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I am computing singular value decomposition at the moment and the book gave us the left singular matrix to be $$U = \begin{pmatrix} \frac{-1}{3} & \frac{2}{3}&\frac{2}{3} \\ \frac{2}{3} & \frac{-1}{3} &\frac{2}{3} \\ \frac{2}{3}& \frac{2}{3} & \frac{-1}{3} \end{pmatrix}$$ for the matrix $$A = \begin{pmatrix} -3 &1 \\ 6&-2 \\ 6&-2 \end{pmatrix}$$

Noticed that the second column and the third column of $U$ are not orthogonal, yet the matrix still works!

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Um, they look orthogonal to me... –  David Mitra Apr 18 '12 at 1:20
    
No they aren't, (2/3)(2/3) + (-1/3)(2/3) + (2/3)(-1/3) = (2/3)(2/3) = 4/9 –  sidht Apr 18 '12 at 1:34
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$(2/3)(2/3)+(-1/3)(2/3)+(2/3)(-1/3)=(4/9)+(-2/3)+(-2/3)=(4/9)+(-4/9)=0$ –  David Mitra Apr 18 '12 at 1:39
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up vote 2 down vote accepted

To answer the title of your question, yes. If you have an orthonormal basis, then not only must each vector in the basis be orthogonal, they must also be of unit length[1]. However, as David Mitra has pointed out, the rows of $U$ that you provided are in fact orthogonal.

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"each vector in the basis be orthogonal"--You mean, each vector much be orthogonal to all of the other vectors. –  Jonas Meyer May 18 '12 at 14:30
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