For matrix A = \(\begin{bmatrix} 2 & 1\quad 1 \\ 1 & 2\quad 1 \\ 1 & 1\quad 2 \end{bmatrix}\) the characteristic polynomial of A is \(p_{ A }{ (X)\quad =\quad (4-x){ (1-x) }^{ 2 } }\). Find the eigenvalues and corresponding eigenspaces of A.
Solution:
Since we are given the characteristic polynomial of A we can find the eigenvalues of A to be \(\lambda\) = 4 and \(\lambda\) = 1. To find the corresponding eigenspaces of A we use the eigenvalues and row reduce the matrix A - \(\lambda\)\(I_{ 3 }\), where I is the identity matrix. Then we express the eigenspace as the null space of A - \(\lambda\)\(I_{ 3 }\),
\(E_{ A }(4)\quad\) = \(A - 4I_{ 3 }\) = \(A - 4\begin{bmatrix} 1 & 0\quad 0 \\ 0 & 1\quad 0 \\ 0 & 0\quad 1 \end{bmatrix}\) = \(\begin{bmatrix} 2 & 1\quad 1 \\ 1 & 2\quad 1 \\ 1 & 1\quad 2 \end{bmatrix} - \begin{bmatrix} 4 & 0\quad 0 \\ 0 & 4\quad 0 \\ 0 & 0\quad 4 \end{bmatrix}\) = \(\begin{bmatrix} -2 & 1\quad 1 \\ 1 & -2\quad 1 \\ 1 & 1\quad -2 \end{bmatrix}\quad\)
Row reduction of \(\begin{bmatrix} -2 & 1\quad 1 \\ 1 & -2\quad 1 \\ 1 & 1\quad -2 \end{bmatrix}\quad\) =
-(1/2) * row 1 = \(\begin{bmatrix} 1 & 1/2\quad 1/2 \\ 1 & -2\quad 1 \\ 1 & 1\quad -2 \end{bmatrix}\quad\)
-1 * row 1 + row 2 to row 2 = \(\begin{bmatrix} 1 & -1/2 & -1/2 \\ 0 & -3/2 & 3/2 \\ 1 & 1 & -2 \end{bmatrix}\)
-1 * row 1 + row 3 to row 3 = \(\begin{bmatrix} 1 & -1/2 & -1/2 \\ 0 & -3/2 & 3/2 \\ 1 & 3/2 & -3/2 \end{bmatrix}\)
-(2/3) * row 2 = \(\begin{bmatrix} 1 & -1/2 & -1/2 \\ 0 & 1 & -1 \\ 1 & 3/2 & -3/2 \end{bmatrix}\)
-(3/2) * row 2 + row 3 to row 3 = \(\begin{bmatrix} 1 & -1/2 & -1/2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}\)
(1/2) * row 2 + row 1 to row 1 = \(\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}\)
\(E_{ A }(4)\quad\) = N(\(A - 4I_{ 3 }\)) = \(<\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}>\)
\(E_{ A }(1)\quad\) = \(A - 1I_{ 3 }\) = \(A - 1\begin{bmatrix} 1 & 0\quad 0 \\ 0 & 1\quad 0 \\ 0 & 0\quad 1 \end{bmatrix}\) = \(\begin{bmatrix} 2 & 1\quad 1 \\ 1 & 2\quad 1 \\ 1 & 1\quad 2 \end{bmatrix} - \begin{bmatrix} 1 & 0\quad 0 \\ 0 & 1\quad 0 \\ 0 & 0\quad 1 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}\)
Row reduction of \(\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}\) =
-1 * row 1 + row 2 to row 2 = \(\begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix}\)
-1 * row 1 + row 3 to row 3 = \(\begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\)
\(E_{ A }(1)\quad\) = N(\(A - 1I_{ 3 }\)) = \(<\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix},\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}>\)
Eigenvalues = \(\lambda\) = 4 and \(\lambda\) = 1
Eigenspaces = \(E_{ A }(4)\quad\) = \(<\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}>\) and \(E_{ A }(1)\quad\) = \(<\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix},\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}>\)