Question C22

Compute the eigenvalues of the matrix by hand \[ \begin{bmatrix} 2 &-1\\ 1 &1 \end{bmatrix} \]

To find the eigenvalues, we want to compute \(\lambda\) where \(\text{det}(A- \lambda I_2)=0\).
\[ \text{det}(A-\lambda I_2)=\begin{vmatrix} 2-\lambda & -2\\ 1 & 1-\lambda \end{vmatrix} \\= (2-\lambda)(1-\lambda)-(-2)(1) \\= 2- \lambda-2\lambda +\lambda^2 +1 \\= \lambda^2 -3\lambda +3 \] If we plot this, we can see that the roots of \(\lambda^2 -3\lambda +3\) are complex:

curve(x^2 -3*x+3, from = -2, to =5)

Thus, the roots are easiest found via the quadratic formula.

\[ \lambda= \frac{3\pm \sqrt{3^2-4(1)(3)} \ }{2} \\= \frac{3 \pm \sqrt{-3}}{2} \\ \text{so} \\ \lambda=\frac{3+i\sqrt{3}}{2}, \lambda=\frac{3-i\sqrt{3}}{2} \]

Checking my work in R:

A<-matrix(c(2,1,-1,1),nrow=2)
e<-eigen(A)
e$values
## [1] 1.5+0.866025i 1.5-0.866025i