My Math Forum base of the orthogonal complement of a subspace

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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+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.
 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 $V_1,V_2$. Thanks from david940

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