Week 3 Discussion

Page 306 question C22

Find eigenvalues of the matrix B:

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

To find the eigenvalues of B we must find where:

\[ det( \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} ) = 0 \]

Simplify:

\[ det( \begin{bmatrix} \lambda - 2 & 1 \\ -1 & \lambda - 1 \end{bmatrix} ) = 0 \]

This is equal to:

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

Now solve the equation: \(\lambda^2-3\lambda+3=0\)

\(\lambda^2-3\lambda+3=0\)

Now solve using the quadratic formula \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

\(\lambda=\frac{3\pm\sqrt{9-(4)(1)(3)}}{2}\)

\(\lambda=\frac{3\pm\sqrt{-3}}{2}\)

This is equal to:

\(\lambda=\frac{3\pm\sqrt{3i}}{2}\)

So the two eigenvalues are:

\(\lambda=\frac{3+\sqrt{3i}}{2}\)

\(\lambda=\frac{3-\sqrt{3i}}{2}\)

Now confirm using R:

B <- matrix(c(2, -1, 1, 1), nrow = 2, byrow = TRUE)

# Compute eigenvalues
eigenvalues <- eigen(B)$values

eigenvalues

## [1] 1.5+0.866025i 1.5-0.866025i