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I'm supposed to show that the Gram-Schmidt process:

$\textbf{a}_j = \left\{ \begin{array}{lr} \textbf{d}_j, \;\;\textbf{if} \;\;\lambda_j = 0\\ \sum_{i=j}^n \lambda_i\textbf{d}_i & \end{array} \right. \rightarrow \;\;\;\;\textbf{b}_j = \left\{ \begin{array}{lr} \textbf{a}_j,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;j=1\\ \textbf{a}_j-\sum_{i=1}^{j-1}\left(\textbf{a}_j^T\tilde{\textbf{d}}_i\right)\tilde{\textbf{d}}_i, \;\;j \geq 2 \end{array} \right.\;\;\; \rightarrow \;\;\tilde{\textbf{d}_j} = \displaystyle\frac{\textbf{b}_j}{||\textbf{b}_j||}.$

Produces orthogonal vectors $\tilde{\textbf{d}_1}$ and $\tilde{\textbf{d}_2}$, where $\textbf{d}_1 = (1,0)$ and $\textbf{d}_2 = (0,1)$ and $\lambda_1, \lambda_2 \neq 0$.

I tried three times to go through this process, but every time I got $\langle \tilde{\textbf{d}_1}, \textbf{b}_2 \rangle \neq 0$, which means that $\langle \tilde{\textbf{d}_1}, \tilde{\textbf{d}_2} \rangle \neq 0$. I don't know where I'm doing something wrong, could someone verify that indeed $\langle \tilde{\textbf{d}_1}, \tilde{\textbf{d}_2} \rangle = 0$?

Thank you for your help!

UPDATE: I got it myself, I needed to use the vectors $\textbf{d}_1 = (1,0)$ and $\textbf{d}_2 = (0,1)$ in this problem. I guess in general this procedure I listed does not hold necessarily? Anyway, when $\textbf{d}_1 = (1,0)$ and $\textbf{d}_2 = (0,1)$ then indeed $\langle \tilde{\textbf{d}_1}, \tilde{\textbf{d}_2} \rangle = 0$.

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    $\begingroup$ If I were you I'd draw the two vectors (on a two dimensional space that is, your sheet per say :) ), and see what is actually happening when applying the gram-schmidt process. You "redress" one vector with regards to the other. So you can find the process intuitively with a drawing rather than applying tedious formulas $\endgroup$
    – mvggz
    Nov 24, 2014 at 10:20
  • $\begingroup$ +1 Thank you for your help :) But I have to show that it indeed is the case, I think drawing isn't enough for my professor :/ $\endgroup$
    – jjepsuomi
    Nov 24, 2014 at 10:22
  • $\begingroup$ Ok, but with the drawing you will find yourself what conditions you need in order to have a suitable process. Gram-schmidt process is built so that recursively you make each new vector orthogonal to sub-space of the previous ones. And you normalize it after to have an orthonormal basis. Let's start with two vectors linearly independant. Without loss of generality you can norm them to 1. Then what you need is to build a new vector orthogonal to the first one, so that you have two orthogonal vectors. So write the fact that the scalar product is zero and after divide it with it's norm $\endgroup$
    – mvggz
    Nov 24, 2014 at 10:33
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    $\begingroup$ Look, it's well explained in here: en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process The drawing is very explicit as to what you should do $\endgroup$
    – mvggz
    Nov 24, 2014 at 10:38
  • $\begingroup$ Thank you @mvggz =) Appreciate it :) $\endgroup$
    – jjepsuomi
    Nov 24, 2014 at 11:07

1 Answer 1

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If $\lambda_1 \neq0$ ja $\lambda_2\neq0$ we get:

$$\textbf{a}_1 = \sum_{i=1}^2\lambda_i\textbf{d}_i,\;\textbf{b}_1=\sum_{i=1}^2\lambda_i\textbf{d}_i,\;\tilde{\textbf{d}}_1 = \frac{\sum_{i=1}^2 \lambda_i\textbf{d}_i}{\left|\left|\sum_{i=1}^2\lambda_i\textbf{d}_i \right|\right|}$$

$$\textbf{a}_2 = \lambda_2\textbf{d}_2,\;\textbf{b}_2 = \lambda_2\textbf{d}_2-\lambda_2(\textbf{d}_2^T\tilde{\textbf{d}}_1)\tilde{\textbf{d}}_1 = \lambda_2\textbf{d}_2-\left(\frac{\lambda_2}{\left|\left| \sum_{i=1}^2\lambda_i\textbf{d}_i\right|\right|}\textbf{d}_2^T\left(\sum_{i=1}^2 \lambda_i\textbf{d}_i\right)\right)\tilde{\textbf{d}}_1$$

$$=\lambda_2\textbf{d}_2-\left(\frac{\lambda_2^2}{\left|\left|\sum_{i=1}^2\lambda_i\textbf{d}_i\right|\right|^2}\right)(\lambda_1\textbf{d}_1+\lambda_2\textbf{d}_2) = \lambda_2\textbf{d}_2-\frac{\sum_{i=1}^2\lambda_i\lambda_2^2\textbf{d}_i}{\left|\left|\sum_{i=1}^2\lambda_i\textbf{d}_i\right|\right|^2} = \textbf{b}_2.$$

Lets denote $c = \left|\left|\sum_{i=1}^2\lambda_i\textbf{d}_i\right|\right|$.

$$\langle \tilde{\textbf{d}}_1, \textbf{b}_2\rangle = \left( \frac{\lambda_1}{c}\textbf{d}_1 + \frac{\lambda_2}{c}\textbf{d}_2 \right)^T\left( \lambda_2\textbf{d}_2 - \frac{\lambda_2^2\lambda_1}{c}\textbf{d}_1 - \frac{\lambda_2^3}{c}\textbf{d}_2\right)$$

$$= -\frac{\lambda_2^2\lambda_1^2}{c^3}+\frac{\lambda_2^2}{c}-\frac{\lambda_2^4}{c^3} = \frac{c^2\lambda_2^2-\lambda_2^2\lambda_1^2-\lambda_2^4}{c^3} = \frac{\lambda_2^2(\lambda_1^2+\lambda_2^2)-\lambda_2^2(\lambda_1^2+\lambda_2^2)}{c^3} = \frac{0}{c^3} = 0.$$

The numerator was zero because:

$\left|\left|\sum_{i=1}^2\lambda_i\textbf{d}_i\right|\right| = \left|\left| \lambda_1(1, 0) + \lambda_2(0,1)\right|\right| = \left|\left| (\lambda_1, \lambda_2)\right|\right| = \sqrt{\lambda_1^2 + \lambda_2^2}.$

From these it follows that:

$$\langle\tilde{\textbf{d}}_1, \tilde{\textbf{d}}_2 \rangle = \langle\tilde{\textbf{d}}_1, \frac{\textbf{b}_2}{||\textbf{b}_2||}\rangle = 0\;\;\;\blacksquare$$

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