Question

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 \(B\).

Short answer

Based on the steps explained above, we calculated the eigenvalues of matrix \(B\) to be \(1, 2i, -2i\). After substituting each eigenvalue back into the equation \((B-\lambda I)X=0\), where X is the eigenvector and I is the identity matrix, we found the corresponding eigenvectors \(X_1 = \left(\begin{array}{c} 1 \\ -2 \\ 2 \end{array}\right), X_2 = \left(\begin{array}{c} 1 \\ i \\ 1-i \end{array}\right), X_3 = \left(\begin{array}{c} 1 \\ -i \\ 1+i \end{array}\right)\).

Step-by-step solution

Calculate the characteristic equation

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

Calculate the eigenvalues

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 \(\lambda_2 = 2i\), \(\lambda_3 = -2i\).

Calculate the eigenvectors for λ = 1

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 \(X_1 = \left(\begin{array}{c} 1 \\ -2 \\ 2 \end{array}\right)\).

Calculate the eigenvectors for λ = 2i and λ = -2i

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: $$ X_1 = \left(\begin{array}{c} 1 \\ -2 \\ 2 \end{array}\right), X_2 = \left(\begin{array}{c} 1 \\ i \\ 1-i \end{array}\right), X_3 = \left(\begin{array}{c} 1 \\ -i \\ 1+i \end{array}\right) $$

Key concepts

Characteristic Equation

The characteristic equation is a crucial concept when finding the eigenvalues of a matrix. To get started, you first need to form the matrix \(B - \lambda I\), where \(B\) is your given matrix and \(\lambda I\) is the diagonal matrix where \(\lambda\) is subtracted from each diagonal entry of \(B\). In mathematical terms, the matrix \(B - \lambda I\) for our example becomes:\[B - \lambda I = \begin{pmatrix} -3-\lambda & -1 & 0 \ 5 & 2-\lambda & 1 \ -5 & 5 & -4-\lambda \end{pmatrix}\]Once the matrix \(B - \lambda I\) is formed, the next step is to calculate its determinant. Setting this determinant equal to zero gives us the characteristic equation. Solving this equation provides the possible values of \(\lambda\), which are referred to as the eigenvalues. For matrix \(B\), the characteristic equation results from:\[\text{det}(B - \lambda I) = (-3-\lambda)((2-\lambda)(-4-\lambda) - 5) + (-1)(5 - 5(\lambda + 4)) = (\lambda-1)(\lambda^2+4)\]From this equation, it's apparent that the eigenvalues are \(\lambda_1 = 1\), \(\lambda_2 = 2i\), and \(\lambda_3 = -2i\). Each eigenvalue contributes to the formation of the final solution.

Complex Conjugates

Complex conjugates arise naturally when dealing with matrices that result in complex eigenvalues. In our example, the eigenvalues \(\lambda_2 = 2i\) and \(\lambda_3 = -2i\) are such complex conjugates.A complex conjugate of a number is achieved by changing the sign of its imaginary part. For example, \(2i\) becomes \(-2i\) when conjugated. It's essential to know that if a matrix has eigenvalues that are complex conjugates, they will always appear in pairs. This property is particularly useful because the behavior of these eigenvalues affects the structure and properties of the matrix.When finding the eigenvectors for complex eigenvalues, once you solve for one, the corresponding eigenvector for the conjugate can be found by taking the complex conjugate of the original eigenvector. For instance, after finding the eigenvector for \(\lambda_2 = 2i\), the eigenvector for \(\lambda_3 = -2i\) can be determined by converting the imaginary parts of each component to their opposite sign.

Eigenvector Calculation

Calculating eigenvectors involves solving a system of linear equations derived from the eigenvalue and the matrix in question. Once you have an eigenvalue \(\lambda\), substitute it back into the equation \((B - \lambda I)X = 0\). This method will produce a homogeneous system of linear equations.A homogeneous system such as this:\[\begin{pmatrix} -4 & -1 & 0 \ 5 & 1 & 1 \ -5 & 5 & -5 \end{pmatrix} \cdot \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}\]By solving this system, you're able to determine the eigenvector. For \(\lambda_1 = 1\), the non-trivial solution gives the eigenvector \(X_1 = \begin{pmatrix} 1 \ -2 \ 2 \end{pmatrix}\).For complex eigenvalues, solving these systems can yield complex vectors. For instance, the vector \(X_2 = \begin{pmatrix} 1 \ i \ 1-i \end{pmatrix}\) for \(\lambda_2 = 2i\). The eigenvector for the conjugate eigenvalue \(\lambda_3 = -2i\) is found by computing the complex conjugate of \(X_2\), leading to \(X_3 = \begin{pmatrix} 1 \ -i \ 1+i \end{pmatrix}\).Detailed computations are crucial in ensuring the correct eigenvectors are associated with each eigenvalue.