Let \(\boldsymbol{u} = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \end{bmatrix}\), and let \(\sigma_{1}, \sigma_{2}, \sigma_{3}, \sigma_{4}, \sigma_{5}\) be the singular values of the matrix \(\boldsymbol{M} = \boldsymbol{u}\boldsymbol{u}^{\boldsymbol{T}}\) (where \(\boldsymbol{u}^{\boldsymbol{T}}\) is the transpose of \(\boldsymbol{u}\)). The value of \(\sum_{i=1}^{5} \sigma_{i}\) is ______.
Answer
55
Explanation
\(\boldsymbol{M} = \boldsymbol{u}\boldsymbol{u}^{T}\) is a symmetric \(5 \times 5\) matrix of rank 1, since every column is a multiple of \(\boldsymbol{u}\). Moreover
\[ \boldsymbol{M}\boldsymbol{u} = \boldsymbol{u}(\boldsymbol{u}^{T}\boldsymbol{u}) = \|\boldsymbol{u}\|^{2}\,\boldsymbol{u}, \qquad \|\boldsymbol{u}\|^{2} = 1 + 4 + 9 + 16 + 25 = 55 \]
while every vector orthogonal to \(\boldsymbol{u}\) is sent to \(\boldsymbol{0}\). So the eigenvalues are 55 and 0 (four times). For a symmetric matrix the singular values are the absolute values of the eigenvalues, here 55, 0, 0, 0, 0.
Hence \(\sum_{i=1}^{5} \sigma_{i} = 55\), which is also the trace of \(\boldsymbol{M}\).
Answer: 55