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What shape does $x+y+z+w=0$ form in $\mathbb{R}^4$?

I suspect that it's a plane in $\mathbb{R}^4$. Is my intuition correct? why?

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  • $\begingroup$ The title is not part of the body for a reason. It's not that your question is very long. You can easily copy it also into the body of your question to make it all more readable. $\endgroup$ – Asaf Karagila Oct 29 '18 at 14:03
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In that case we refer to a hyperplane which is a subspace whose dimension is one less than that of its ambient space.

Indeed in that case the given equation $x+y+z+w=0$ define a subspace $\subseteq \mathbb{R}^4$ with dimension $3$.

To show this rigorously we need

  • to check that it is a subspace
  • to check that its dimension is $3$

Can you show these two fact?

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  • $\begingroup$ Awesome thanks! It's a subspace because it's closed under addition and scalar multiplication and I think that the dimension is 3 because it's spanned by three independent vectors. $\endgroup$ – Jake Oct 29 '18 at 13:57
  • $\begingroup$ @Josh Exactly! Perfect explanation and can you give three spanning vectors? $\endgroup$ – gimusi Oct 29 '18 at 14:00
  • $\begingroup$ I know how to show closure under addition and multiplication but am not sure how to show that the smallest spanning set consists of three independent vectors -- Is it because we have x=-y-z-w? $\endgroup$ – Jake Oct 29 '18 at 14:01
  • $\begingroup$ sure (1,1,1,-3), (-1,1,0,0), (-1,0,1,0) $\endgroup$ – Jake Oct 29 '18 at 14:02
  • $\begingroup$ @amWhy "Rigorously", I took note of it! Thanks $\endgroup$ – gimusi Oct 29 '18 at 14:02

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