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June 27th, 2014, 01:17 AM  #1 
Newbie Joined: Jun 2014 From: italy Posts: 9 Thanks: 0  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+2yz &=0 \\ x+yz &=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. 
June 27th, 2014, 07:46 AM  #2 
Senior Member Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108 
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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base, complement, orthogonal, subspace 
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