GATE 2026 IN Q.11: solution and answer

Official answer: CSource: Official answer key, GATE 2026 IN (IIT Guwahati)

1. Eigenvalues satisfy any polynomial that annihilates the matrix. If $P\mathbf{x}=\lambda \mathbf{x}$ with $\mathbf{x}\ne 0$, then $P^k\mathbf{x}={\lambda }^k\mathbf{x}$ for every $k$. Applying the given identity to $\mathbf{x}$: $$\left(P^3-4P^2+5P-2I\right)\mathbf{x}=\left({\lambda }^3-4{\lambda }^2+5\lambda -2\right)\mathbf{x}=0$$ Since $\mathbf{x}\ne 0$, every eigenvalue must satisfy ${\lambda }^3-4{\lambda }^2+5\lambda -2=0$.

2. Find a root by inspection. At $\lambda =1$: $1-4+5-2=0$, so $(\lambda -1)$ is a factor.

3. Factorise. $${\lambda }^3-4{\lambda }^2+5\lambda -2=(\lambda -1)({\lambda }^2-3\lambda +2)=(\lambda -1{)}^2(\lambda -2)$$

4. Read off the candidates. The only roots are $1$ and $2$ (both real), so every eigenvalue of $P$, real or complex, must be $1$ or $2$. The value $3$ can never occur, which rules out option (A).

5. Both values actually occur. Take $P=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 2\end{array}\right]$. Then $P-I=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right]$ and $P-2I=\left[\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& 0\end{array}\right]$, so $(P-I{)}^2(P-2I)=0$, which is the given identity. This $P$ has both eigenvalues $1$ and $2$, so neither $\{1\}$ nor $\{2\}$ alone is the full set. The set of all possible eigenvalues is $\{1,2\}$, option (C).