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Consider the matrix

\[ A = \left( \begin{array}{ccccc} 1 & 0 & 0 & 0 & 2 \\0 & 0 & 3 & 0 & 0 \\0 & 0 & 0 & 0 & 0 \\0 & 4 & 0 & 0 & 0 \\\end{array} \right). \]

A singular value decomposition of this matrix is given by \( A = U \tilde{\mathit{S}} V^T \), with

\[ U = \left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ \end{array} \right) , \quad \tilde{\mathit{S}} = \left( \begin{array}{ccccc} 4 & 0 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 & 0 \\ 0 & 0 & \sqrt{5} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right) , \quad V^T = \left( \begin{array}{ccccc} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ \sqrt{0.2} & 0 & 0 & 0 & \sqrt{0.8} \\ 0 & 0 & 0 & 1 & 0 \\ – \sqrt{0.8} & 0 & 0 & 0 & \sqrt{0.2} \\ \end{array} \right) . \]

Notice above that \( \tilde{\mathit{S}} \) has non-zero values only in its diagonal, and can be written as

\[ \tilde{\mathit{S}} = \mathbf{diag}(\mathit{S}, 0, 0), \quad \mathit{S} := \mathbf{diag}(\sigma_1,\sigma_2,\sigma_3), \]

with \( \sigma_1 = 4, \sigma_2 = 3, \sigma_3 = \sqrt{5} \). The rank of \( A \) (which is the number of non-zero elements on the diagonal matrix \( \tilde{\mathit{S}} \)) is thus \( r = 3 < \min(m,n) \). We can check that \( V^TV = VV^T = I_5 \), and \( UU^T = U^TU= I_4 \).