GATE 2024 DA Q.61: solution and answer

Official answer: 55Source: Official answer key, GATE 2024 DA (IISc Bengaluru)

1. $\mathbf{M}=\mathbf{u}{\mathbf{u}}^T$ is the outer product of the nonzero vector $\mathbf{u}$ with itself, so every column of $\mathbf{M}$ is a multiple of $\mathbf{u}$ and $\mathrm{rank}(\mathbf{M})=1$.

2. A matrix of rank 1 has exactly one nonzero singular value; the other four singular values are $0$. So ${\sum }_{i=1}^5{\sigma }_i={\sigma }_1$.

3. $\mathbf{M}$ is symmetric, and for any $\mathbf{x}$ we have ${\mathbf{x}}^T\mathbf{Mx}=({\mathbf{u}}^T\mathbf{x}{)}^2\ge 0$, so it is positive semidefinite. For such a matrix the singular values equal the eigenvalues.

4. $\mathbf{Mu}=\mathbf{u}({\mathbf{u}}^T\mathbf{u})=({\mathbf{u}}^T\mathbf{u})\mathbf{u}$, so $\mathbf{u}$ is an eigenvector with eigenvalue ${\mathbf{u}}^T\mathbf{u}$. This is the nonzero eigenvalue, since the rank is 1.

5. ${\mathbf{u}}^T\mathbf{u}=1^2+2^2+3^2+4^2+5^2=1+4+9+16+25=55$.

6. Hence ${\sigma }_1=55$ and ${\sigma }_2=\cdots ={\sigma }_5=0$, giving ${\sum }_{i=1}^5{\sigma }_i=55$. Check: $\mathrm{trace}(\mathbf{M})={\sum }_i{u}_i^2=55$, which equals the sum of the eigenvalues.