Question

If the matrix \[ A = \begin{pmatrix} 3 & -1 & 1 \\ -1 & 5 & -1 \\ 1 & -1 & 3 \end{pmatrix} \] has three distinct eigenvalues, and one of its eigenvectors is \[ \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \] then which of the following can be another eigenvector of \( A \)?

• A. \( \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \)
• B. \( \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} \)
• C. \( \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \)
• D. \( \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \)

Hint:

To verify if a vector is an eigenvector, multiply it with the matrix and see if the result is a scalar multiple of the original vector.

Answer 1

The Correct Option is D

We are given that matrix \( A \) has three distinct eigenvalues, which means it has three linearly independent eigenvectors.

One known eigenvector is: \[ v_1 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \]
We are to determine which among the given options can be another eigenvector. Since eigenvectors corresponding to distinct eigenvalues are linearly independent, any other eigenvector must be linearly independent of \( v_1 \).

Check each option:

(1) \( \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \)
This is a scalar multiple of \( v_1 \) (specifically, \( -1 \cdot v_1 \)), so it is not linearly independent. Hence, not acceptable.

(2) \( \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} \)
This vector is not a scalar multiple of \( v_1 \), but let’s test better options.

(3) \( \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \)
Compare with \( v_1 \). Subtract \( v_1 \) from this: \[ \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \] Clearly not a scalar multiple. Still not confirmed as eigenvector without matrix multiplication.

(4) \( \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \)
Now test this by checking if \( A \cdot v = \lambda v \) for some scalar \( \lambda \).

Compute \( A \cdot \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \): \[ A \cdot \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 3(1) + (-1)(-2) + 1(1) \\ -1(1) + 5(-2) + (-1)(1) \\ 1(1) + (-1)(-2) + 3(1) \end{pmatrix} = \begin{pmatrix} 6 \\ -12 \\ 6 \end{pmatrix} \] This equals \( 6 \cdot \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \), which confirms it is an eigenvector.

Hence, option (4) is a valid eigenvector corresponding to eigenvalue \( \lambda = 6 \).