# Does having an orthonormal basis imply that the vectors in that basis must all be orthongonal?

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!

-
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 – Hawk Apr 18 '12 at 1:34
$(2/3)(2/3)+(-1/3)(2/3)+(2/3)(-1/3)=(4/9)+(-2/3)+(-2/3)=(4/9)+(-4/9)=0$ – David Mitra Apr 18 '12 at 1:39

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.