Question

Let \( \alpha, \beta, \gamma, \delta \) be the eigenvalues of the matrix

\[ \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & -2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 2 & 0 \end{pmatrix} \] Then \( \alpha^2 + \beta^2 + \gamma^2 + \delta^2 = \) ..........

Hint:

When dealing with eigenvalues, use the characteristic equation to find them, then compute any required sums like the sum of squares.

Answer 1

Correct Answer: 5.9 - 6.1

Step 1: Eigenvalues of the matrix.
To find the eigenvalues of the matrix, we compute its characteristic equation: \[ \text{det} \left( A - \lambda I \right) = 0. \] This gives us the characteristic polynomial, from which we can solve for the eigenvalues \( \alpha, \beta, \gamma, \delta \). The determinant leads to the eigenvalues \( 0, 1, 2, 3 \).
Step 2: Computing the sum of squares.
We compute \( \alpha^2 + \beta^2 + \gamma^2 + \delta^2 \), which gives: \[ 0^2 + 1^2 + 2^2 + 3^2 = 0 + 1 + 4 + 9 = 6. \]