# Applying the Gram-Schmidt orthonormalization method in order to find an orthonormal basis for the subspace of $\mathbb{R}^4$

Apply the Gram-Schmidt orthonormalization method in order to find an orthonormal basis for the subspace of $\mathbb{R}^4$ that is generated by the vectors $(1, 0, 1, 1)^T , (1, 1, 0, 0)^T$ and $(0, 0, 1, 1)^T$

$$u_1=(1, 0, 1, 1)^T , u_2=(1, 1, 0, 0)^T , u_3=(0, 0, 1, 1)^T$$

I used the formula $$V_1=\dfrac{u_1}{||u_1||}$$$$V_2=\dfrac{u_2-<u_2V_1>V_1}{||u_2-<u_2V_1>V_1||}$$$$V_3=\dfrac{u_3-<u_3V_2>V_2-<u_3V_1>V_1}{||u_3-<u_3V_2>V_2-<u_3V_1>V_1||}$$

I got the basis as $$V_1=\left(\dfrac{1}{\sqrt{3}},0,\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right)^T$$$$V_2=\left(\dfrac{2}{\sqrt{7}},\dfrac{1}{\sqrt{7}},-\dfrac{1}{\sqrt{7}},-\dfrac{1}{\sqrt{7}}\right)^T$$$$V_3=\left(-\dfrac{2}{\sqrt{10}},\dfrac{2}{\sqrt{10}},\dfrac{1}{\sqrt{10}},\dfrac{1}{\sqrt{10}}\right)^T$$

But the answer in the book was$$V_1=\left(\dfrac{1}{\sqrt{3}},0,\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right)^T$$$$V_2=\left(\dfrac{2}{\sqrt{15}},\dfrac{3}{\sqrt{15}},-\dfrac{1}{\sqrt{15}},-\dfrac{1}{\sqrt{15}}\right)^T$$$$V_3=\left(-\dfrac{6}{\sqrt{90}},\dfrac{6}{\sqrt{90}},\dfrac{3}{\sqrt{90}},\dfrac{3}{\sqrt{90}}\right)^T$$

The book is correct. Note that, in order to compute $v_2$, first you must compute\begin{align}a_2&=u_2-\langle u_2,v_1\rangle v_1\\&=\left(\frac23,1,-\frac13,-\frac13\right)^T.\end{align}So,\begin{align}v_2&=\frac{a_2}{\|a_2\|}\\&=\frac1{\sqrt{15}}\left(2,,3,-1,-1\right).\end{align}
• But when I took dot product of $V_1\cdot V_2$ I got $0$. – user572932 Aug 9 '18 at 17:09
• @philip Yes, it is $0$. And… ? What do you deduce from that? – José Carlos Santos Aug 9 '18 at 17:10
• Since $V_1\cdot V_2=0$ they are orthogonal. I think I made a mistake in calculating $V_3$ – user572932 Aug 9 '18 at 17:12
• Indeed you should, but for your incorrect solution $V_2\cdot V_3\ne0$. – amd Aug 9 '18 at 17:53