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Given a triangle with the edges $e_1$, $e_2$, $e_3$, it seems (from numerical evidence) that there are coefficients $\alpha_i$ such that $$ u^Hv = \sum_{i=1}^3 \alpha_i \, (u^He_i)\, (e_i^H v) $$ holds for all vectors $u, v\in\mathbb{C}^2$.

There is of course the brute-force way of taking the edge coefficients in terms of their $x$- and $y$-coordinates to show this is true, but the simplicity of the statement makes me think there should be a more elegant way.

I looked at a bunch of geometry text books and couldn't find it. I'm thinking to maybe take off my geometry hat and put on my linear algebra hat. What's your take on the problem; where do you think are similar results to be found?

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Try showing that $\sum \alpha_i e_i e_i^H = I$, for some $\alpha_i$. Maybe the $e_i e_i^H$ spans the set of hermitian matrices? – copper.hat Jul 29 '12 at 18:08
The symmetric matrices, not the hermitian matrices (if the $e_i$ are real). – Robert Israel Jul 29 '12 at 18:31
Try to show that $\alpha_i e_i e_i^H$ is the oblique projection onto the line of $e_i$ – chaohuang Jul 29 '12 at 18:39
up vote 2 down vote accepted

What you're saying is that the $2 \times 2$ identity matrix is in the linear span of the three matrices $M_i = e_i \otimes e_i^H$, where $e_i$ are the (column) vectors corresponding to the edges of the triangle, and thus $e_1 + e_2 + e_3 = 0$. In fact the three matrices are linearly independent, and thus form a basis of the symmetric $2 \times 2$ matrices, as long as two of the vectors are linearly independent.

Note that if you replace $e_i$ by $f_i = U e_i$ where $U$ is any invertible $2 \times 2$ matrix, you replace $M_i$ by $U M_i U^H$, and these are linearly independent iff the original $M_i$ are. By taking a suitable $U$, we may assume $e_1 = \pmatrix{1\cr 0\cr}$ and $e_2 = \pmatrix{0\cr 1\cr}$, so $M_1 = \pmatrix{1 & 0\cr 0 & 0\cr}$, $M_2 = \pmatrix{0 & 0\cr 0 & 1\cr}$, $M_2 = \pmatrix{1 & 1\cr 1 & 1\cr}$, which are clearly linearly independent.

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