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Given four 2D planes in the 4D space, how can I specify if they lie on a 3D hyperplane or not. Assume $\langle x_i, y_i, z_i, w_i \rangle$ for $i=1,\cdots,4$ be the normal vectors of the planes. Is that enough to show that they are not linearly independent, i.e., the following determinant is zero?

$$ \begin{vmatrix} x_1& y_1& z_1& w_1 \\ x_2& y_2& z_2& w_2 \\ x_3& y_3& z_3& w_3 \\ x_4& y_4& z_4& w_4 \end{vmatrix} $$

What about more than 4 dimensions? If we want to show that in $n$ dimensional space, $n$ hyperplanes of dimension $n-2$ lie on a hyperplane of dimension $n-1$, is that enough to show that the normal vectors of the planes are not linearly independent?

Ben Grossmann
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Helium
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    Note that 2D planes do not have unique normal vectors in 4D space. Your choice of normal vectors might change the result you get from this process – Ben Grossmann Aug 21 '13 at 20:07

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Your method doesn't really work. Here's something that does work no matter what your subspaces look like, in any number of dimensions.

The general problem we are solving is as follows: what is the dimension of the smallest subspace containing subspaces $S_1,S_2,\dots,S_m\subset \mathbb R^n$?

Here's the process: find a basis for each subset $S_k$ consisting of the vectors $v_{k1},v_{k2},\dots,v_{kd_k}$. Take all of these vectors, and place them as column vectors in the matrix $$ A = \begin{bmatrix} | & | & \, & | & | & \, & |\\ v_{11} & v_{12} & \cdots & v_{1d_1} & v_{21} & \cdots & v_{md_m}\\ | & | & \, & | & | & \, & | \end{bmatrix} $$

Put $A$ in reduced row echelon form. The number of pivots (row-starting $1$s) gives you the dimension of the resulting union, and the column vectors corresponding to the pivots give you a basis for this space.

An example where your method fails:

Consider the $xy,yz,zw$ and $wx$ planes in $4$-space. Choose the following normal-vectors, respectively: $$ \pmatrix{0\\0\\1\\0}\quad\pmatrix{1\\0\\0\\0}\quad\pmatrix{1\\0\\0\\0}\quad\pmatrix{0\\0\\1\\0} $$

Ben Grossmann
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  • Thanks for your answer. What do you mean by "find a basis for each subset $S_k$ consisting ..."? Selecting any arbitrary basis for the vectors results in the same output? – Helium Aug 21 '13 at 21:22
  • Selecting any arbitrary basis gives you the same number of $1$s in the resulting rref matrix. So yes. – Ben Grossmann Aug 21 '13 at 21:30
  • I asked this question with the hope to solve another problem that I have already asked here: http://math.stackexchange.com/questions/387380/rank-of-a-n-times-n-matrix but I think that the method you described, while correct, is not so useful to solve the other problem. The other problem has an open bounty now. Could you please have a look at the other problem and see if you can help. – Helium Aug 22 '13 at 00:31
  • I think the key is that you don't save any effort when you look at the perpendicular spaces in dimensions 4 and higher, because the space perpendicular to a plane has higher or the same dimension than the plane itself. I'll give your question a shot – Ben Grossmann Aug 22 '13 at 00:41