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Problem 8 in problem set 2.1 in Gilbert Strang's Linear Algebra says

"Normally 4 planes in a 4-dimensional space meet at a ____."

Let the four dimensions be $x,y,z,w$. In three dimensions, a plane is defined to be a two dimensional surface that extends infinitely. A plane satisfies some equation of the form $ax+by+cz=n$. Abstractly extending this definition to $4$D space, a plane would satisfy some equation of the form $ax+by+cz+dw=n$. First off, is this correct? Secondly, I'm not sure where $4$ of such planes would meet.

Could someone please help?

Saksham Sethi
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    The equation $ax+by+cz=n$ defines a plane through its normal vector $(a,b,c)$. (If that vector has length $1$, $n$ is the signed distance of the plane from the origin) If you carry this over formally to $\mathbb{R}^4$ the way you did you get a three dimensional subspace, which is orthogonal to $(a,b,c,d)$. From the question it is not clear whether this is what you want, or whether you want to consider two dimensional subspaces. In other words: what is the (general) definition of a plane in that book? – Thomas Sep 19 '22 at 17:59
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    Doesn’t Strang refer to hyperplanes in dimensions >3? – Ted Shifrin Sep 19 '22 at 18:23

2 Answers2

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Not quite. What you are trying to define is called a hyperplane, not a plane. Basically, in an $n$-dimensional space, like $\mathbb{R}^n$, a hyperplane is derived by imposing one constraint on the space, like $$ c_1 x_1 + c_2 x_2 + \ldots + c_n x_n = c_0, $$ which makes hyperplane have dimension $n-1$. Then, intersecting 2 hyperplanes imposes another constraint, so the result will be $n-2$ dimensional, and the intersection of $n$ hyperplanes would have dimension $0$, i.e. be a single point.

The real question is, however, if this is what Gilbert Strang really means in the question. He is saying he is operating on planes (not hyperplanes). Planes in any space are defined as 2D objects, so even though in $\mathbb{R}^3$ (which is 3-D), the plane and the hyperplane are the same notion (which explains where the name hyperplane comes from), in higher dimensions that is no longer the case. In particular, in $\mathbb{R}^4$, plane is still 2D, so requires 2 constraints, not 1, as a hyperplane would.

Two planes usually intersect in a line, but 4 planes is a different story...

gt6989b
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    Indeed. And, in any case, two 2-dimensional planes in 4-space typically intersect in a point. :) ... and four 2-dimensional planes in 4-spaces typically have empty intersection... – paul garrett Sep 19 '22 at 18:24
  • @paulgarrett The solution manual disagrees with your answer. See problem 8's solution in problem set 2.1 here: https://math.mit.edu/~gs/linearalgebra/ila_sol5_ch02.pdf – Saksham Sethi Sep 19 '22 at 18:28
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    @SakshamSethi, well, referring directly to reality, the 2D planes $x=0=y$ and $z=0=w$ intersect just at ${0}$. Also, the "constraint-counting" argument given in this answer gives the same outcome for "generic" intersections. Sure, four 3D-planes (=hyperplanes) in 4-space generically intersect in a point. It's not good if Strang inadvertently used "plane" for "hyperplane"... – paul garrett Sep 19 '22 at 18:33
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    @paulgarrett Yes, most likely Strang used "plane" for "hyperplane" – Saksham Sethi Sep 19 '22 at 18:48
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Assume Strang meant hyperplane. A hyperplane in $\mathbb R^n$ is generally defined as all points $\mathbf x$ satisfying $\mathbf n \cdot \mathbf x = c$, where $\mathbf n$ is the normal vector and $c$ is a constant real number. If $n-1$ coordinates are known, the last is determined, so the hyperplane has dimension $n-1$.

In this case our intersection can be represented as 4 simultaneous equations: $$ \begin{align} \mathbf n_1 \cdot \mathbf x &= c_1 \\ \mathbf n_2 \cdot \mathbf x &= c_2 \\ \mathbf n_3 \cdot \mathbf x &= c_3 \\ \mathbf n_4 \cdot \mathbf x &= c_4 \\ \end{align} $$

or a matrix equation $$ \begin{bmatrix} \mathbf n_1 \\ \mathbf n_2 \\ \mathbf n_3 \\ \mathbf n_4 \end{bmatrix} \mathbf x = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \end{bmatrix} $$

If the normals are picked to point in uniformly random directions, with probability 1 all are linearly independent (see Probability that $n$ vectors drawn randomly from $\mathbb{R}^n$ are linearly independent), so the matrix of normals is full rank and there is one solution for $\mathbf x$ at a point. You don't have to use matrices to solve this, but this is how you would get a computer to solve given actual values.

qwr
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