Question

Select the correct pair of eigenvectors for the following symmetric matrix.
\[ A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \]

• A. \(\begin{bmatrix}1\\2\end{bmatrix}\) and \(\begin{bmatrix}-2\\1\end{bmatrix}\)
• B. \(\begin{bmatrix}1\\2\end{bmatrix}\) and \(\begin{bmatrix}2\\1\end{bmatrix}\)
• C. \(\begin{bmatrix}-1\\-2\end{bmatrix}\) and \(\begin{bmatrix}2\\1\end{bmatrix}\)
• D. \(\begin{bmatrix}-1\\2\end{bmatrix}\) and \(\begin{bmatrix}2\\-1\end{bmatrix}\)

Hint:

Find the eigenvalues from \(\det(A-\lambda I)=0\) first, then solve \((A-\lambda I)v=0\) for each \(\lambda\); for a symmetric matrix, the two eigenvectors must be perpendicular.

Answer 1

The Correct Option is A

Step 1: Set up the characteristic equation.
For a matrix \(A\), the eigenvalues \(\lambda\) satisfy \(\det(A-\lambda I)=0\). Here
\[ A-\lambda I = \begin{bmatrix} 1-\lambda & 2 \\ 2 & 4-\lambda \end{bmatrix} \]

Step 2: Expand the determinant.
\[ \det(A-\lambda I) = (1-\lambda)(4-\lambda) - (2)(2) \] \[ = 4 - 5\lambda + \lambda^2 - 4 = \lambda^2 - 5\lambda \] Setting this to zero,
\[ \lambda(\lambda-5) = 0 \;\Rightarrow\; \lambda = 0 \text{ or } \lambda = 5 \]

Step 3: Find the eigenvector for \(\lambda=5\).
Solve \((A-5I)v=0\):
\[ \begin{bmatrix} -4 & 2 \\ 2 & -1 \end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix} \] The first row gives \(-4x+2y=0\), so \(y=2x\). Taking \(x=1\) gives the eigenvector
\[ v_1 = \begin{bmatrix}1\\2\end{bmatrix} \]

Step 4: Find the eigenvector for \(\lambda=0\).
Solve \(Av=0\):
\[ \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix} \] The first row gives \(x+2y=0\), so \(x=-2y\). Taking \(y=1\) gives the eigenvector
\[ v_2 = \begin{bmatrix}-2\\1\end{bmatrix} \]

Step 5: Verify orthogonality (a check for symmetric matrices).
Since \(A\) is symmetric, eigenvectors from different eigenvalues must be perpendicular.
\[ v_1 \cdot v_2 = (1)(-2) + (2)(1) = -2 + 2 = 0 \] The dot product is zero, confirming the pair is consistent.

Step 6: Match with the options.
The pair \(\begin{bmatrix}1\\2\end{bmatrix}\) and \(\begin{bmatrix}-2\\1\end{bmatrix}\) matches option (A). Testing option (B)'s second vector \(\begin{bmatrix}2\\1\end{bmatrix}\):
\[ A\begin{bmatrix}2\\1\end{bmatrix} = \begin{bmatrix}4\\8\end{bmatrix} \] This is not a scalar multiple of \(\begin{bmatrix}2\\1\end{bmatrix}\) (it would need to be \(\begin{bmatrix}8\\4\end{bmatrix}\) for that), so \(\begin{bmatrix}2\\1\end{bmatrix}\) is not an eigenvector, ruling out (B) and (C).

Step 7: Final conclusion.
\[ \boxed{\begin{bmatrix}1\\2\end{bmatrix} \text{ and } \begin{bmatrix}-2\\1\end{bmatrix}} \]
Hence, the correct option is (A).