Is there any short cut method to find the characteristic and minimal polynomial of $A$? Given $A= \begin{bmatrix} 0 &1&0&0\\0&0&1&0\\0&0&0&1\\1&0&0&0 \ \end{bmatrix}$
Is  there  any short cut method  to find the  characteristic and minimal polynomial  of $A$ ?
 A: Actually, there is. You can use the cofactor procedure. Note that if you choose eliminate the first row and the first column and then proceed eliminating the second, third and fourth row, you will have to deal with determinants of either upper triangular or diagonal matrices. 
A: It helps to note that $A$ is a permutation matrix. In particular, we can conclude that there must be an integer $n$ such that $A^n = I$.  In this case, we find that $A^4 = I$.  Thus, the minimal polynomial of $A$ must divide $x^4 - 1$.  In fact, we find in this case that the minimal polynomial of $A$ is $x^4 - 1$, since $I,A,A^2,A^3$ are linearly independent.
Thus, the characteristic and minimal polynomials of $A$ are $x^4 - 1$.
More generally, we can deduce that the minimal and characteristic polynomials of a permutation matrix by using the cycle-decomposition of the permutation.
Another approach is to recognize $A$ (or $A^T$ depending on your definition) as the companion matrix to the polynomial $x^4 - 1$.
Yet another approach is to recognize $A$ as a circulant matrix.
A: Any matrix is similar to its transpose, and the transpose is seen to be the companion matrix of $x^4-1$.
A: We can use the minor expansion formula to find the characteristic polynomial:
\begin{aligned}\det(A-\lambda I)
&= \begin{vmatrix}-\lambda&1\\&-\lambda&1\\&&-\lambda&1\\1&&&-\lambda\end{vmatrix} \\
&= -\lambda\cdot\begin{vmatrix}-\lambda&1\\&-\lambda&1\\&&-\lambda\end{vmatrix} - 1\cdot\begin{vmatrix}1\\-\lambda&1\\&-\lambda&1\end{vmatrix} \\
&= -\lambda\cdot (-\lambda)^3 - 1 \cdot 1^3 \\
&= \lambda^4-1
\end{aligned}
The roots are the 4th roots of unity: $\{\pm 1,\pm i\}$. 
Since they are distinct, the minimal polynomial is the same.
