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It's a problem from MIT 18.06 14 Spring Exam1. The vectors are all 4 dimensional. I suppose that there are infinitely many such space orthogonal to that fixed subspace spanned by those 2 independent vectors (let's call them v and w). But 2 dimensional subspace that contains all and only vectors orthogonal to those 2 vectors is a different thing. It's hard to visualize this problem in 4 dimensional space so I tried reasoning in 3 dimensional space. There are infinitely many 2 dimensional plane orthogonal to the subspace spanned by z-axis. But only the x-y plane contains all vectors orthogonal to z-axis. I believe it's also true for higher dimension (such space is unique).

OK, but I don't know how to extend this 3 dimensional case to 4 dimensional space to convince myself. Any help?

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A vector $\mathbf{a}$ belonging to that space satisfy the two equations $$ \mathbf{a}\cdot\mathbf{v}=0\\ \mathbf{a}\cdot\mathbf{w}=0 $$ If you have a basis, and a representation of the inner product in this basis, then that is a homogeneous linear system of two equations in four unknown (the components of $\mathbf{a}$), and the set of its solutions is a subspace of dimension $2$ (given that the two equations are independent because the two vector are independent).

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  • $\begingroup$ Right. How could I miss that. Thank you. $\endgroup$ – Xiaohong Deng Aug 14 '14 at 1:32

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