Question

For the matrix $\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$ which of the following is not an Eigen Vector

• A. \( (1,0,0) \)
• B. \( (0,1,0) \)
• C. \( (0,0,1) \)
• D. \( (0,0,0) \)

Hint:

Eigenvectors must be non-zero vectors. The zero vector never satisfies the eigenvector equation in a meaningful way.

Answer 1

The Correct Option is D

Let the given matrix be $B = \begin{bmatrix} 2 & 0 & 0
0 & 2 & 0
0 & 0 & 2 \end{bmatrix}$. An eigenvector $v$ of a matrix $B$ is a non-zero vector such that $Bv = \lambda v$ for some scalar $\lambda$ (the eigenvalue). For option (A): $B \begin{bmatrix} 1
0
0 \end{bmatrix} = \begin{bmatrix} 2 \times 1 + 0 \times 0 + 0 \times 0
0 \times 1 + 2 \times 0 + 0 \times 0
0 \times 1 + 0 \times 0 + 2 \times 0 \end{bmatrix} = \begin{bmatrix} 2
0
0 \end{bmatrix} = 2 \begin{bmatrix} 1
0
0 \end{bmatrix}$. So, $(1,0,0)$ is an eigenvector with eigenvalue $\lambda = 2$. For option (B): $B \begin{bmatrix} 0
1
0 \end{bmatrix} = \begin{bmatrix} 2 \times 0 + 0 \times 1 + 0 \times 0
0 \times 0 + 2 \times 1 + 0 \times 0
0 \times 0 + 0 \times 1 + 2 \times 0 \end{bmatrix} = \begin{bmatrix} 0
2
0 \end{bmatrix} = 2 \begin{bmatrix} 0
1
0 \end{bmatrix}$. So, $(0,1,0)$ is an eigenvector with eigenvalue $\lambda = 2$. For option (C): $B \begin{bmatrix} 0
0
1 \end{bmatrix} = \begin{bmatrix} 2 \times 0 + 0 \times 0 + 0 \times 1
0 \times 0 + 2 \times 0 + 0 \times 1
0 \times 0 + 0 \times 0 + 2 \times 1 \end{bmatrix} = \begin{bmatrix} 0
0
2 \end{bmatrix} = 2 \begin{bmatrix} 0
0
1 \end{bmatrix}$. So, $(0,0,1)$ is an eigenvector with eigenvalue $\lambda = 2$. For option (D): The zero vector $(0,0,0)$ cannot be an eigenvector because eigenvectors must be non-zero vectors by definition.