split over$~\R$); therefore point c. is true. 
