Let $R$ be a commutative ring with $1 \neq 0$.

It is known that $R^{n} \hookrightarrow R^{m}$ implies $n \leq m$ and $R^{n} \twoheadrightarrow R^{m}$ implies $n \geq m$, and I might use this without mentioning (though I don't think I did). Isomorphisms are $R$-module maps if not mentioned otherwise.

Let $m \geq n$ and consider $R^{n} = R^{n} \times 0 \times \cdots \times 0 \subseteq R^{m}$. We have $\text{Hom}_{R}(R^{n}, R^{m}) \simeq M_{m \times n}(R)$ via $\phi \mapsto M(\phi) = (c_{ij})$ where $\phi(e_{j}) = \sum_{i=1}^{m}c_{ij}e_{i}$. Take any $R$-linearly independent $y_{1}, \cdots, y_{n} \subseteq R^{m}$ and define an $m \times n$ matrix $A = (a_{ij})$ via $y_{j} = b_{1j}e_{1} + \cdots + b_{mj}e_{m}$.

If we understand $e_{j}$ with $1 \leq j \leq n$ as $n \times 1$ column matrix in $R^{n}$ when we multiply an $m \times n$ matrix to it on its left, then $Ae_{j} = y_{j}$ for $1 \leq j \leq n$. Thus, we can understand this matrix as an injection $A : R^{n} \hookrightarrow R^{m}$.

By definition $A$ is an $m \times n$ matrix, but if what I wrote is correct, then $A|_{R^{n}}$ must be an isomorphism $R^{n} \simeq \bigoplus_{j=1}^{n} Ry_{j}$. I am not trying to find an invertible matrix to $A$, but is it okay to think of it as an embedding like how I described?

  • $\begingroup$ I don't get the question. If $m \geq n$, then of course there is a (split) monomorphism from $R^n$ to $R^m$. $\endgroup$ – Martin Brandenburg Jul 8 '13 at 7:48
  • $\begingroup$ I was wondering if writing the $m \times n$ matrix representation $A$ of a monomorphism $\phi : R^{n} \hookrightarrow R^{m}$ and (using the basis of $R^{n} \subseteq R^{m}$) considering the matrix the matrix $A$ as the monomorphism by understanding $A(x) = Ax$, the multiplication by $n \times 1$ column matrix $x \in R^{n} \subseteq R^{m}$. And the question came from the fact that I was unfamiliar with thinking about non-square matrix as an injection. (I guess $A$ is essentially $m \times m$ matrix with zeros on the rows at the bottom, but I want to know how others think of this situation.) $\endgroup$ – Gil Jul 8 '13 at 14:32

To address the title question: normally, an element $A$ is invertible if $AB=BA=I$ where $A,B,I$ all live in the same algebraic system, and $I$ is the identity for that system.

In this case, where $A$ and $B$ are matrices of different sizes, they don't really have a common algebraic system. If you put the $m\times n$ matrices and $n\times m$ matrices together into a single set, then when you multiply with matrix operations you get $n\times n$ and $m\times m$ square matrices. If you throw those square matrices into the set, then you find that sometimes you can't multiply two elements of the set because their dimensions don't match up. So, you can see the $A$ in your example isn't really invertible in this sense.

However, matrices can and do have one-sided inverses. We usually say that $A$ is left invertible if there is $B$ such that $BA=I_n$ and right invertible if there is $C$ such that $AC=I_m$. In a moment we'll see how the body of your question was dealing with a left inverible homomorphism.

To address the body of the question:

Sure: any homomorphism between free modules including the one you described can be identified with a matrix, if you have picked bases. You chose $A$ to represent your monomorphism in those particular bases. Since the $e_i$ form a basis for $R^n$ and $A$ is 1-1, then the images $Ae_i$ form a basis for $Im(A)$ which is isomorphic to $R^n$.

Now, take the $y_i=Ae_i$ for $1\leq i\leq n$ and extend them to a basis of $R^m$ by finding $y_i$, $n<i\leq m$. Map the $y_i$ back to $e_i$ for $i\leq n$, and send the rest of the $y_i$ to zero. This must give rise to a homomorphism $B$ from $R^m$ to $R^n$, and you can check that $BA$ is the identity on $R^n$. (Of course this $B$ has a nonzero kernel, so it can't be left invertible. It is, evidentally, right invertible!)

While this is all perfectly fine, it is known that $m\times n$ matrices (where $m\neq n$) over commutative rings with 1 can't be two-sided invertible. Such a pair would amount to an isomorphism between $R^n$ and $R^m$, which isn't possible because commutative rings have the IBN.

It's interesting to know though that for noncommutative rings, you can have right-and-left invertible matrices with $m\neq n$! The full linear ring is a good example for such a ring $R$, because $R^m\cong R^n$ for every pair of positive integers $m,n$.

  • $\begingroup$ Thanks for your thoughtful answer! $\endgroup$ – Gil Jul 20 '13 at 1:59

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