Suppose I have $v_1,\ldots,v_n$ vectors in $\mathbb{Z}^n$. Let $M$ be the matrix whose columns are $v_1,\ldots,v_n$. I would like to know if, as it happens with a vector space over a field,

  1. $M$ is ivertible if and only if $v_1,\ldots,v_n$ are linearly independent.

  2. If $v_1,\ldots,v_n$ are linearly independent, they form a basis.

Are there any notes on the Internet where I can find results of something like ``vector spaces over a ring''?

  • 2
    $\begingroup$ A vector space over a ring is called a module: en.wikipedia.org/wiki/Module_%28mathematics%29 $\endgroup$ – roman May 4 '16 at 8:54
  • 1
    $\begingroup$ "Vector spaces over a ring" are called modules. To be more precise, a vector space is a module over a field. There is plenty of literature about it. Both claims you have are not true, consider $2I_n$, where $I_n$ is the identity matrix of dimension $n$. $\endgroup$ – sebigu May 4 '16 at 8:55

Let's expand on question one a little bit. Here is something that you can say for a matrix over a general (commutative) ring.

For any matrix $M$, there is always a matrix $M^{adj}$, the adjugate of $M$, which satisfies the following equations. $$ MM^{adj} = \det(M) \cdot I \qquad\qquad M^{adj}M = \det(M) \cdot I $$ So this is a matrix that can be constructed that is almost an inverse to $M$. One can construct it exactly as one constructs the inverse; the only thing that we do not do is divide by the determinant of $M$.

So what does this provide? This tells us that if $M$ is a matrix over $R$ such that $\det(M)$ is a unit in $R$, then $M$ will be invertible, since we can then define $M^{-1} = (\det M)^{-1} \cdot M^{adj}$.

In particular, over $\mathbb{Z}$, the only units are $\pm1$. This tells us that the matrix $M$ you have constructed will have an inverse if and only if its determinant is $\pm1$.


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