# Gram-Schmidt orthornormalization , $\dim(S)$.

By means of Gram-Schmidt orthonormalization, find an orthonormal basis in

$$S=\{v=\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{pmatrix}: x_{1}-x_{2}+2x_{3}-3x_{4}=0\}$$

subspace of $\left(\mathbb{R}^{4},\langle\,,\,\rangle\right)$.

What I do:

$x_{2}=\alpha, x_{3}=\beta, x_{4}=\gamma$ so $x_{1}=\alpha-2\beta+3\gamma$ and :

$$\left(\alpha-2\beta+3\gamma,\alpha,\beta,\gamma\right)=\left(\alpha, \alpha,0,0\right)+\left(-2\beta,0, \beta,0\right)+\left(3\gamma,0,0,\gamma\right).$$

So a base is : $\{(1,1,0,0),(-2,0,1,0),(3,0,0,1)\}$ and I have to apply Gram-Schmidt for this base?

Another question, what is the $\dim(S)$? $3$ or $4$ ?

because I can find canonical base in $\mathbb{R}^{4}$ to write $\left(\alpha-2\beta+3\gamma,\alpha,\beta,\gamma\right)$.

Thanks :)

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There are infinitely many orthonormal basis for S, but best way to find one is as you did. And Dim S=3 clearly – Detectives Dec 2 '12 at 18:03
@Detectives Why $dim(S)$ can not be $2$ ? thanks :) – Iuli Dec 2 '12 at 18:10
@luli since GS process preserves linearly independent set, you should obtain 3 orthonormal vectors after processing – Detectives Dec 2 '12 at 18:13
$\left(\alpha-2\beta+3\gamma,\alpha,\beta,\gamma\right)$=$(\alpha-2\beta,\alpha,‌​0, \gamma)+(3\gamma,0,\beta,0)$? can I form from here a base ? thanks :) – Iuli Dec 2 '12 at 18:20
Since every vector in S is written as what you wrote, alpha, beta, gamma can be arbitrary. That makes the reasoning for dim S =3. no way to express soution as two free variables.. – Detectives Dec 2 '12 at 18:32

The question was basically answered in comments, so I am posting a CW-answer, so that it is not left unanswered.

Since you exhibited a basis for $S$ which has 3 elements, you get that $\operatorname{dim}(S)=3$. Using Gram-Schmidt process you will obtain from this basis an orthonormal basis of the same subspace.

If you have any doubts whether your three vectors $(1,1,0,0)$, $(-2,0,1,0)$ and $(3,0,0,1)$ are linearly independent, you just need to have a look at the last three coordinates. If $c_1(1,1,0,0)+c_2(-2,0,1,0)+c_3(3,0,0,1)=(\dots,c_1,c_2,c_3)=(0,0,0,0)$, then obviously $c_1=c_2=c_3=0$.

This is similar to the argument which is used to show that non-zero rows of matrix in the reduced row echelon form are linearly independent. For a slightly more general claim, see this question.

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So the orthonormal basis vectors of the nullspace given in this post are as follows (using the Gram-Schmidt process):

\begin{eqnarray} q_1^T &=& (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \, 0, \, 0)\, , \\ q_2^T &=& (\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} , 0)\, , \\ q_3^T &=& (\frac{1}{\sqrt{10}}, \frac{-1}{\sqrt{10}}, \frac{2}{\sqrt{10}} , \frac{2}{\sqrt{10}})\, . \end{eqnarray}

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