Problem-2

Question

What is the product of the non-zero eigenvalues of the following matrix: \[ \begin{bmatrix} 1 & 0 & 0 & 0 & 1\\ 0 & 1 & 1 & 1 & 0\\ 0 & 1 & 1 & 1 & 0\\ 0 & 1 & 1 & 1 & 0\\ 1 & 0 & 0 & 0 & 1 \end{bmatrix} \]

Solution

The product of the eigenvalues of a matrix is equal to its determinant. But the determinant of this matrix is zero and we are asked to compute the product of the non-zero eigenvalues. We should therefore take recourse to the characteristic polynomial: \[ D = \begin{vmatrix} 1 - \lambda & 0 & 0 & 0 & 1\\ 0 & 1 - \lambda & 1 & 1 & 0\\ 0 & 1 & 1 - \lambda & 1 & 0\\ 0 & 1 & 1 & 1 - \lambda & 0\\ 1 & 0 & 0 & 0 & 1 - \lambda \end{vmatrix} \] Though this determinant looks forbidding, computing it is not all that cumbersome if we choose the right row/column along which to expand. Let us first break this down into the following sum by expanding along the first row: \[ D = (1 - \lambda) \begin{vmatrix} 1 - \lambda & 1 & 1 & 0\\ 1 & 1 - \lambda & 1 & 0\\ 1 & 1 & 1 - \lambda & 0\\ 0 & 0 & 0 & 1 - \lambda \end{vmatrix} + \begin{vmatrix} 0 & 1 - \lambda & 1 & 1\\ 0 & 1 & 1 - \lambda & 1\\ 0 & 1 & 1 & 1 - \lambda\\ 1 & 0 & 0 & 0 \end{vmatrix} \] The two determinants look quite similar to each other. We can expand along the last row for both of them: \[ D = \left[(1 - \lambda)^2 - 1 \right] \begin{vmatrix} 1 - \lambda & 1 & 1\\ 1 & 1 - \lambda & 1\\ 1 & 1 & 1 - \lambda\\ \end{vmatrix} \] Let us first compute the \(3 \times 3\) determinant: \[ \begin{vmatrix} 1 - \lambda & 1 & 1\\ 1 & 1 - \lambda & 1\\ 1 & 1 & 1 - \lambda\\ \end{vmatrix} = - \lambda^2 (\lambda - 3) \] Putting it all together: \[ D = -\lambda^3(\lambda - 2)(\lambda - 3) \] We see that there are five eigenvalues: \(0, 0, 0, 2, 3\). The product of the non-zero eigenvalues is therefore \(\boxed{6}\).