GATE 2018 IN Q.1: solution and answer

Official answer: ASource: Official answer key, GATE 2018 IN (IIT Guwahati)

1. Let $\lambda$ be any eigenvalue of $N$, with eigenvector $v\ne 0$ (real or complex), so $Nv=\lambda v$.

2. Apply $N$ again: $N^{2}v=N(\lambda v)=\lambda (Nv)={\lambda }^{2}v$. In general, if $\lambda$ is an eigenvalue of $N$, then ${\lambda }^k$ is an eigenvalue of $N^k$.

3. Since $N^2=0$, the left side is $0$, so ${\lambda }^{2}v=0$. Because $v\ne 0$, this forces ${\lambda }^2=0$, so $\lambda =0$.

4. This holds for every eigenvalue, real or complex. The characteristic polynomial of the $3\times 3$ matrix $N$ has degree 3, so it has three roots counted with multiplicity, and each of them is $0$. The eigenvalues are therefore $0,0,0$. (Equivalently: $x^2$ annihilates $N$, so the minimal polynomial divides $x^2$, and its only possible root is $0$.)

5. Check with an example: $N=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$ satisfies $N^2=0$. The matrix $N-\lambda I$ is upper triangular, so $\det (N-\lambda I)=(-\lambda {)}^3=-{\lambda }^3$, and the eigenvalues are $0,0,0$.

6. An eigenvalue of $1$ would mean $1^2=1$ is an eigenvalue of $N^2=0$, which is impossible because the zero matrix has only the eigenvalue $0$. This rules out options (B), (C) and (D). The answer is (A).