Singular value decomposition of a 4×5 matrix

Consider the matrix

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

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

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

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

    \[\tilde{S}=\operatorname{diag}(S, 0,0), \quad S:=\operatorname{diag}\left(\sigma_1, \sigma_2, \sigma_3\right),\]

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{S} ) is thus r=3<\min (m, n). We can check that V^T V=V V^T=I_5, and U U^T=U^T U=I_4.

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Hyper-Textbook: Optimization Models and Applications Copyright © by L. El Ghaoui. All Rights Reserved.

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