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Consider $\mathbb R^3$ with the standard inner product. Let $W$ be the subspace of $\mathbb R^3$ spanned by $(1,0, -1)$. Which of the following is a basis for the orthogonal complement of $W$?

  1. $\{ ( 1, 0, 1), ( 0, 1, 0)\}$
  2. $\{(1,2,1),(0,1,1)\}$
  3. $\{(2,1,2),(4,2,4)\}$
  4. $\{(2,-1,2),(1,3,1),(-1,-1,-1)\}$

only first set is orthogonal, so it should be correct option. but what is the correct method to solve this problem.

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  • $\begingroup$ LaTeX + FAQ section = writing properly mathematics in this site $\endgroup$
    – DonAntonio
    Jun 26, 2013 at 6:43
  • $\begingroup$ The basis for the orthogonal complement does not need to consist of orthogonal vectors. Hence your reasoning that for the first set is invalid. E.g. a basis for the orthogonal complement of the singleton $(0,0,0)$ could be $(1,0,0), (1,1,0), (1,1,1)$. $\endgroup$
    – Calvin Lin
    Jun 26, 2013 at 6:44
  • $\begingroup$ Where did you get the idea that everybody would understand that [R!.3 means $\mathbb R^3$?? $\endgroup$ Jun 26, 2013 at 6:44
  • $\begingroup$ Actually i have copied from pdf formet. $\endgroup$
    – Alka Goyal
    Jun 26, 2013 at 6:48

1 Answer 1

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As $\dim W=1$, you know $\dim W^\perp = 3-1=2$, so $4$ is wrong. The vectors must be linearly independant, so $3$ is wrong. Each of the vectors must be orthogonal to each element of $W$ (it suffices to check against a basis of $W$, here only against $(1,0,-1)$), so $2$ is wrong and $1$ is correct (it does not matter if the two vectors in $1$ are othogonal to each other).

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