Let \( A=\left[\begin{array}{ccc}1 & -1 & -1 \\ 1 & -1 & 0 \\ 1 & 0 & -1\end{array}\right] \), Find the eigenvalues of \( A \). a) Find the characteristic equation of \( A \) bi) Find the eigenvectors of \( A \).

To find the eigenvalues of the matrix \( A=\left[\begin{array}{ccc}1 & -1 & -1 \\ 1 & -1 & 0 \\ 1 & 0 & -1\end{array}\right] \), we first derive the characteristic equation by computing the determinant of \( A - \lambda I \), where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalues. The characteristic polynomial is given by \( \text{det}(A - \lambda I) = 0 \). Let's compute \( A - \lambda I \): \[ A - \lambda I = \left[\begin{array}{ccc} 1-\lambda & -1 & -1 \\ 1 & -1-\lambda & 0 \\ 1 & 0 & -1-\lambda \end{array}\right] \] Next, we calculate the determinant: \[ \text{det}(A - \lambda I) = (1-\lambda)((-1-\lambda)(-1-\lambda) - 0) - (-1)(1(-1-\lambda) - 0) - (-1)(1(0) - (-1)(1)) \] Calculating the deteminant step-by-step: \[ = (1-\lambda)((-1-\lambda)^2) + (1)(1+\lambda) \] \[ = (1-\lambda)(\lambda^2 + 2\lambda + 1) + 1 + \lambda \] \[ = (1-\lambda)(\lambda^2 + 2\lambda + 1) + \lambda + 1 \] Now expanding \( (1-\lambda)(\lambda^2 + 2\lambda + 1) \): \[ = \lambda^2 + 2\lambda + 1 - \lambda^3 - 2\lambda^2 - \lambda \] \[ = -\lambda^3 - \lambda^2 + \lambda + 1 \] Thus: \[ \text{det}(A - \lambda I) = -\lambda^3 - \lambda^2 + 2\lambda + 2 = 0 \] This can be rewritten as: \[ \lambda^3 + \lambda^2 - 2\lambda - 2 = 0 \] Now, factoring or using the Rational Root Theorem, we can find rational roots. Testing \(\lambda = -1\): \[ (-1)^3 + (-1)^2 - 2(-1) - 2 = -1 + 1 + 2 - 2 = 0 \] Thus, \(\lambda = -1\) is a root. We can factor \((\lambda + 1)\) out of \(\lambda^3 + \lambda^2 - 2\lambda - 2\). Using synthetic division: \[ 1 | 1 1 -2 -2 \\ | -1 0 2 \\ \hline | 1 0 -2 0 \] Resulting in: \[ \lambda^2 - 2 \] Thus: \[ \lambda^3 + \lambda^2 - 2\lambda - 2 = (\lambda + 1)(\lambda^2 - 2) = 0 \] Setting each factor to zero: \[ \lambda + 1 = 0 \implies \lambda = -1 \] \[ \lambda^2 - 2 = 0 \implies \lambda = \pm\sqrt{2} \] Thus, the eigenvalues are: \[ \lambda_1 = -1, \ \lambda_2 = \sqrt{2}, \ \lambda_3 = -\sqrt{2} \] For part bi, we now find the eigenvectors corresponding to each eigenvalue. 1. **Eigenvector for \( \lambda_1 = -1 \)**: We solve \((A - (-1)I)\mathbf{v} = 0\): \[ \left[\begin{array}{ccc} 2 & -1 & -1 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{array}\right]\begin{bmatrix} x\\y\\z \end{bmatrix}= \begin{bmatrix} 0\\0\\0 \end{bmatrix} \] This results in: \[ \begin{cases} 2x - y - z = 0 \\ x = 0 \\ x = 0 \end{cases} \] Choosing \( z = 1, y