Question-61

Let \(\mathbf{u} = \begin{bmatrix}1\\2\\3\\4\\5\end{bmatrix}\) and let \(\sigma_1, \sigma_2, \sigma_3, \sigma_4, \sigma_5\) be the singular values of the matrix \(\mathbf{M} = \mathbf{u} \mathbf{u}^T\) (where \(\mathbf{u}^T\) is the transpose of \(\mathbf{u}\)). The value of \(\sum \limits_{i = 1}^{5} \sigma_i\) is ________.

Answer

\(55\)

Solution

Some useful results:

• A matrix \(A\) of rank \(r\) has \(r\) positive singular values.
• The first \(r\) eigenvalues of \(A^TA\) are the squares of the first \(r\) singular values.
• A matrix of the form \(u u^T\) is unit rank.
• \((||u||^2, u)\) is an eigenpair of \(uu^T\). In fact, \(||u||^2\) is the only eigenvalue for this matrix.
• If \(\lambda\) is an eigenvalue of \(A\), \(n\lambda\) is an eigenvalue of \(nA\).

With these results, we see that \(M\) has exactly one non-zero singular value, namely \(\sigma_1\), assuming it is the largest of the lot. The rest are zero. The largest eigenvalue of \(M^TM\) is therefore \(\sigma_1^2\). Since \(M = uu^T\), \(M^TM = ||u||^2 M\). From the RHS, we gather that the largest singular value of \(M^TM\) is \(||u||^4\). This leads us to: \[ \sigma_1^2 = ||u||^4 \implies \sigma_1 = ||u||^2 \] Since \(||u||^2 = 55\), \(\sum \limits_{i = 1}^{5} \sigma_i = 55\).