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Ilya Bogdanov
Ilya Bogdanov
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Surely, the Smith normal form does it all. But if you need a more concrete condition, here is one.

Let $X$ be the set of all $n\times n$ minors of $A$, and let $Y$ be the set of all $n\times n$ minors of $(A|B)$$(A\,| B)$. Then the equivalent condition is that $\gcd(X)=\gcd(Y)$. Indeed, this condition is equivalent if the system is in the Smith form, and moreover it is preserved by $SL(n,\mathbb Z)$-transforms.

Surely, the Smith normal form does it all. But if you need a more concrete condition, here is one.

Let $X$ be the set of all $n\times n$ minors of $A$, and let $Y$ be the set of all $n\times n$ minors of $(A|B)$. Then the equivalent condition is that $\gcd(X)=\gcd(Y)$. Indeed, this condition is equivalent if the system is in the Smith form, and moreover it is preserved by $SL(n,\mathbb Z)$-transforms.

Surely, the Smith normal form does it all. But if you need a more concrete condition, here is one.

Let $X$ be the set of all $n\times n$ minors of $A$, and let $Y$ be the set of all $n\times n$ minors of $(A\,| B)$. Then the equivalent condition is that $\gcd(X)=\gcd(Y)$. Indeed, this condition is equivalent if the system is in the Smith form, and moreover it is preserved by $SL(n,\mathbb Z)$-transforms.

Source Link
Ilya Bogdanov
Ilya Bogdanov
  • 23.7k
  • 54
  • 92

Surely, the Smith normal form does it all. But if you need a more concrete condition, here is one.

Let $X$ be the set of all $n\times n$ minors of $A$, and let $Y$ be the set of all $n\times n$ minors of $(A|B)$. Then the equivalent condition is that $\gcd(X)=\gcd(Y)$. Indeed, this condition is equivalent if the system is in the Smith form, and moreover it is preserved by $SL(n,\mathbb Z)$-transforms.