Finding the Jordan Form of a transformation defined by $T(X)=AX$ when $A,X \in M_{4\times 4}(\mathbb C)$ 
Given $A=\left(\matrix{0&1&0&0\\1&0&0&0\\0&0&1&0\\0&0&0&-1}\right)$ and define $T(X)=AX$. when $A,X \in M_{4\times 4}(\mathbb C)$
Find the Jordan Form of T

I found that the minimal polynomial is $(t-1)(t+1)$ therefore $T$ is diagonalizable. However, I'm not sure how to find the amount of each eigenvalue to put in the diagonal without explicity finding the $16\times16$ representative matrix of T.

I identify a $4\times 4$ matrix by a vector of length $16$ by reading it from top to bottom, from left to right. So the first four coordinates are from the first column, etc. 
Then $T$ is diagonal in the last $8$ coordinates: identical on coordinates $9-12$, and negative of the identity on $13-16$, so that part is covered. 
Furthermore, $T$ switches the first four coordinates by the second four as a product of four transpositions. You can view these transpositions separately. The matrix of a transposition of two coordinates is $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, whose Jordan normal form is $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$. 
So the Jordan normal form of $T$ is diagonal with eigenvalues: 
$$1,-1,1,-1,1,-1,1,-1,1,1,1,1,-1,-1,-1,-1$$
