Recently I was surprised to learn that the only symmetric 0-1 matrix with all eigenvalues positive is the identity matrix. Here’s a very nice proof of this fact given in this answer of Robert Israel’s:
Let A be an symmetric 0-1 matrix with all eigenvalues positive. Since A is symmetric with all eigenvalues positive, A is positive definite. The positive definiteness of A implies that any submatrix consisting of rows and columns i and j (for ) is also positive definite. Thus the determinant of this submatrix is positive. Therefore, . The only way this is possible for a symmetric 0-1 matrix is to have and . Thus A has all 1’s for its diagonal elements and all o’s for its off-diagonal elements.
I feel like there should be a more high-level explanation of this fact, though – a bigger-picture answer than one that operates on the level of the individual elements in the matrix. For instance, if a matrix being symmetric implies X about the underlying linear transformation, a matrix being 0-1 implies Y about the underlying linear transformation, and a matrix having all eigenvalues positive implies Z about the underlying linear transformation, then there ought to be some intuitive argument as to why the only linear transformation that has properties X, Y, and Z simultaneously is the identity transformation. I asked this question on math.SE (in fact, Robert Israel’s argument above comes from there), but I haven’t yet received such a big-picture answer.
Added (January 25, 2013): I wrote a follow-up post where I give two bigger-picture proofs.