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I have two planes in an $n$-dimensional space, and a parallelogram in one of the planes, defined by two vectors ($v_1$, $v_2$), with area $S$. I want to find the area of the parallelogram defined by the projections of the vectors onto the other plane ($v'_1$, $v'_2$).

I can obviously just obtain the projections and compute the area, but I'm after a "simpler" result. For example, in 3-dimensional space a single angle $\theta$ can be defined between the two planes, and the projected area is then $S'=S\cos\theta$. But in an $n$-dimensional space there are, in general, two principal angles between the planes, is there a similarly simple relation in this case? Intuitively, I believe that if the principal angles are $\theta_1$ and $\theta_2$, maybe the area is $S'=S\cos\theta_1\cos\theta_2$ (which would still be applicable to 3 dimensions, since then $\theta_2=0$). Am I right? Where can I find a reference for this (or the correct) result?

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$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$Denote your planes by $P$ and $P'$, let $B = (\Basis_{1}, \Basis_{2})$ and $B' = (\Basis_{1}', \Basis_{2}')$ be orthonormal (ordered, oriented) bases for the respective planes, and put $$ a_{ij} = \Basis_{i} \cdot \Basis_{j}',\quad i, j = 1, 2. $$ (The coefficient $a_{ij}$ is, if it matters, the cosine of the angle between $\Basis_{i}$ and $\Basis_{j}'$.)

Orthogonal projection $\Reals^{n} \to P'$ defines a map $\Pi:P \to P'$ given by \begin{align*} \Pi(\Basis_{j}) &= (\Basis_{j} \cdot \Basis_{1}')\, \Basis_{1}' + (\Basis_{j} \cdot \Basis_{2}')\, \Basis_{2}' \\ &= a_{j1} \Basis_{1}' + a_{j2} \Basis_{2}'. \end{align*} The scale factor for area under $\Pi$ is $$ \det(a_{ij}) = (\Basis_{1} \cdot \Basis_{1}')(\Basis_{2} \cdot \Basis_{2}') - (\Basis_{1} \cdot \Basis_{2}')(\Basis_{2} \cdot \Basis_{1}') $$ (or its absolute value if you ignore orientation): Indeed, the (unique) linear map $\sigma:P' \to P$ satisfying $\sigma(\Basis_{i}') = \Basis_{i}$ obviously preserves area, and the composition $\Pi\sigma:P' \to P'$ has matrix $(a_{ij})$ with respect to $B'$.

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    $\begingroup$ +1. It's a bit confusing that you seem to have switched the primed and unprimed planes (with respect to the question). $\endgroup$ – joriki May 19 '16 at 13:33
  • $\begingroup$ @joriki: Thanks. :) $\endgroup$ – Andrew D. Hwang May 19 '16 at 13:49
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    $\begingroup$ Note that you can perform a singular value decomposition on $A$ to find bases for the two planes such that $A$ is diagonal. Presumably the singular values on the diagonal are related to the angles mentioned in the question? Perhaps the cosines? $\endgroup$ – joriki May 19 '16 at 14:14
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    $\begingroup$ Yes, you're right: On geometric grounds, they're the minimum and maximum cosines of angles between $P$ and $P'$. These two planes span a space of dimension at most four, and without loss of generality, $P'$ is the copy of $\Reals^{2}$ in the first two coordinates. The unit circle in $P$ projects to an ellipse in $P'$ (possibly degenerate), whose semi-axes are these cosines. The product of the semi-axes is the scale factor for area. $\endgroup$ – Andrew D. Hwang May 19 '16 at 14:27
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    $\begingroup$ Funny -- I was writing an answer based on that embedding when you posted yours -- and I thought: Ah, much more elegant independent of coordinates :-) $\endgroup$ – joriki May 19 '16 at 14:28

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