Answer to Question #217997 in Linear Algebra for sabelo Zwelakhe Xu

Question

suppose u is a subspace of r^4 defined by u=span((1,2,3,-4),(-5,4,3,2)). find an orthonormal basis of u and an orthonormal basis of u^⊥

Answer 1

Notice that (1,2,3-4) and (-5,4,3,2) are linearly independent since neither vetor is a scalar multiple of the other. Thus the basis is

(1,2,3,-4),(-5,4,3,2),(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)

So basis of $\R^4$ is

(1,2,3,-4),(-5,4,3,2),(1,0,0,0),(0,1,0,0)

$e_1=\frac{(1,2,3,-4)}{\|(1,2,3,-4)\|}\\ e_1=(\frac{1}{\sqrt{30}},\sqrt{\frac{2}{15}},\sqrt{\frac{3}{10}},-2\sqrt{\frac{2}{15}})\\ e_2=\frac{(-5,4,3,2)-<(-5,4,3,2),e_1>e_1}{\|(-5,4,3,2)-<(-5,4,3,2),e_1>e_1\|}\\ e_2=(-\frac{77}{\sqrt{12030}},28\sqrt{\frac{2}{6015}},13\sqrt{\frac{3}{4010}},19\sqrt{\frac{2}{6015}})\\ e_3=\frac{(1,0,0,0)-<(1,0,0,0),e_1>e_1- <(1,0,0,0),e_2>e_2}{\|(1,0,0,0)-<(1,0,0,0),e_1>e_1- <(1,0,0,0),e_2>e_2\|}\\ e_3=(\sqrt{\frac{190}{401}},\frac{117}{\sqrt{76190}},6\sqrt{\frac{10}{7619}},\frac{151}{\sqrt{76190}})\\ e_4=\frac{(0,1,0,0)-<(0,1,0,0),e_1>e_1- <(0,1,0,0),e_2>e_2-<(0,1,0,0),e_3>e_3}{\|(0,1,0,0)-<(0,1,0,0),e_1>e_1- <(0,1,0,0),e_2>e_2-<(0,1,0,0),e_3>e_3\|}\\ e_4=(0,\frac{9}{\sqrt{190}},-\sqrt{\frac{10}{19}},-\frac{3}{\sqrt{190}})$ Orthogonal basis of u is

$(\frac{1}{\sqrt{30}},\sqrt{\frac{2}{15}},\sqrt{\frac{3}{10}},-2\sqrt{\frac{2}{15}})$ and $(-\frac{77}{\sqrt{12030}},28\sqrt{\frac{2}{6015}},13\sqrt{\frac{3}{4010}},19\sqrt{\frac{2}{6015}})$ while the orthogonal basis of $u^\bot$ $(\sqrt{\frac{190}{401}},\frac{117}{\sqrt{76190}},6\sqrt{\frac{10}{7619}},\frac{151}{\sqrt{76190}})$

and $(0,\frac{9}{\sqrt{190}},-\sqrt{\frac{10}{19}},-\frac{3}{\sqrt{190}})$