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Let $i := \sqrt{-1}$ . Consider $W \subseteq \mathbb{C}^3$ defined by $W := \{(1, 0, i),(1, 2, 1)\} $. Find $W^\perp$.

My biggest issue with this problem is not knowing how to extend the basis of $W$ to a basis of $\mathbb{C}^3$. After that, I know that I'd have to do the Gram-Schmidt process.

Andrew D. Hwang
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Chad
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    You deleted your question where we had a discussion, so I cannot post my comments there. This is why I am contacting you over this question to tell you this: what you did is very much frowned upon. It is wrong. You asked a question, and I wanted to try to help you. Your response was to delete the question with no comments. At the very least, it is rude. – 5xum Jun 02 '15 at 06:36
  • He also deselects answers he previously chose. @5xum – David South Jun 02 '15 at 07:10
  • @DavidSouth Thanks for sharing. If he keeps that up, soon, nobody will answer his questions. – 5xum Jun 02 '15 at 08:20
  • @5xum u were being extremely rude – Chad Jun 03 '15 at 01:40
  • @DavidSouth i have no idea who you are. i apologize if i 'deselected an answer' however i dont recall doing that. and to you both, please leave me alone and lets end this on good terms – Chad Jun 03 '15 at 01:41
  • @Chad (1) It was not my intention to be rude. If I sounded rude, I apologize. (2) If you think I am rude, you can either flag my comment or tell me in person. Me being rude does not change the fact that you should not delete your questions when you already get some feedback. – 5xum Jun 03 '15 at 05:27
  • @5xum u gave no feedback. feel free to reply, but im done with u – Chad Jun 03 '15 at 05:30
  • @Chad I told you where you made your mistake. Then I aksed you how you got to the answer (which was wrong). If you explained that, I would explain to you what exactly you did wrong. Your response was to simply delete your whole question. – 5xum Jun 03 '15 at 06:07

2 Answers2

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Method 1: To directly answer your question, the orthogonal complement of the span of the two dimensional subspace spanned by $(1,0,i)$ and $(1,2,1)$ is a one-dimensional subspace. Assume that this subspace is spanned by the vector $(a,b,c)$. For $(a,b,c)$ to be orthogonal to both $(1,0,i)$ and $(1,2,1)$, the inner (dot) products must be zero. In other words, $(a,b,c)$ must satisfy $$ a-ci=0\qquad\text{and}\qquad a+2b+c=0. $$ A nonzero solution to this system is a basis for the orthogonal complement.

Method 2: To get a basis, it is easier to start with a spanning set and reduce it to a basis than to get the basis directly. Observe that $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ form a basis for $\mathbb{C}^3$. Therefore, the vectors of $\{(1,0,i),(1,2,1),(1,0,0),(0,1,0),(0,0,1)\}$ form a spanning subset of $\mathbb{C}^3$. Consider adding $(1,0,0)$ to $\{(1,0,i),(1,2,1)\}$, either the vectors in this new set $\{(1,0,i),(1,2,1),(1,0,0)\}$ are linearly independent or linearly dependent. If they are independent, then you have a basis. If they are dependent, then the span of the vectors in $\{(1,0,i),(1,2,1)\}$ is the same as the span of the vectors $\{(1,0,i),(1,2,1),(1,0,0)\}$, therefore, throw $(1,0,0)$ out and continue with $(0,1,0)$. You will eventually have a basis because the original set was spanning.

Michael Burr
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The cross product is well suited to the case of finding a vector orthogonal to two others in a three-dimensional space.

Edit: As pointed out below, the complex cross product differs from the usual real-valued version in that it requires conjugation. See this stackexchange question.

Community
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cxseven
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  • Oh, this is true. I hadn't thought of that. Would it work with complex numbers, too, though? – Chad May 31 '15 at 02:18
  • Yes. The proof I keep in mind uses properties of the determinant that hold in any field: http://people.sju.edu/~pklingsb/crossproduct.pdf – cxseven May 31 '15 at 02:27
  • Beware the cross product depends very much on the definition of the inner product. So in $\mathbb{R}^3$ where $v\cdot w=\sum v_iw_i$ we have the traditional cross product. What would be the cross product in $\mathbb{C}^3$ where the inner product is $v\cdot w=\sum v_i\bar{w_i}$ ? – marwalix May 31 '15 at 03:13
  • Good point. Conjugation should be added to the traditional cross product. There was more discussion about this at http://math.stackexchange.com/q/129227/1174 – cxseven May 31 '15 at 04:33