Find the eigenvalues, eigenspaces, algebraic and geometric multiplicities for
\[ A = \begin{bmatrix} 1 & -1 & 1\\ -1 & 1 & -1\\ 1 & -1 & 1\\ \end{bmatrix}\\ \]
\[ \det(A - \lambda I) = 0\\ \det( \begin{bmatrix} 1 & -1 & 1\\ -1 & 1 & -1\\ 1 & -1 & 1\\ \end{bmatrix} - \begin{bmatrix} \lambda & 0 & 0\\ 0 & \lambda & 0\\ 0 & 0 & \lambda\\ \end{bmatrix}) = 0 \\ \det( \begin{bmatrix} 1 - \lambda & -1 & 1\\ -1 & 1 - \lambda & -1\\ 1 & -1 & 1 - \lambda\\ \end{bmatrix}) = 0\\ (1-\lambda) \det(\begin{bmatrix} 1 - \lambda & -1\\ -1 & 1 -\lambda\\ \end{bmatrix}) - (-1) \det( \begin{bmatrix} -1 & 1\\ -1 & 1 - \lambda\\ \end{bmatrix}) + (1) \det( \begin{bmatrix} -1 & 1 \\ 1 - \lambda & -1\\ \end{bmatrix}) = 0\\ (1 - \lambda)((1-\lambda) \times (1-\lambda)-(-1 \times -1)) + ((-1 \times (1 - \lambda))-(-1 \times 1)) + ((-1 \times -1) - (1 \times (1-\lambda)) = 0\\ (1 - \lambda)(-2\lambda + \lambda ^2) + \lambda + \lambda = 0\\ -\lambda^3 +2\lambda^2 - 2\lambda + \lambda ^2 + 2\lambda = 0\\ -\lambda ^3 +3\lambda^2 = 0\\ -\lambda^2 (\lambda - 3) = 0\\ \lambda_1 = 0, \lambda_2 = 3\\ \]
\(\lambda_1 = 0\) with multiplicity 2 and \(\lambda_2 = 3\).
First, we’ll find the eigenvector for \(\lambda_1 = 0\).
\[ (A-\lambda I)x = 0\\ \begin{bmatrix} 1 & -1 & 1\\ -1 & 1 & -1\\ 1 & -1 & 1\\ \end{bmatrix} - \begin{bmatrix} \lambda_1 & 0 & 0\\ 0 & \lambda_1 & 0\\ 0 & 0 & \lambda_1\\ \end{bmatrix})x = 0 \\ (\begin{bmatrix} 1 & -1 & 1\\ -1 & 1 & -1\\ 1 & -1 & 1\\ \end{bmatrix} - \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\\ \end{bmatrix})x = 0 \\ (\begin{bmatrix} 1 & -1 & 1\\ -1 & 1 & -1\\ 1 & -1 & 1\\ \end{bmatrix}) x = 0\\ \]
Reduced Echelon Form:
\[ \begin{bmatrix} 1 & -1 & 1\\ -1 & 1 & -1\\ 1 & -1 & 1\\ \end{bmatrix} \rightarrow \text{R2 = R1 + R2} \rightarrow \begin{bmatrix} 1 & -1 & 1\\ 0 & 0 & 0\\ 1 & -1 & 1\\ \end{bmatrix} \rightarrow \text{R3 = R3 - R1} \rightarrow \begin{bmatrix} 1 & -1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}\\ \] Find the Null Space:
\[ \begin{bmatrix} 1 & -1 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0\\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}\\ x_1 - x_2 + x_3 = 0 \\ \text{Let's let }x_2 = t, x_3 = s, x_1 = -s +t\\ \overrightarrow{x} = \begin{bmatrix} - s + t\\ t\\ s\\ \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \\ \end{bmatrix} t + \begin{bmatrix} -1 \\ 0 \\ 1\\ \end{bmatrix}s \\ \] The eigenvectors are \(v_1 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}\) and \(v_2 = \begin{bmatrix} -1 \\ 0 \\ 1 \\ \end{bmatrix}\) for corresponding eigenvalue \(\lambda_1 = 0\) with multiplicity 2.
