The eigen vector(s) of the matrix [ 0 0 alpha ] [ 0 0 0 ] [ 0 0 0 ] where alpha != 0, is (are) A. (0, 0, alpha) B. (alpha, 0, 0) C. (0, 0, 1) D. (0, alpha, 0)
GATE 1993 · Engineering Mathematics · Eigen Value · easy
Answer: B=(alpha,0,0) and D=(0,alpha,0) are eigenvectors.
- Apply A to each candidate: A*(0,0,alpha)^T = (alpha*alpha,0,0) = (alpha^2,0,0). Is (alpha^2,0,0) = lambda*(0,0,alpha)? That requires alpha^2=0 and lambda*alpha=0, but alpha!=0, contradiction. NOT an eigenvector. A*(alpha,0,0)^T = (0,0,0) = 0*(alpha,0,0). YES, eigenvector with lambda=0. A*(0,0,1)^T = (alpha,0,0). Is (alpha,0,0) = lambda*(0,0,1)? Requires alpha=0, contradiction. NOT an eigenvector. A*(0,alpha,0)^T = (0,0,0) = 0*(0,alpha,0). YES, eigenvector with lambda=0.
- Confirm the null space condition: B=(alpha,0,0): third component=0, in null space. D=(0,alpha,0): third component=0, in null space. A=(0,0,alpha): third component=alpha!=0, not in null space. C=(0,0,1): third component=1!=0, not in null space.