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I have a problem that says: "By using the Gram Schmidt process(if you need it), find an orthonormal basis B in the subspace: $A=\{ u \in R^3 \ | \ x_1-4x_2-x_3=0 \}$

I chose as the basis, $B=\{[1 \ -4 \ -1]\}$ and normed it by dividing each entry with $\sqrt{16}$. Is it correct? How could I use Gram Schmidt? I know how to use it when I have 2+ basis vectors.

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    @Andrew: Yes it is. Gram-Schmidt consists of two parts: the *orthogonalization part*, and the *normalization part*. The orthogonalization part is trivial (has no steps) when the dimension is $0$ or $1$, since a set with 1 or fewer elements is always orthogonal. But the *normalization part* is still relevant.2012-01-17

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The set $A$ describes a plane in $\Bbb R^3$; thus $A$ has dimension two. If you intend to use Gram-Schmidt to obtain an orthonormal basis of of $A$, then all you need to start are two independent vectors in $A$.

A cursory examination of the equation $x_1-4x_2-x_3=0$ readily supplies these (for instance $(1,0,1)$ and $(0,1,-4 )$).

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Your basis is incorrect. The subspace $A$ has dimension $2$, not $1$, so you need to start with two basis vectors, not one.

In addition, your vector $B$ is not even in the subspace: if you plug in $x_1=1$, $x_2=-4$, $x_3=-1$ into $x_1-4x_2-x_3$, you do not get $0$.

So, first write out the solutions to $x_1-4x_2-x_3=0$, e.g., in parametric form. This will give you the two basis vectors. Then apply Gram-Schmidt to the basis you got.

(In the special case where the vector space is $1$-dimensional, the orthogonalization part of Gram-Schmidt is empty, so all you need to do is normalize a single basis vector).

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    @Andrew: Re the first sentence of this answer: What you wrote down happens to be a basis for $A^{\perp}$, not $A$. It seems you just read off the coefficients from the equation $x_1 - 4x_2 - x_3 = 0$, instead of *solving* it.2012-01-17