Now we’ll find the eigenvector for eigenvalue \(\lambda_1 = 3\).
\[ (A-\lambda I)x = 0\\ (\begin{bmatrix} 1 & -1 & 1\\ -1 & 1 & -1\\ 1 & -1 & 1\\ \end{bmatrix} - \begin{bmatrix} \lambda_1 & 0 & 0\\ 0 & \lambda_1 & 0\\ 0 & 0 & \lambda_1\\ \end{bmatrix})x = 0 \\ (\begin{bmatrix} 1 & -1 & 1\\ -1 & 1 & -1\\ 1 & -1 & 1\\ \end{bmatrix} - \begin{bmatrix} 3 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 3\\ \end{bmatrix})x = 0 \\ (\begin{bmatrix} 1 - 3 & -1& 1 \\ -1 & 1-3 & -1 \\ 1 & -1 & 1-3 \\ \end{bmatrix})x = 0 \\ (\begin{bmatrix} -2 & -1 & 1 \\ -1 & -2 & -1 \\ 1 & -1 & -2 \\ \end{bmatrix})x = 0\\ \] Reduced Echelon Form
\[ \begin{bmatrix} -2 & -1 & 1 \\ -1 & -2 & -1 \\ 1 & -1 & -2 \\ \end{bmatrix} \rightarrow \text{R1 = - 1/2 R1} \rightarrow \begin{bmatrix} 1 & 1/2 & -1/2\\ -1 & -2 & -1 \\ 1 & -1 & -2 \\ \end{bmatrix} \rightarrow \text{R2 = R2 + R1} \rightarrow \begin{bmatrix} 1 & 1/2 & -1/2\\ 0 & -3/2 & -3/2\\ 1 & -1 & -2 \\ \end{bmatrix} \rightarrow\\ \rightarrow \text{R3 = R3 - R1} \rightarrow \begin{bmatrix} 1 & 1/2 & -1/2\\ 0 & -3/2 & -3/2\\ 0 & -3/2 & -3/2\\ \end{bmatrix} \rightarrow \text{R2 = -2/3 R2} \rightarrow \begin{bmatrix} 1 & 1/2 & -1/2\\ 0 & 1 & 1\\ 0 & -3/2 & -3/2\\ \end{bmatrix} \rightarrow \text{R3 = R3 + 3/2 R2} \rightarrow\\ \begin{bmatrix} 1 & 1/2 & -1/2\\ 0 & 1 & 1\\ 0 & 0 & 0\\ \end{bmatrix} \rightarrow \text{R1 = R1 - R2 / 2} \rightarrow \begin{bmatrix} 1 & 0 & -1\\ 0 & 1 & 1\\ 0 & 0 & 0\\ \end{bmatrix} \\ \]
Find the Null Space.
\[ \begin{bmatrix} 1 & 0 & -1\\ 0 & 1 & 1\\ 0 & 0 & 0\\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix}\\ \text{SOLVE}\\ x_1 - x_3 = 0 \rightarrow x_1 = x_3 \\ x_2 + x_3 = 0 \rightarrow x_2 = -x_3 = -x_1\\ x_1 = t, x_3 = t, x_2 = -t\\ \text{THUS}\\ \overrightarrow{x} = \begin{bmatrix} t \\ -t \\ t\\ \end{bmatrix} = \begin{bmatrix} 1\\ -1\\ 1\\ \end{bmatrix} t \]
The eigenvector corresponding to eigenvalue \(\lambda_2 = 3\) is \(v_3 = \begin{bmatrix} 1 \\ -1 \\ 1 \\ \end{bmatrix}\).
In conclusion, the eigenvalues associated with matrix \(A\) are \(\lambda_1 = 0\) with geometric multiplicity 2 and algebraic multiplicity 2 and \(\lambda_2 = 3\) with geometric multiplicity 1 and algebraic multiplicity 1.
The eigenvectors corresponding to \(\lambda_1 = 0\) are \(v_1 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}\) and \(v_2 = \begin{bmatrix} -1 \\ 0 \\ 1 \\ \end{bmatrix}\).
The eigenvector corresponding to \(\lambda_2 = 3\) is \(v_3 = \begin{bmatrix} 1 \\ -1 \\ 1 \\ \end{bmatrix}\).