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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
up vote 3 down vote accepted

The first idea that came to my mind was to perform row/column elimination (try to put it an a row/column echelon form) and conclude something about oddness & evenness of entries.

Then I realized that it's easier to just use the determinant formula.

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Evaluate mod $2$. Answer is $1$. – André Nicolas Jan 22 '13 at 8:43
@Tunococ : While I am not dextrous enough with Leibniz's formula to see why it leads to the answer, your answer suggested to me a more pedestrian approach: Show that the determinant of every leading principal submatrix is nonzero. Do this by induction on the size of the submatrix, using the fact that in the expression for each such determinant there is exactly one one odd term. Thank you! – Math Noob Jan 22 '13 at 9:09
@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
@MathNoob You knew exactly how I thought! I thought about Leibniz's formula. There's only one term, which is the product of diagonal entries, that will be odd. Other terms will have some off-diagonal factors. – Tunococ Jan 22 '13 at 11:34

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