“Without using a calculator, find the eigenvalues of the matrix \(B\).”
\(B=\begin{bmatrix}2 & -1\\1 & 1\end{bmatrix}\)
\(det(\begin{bmatrix}2 & -1\\1 & 1\end{bmatrix}-\begin{bmatrix}\lambda & 0\\0 & \lambda\end{bmatrix})=0\)
\(det(\begin{bmatrix}2-\lambda & -1\\1 & 1-\lambda\end{bmatrix})=0\)
\([(2-\lambda)(1-\lambda)]-[(1)(-1)]=0\)
\(2-3\lambda+\lambda^2+1=0\)
\(\lambda^2-3\lambda+3=0\)
We can now use the quadratic formula to find the eigenvalues:
\(\lambda=\frac{3\pm\sqrt{9-4(1)(3)}}{2}\)
\(\lambda=\frac{3\pm\sqrt{-3}}{2}\)
So the two eigenvalues for matrix \(B\) are \(\lambda_1=\frac{3+\sqrt{3}i}{2}\) and \(\lambda_2=\frac{3-\sqrt{3}i}{2}\)
Testing that this solution is correct:
matrix_B <- matrix(c(2,1,-1,1), nrow = 2)
matrix_B## [,1] [,2]
## [1,] 2 -1
## [2,] 1 1eigen(matrix_B)## eigen() decomposition
## $values
## [1] 1.5+0.866025i 1.5-0.866025i
##
## $vectors
## [,1] [,2]
## [1,] -0.7071068+0.0000000i -0.7071068+0.0000000i
## [2,] -0.3535534+0.6123724i -0.3535534-0.6123724i