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^π

Answer 1

Let $v_1=(1,2,3,-4)$, $v_2=(-5,4,3,2)$, $v_3=(0,0,1,0)$, $v_4=(0,0,0,1)$. This is a basis of the vector space ${\mathbb R}^4$, $v_1$ and $v_2$ form a basis of $U$. Let's apply the Gram-Schmidt's method of orthogonalization to this vector system.

Let $w_1=v_1=(1,2,3,-4)$, then:

$\langle{w_1,w_1}\rangle=1^2+2^2+3^2+(-4)^2=30$.

$\langle{v_2,w_1}\rangle=-5\cdot 1+4\cdot 2+3\cdot 3+2\cdot(-4)=4$.

$\langle{v_3,w_1}\rangle=3$.

$\langle{v_4,w_1}\rangle=-4$.

Calculate $v_2-\frac{\langle{v_2,w_1}\rangle}{\langle{w_1,w_1}\rangle} w_1=v_2-\frac{4}{30}w_1=$

$=(-5,4,3,2)-\frac{2}{15}(1,2,3,-4)=\frac{1}{15}(-77,56,39,38)$.

Let $w_2=(-77,56,39,38)$, then:

$\langle{w_1,w_2}\rangle=0$.

$\langle{w_2,w_2}\rangle=(-77)^2+56^2+39^2+38^2=12030$

$\langle{v_3,w_2}\rangle=39$.

$\langle{v_4,w_2}\rangle=38$.

Calculate $v_3-\frac{\langle{v_3,w_1}\rangle}{\langle{w_1,w_1}\rangle} w_1-\frac{\langle{v_3,w_2}\rangle}{\langle{w_2,w_2}\rangle} w_2=v_3-\frac{3}{30}w_1-\frac{39}{12030}w_2$

$=\frac{1}{4010}(4010v_3-401w_1-13w_2)=\frac{1}{4010}(600,-1530,2300,1110)$

$=\frac{1}{401}(60,-153,230,111)$.

Let $w_3=(60,-153,230,111)$. Then

$\langle{w_1,w_3}\rangle=0$

$\langle{w_2,w_3}\rangle=0$

$\langle{w_3,w_3}\rangle=60^2+(-153)^2+230^2+111^2=92230$

$\langle{v_4,w_3}\rangle=111$.

Calculate $v_4-\frac{\langle{v_4,w_1}\rangle}{\langle{w_1,w_1}\rangle} w_1-\frac{\langle{v_4,w_2}\rangle}{\langle{w_2,w_2}\rangle} w_2-\frac{\langle{v_4,w_3}\rangle}{\langle{w_3,w_3}\rangle} w_3$

$=v_4+\frac{4}{30}w_1-\frac{38}{12030}w_2-\frac{111}{92230}w_3$

$=\frac{1}{276690}(276690v_4+27669w_1-874w_2-333w_3)$

$=\frac{1}{276690}(74987,57343,-27669,95839 )$.

Let $w_4=(74987,57343,-27669,95839)$. Then

$\langle{w_1,w_4}\rangle=0$

$\langle{w_2,w_4}\rangle=0$

$\langle{w_3,w_4}\rangle=0$

$\langle{w_4,w_4}\rangle=18861957300$.

We have constructed an orthogonal basis $w_1,\,w_2,\,w_3,\,w_4$ of ${\mathbb R}^4$. The vectors $w_1,\,w_2$ form a basis of U, because $w_1,w_2\in Span\{v_1,v_2\}=U$ and $\dim U=2$.

Since the vectors $w_3,\,w_4$ are orthogonal to $w_1,w_2$, they form a basis of $U^{\perp}$.

To obtain orthonormal bases, one should normalize these vectors and take $\frac{1}{\sqrt{30}}w_1$, $\frac{1}{\sqrt{12030}}w_2$, $\frac{1}{\sqrt{92230}}w_3$ and $\frac{1}{\sqrt{18861957300}}w_4$ instead of them.