Question-27
What is the product of the non-zero eigenvalues of the following matrix: \[ \begin{bmatrix} 1 & 0 & 0 & 0 & 1\\ 0 & 1 & 1 & 1 & 0\\ 0 & 1 & 1 & 1 & 0\\ 0 & 1 & 1 & 1 & 0\\ 1 & 0 & 0 & 0 & 1 \end{bmatrix} \]
\(6\)
Thanks to Vivek, Sherry and Aniruddha for this elegant solution.
We have:
\[ \begin{equation*} A=\begin{bmatrix} 1 & 0 & 0 & 0 & 1\\ 0 & 1 & 1 & 1 & 0\\ 0 & 1 & 1 & 1 & 0\\ 0 & 1 & 1 & 1 & 0\\ 1 & 0 & 0 & 0 & 1 \end{bmatrix} \end{equation*} \]
Given the symmetry in the problem, by inspection, we see that \(\displaystyle A\) has the following eigenpairs:
\[ \begin{gather*} ( 2,( 1,0,0,0,1))\\ \\ ( 3,( 0,1,1,1,0)) \end{gather*} \]
We also see that the rank of \(\displaystyle A\) is \(\displaystyle 2\). For a symmetric matrix (a diagonalizable matrix, to be more specific), the rank is equal to the number of non-zero eigenvalues (including the multiplicity). Therefore \(2\) and \(3\) are the only non-zero eigenvalues and each of them occurs just once. The required product is \(\boxed{6}\).
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 \(\boxed{6}\).