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.