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

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?

• 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. – Asaf Karagila Oct 29 '18 at 14:03

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?

• 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. – Jake Oct 29 '18 at 13:57
• @Josh Exactly! Perfect explanation and can you give three spanning vectors? – gimusi Oct 29 '18 at 14:00
• 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? – Jake Oct 29 '18 at 14:01
• sure (1,1,1,-3), (-1,1,0,0), (-1,0,1,0) – Jake Oct 29 '18 at 14:02
• @amWhy "Rigorously", I took note of it! Thanks – gimusi Oct 29 '18 at 14:02