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I have a matrix in mathematica that looks like

$m= \left( \begin{array}{cccc} -i & 0 & 1 & 0 \\ 0 & -i & 0 & -1 \\ 1 & 0 & -i & 0 \\ 0 & -1 & 0 & -1. i \\ \end{array} \right) $

Note the last element I explicitly use a numerical value of $i$. Now if I use SVD without dropping singular values, I should be able to recover the original matrix:

{U, W, V} = SingularValueDecomposition[m]
U.W.Transpose[V]

The result however is $ \left( \begin{array}{cccc} 0.\, +1. i & 0 & 1. & 0 \\ 0 & 0.\, -1. i & 0 & 1. \\ -1. & 0 & 0.\, -1. i & 0 \\ 0 & -0.6+0.8 i & 0 & 0.8\, +0.6 i \\ \end{array} \right) $

If I replace the $1.i$ with the accurate symbol of $i$ there is no problem with recovering the matrix.

What is the reason behind it? Observation shows the matrix $m$ has a non-zero determinant and a close-to-one condition number. I would like to know the properties of such matrix that has issue.

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closed as off-topic by Henrik Schumacher, Daniel Lichtblau, Michael E2, m_goldberg, b.gates.you.know.what Jan 11 at 8:41

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  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Henrik Schumacher, Daniel Lichtblau, Michael E2, m_goldberg, b.gates.you.know.what
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  • $\begingroup$ (1) Best to post full code. (2) Also should check the documentation-- there is no actual issue here. $\endgroup$ – Daniel Lichtblau Jan 10 at 22:34
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The reason is that you have to use ConjugateTranspose instead of Transpose:

m = {
   {-I, 0, 1, 0},
   {0, -I, 0, -1},
   {1, 0, -I, 0},
   {0, -1, 0, -1. I}
   };
{U, W, V} = SingularValueDecomposition[m];
Max[Abs[U.W.ConjugateTranspose[V] - m]]

4.74287*10^-16

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  • $\begingroup$ Thanks a lot! I was fooled by the numerical accuracy stuff and did not think about the complex values. $\endgroup$ – Peter Zhang Jan 10 at 22:52

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