Find the eigenvalues for A

\[ A = \begin{bmatrix} 1 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} \]

We use the following equations to find the eigenvalues: \[ det(A - \lambda I) = 0 \]

\[ det \begin{pmatrix} 1-\lambda & -1 & 1 \\ -1 & 1-\lambda & -1 \\ 1 & -1 & 1-\lambda \end{pmatrix} = 0 \]

Exanding the determinant (Rule of Sarrus) we get:

\[(1-\lambda )(1-\lambda )(1-\lambda ) + (-1)(-1)(1) + (-1)(-1)(1) - (-1)(-1)(1-\lambda) - (1-\lambda )(-1)(-1) - (1)(1-\lambda )(1) = 0\]

This factors to…. \[-\lambda * \lambda * (\lambda - 3) \]

we find the 2 eigenvalues to be:

  1. \(\lambda_1 = 0\) with an algebraic multiplicity \(\alpha _{B}(0) = 1\)
  2. \(\lambda_2 = 3\) with an algebraic multiplicity \(\alpha _{B}(3) = 1\)

Find the eigenspace

  1. \(\lambda = 0\)

\[ \left[ \begin{array}{ccc|c} 1-0 & -1 & 1 & 0 \\ -1 & 1-0 & -1 & 0 \\ 1 & -1 & 1-0 & 0 \\ \end{array} \right] \longrightarrow \left[ \begin{array}{ccc|c} 1 & -1 & 1 & 0 \\ -1 & 1 & -1 & 0 \\ 1 & -1 & 1 & 0 \\ \end{array} \right] \]

Now to solve with gaussian elimination $$ (R_2+R_1) ((R_3-R_1)

$$

This gives Var1 - Var2 + Var3 = 0 (row 1 equation) Var1 = Var2 - Var3 , Var2 = Var2 (row 2 equation) and Var3 = Var3

\[X = \begin{bmatrix} Var2 - Var3 \\ Var2 \\ Var3 \end{bmatrix}\]

There are two solutions, one of Var2 = 1 \[v_1 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}\]

The other if Var3 = 1 \[v_1 = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}\]

This gives the Eigenspace for \(\lambda = 0\) \[\mathcal{E}_{\lambda=0} = \left\langle \left\{ \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \right\} \right\rangle\] The dimension = 2, therefore the geometric multiplicity, \(\gamma _{B}(0)=2\)

  1. \(\lambda = 3\)

\[ \left[ \begin{array}{ccc|c} 1-3 & -1 & 1 & 0 \\ -1 & 1-3 & -1 & 0 \\ 1 & -1 & 1-3 & 0 \\ \end{array} \right] \longrightarrow \left[ \begin{array}{ccc|c} -2 & -1 & 1 & 0 \\ -1 & -2 & -1 & 0 \\ 1 & -1 & -2 & 0 \\ \end{array} \right] \]

Now to solve with gaussian elimination \[ \left[ \begin{array}{ccc|c} -2 & -1 & 1 & 0 \\ -1 & -2 & -1 & 0 \\ 1 & -1 & -2 & 0 \\ \end{array} \right] \rightarrow (R_1/-2) \rightarrow \left[ \begin{array}{ccc|c} 1 & \frac{1}{2} & \frac{-1}{2} & 0 \\ -1 & -2 & -1 & 0 \\ 1 & -1 & -2 & 0 \\ \end{array} \right] \rightarrow ((R_2+R_1) \rightarrow \left[ \begin{array}{ccc|c} 1 & \frac{1}{2} & \frac{-1}{2} & 0 \\ 0 & \frac{-3}{2} & \frac{-3}{2} & 0 \\ 1 & -1 & -2 & 0 \\ \end{array} \right] \rightarrow (R_3-R_1) \rightarrow \left[ \begin{array}{ccc|c} 1 & \frac{1}{2} & \frac{-1}{2} & 0 \\ 0 & \frac{-3}{2} & \frac{-3}{2} & 0 \\ 0 & \frac{-3}{2} & \frac{-3}{2} & 0 \\ \end{array} \right] \rightarrow (R_2/(\frac{-3}{2})) \rightarrow \] \[ \left[ \begin{array}{ccc|c} 1 & \frac{1}{2} & \frac{-1}{2} & 0 \\ 0 & 1 & 1 & 0 \\ 0 & \frac{-3}{2} & \frac{-3}{2} & 0 \\ \end{array} \right] \rightarrow (R_3-\frac{-3}{2}R_2) \rightarrow \left[ \begin{array}{ccc|c} 1 & \frac{1}{2} & \frac{-1}{2} & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right] \rightarrow (R_1 - \frac{1}{2}R_2) \rightarrow \left[ \begin{array}{ccc|c} 1 & 0 & -1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right] \]

This give us Var1 - Var3 = 0 (from row1), Var2 + Var3 = 0 (from row2) Var1 = -Var3, Var2 = Var3 & Var3 = Var3 which we can express as: \[X = \begin{bmatrix} Var3 \\ -Var3 \\ Var3 \end{bmatrix}\]

setting \(Var3=0\) the next eigenvector: \[v_2 = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}\]

This gives the Eigenspace for \(\lambda = 3\) \[\mathcal{E}_{\lambda=0} = \left\langle \left\{ \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} \right\} \right\rangle\] The dimension = 1, therefore the geometric multiplicity, \(\gamma _{B}(3)=1\)