Find a set of vectors that spans the subspace

mateomy
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Find a set of vectors in \mathbb{R}^3 that spans the subspace
<br /> S\,=\,\{\,u\,\in\,\mathbb{R}^3\,|\,u\cdot v\,=\,0\,\}<br />
where v=<1,2,3>


Maybe 12 hours of studying is too much and I'm fried or, maybe I'm looking for excuses. Either way...

To solve this I'm trying to set up a matrix multiplication and augment it at zero. But, I just get a single linear equation which tells me that the only way I can have a span of this subspace is if my other vector is the zero vector <0,0,0>. I don't think that's right.

<br /> <br /> \begin{bmatrix}<br /> a &amp; b &amp; c<br /> \end{bmatrix} <br /> <br /> *<br /> <br /> \begin{bmatrix}<br /> 1\\2\\3<br /> \end{bmatrix}<br /> <br /> =<br /> <br /> \mathbf{0}<br /> <br /> <br />

Getting a+2b+3c=0

Where's my issue?


Thanks.
 
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Perhaps you should stop and think what it is you're calculating here. You are looking for a subspace which is perpendicular to a vector, so it's a plane. Maybe that will help you interpret your result.
 
From a+ 2b+ 3c= 0, you have a= -2b- 3c so <a, b, c>= <-2b- 3c, b, c>.
 
HallsofIvy said:
From a+ 2b+ 3c= 0, you have a= -2b- 3c so <a, b, c>= <-2b- 3c, b, c>.

Thanks. I was looking at that and didn't take the next step of actually putting it into a vector form. Again, thanks.
 
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