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