Find a set of vectors that spans the subspace

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Homework Help Overview

The original poster is attempting to find a set of vectors in \(\mathbb{R}^3\) that spans a subspace defined by the condition \(u \cdot v = 0\), where \(v = \langle 1, 2, 3 \rangle\). The discussion revolves around understanding the implications of this condition in terms of linear algebra and vector spaces.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster discusses setting up a matrix multiplication to find vectors that satisfy the equation but expresses confusion over obtaining only the zero vector as a solution. Some participants suggest reconsidering the geometric interpretation of the problem, noting that the subspace is a plane perpendicular to the vector \(v\).

Discussion Status

Participants are exploring the implications of the equation derived from the dot product and discussing how to express the solution in vector form. There is acknowledgment of the need to interpret the results geometrically, and some guidance has been provided regarding the structure of the solution.

Contextual Notes

The original poster mentions feeling overwhelmed after extensive study, which may influence their approach to the problem. There is an indication of a potential misunderstanding regarding the nature of the subspace being sought.

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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