GATE 2018 EC Q.11: solution and answer

Official answer: CSource: Official answer key, GATE 2018 EC (IIT Guwahati)

1. Theorem used: eigenvectors that belong to distinct eigenvalues are linearly independent. Suppose $\mathbf{M}$ has 4 distinct eigenvalues ${\lambda }_1,\dots ,{\lambda }_4$ and we pick one eigenvector for each. By the theorem these 4 eigenvectors are linearly independent, so S2 $\Rightarrow$ S1 and option (C) holds. (For a real matrix some of the ${\lambda }_i$ may be complex. The theorem still applies, with the eigenvectors in ${\mathbb{C}}^4$.)

2. (A) is false. Take $\mathbf{M}=I_4$. Every nonzero vector is an eigenvector, so the 4 standard basis vectors are 4 linearly independent eigenvectors and S1 holds. But the only eigenvalue is $\lambda =1$, repeated 4 times, so S2 fails.

3. (B) is false. Take $\mathbf{M}=0_{4\times 4}$. The 4 standard basis vectors are linearly independent eigenvectors, all with $\lambda =0$, so S1 holds. But $\det \mathbf{M}=0$, so $\mathbf{M}$ is singular and S3 fails.

4. (D) is false. $\mathbf{M}=I_4$ is invertible because $\det I_4=1\ne 0$, so S3 holds. Its eigenvalues $1,1,1,1$ are not distinct, so S2 fails.

5. How S3 relates to the eigenvalues: $\det \mathbf{M}={\lambda }_1{\lambda }_2{\lambda }_3{\lambda }_4$, with eigenvalues counted by multiplicity. Invertibility therefore only says that no eigenvalue is zero. It says nothing about whether the eigenvalues are distinct or whether $\mathbf{M}$ is diagonalisable, and diagonalisability says nothing about invertibility.

6. Only (C) survives, and it holds for every matrix, not just for one example: distinct eigenvalues are sufficient (though not necessary) for $\mathbf{M}$ to be diagonalisable. Answer: (C)