# Finding an orthonormal basis in a subspace

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.

-
math.stackexchange.com/questions/88469/… – user38268 Jan 17 '12 at 18:00
@BenjaminLim I guess that's not applicable to one vector basis. – Andrew Jan 17 '12 at 18:01
I don't think so. $A$ has dimension 2 (the vectors $(2,1,-2)$ and $(4,1,0)$ are in $A$). Apply G-S to two independent vectors in $A$. – David Mitra Jan 17 '12 at 18:03
@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. – Arturo Magidin Jan 17 '12 at 18:04

## 2 Answers

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 $(0,1,-4 )$).

-

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).

-
@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. – Srivatsan Jan 17 '12 at 20:02