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.

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Comments

  • Yun  On October 30, 2011 at 5:58 am

    I’m currently working through AA:DF also, just for the hell of it, and I think this project is great, keep up the good work!

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