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Let

A = $\begin{bmatrix}1&7&6\\2&6&4\\4&3&-1\end{bmatrix}$

and U = Col(A). Find the orthonormal basis for U.

Am I supposed to designate vectors $x_1, x_2, x_3$ as the columns of A, and then, perform the Gram-Schmidt algorithm using those vectors? I tried that but ended up with a 0 vector, and that wouldn't form a basis. What am I doing wrong?

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  • $\begingroup$ I’ve got a slight quibble with the wording of the problem: there is an infinite number of orthonormal bases for $U$—asking for the basis isn’t well-defined. $\endgroup$ – amd Mar 2 '16 at 3:39
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You haven’t done anything wrong. You’ve just discovered that the column space is only two-dimensional. It’s pretty easy to see that any one column of the matrix is a linear combination of the other two.

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  • $\begingroup$ So do I pick two of the columns that are linearly independent with each other and use the Gram-Schimdt algorithm on them? Or just take out the zero vector? $\endgroup$ – sucksatmath Mar 2 '16 at 3:29
  • $\begingroup$ @sucksatmath That’s right. You’ve effectively done that already. When you fed the third vector into G-S, it got “absorbed” into the first two, so discard the zero vector that resulted. $\endgroup$ – amd Mar 2 '16 at 3:31

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