Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 :)

share|improve this question
1  
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
add comment

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer
add comment

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}

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.