I was trying to solve some problems related to Gram-Schmidt orthogonalisation when I came across this question

Use Gram Schmidt process to obtain an orthonormal set of vectors from the vectors (1,0,1),(0,1,1,),(1,-1,3)

My attempt:
I verified that the three vectors are linearly independent.
Let $$\vec u_1=(1,0,1)$$ $$\vec {u_2}=(0,1,1,)-\frac{(1,0,1)'(0,1,1)}{2}*(0,1,1)$$ $$\Rightarrow\vec u_2=(0,\frac12 ,\frac 12)$$ Straight away I can tell that the two vectors are not perpendicular. What am I doing wrong over here?

  • 1
    $\begingroup$ You first need to normalize $\overline{u_1}$ so that its norm is 1. $\endgroup$ – Prahlad Vaidyanathan Apr 20 '14 at 16:21
  • $\begingroup$ $u_2 = v_1 - Proj_{v_2}v_1$. You have $u_2 = v_2 - Proj_{v_2}v_1$. $\endgroup$ – Braindead Apr 20 '14 at 16:22
  • $\begingroup$ @Braindead ah got it. Thank you for the quick reply. $\endgroup$ – GTX OC Apr 20 '14 at 16:25

Put (assuming the usual Euclidean inner product)

$$\begin{align*}u_1:=&\frac{(1,0,1)}{||(1,0,1)||}=\frac1{\sqrt2}(1,0,1)\\ v_2:=&(0,1,1)-\langle\; (0,1,1)\,,\,u_1\;\rangle \,u_1=(0,1,1)-\frac12(1,0,1)=\left(-\frac12\,,\,1\,,\,\frac12\right)\\ u_2:=&\frac{v_2}{||v_2||}=\frac1{\sqrt{6}}(-1,2,1)\\{}\\ v_3:=&(1,-1,3)-\langle\;1,-1,3)\,,\,u_1\;\rangle \,u_1-\langle\;(1,-1,3)\,,\,u_2\;\rangle \,u_2\\ u_3:=&\frac{v_3}{||v_3||}\end{align*}$$

and etc.


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