## C12 Page 298

### Find the characteristic polynomial of the matrix:

$$A = \left[ \begin{array}{cccc} 1 & 2 & 1 & 0 \\ 1 & 0 & 1 & 0 \\ 2 & 1 & 1 & 0 \\ 3 & 1 & 0 & 1 \\ \end{array} \right] \\$$

The characteristic polynomial of square matrix $$A$$ is defined as $$f_A(\lambda)=det(A−\lambda I)$$

$$(A−\lambda I)$$ = $$A = \left[ \begin{array}{cccc} 1 & 2 & 1 & 0 \\ 1 & 0 & 1 & 0 \\ 2 & 1 & 1 & 0 \\ 3 & 1 & 0 & 1 \\ \end{array} \right] \\$$ - $$\lambda$$ $$\left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right] \\$$ = $$\left[ \begin{array}{cccc} 1-\lambda & 2 & 1 & 0 \\ 1 & -\lambda & 1 & 0 \\ 2 & 1 & 1-\lambda & 0 \\ 3 & 1 & 0 & 1-\lambda \\ \end{array} \right] \\$$

Therefore $$det(A−\lambda I) = det$$ $$\left[ \begin{array}{cccc} 1-\lambda & 2 & 1 & 0 \\ 1 & -\lambda & 1 & 0 \\ 2 & 1 & 1-\lambda & 0 \\ 3 & 1 & 0 & 1-\lambda \\ \end{array} \right] \\$$

After reducing the matrix to row echelon and simplifyin all expression, we’ll obtain:

$$det(A−\lambda I) = \lambda^4-3\lambda^3-2\lambda^2+2\lambda+2$$

Hence: $$f_A(\lambda)= \lambda^4-3\lambda^3-2\lambda^2+2\lambda+2$$

### If in doubt, let’s get R opinion using the pracma library

library(pracma)

A <- matrix(data = c(1,2,1,0,
1,0,1,0,
2,1,1,0,
3,1,0,1), nrow=4, byrow=TRUE)

A
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    1    0
## [2,]    1    0    1    0
## [3,]    2    1    1    0
## [4,]    3    1    0    1
#determinind the characteristic polynomial of A

charpoly(A, info = FALSE)
## [1]  1 -3 -2  2  2

The result is interpreted as: $$f_A(\lambda)= \lambda^4-3\lambda^3-2\lambda^2+2\lambda+2$$