The matrix \(B\) is defined by $$ B=\left(\begin{array}{ccc} -3 & -1 & 0 \\ 5 & 2 & 1 \\ -5 & 5 & -4 \end{array}\right) $$ (a) Calculate the eigenvalues of \(B\). (b) Calculate the eigenvectors of \(B\).

To find the eigenvalues, we must first calculate the characteristic equation given by the determinant of \((B-\lambda I)\), which is $$ \text{det}(B-\lambda I) = \text{det}\left(\begin{array}{ccc} -3-\lambda & -1 & 0 \\\ 5 & 2-\lambda & 1 \\\ -5 & 5 & -4 -\lambda \end{array}\right) $$

Now, we find the determinant of the resulting matrix to form a polynomial equation in terms of \(\lambda\): $$ \text{det}(B-\lambda I) = (-3-\lambda)((2-\lambda)(-4-\lambda) - 5) + (-1)(5 - 5(\lambda + 4)) = (\lambda-1)(\lambda^2+4) $$ It can be seen that the eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = 2i\), \(\lambda_3 = -2i\).

We substitute the eigenvalue \(\lambda_1 = 1\) back into the equation \((B-\lambda I)X=0\): $$ \left(\begin{array}{ccc} -3-1 & -1 & 0 \\\ 5 & 2-1 & 1 \\\ -5 & 5 & -4-1 \end{array}\right) \cdot \left(\begin{array}{c} x_{1} \\\ x_{2} \\\ x_{3} \end{array}\right) = \left(\begin{array}{c} 0 \\\ 0 \\\ 0 \end{array}\right) $$ $$ \left(\begin{array}{ccc} -4 & -1 & 0 \\\ 5 & 1 & 1 \\\ -5 & 5 & -5 \end{array}\right) \cdot \left(\begin{array}{c} x_{1} \\\ x_{2} \\\ x_{3} \end{array}\right) = \left(\begin{array}{c} 0 \\\ 0 \\\ 0 \end{array}\right) $$ Now we solve this system of linear equations to find one non-trivial eigenvector associated with \(\lambda_1 = 1\). We can get the eigenvector \(X_1 = \left(\begin{array}{c} 1 \\ -2 \\ 2 \end{array}\right)\).

Since the other eigenvalues are complex conjugates (\(\lambda_2 = 2i\) and \(\lambda_3 = -2i\)), we can find one eigenvector for the complex eigenvalue and then take its complex conjugate to find the eigenvector for the conjugate eigenvalue. We substitute the eigenvalue \(\lambda_2 = 2i\) back into the equation \((B-\lambda I)X=0\): $$ \left(\begin{array}{ccc} -3-2i & -1 & 0 \\\ 5 & 2-2i & 1 \\\ -5 & 5 & -4 - 2i \end{array}\right) \cdot \left(\begin{array}{c} x_{1} \\\ x_{2} \\\ x_{3} \end{array}\right) = \left(\begin{array}{c} 0 \\\ 0 \\\ 0 \end{array}\right) $$ Now we solve this system of linear equations to find one eigenvector associated with \(\lambda_2 = 2i\). We can find the eigenvector \(X_2 = \left(\begin{array}{c} 1 \\ i \\ 1-i \end{array}\right)\). To find the eigenvector associated with \(\lambda_3 = -2i\), we take the complex conjugate of \(X_2\): \(X_3 = \left(\begin{array}{c} 1 \\ -i \\ 1+i \end{array}\right)\). The eigenvectors of matrix \(B\) are: $$ X_1 = \left(\begin{array}{c} 1 \\ -2 \\ 2 \end{array}\right), X_2 = \left(\begin{array}{c} 1 \\ i \\ 1-i \end{array}\right), X_3 = \left(\begin{array}{c} 1 \\ -i \\ 1+i \end{array}\right) $$