Question C22 - Without using a calculator, find the eigenvalues of the matrix \(B\).

\[ B = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} \]

Solve to get the characteristic equation for the eigenvalues:

\(B\) is the original matrix and \(\lambda I\) is \(\lambda\) * \(\textbf{The Identity Matrix}\)

\[ \text{det}(B - \lambda I) = 0 \]

The matrix \(B - \lambda I\) is:

\[ B - \lambda I = \begin{bmatrix} 2 - \lambda & -1 \\ 1 & 1 - \lambda \end{bmatrix} \]

The determinant is:

\[ \text{det}(B - \lambda I) = (2 - \lambda)(1 - \lambda) - (-1)(1) \]

Solving for \(\lambda\):

\[ (2 - \lambda)(1 - \lambda) + 1 = 0 \]

Finally I get the the characteristic polynomial function: The problem here is it can not be factored like other normal instances in the characteristic polynomial function.

\[ \lambda^2 - 3\lambda + 3 = 0 \] Since n=3 this is a quadratic polynomial and I can try and use the quadratic formula to solve:

\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \(a = 1\), \(b = -3\), and \(c = 3\) :

\[ \lambda = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(3)}}{2(1)} \]

\[ \lambda = \frac{3 \pm \sqrt{-3}}{2} \]

So, the eigenvalues are complex numbers:

\[ \lambda_1 = \frac{3}{2} + \frac{\sqrt{3}i}{2} \]

\[ \lambda_2 = \frac{3}{2} - \frac{\sqrt{3}i}{2} \]

The operations I did by hand can now be verified by using the eigen(\(B\)) function from base R.

B = matrix(c(2,-1,1,1), nrow = 2, byrow = T)
eigen_values <- eigen(B)$values
print(eigen_values)
## [1] 1.5+0.866025i 1.5-0.866025i