Discussion 3

Exercises C10

Find the characteristic polynomial of the matrix

\[ A= \left[ \begin{array}{cccc} 1 & 2 \\ 3 & 4 \end{array} \right] \]

Answer

library(pracma)
A <- matrix(c(1, 3, 2, 4), nrow = 2)
A

##      [,1] [,2]
## [1,]    1    2
## [2,]    3    4

A <- charpoly(A, info = TRUE)

## Error term: 4

A$cp

## [1]  1 -5 -2

The characteristic polynomial or characteristic equation is:

\[ Det(\lambda I - A) = 0 \]

\[ \lambda I = \lambda \left[ \begin{array}{cccc} 1 & 0 \\ 0 & 1 \end{array} \right] \]

\[ \lambda I = \left[ \begin{array}{cccc} \lambda & 0 \\ 0 & \lambda \end{array} \right] \]

\[ \left[ \begin{array}{cccc} \lambda & 0 \\ 0 & \lambda \end{array} \right] - \left[ \begin{array}{cccc} 1 & 2 \\ 3 & 4 \end{array} \right] \]

\[ \left[ \begin{array}{cccc} \lambda -1 & 0 - 2 \\ 0 - 3 & \lambda - 4 \end{array} \right] \]

\[ \left[ \begin{array}{cccc} \lambda -1 & - 2 \\ - 3 & \lambda - 4 \end{array} \right] \]

Determinant of a 2x2 matrix

\[ \lambda -1 \times \lambda -4 - -2 \times -3 \]

\[ \lambda^2 - 5\lambda -2 = 0 \]