Question

Check that the vectors u =(3/5, 4/5, 0) v=(-4/5, 3/5, 0)
and w = (0;0;1) are orthonormal.
Further, write the vector a = (1;-1;2) as a linear combination of the vectors.

Answer 1

Answer on Question #44584 – Math – Linear Algebra

Question 1.

Check that the vectors $u=(3/5,4/5,0)$, $v=(-4/5,3/5,0)$ and $w=(0,0,1)$ are orthonormal. Further, write the vector $a=(1,-1,2)$ as a linear combination of the vectors.

Solution. We have

$u\cdot u$ $=(3/5)^{2}+(4/5)^{2}+0^{2}=9/25+16/25+0=25/25=1;$

$u\cdot v$ $=(3/5)(-4/5)+(4/5)(3/5)+0\cdot 0=-12/25+12/25+0=0;$

$u\cdot w$ $=(3/5)0+(4/5)0+0\cdot 1=0+0+0=0;$

$v\cdot v$ $=(-4/5)^{2}+(3/5)^{2}+0^{2}=16/25+9/25+0=25/25=1;$

$v\cdot w$ $=(-4/5)0+(3/5)0+0\cdot 1=0+0+0=0;$

$w\cdot w$ $=0^{2}+0^{2}+1^{2}=1,$

so, $u,v$ and $w$ are orthonormal.

Let $a=\alpha u+\beta v+\gamma w$, that is

$(1,-1,2)$ $=\alpha(3/5,4/5,0)+\beta(-4/5,3/5,0)+\gamma(0,0,1)$

$=((3/5)\alpha-(4/5)\beta,(4/5)\alpha+(3/5)\beta,\gamma).$

This gives

$(3/5)\alpha-(4/5)\beta$ $=1,$

$(4/5)\alpha+(3/5)\beta$ $=-1,$

$\gamma$ $=2.$

Multiplying the first equation by 20 and the second one by $-15$, we get

$12\alpha-16\beta$ $=20,$

$-12\alpha-9\beta$ $=15.$

Adding the equations, we obtain $-25\beta=35$, so $\beta=-7/5$. Then

$\alpha=(5/3)(1+(4/5)\beta)=(5/3)(1-28/25)=(5/3)(-3/25)=-1/5.$

Thus $a=(-1/5)u+(-7/5)v+2w$. $\Box$