\[ A = \begin{bmatrix} 1 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} \]
We are looking for the values of \(\lambda\) that makes \(det(\lambda I_n -A) = 0\) true. In order to solve this, we build the following matrix:
\[ \lambda I_n -A = \begin{bmatrix} \lambda-1 & 1 & -1 \\ 1 & \lambda-1 & 1 \\ -1 & 1 & \lambda-1 \end{bmatrix} \] and find the roots of its characteristic polynomial by setting its determinant to 0: \[ det(\lambda I_n -A) = (\lambda-1)[(\lambda-1)(\lambda-1)-1] - (1)[(\lambda-1)+1]+(-1)[1+\lambda01] = 0 \\ (\lambda-1)^3 - 3(\lambda-1) - 2 = 0 \\ (\lambda-1)(\lambda^2-2\lambda-2) - 2 = 0 \\ \lambda^3-3\lambda^2 = 0 \\ \lambda^2(\lambda-3) = 0 \\ \lambda = 0 \ or\ \lambda = 3 \]
We found two eigenvalues \(\lambda = 0, \ 3\) that we now plug into the matrix \(\lambda I_n -A\) and row reduce:
\[ \lambda = 0 \ \ \ \ \begin{bmatrix} -1 & 1 & -1 \\ 1 & -1 & 1 \\ -1 & 1 & -1 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & -1 & 1 \\ 1 & -1 & 1 \\ 1 & -1 & 1 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & -1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]
We now solve for values of \(v_i\) that make the following system true. Arbitrary values are chosen for free variables \(v_2 \ and \ v_3\).
\[ \begin{bmatrix} 1 & -1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]
\[ v_3 = b \\ v_2 = a \\ v_1-v_2+v_3=0 \longrightarrow v_1 = v_2-v_3 \longrightarrow v_1 = a-b \]
We now have the eigenspace for the matrix A with associated eigenvalue \(\lambda = 0\)
\[ E_{\lambda=0} = \Bigg\{ \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = a \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + b \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \Bigg| \ a, b \in \Re \Bigg\} \]
\[ E_{\lambda=0} = span \Bigg( \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \Bigg) \]
We conclude that the algebraic multipicity is 2 as 0 was a repeated root of the characteristic polynomial. Two vectors span the space so the geometric multiplicity is also 2.
\[ \alpha_{0} = 2, \ \gamma_{0} = 2 \]
Repeat the process for the eigenvalye \(\lambda = 3\) \[ \lambda = 3 \ \ \ \begin{bmatrix} 2 & 1 & -1 \\ 1 & 2 & 1 \\ -1 & -1 & 2 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & -1 \\ -1 & -1 & 2 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \]
\[ \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]
\[ v_3 = a \\ v_2 + v_3 = 0 \longrightarrow v_2 = -v_3 \longrightarrow v_2 = -a \\ v_1 + 2v_2 + v_3 = 0 \longrightarrow v_1 = -v_3 - 2v_2 \longrightarrow v_1 = a \]
\[ E_{\lambda=3} = \Bigg\{ \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = a \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} \Bigg| \ \ a \in \Re \Bigg\} \]
\[ E_{\lambda=3} = span \Bigg( \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} \Bigg) \]
We then conclude that the algebraic multiplicity is 1 (one root) and that the geometric multiplicity is also 1 (one vector spans the space).
\[ \alpha_{3} = 1, \ \gamma_{3} = 1 \]
The eigen() function returns both eigenvalues and eigenvectors. The middle eigenvalue returns is not exactly equal to 0, but with -16 as an exponent we can also consider it to be zero.
## eigen() decomposition
## $values
## [1] 3.000000e+00 8.881784e-16 0.000000e+00
##
## $vectors
## [,1] [,2] [,3]
## [1,] 0.5773503 -0.8164966 0.0000000
## [2,] -0.5773503 -0.4082483 0.7071068
## [3,] 0.5773503 0.4082483 0.7071068At first glance, it appears that the returned eigenvectors don’t match the vectors found above. The eigen() function returns vectors that are normalized to unit length. So we can divide each column by it’s smallest entry to get integers.
## [,1] [,2] [,3]
## [1,] 1 -2 0
## [2,] -1 -1 1
## [3,] 1 1 1We recognize the first eigenvector from \(\lambda = 3\) but the following two are different for \(\lambda = 0\). Upon investigation we find that the two vectors are linear combinations of the eigenvectors found above with \(a=-1, b=1\) and \(a=1, b=1\)