Question

Which one of the following vectors is an eigenvector corresponding to the eigenvalue \(\lambda = 1\) for the matrix \[ A = \begin{bmatrix} 1 & 1 & 0 \\ 1 & -1 & 0 \\ 1 & -1 & 1 \end{bmatrix}? \]

• A. \([1\ 0\ 1]^T\)
• B. \([1\ 1\ 0]^T\)
• C. \([0\ 1\ 0]^T\)
• D. \([0\ 0\ 1]^T\)

Hint:

To find eigenvectors for eigenvalue \(\lambda\), solve \((A-\lambda I)v=0\). Free variables correspond to eigenvector directions.

Answer 1

The Correct Option is D

Eigenvector corresponding to \(\lambda = 1\)

Step 1: Solve \((A - I)v = 0\). \[ A - I = \begin{bmatrix} 0 & 1 & 0 \\ 1 & -2 & 0 \\ 1 & -1 & 0 \end{bmatrix}. \]

Step 2: From the row equations: \[ y = 0, \quad x - 2y = 0 \;\Rightarrow\; x = 0, \quad x-y=0 \;(\text{satisfied}). \] Thus \(z\) is free.

Step 3: Therefore eigenvectors are of the form: \[ v = \begin{bmatrix}0 \\ 0 \\ z\end{bmatrix}, \quad z \neq 0. \] i.e., multiples of \([0,0,1]^T\).

Hence the correct option is: \[ \boxed{\text{(D)}} \]