Find the Jordan Canonical Form of the transformation $L$. 
Let $V$ be the vector space of all 4x4 matrices over $\mathbb{R}$ and let $L$ be the linear operator defined by $L(A)= 4A^T.$

First I found the minimal polynomial of $L$ to be $m_L(x)=x^2-16$. Because the lowest power of $L$ that is linearly dependent on other powers of $L$ is $L^2 = 16A,$ so $m_L(L) = L(A)^2 - 16id(A) =0$. This tells us that the eigenvalues are 4, -4 and the largest size Jordan block of each is one.

Finding the characteristic polynomial was trickier and this is what I am less confident on.

The matrix representation of $L$ is going to be 16x16, so I did not want to take the determinant of that, however just building up a few rows of the matrix, I am fairly certain the characteristic polynomial, $p(x)$ will be $p(x) = (4-x)^{15}(x+4).$ So our matrix in Jordan form will be a 16x16 diagonal matrix with fifteen $4$'s and one $-4$ on the diagonal.
 A: As you have found, the minimal polynomial is $m(x) = x^2 - 16 = (x-4)(x+4)$. Because each (linear) factor has exponent $1$, we know that $L$ will be diagonalizable with eigenvalues $\pm 4$. It follows that the exponent of $(x - \lambda)$ in the characteristic polynomial is equal to the dimension of $E_\lambda(L)$ (the eigenspace of $\lambda$), i.e. the dimension of the set of solutions to the equation to $L(A) = \lambda A$.
To begin, let's look for the dimension of $E_{-4}(L)$. We want the dimension of the space of solutions to
$$
L(A) = -4A \implies 4A^T = -4A \implies A = -A^T.
$$
That is, we want the dimension of the space of skew-symmetric $4\times 4$ matrices. The dimension of this space is $1 + 2 + 3 = 6$, can you see why?
We could do the same for the other eigenvalue, but the easier approach is to note that the characteristic polynomial has degree $16$, which means that the exponent of $(x-4)$ be $16-6 = 10$. Conclude that the characteristic polynomial is given by
$$
p(x) = (x+4)^6(x-4)^{10}.
$$
