Normalized Eigenvector Problem
For the matrix \( A = \begin{bmatrix} 5 & 3 \\ 1 & 3 \end{bmatrix} \), find one of the normalized eigenvectors.
Step 1: Find Eigenvalues
Solve \( \det(A - \lambda I) = 0 \)
\[ \det\begin{pmatrix} 5-\lambda & 3 \\ 1 & 3-\lambda \end{pmatrix} = 0 \]
\[ (5-\lambda)(3-\lambda) - 3 = 0 \]
\[ 15 - 8\lambda + \lambda^2 - 3 = 0 \]
\[ \lambda^2 - 8\lambda + 12 = 0 \]
\[ (\lambda - 6)(\lambda - 2) = 0 \]
\[ \boxed{\lambda_1 = 6, \quad \lambda_2 = 2} \]
Step 2: Find Eigenvectors
For \( \lambda = 6 \):
\[ A - 6I = \begin{pmatrix} -1 & 3 \\ 1 & -3 \end{pmatrix} \]
Row reduce → \( x_1 = 3x_2 \), so eigenvector = \( (3,\ 1) \)
For \( \lambda = 2 \):
\[ A - 2I = \begin{pmatrix} 3 & 3 \\ 1 & 1 \end{pmatrix} \]
Row reduce → \( x_1 = -x_2 \), so eigenvector = \( (1,\ -1) \)
Step 3: Normalize the Eigenvectors
Normalize \( (3,\ 1) \):
\[ \|(3,1)\| = \sqrt{9+1} = \sqrt{10} \]
\[ \text{Normalized} = \begin{pmatrix} \dfrac{3}{\sqrt{10}} \\[10pt] \dfrac{1}{\sqrt{10}} \end{pmatrix} \]
Normalize \( (1,\ -1) \):
\[ \|(1,-1)\| = \sqrt{1+1} = \sqrt{2} \]
\[ \text{Normalized} = \begin{pmatrix} \dfrac{1}{\sqrt{2}} \\[10pt] \dfrac{-1}{\sqrt{2}} \end{pmatrix} \]
Answer: Option B
\[ \boxed{ \begin{pmatrix} \dfrac{1}{\sqrt{2}} \\[10pt] \dfrac{-1}{\sqrt{2}} \end{pmatrix} } \]
This is the normalized eigenvector corresponding to \( \lambda = 2 \).
