In the third term of formula (4) two squares are missing.

The basis for the subspace U in example 1 is { (3,1,0,0,0), (0,0,7,1,0), (0,0,0,0,1)} since
3x_2+x_2+7x_4+x_4+x_5 = x_2 (3,1,0,0,0) + x_4 (0,0,7,1,0) + x_5 (0,0,0,0,1)
and the three vectors in the basis are linearly independent.

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Title. Tested and confirmed with several dozen data points to back up our conclusion. Please fix ASAP.

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