## Symmetric 0-1 Matrices with All Eigenvalues Positive

Recently I was surprised to learn that the only $n \times n$ symmetric 0-1 matrix with all eigenvalues positive is the $n \times n$ identity matrix.  Here’s a very nice proof of this fact given in this answer of Robert Israel’s:

Let A be an $n \times n$ 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 $2 \times 2$ submatrix consisting of rows and columns i and j (for $i \neq j$) is also positive definite.  Thus the determinant of this submatrix is positive.  Therefore, $a_{ii} a_{jj} - a_{ij} a_{ji} > 0$.  The only way this is possible for a symmetric 0-1 matrix is to have $a_{ii} = a_{jj} = 1$ and $a_{ij} = a_{ji} = 0$.  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 XY, 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.