Finding the orthonormal basis of $\Bbb R^3$ using the Gram-Schmidt algorithm

Question

I'm relatively new to inner product and I was asked to obtain an orthonormal basis of $\Bbb R^3$ using

$$ \begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix} , \begin{pmatrix} 1 \\ 0 \\ 1 \\ \end{pmatrix} , \begin{pmatrix} 0 \\ 1 \\ 2 \\ \end{pmatrix} $$

by applying the Gram Schmidt algorithm which I have found to be

$$ (\frac{1}{\sqrt 3})\begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix} , (\frac{\sqrt 6}{2})\begin{pmatrix} 1/3 \\ -2/3 \\ 1/3 \\ \end{pmatrix} , (\frac{1}{\sqrt 2})\begin{pmatrix} -1 \\ 0 \\ 1 \\ \end{pmatrix} $$

In addition to that my inner product was defined to be $\langle u,w\rangle = u_1w_1+ 2u_2w_2+ 3u_3w_3 $.

I'm not sure what I'm supposed to do with the defined innerproduct as such. What do i have to do to solve this question or have i gotten the concept all wrong?

Answer 1

Note that the inner product is an abstract concept. Check out the Wikipedia article. You can use the GS process with any inner product, just use your inner product definition to calculate $\langle\cdot,\cdot\rangle$ where it appears!

Answer 2

Which inner product did you use for your orthogonal projections during Gram-Schmidt ? Make sur you stay consistent with using the same inner product all the time. Recall : the orthogonal projection of $u$ onto $v$ is given by $\text{proj}_vu=\frac{\langle u,v\rangle}{\langle v,v \rangle}v$, and the norm of a vector $w$ as defined is this context is equal to $||w||=\sqrt{\langle w,w \rangle}$

Answer 3

You are supposed to use the inner product that you were given. With respect to that inner product, $\|(1,1,1)\|=\sqrt6$ and therefore the first vector that you get when you apply Gramm-Schmidt is $e_1=\frac1{\sqrt6}(1,1,1)$. Then, let$$a_2=(1,0,1)-\langle(1,0,1),e_1\rangle e_1=\left(\frac13,-\frac23,\frac13\right).$$It norm is $\sqrt{\frac43}$ and so you take $e_2=\frac1{2\sqrt3}(1,-2,1)$. Finally, you take$$a_3=(0,1,2)-\langle(0,1,2),e_1\rangle e_1+\langle(0,1,2),e_2\rangle e_2=\left(-\frac32,0,\frac12\right).$$Its norm is $\sqrt3$ and so you take $e_3=\frac1{\sqrt3}\left(-\frac32,0,\frac12\right)$.