# Is a symmetric non-negative integral matrix with odd diagonal entries and even non-diagonal entries full rank over $\mathbb{R}$?

Let $A$ be an $n\times{}n$ matrix which satisfies the following properties:

1. The elements of $A$ are non-negative integers.
2. The diagonal elements of $A$ are all odd.
3. The non-diagonal elements of $A$ are all even.

Is it true that $A$ has full rank over the field of real numbers? How does one prove this?

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Symmetry is irrelevant. Non-negative is irrelevant. –  André Nicolas Jan 22 '13 at 8:40

Evaluate mod $2$. Answer is $1$. –  André Nicolas Jan 22 '13 at 8:43
@MathNoob: It is a fundamental property of the determinant that it is compatible with ring homomorphisms (because it it just a polynomial expression): applying a homomorphism to all entries results in the same homomorphism being applied to the determinant. Calculating the determinant of the matrix in $\Bbb Z$ and then applying the morphism of reducing modulo $2$ we may conclude that the determinant is odd, and therefore nonzero. –  Marc van Leeuwen Jan 22 '13 at 10:08