A First Course in Linear Algebra, Section EEM: Eigenvalues and Eigenvectors of a Matrix
\[ B=\begin{bmatrix} 2 & -1\\ 1 & 1\\ \end{bmatrix}. \]
The characteristic polynomial of a matrix \(B\) can be found using
\[ \text{det}(B-\lambda\ I) = 0, \]
where \(\lambda\) is some scalar and \(I\) is an identity matrix. The result of this equation will be a polynomial whose roots are the eigenvalues of \(B\).
The product of \(\lambda\) and the identity matrix will result in a matrix with \(\lambda\) down the diagonal.
\[ \lambda\ I = \begin{bmatrix} \lambda & 0\\ 0 & \lambda\\ \end{bmatrix} \]
The difference of \(\lambda\ I\) and \(B\) results in:
\[ B-\lambda\ I = \begin{bmatrix} 2 & -1\\ 1 & 1\\ \end{bmatrix} - \begin{bmatrix} \lambda & 0\\ 0 & \lambda\\ \end{bmatrix} = \begin{bmatrix} 2-\lambda & -1-0\\ 1-0 & 1-\lambda\\ \end{bmatrix} = \begin{bmatrix} 2-\lambda & -1\\ 1 & 1-\lambda\\ \end{bmatrix} \]
The determinant of \(B-\lambda\ I\) is:
\[ \text{det}(B-\lambda\ I) = (2 - \lambda)(1-\lambda) - 1(-1) \]
Which gives us the characteristic polynomial:
\[ \lambda^2 - 3\lambda + 3 \]
To find the solutions of this polynomial, and the corresponding eigenvalues, we need to apply the quadratic formula.
\[ \lambda = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
Plugging in the coefficients of the characteristic polynomial, we get a complex solution:
\[ \lambda = \frac{3 \pm \sqrt{3^2-4(1)(3)}}{2(1)} = \frac{3 \pm \sqrt{-3}}{2} = \frac{3 \pm \sqrt{3i}}{2} \]
Thus, eigenvalues of the matrix \(B\) are:
\[ \lambda = \frac{3 + \sqrt{3i}}{2} \quad \text{and} \quad \frac{3 - \sqrt{3i}}{2} \]