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Problem Statement

The problem C11, selected page 388,
Find the characteristic polynomial of the matrix.

##      [,1] [,2] [,3]
## [1,]    3    2    1
## [2,]    0    1    1
## [3,]    1    2    0

Solution using R

## [1]  1 -4  0  5

Which gives us the characteristic polynomial as

\({ PA(x)\quad =\quad λ }^{ 3 }-4{ λ }^{ 2 }+5\)

Solution by hand

Let’s find the characteristic polynomial of a 3x3 matrix. Let

\(A=\begin{vmatrix} 3 & 2 & 1 \\ 0 & 1 & 1 \\ 1 & 2 & 0 \end{vmatrix}\)

To get the eigenvalues, we need to solve

\(det(\begin{vmatrix} λ & 0 & 0 \\ 0 & λ & 0 \\ 0 & 0 & λ \end{vmatrix}-\begin{vmatrix} 3 & 2 & 1 \\ 0 & 1 & 1 \\ 1 & 2 & 0 \end{vmatrix})\quad =\quad 0\)

That is

\(det(\begin{vmatrix} λ-3 & -2 & -1 \\ 0 & λ-1 & -1 \\ -1 & -2 & λ \end{vmatrix})\quad =\quad 0\)

Solving determinant
\((λ-3)\begin{vmatrix} λ-1 & -1 \\ -2 & λ \end{vmatrix}-(-2)\begin{vmatrix} 0 & -1 \\ -1 & λ \end{vmatrix}+(-1)\begin{vmatrix} 0 & λ-1 \\ -1 & -2 \end{vmatrix}\quad =\quad 0\)

\((λ-3)\left[ ({ λ }^{ 2 }-λ)-2 \right] +2(λ-1)+2+(-λ+1)\quad =\quad 0\\ { λ }^{ 3 }-{ λ }^{ 2 }-2λ-3{ λ }^{ 2 }+3{ λ }+6+2{ λ }-2-{ λ }+1\quad =\quad 0\\ { λ }^{ 3 }-4{ λ }^{ 2 }+5\quad =\quad 0\)

Which gives us the characteristic polynomial as

\({ PA(x)\quad =\quad λ }^{ 3 }-4{ λ }^{ 2 }+5\)