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: Test each candidate vector directly instead of deriving eigenvalues first.
An eigenvector $v$ of matrix $A$ must satisfy $Av=\lambda v$ for some scalar $\lambda$, meaning $Av$ comes out parallel to $v$ itself. We can check each option's vectors against $A=\begin{bmatrix}1&2\\2&4\end{bmatrix}$ by direct multiplication.

Step 2: Test $\begin{bmatrix}1\\2\end{bmatrix}$.
\[ A\begin{bmatrix}1\\2\end{bmatrix} = \begin{bmatrix}1(1)+2(2)\\2(1)+4(2)\end{bmatrix} = \begin{bmatrix}5\\10\end{bmatrix} = 5\begin{bmatrix}1\\2\end{bmatrix} \]
This comes back as $5$ times the same vector, so $\begin{bmatrix}1\\2\end{bmatrix}$ is a genuine eigenvector, with eigenvalue $5$.

Step 3: Test $\begin{bmatrix}-2\\1\end{bmatrix}$.
\[ A\begin{bmatrix}-2\\1\end{bmatrix} = \begin{bmatrix}1(-2)+2(1)\\2(-2)+4(1)\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix} = 0\begin{bmatrix}-2\\1\end{bmatrix} \]
This comes back as the zero vector, which is $0$ times itself, so $\begin{bmatrix}-2\\1\end{bmatrix}$ is also a genuine eigenvector, with eigenvalue $0$.

Step 4: Test the rival vector $\begin{bmatrix}2\\1\end{bmatrix}$ used in options (B) and (C).
\[ A\begin{bmatrix}2\\1\end{bmatrix} = \begin{bmatrix}4\\8\end{bmatrix} \]
For this to be an eigenvector we would need $\begin{bmatrix}4\\8\end{bmatrix}=k\begin{bmatrix}2\\1\end{bmatrix}$ for some $k$, but $4=2k$ gives $k=2$ while $8=k$ gives $k=8$. These do not agree, so $\begin{bmatrix}2\\1\end{bmatrix}$ fails and options (B) and (C) are both wrong.

Step 5: Conclude.
Only $\begin{bmatrix}1\\2\end{bmatrix}$ and $\begin{bmatrix}-2\\1\end{bmatrix}$ pass the direct test.
\[ \boxed{\begin{bmatrix}1\\2\end{bmatrix} \text{ and } \begin{bmatrix}-2\\1\end{bmatrix}} \]
This is option (A).