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  • 1 Post By Evgeny.Makarov
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June 27th, 2014, 01:17 AM   #1
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base of the orthogonal complement of a subspace

In $\displaystyle R^3 $ the subspace $\displaystyle V_1 $ is generated by the vector $\displaystyle v_1 = (1,2,-1) $ and the subspace $\displaystyle V_2 $ is generated by the vector $\displaystyle v_2 = (1,1,-1) $

Find a base of $\displaystyle (V_1 + V_2)^{\bot} $ and a base of $\displaystyle V_1^{\bot}\cap V_2^{\bot} $


$\displaystyle V_1+V_1 = \langle (1,2,-1),(1,1,-1)\rangle $

Let $\displaystyle v= (x,y,z) \in R^3 $ then $\displaystyle v \in (V_1 + V_2)^{\bot} $ if

$\displaystyle \left\{
\begin{aligned}
v * (1,2,-1) &=0 \\
v * (1,1,-1) &=0 \
\end{aligned}
\right. $
$\displaystyle = \left\{
\begin{aligned}
x+2y-z &=0 \\
x+y-z &=0 \
\end{aligned}
\right. $

the solution is:
$\displaystyle \left\{
\begin{aligned}
x &= z \\
y &=0 \
\end{aligned}
\right. $

So $\displaystyle (x,y,z)=(z,0,z)= z(1,0,1) $

Thus
$\displaystyle (V_1 + V_2)^{\bot} = \langle(1,0,1)\rangle $

$\displaystyle V_1 = \langle (1,2,-1)\rangle $
$\displaystyle V_2 = \langle (1,1,-1)\rangle $

So $\displaystyle v \in V_1^{\bot}\cap V_2^{\bot} $ if

$\displaystyle \left\{
\begin{aligned}
v*(1,2,-1) &= 0 \\
v*(1,1,-1) &= 0 \
\end{aligned}
\right. $
So it is the same as before thus

$\displaystyle V_1^{\bot}\cap V_2^{\bot} = (V_1 + V_2)^{\bot} = \langle(1,0,1)\rangle $

I would like to know if it is solved in the right way.
david940 is offline  
 
June 27th, 2014, 07:46 AM   #2
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You solved it correctly. This problem is probably supposed to illustrate the fact that $(V_1+V_2)^\perp=V_1^\perp\cap V_2^\perp$ for all $V_1,V_2$.
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