Diagonalisation and characteristic polynomial

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.

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.

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.