Here are a few arguments that are possibly useful to prove things when one hasn’t set up all the usual machinery of vector spaces, subspaces and dimensions, presented in decreasing order of how seriously that could be used in a first course on matrix algebra. Since it fits well with the syllabus for the first year course I’m lecturing this term, I’m going to assume reduced row-echelon form is available. Throughout all matrices have coefficients in a field .

### Invertible if and only if injective by row operations

Given a matrix in reduced row-echelon form it is almost trivial to write down all solutions to . There are non-zero solutions if and only if not every column of contains a pivot. For example, the solutions to

are

the pivot column variables are determined by the non-pivot variables, which may be chosen freely.

Let be an invertible linear map represented by the matrix . Obviously if is invertible then is injective. Conversely, suppose that is injective and that has reduced row-echelon form , so for some invertible matrix . If has a zero row then it has at most pivots, so by the remark above, has a non-zero kernel. Therefore has pivots and so is the identity matrix. Hence . Now this only shows that has a left inverse, but since is a product of elementary matrices (corresponding to row operations), and each such elementary matrix is easily seen to be invertible (in the usual sense), is invertible and so is invertible.

Very similarly one can show that is invertible if and only if is surjective.

### Linearly dependent if too many vectors

If is the matrix whose rows are the components of vectors in (expressed in the standard basis) then the reduced row echelon form of has at most pivot columns, and so the bottom row is zero. Correspondingly an -linear combination of the rows of is zero.

### Invertible if and only if injective by disguised minimal polynomial and Primary Decomposition Theorem

The vector space of all linear endomorphisms of is dimensional. By the previous result there is a non-trivial linear dependency between the powers of . Let be a relation with minimal. If then formally multiplying through by gives

as a plausible (and clearly correct) guess for the inverse of . If then where and has degree . By minimality of , we have ; if then , so, choosing least such that , we find that . So either is invertible, or , and so is not injective.

### Row rank is column rank

William P. Wardlaw published a very short proof in Mathematics Magazine (2005) **78** 316—318. Let be a non-zero matrix. Let be minimal such that there exists an matrix and an matrix such that . By minimality the rows of are linearly independent, and clearly the row span of contains the row span of . Hence is the row rank of . Dually, is the column rank of .

### Linear dependencies by pigeonhole principle

Take vectors in . There are linear combinations of these vectors, but only elements of , so two linear combinations with different coefficients are equal.

As I learned from this MathOverflow answer, one can make this argument work in , and so in by scaling to remove denominators. Let non-zero elements of be given, and let be the largest component in absolute value. There are -linear combinations with coefficients in , each of which has components in . The pigeonhole principle strikes as soon as ; clearly this is true for large enough . For instance, since and , it is sufficient if .

**Question.** Is there a trick to extend this argument to arbitrary fields?

### Modules over general rings

Yiftach Barnea pointed out to me that if is an matrix with entries in a principal ideal domain then, regarded as a linear map , if is surjective then is injective. One proof (to my eyes, essentially the same as using the structure theorem for modules over a PID) is to use row and column operations to reduce to the case where is a diagonal matrix: then all the diagonal entries of must be invertible in , so is itself invertible, and so injective. Clearly the converse holds if and only if is a field.

Suppose that is a unital ring containing a non-zero element with a left inverse. Then multiplication by , i.e. , is an injective linear map. This map is surjective if and only if has a right inverse. So for general rings, neither implication holds.

### Question.

Is there a nice characterization of rings such that if is a surjective -linear map then is injective?

### Partial answer.

My Ph.D student Bill O’Donovan told me about Hopfian objects. By definition, an -module is *Hopfian* if and only if it has the property in the question, i.e. any surjective -module endomorphism of is injective.

A related property is that is *strongly Hopfian* if and only if whenever is an -module homomorphism, the kernel chain

stabilises. The relevant implication follows from the standard proof of Fitting’s lemma.

**Claim.** *Strongly Hopfian implies Hopfian.*

**Proof.** Suppose that . Since restricts to a map for any , we see that preserves , and acts on this space as a nilpotent map of degree . Let . Since is surjective, for some , and by hypothesis, . Hence . But , so , hence . Therefore and is injective.

In particular, any Noetherian ring is Hopfian as a module for itself. There are Hopfian modules that are not strongly Hopfian modules, for example Proposition 2.7(4) in this paper by A. Hmaimou, A. Kaidi and E. Sánchez Campos gives the example of the -module , where is the th prime; the kernel chain for the endomorphism acting nilpotently with degree on the th summand does not stabilise, but since any endomorphism preserves the summands, the module is Hopfian.

I asked on MathOverflow is there was a ring and a Hopfian -module such that was not Hopfian, and my worst suspicions were confirmed.