## Compute the eigenvalues of a given matrix

Find the eigenvalues of the following matrix.

 $A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix}$

Recall that the eigenvalues of a matrix $A$ are the roots of the polynomial $c_A(x) = \mathsf{det}(xI - A)$. Now in this case, we have

 $xI - A = [a_{i,j}] = \begin{bmatrix} x & -1 & 0 & 0 \\ 0 & x & -1 & 0 \\ 0 & 0 & x & -1 \\ -1 & 0 & 0 & 0 \end{bmatrix}.$

Using the Cofactor Expansion Formula for the determinant along the 4th row, (Theorem 29 on page 439 of D&F, with $i = 4$), we have that $\mathsf{det}(xI - A) = (-1)^5 a_{4,1} \mathsf{det}(A_{4,1}) + (-1)^6 a_{4,2} \mathsf{det}(A_{4,2})$ $+ (-1)^7 a_{4,3} \mathsf{det}(A_{4,3}) + (-1)^8 a_{4,4} \mathsf{det}(A_{4,4})$. Since $a_{4,2} = a_{4,3} = 0$, this simplifies to

 $\mathsf{det}(xI - A)$ = $\mathsf{det}(A_{4,1}) + x \mathsf{det}(A_{4,4})$ = $\mathsf{det} \begin{bmatrix} -1 & 0 & 0 \\ x & -1 & 0 \\ 0 & x & -1 \end{bmatrix} + x \mathsf{det} \begin{bmatrix} x & -1 & 0 \\ 0 & x & -1 \\ 0 & 0 & x \end{bmatrix}$.

Note that both of these minor matrices are diagonal, and recall that the determinant of an upper- or lower-triangular matrix is the product of the diagonal entries. (To see this, consider the Leibniz formula for the determinant: $\mathsf{det}([a_{i,j}]) = \sum_{\sigma \in S_n} \varepsilon(\sigma) \prod_{i=1}^n a_{\sigma(i),i}$. If $[a_{i,j}]$ is upper (lower) triangular, then for every $\sigma$, if $\sigma$ moves at least one element in $\{1,\ldots,n\}$, then it must move some element to a strictly larger (smaller) one, in which case $\prod a_{\sigma(i),i}$ vanishes since some factor is zero. What remains is merely $\prod_i a_{i,i}$.)

So we have $c_A(x) = -1 + x x^3 = x^4 - 1$. Clearly the roots of this polynomial, hence the eigenvalues of $A$, are $\pm 1$ and $\pm i$.