Q:

Diagonalisable real matrix.  Conclusion

A matrix $A\in\mathbb{R}^{n,n}$ is diagonalisable on $\mathbb{R}$. Then:  
a.  characteristic polynomial of $A$ is product of polynomials degree $1$ with real coefficients
b. matrix $A^2$ is also diagonalisable  
c. $A=A^T$  .  
b. is obviously true, we know that $A^2=(PDP^{-1})^2 = PD^2P^{-1}$  Due to $D^2$ is still diagonal then b. is true. 
a.
We know that $A$ is similar to some diagonal matrix $D$. Therefore, $A$ has the same characteristic polynomial as $D$, so it is product of $(x-\lambda)$ where $\lambda$ is any eigenvalue.  Lets consider:
$A=\left( \begin{array}{ccc}
2 & 0  \\
0 & 2 \\
\end{array} \right)=\left( \begin{array}{ccc}
1 & 0 \\
0 & 1 \\
\end{array} \right) \left( \begin{array}{ccc}
2& 0 \\
0 & 2 \\
\end{array} \right)\left( \begin{array}{ccc}
0& 1  \\
1 & 0 \\
\end{array} \right)$ is diagonalisable, but $P_A(x)=(x-2)^2$.  So it is not prodcut of degree 1 polynomials.  Hence, a. is not true.   
What about these solutions ?   Can you help me solve c. ?  

A:

For $c)$ , just take the matrix $\pmatrix{1&2\\3&4}$ which is not symmetric , but real diagonalizable.