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Gram-Schmidt process is applied to the ordered basis {$i + j + k, i + j, i$} in $R^3$ . Find the resulting orthonormal basis.

My attempts : Im converting them into vectors {(1,1,1), (1,1,0), (1,0,0)}

Now $u_1 = (1,1,1)$

$ u_2 =(1,1,0)$

$ u_3 = (1,0,0)$

Now $u_1 = (1,1,1)$, then $v_1 = \frac {(1,1,1)} {\sqrt 3}$

Now $u_2 =(1,1,0)$,then $ w_2 = u_2 -<(u_2,v_1)>.v_1$

$w_2 =(1,1,0) -(\frac{2}{3},\frac{2}{3},\frac{2}{3})=(\frac{1}{3},\frac{1}{3},-\frac{2}{3})$ so $v_2 = \frac{3}{ \sqrt 6}(\frac{1}{3},\frac{1}{3},-\frac{2}{3})$

Now again $u_3 =(1,0,0)$,then $w_3 = u_3 -<(u_3,v_1)>.v_1 -<(u_3,v_2).v_2$

$w_3 =(1,0,0) -\frac{(1,1,1)} { 3}- \frac{3}{ 6}(\frac{1}{3},\frac{1}{3},-\frac{2}{3})=(\frac{1}{6},\frac {-1}{2},\frac{-2}{3})$

Now $v_3 = \frac {6}{\sqrt26}(\frac{1}{6},\frac {-1}{2},\frac{-2}{3})$

so required resulting orthonormal basis are $ v_1,v_2,v_3$

$v_1 = \frac {(1,1,1)} {\sqrt 3}$

$v_2 = \frac{3}{ \sqrt 6}(\frac{1}{3},\frac{1}{3},-\frac{2}{3})$

$v_3 = \frac {6}{\sqrt26}(\frac{1}{6},\frac {-1}{2},\frac{-2}{3})$

Is my answer is correct or not ???Pliz check its,,

if not correct then any hints/solution will be appreciated

thanks u

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If the answer is correct, we must have $v_1 \cdot v_3=0$,

but $\frac16-\frac12-\frac23 =\frac16-\frac36-\frac46=-\frac16 \ne 0$.

Your idea is correct, just a careless mistake.

$w_3 =(1,0,0) -\frac{(1,1,1)} { 3}- \frac{3}{ 6}(\frac{1}{3},\frac{1}{3},-\frac{2}{3})=(\frac{1}{2},\frac {-1}{2},0)$

$v_3 = \frac1{\sqrt2}(1, -1, 0)$

Also note that $v_2$ can be simplified as $\frac{1}{\sqrt{6}}(1,1,-2)$.

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  • $\begingroup$ thanks @siong Thy Goh,,,,i will rechecked it $\endgroup$ – jasmine May 3 '18 at 2:03

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