= Solution
A <Jordan block> is $J_m(\alpha)=\alpha I+N$, where $N$ has ones on the superdiagonal and zeros elsewhere. Since $N^2=0$ for $m=2$, Taylor expansion gives
$$
p(J_2(\alpha))=p(\alpha)I+p'(\alpha)N
=\begin{pmatrix}p(\alpha)&p'(\alpha)\\0&p(\alpha)\end{pmatrix}.
$$
The <Jordan normal form> is the block-diagonal <matrix>, unique up to block order, to which $A$ is similar and whose blocks are Jordan blocks.
A block of size at least two for <eigenvalue> $\lambda$ contributes a <generalized eigenvector> in $\ker(A-\lambda I)^2\setminus\ker(A-\lambda I)$. Thus the displayed equality for every $\lambda$ is equivalent to every block having size one, which is diagonalizability.
The transpose of each Jordan block is similar to it by reversing its <basis>. Hence $A^T$ has the same Jordan form as $A$.
Write $A=PJP^{-1}$. Let $R$ be block diagonal with the reversal <matrix> on each Jordan block. Both $R$ and $RJ$ are symmetric and $R$ is invertible. Therefore
$$
A=(PRP^T)\{P^{-T}(RJ)P^{-1}\}=CD,
$$
where $C,D$ are symmetric and $C$ is nonsingular.
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