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I'm wondering if the following basic facts about matrix inversion are correct:

1) For any matrix $A$, if $A$ has a right and left inverse, these are equal, unique and called the inverse of $A$

2) If $A$ is square, then any left inverse is a right inverse and vice versa

3) If $A$ is $m$x$n$ with $n > m$, then $A$ can only have a right inverse, and if it does, then it has infinitely many (question: it has infinitely many if and only if the characteristic of the underlying field $\mathbb{F}$ is $0$?)

4) If $A$ is $m$x$n$ with $n < m$ we have the exact same conditions as 3 with "right" replaced by "left".

Typically 2 is proven using elementary matrices, is there a way to do this using the finite dimensional linear operator point of view rather than the matrix point of view. For example, for $A$ $m$x$n$ with $n > m$ you can argue that $Ax = 0$ has non-trivial solutions either from considerations concerning free variables or by using the dimension formula.

Similarly, for 3, you can argue that if $B$ is a right inverse of $A$ then since $Ax = 0$ has non-trivial solutions, we can find another matrix $B'$ s.t. $AB' = 0$ (by taking for the columns of $B'$, scalar multiples of $x_0$, i.e. $b_j = c_j \cdot x_0$ for some non-trivial solution $x_0$ to $Ax=0$) and we see that the family $B + B'$ gives infinitely many right inverses of $A$. Is there a good way to see this from the linear operator point of view?

Finally, are there good rules of thumb or heuristics for identifying the types of problems where one point of view may be more appropriate, efficient, or insightful?

Coriolanus
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2 Answers2

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Statement 1 is correct. If $AB=CA=I$, then $B=(CA)B=C(AB)=C$.

Statement 2 is also correct. It can be proved by some sort of dimension arguments. See the question If $AB = I$ then $BA = I$ and Martin Brandenburg's or Bill Dubuque's answers in particular.

The first part of statement 3 is correct. Let $A\in L(V,W)$ with $n=\dim V>\dim W=m$. Then every $B\in L(W,V)$ is not surjective (because $\dim BW\le \dim W<\dim V$) and hence $BA$ cannot possibly be equal to $I$.

question: it has infinitely many if and only if the characteristic of the underlying field $\mathbb{F}$ is $0$?

Not exactly. Provided that $n>m\ge1$, $A$ has infinitely many right inverses if and only if the underlying field $F$ is infinite. (Do not confuse an infinite field with a field of characteristic zero. While every field of characteristic zero is infinite and every finite field has a nonzero characteristic, a field of nonzero characteristic can be infinite.)

When $F$ is finite, the number of right inverses is finite because there are only finitely many $F$-linear maps between two finitely dimensional vector spaces over $F$.

When $F$ is infinite, let $v$ be any nonzero vector in the null space of $A$ (which is necessarily a nontrivial subspace of $V$ because $\dim V>\dim W$) and $f$ be any nonzero linear function in $W^\ast=L(W,F)$ (note that $f$ can be chosen to be nonzero because $\dim W=m\ge1$). Then $w\mapsto Bw+\alpha f(w)v$ is a right inverse of $A$ for every $\alpha\in F$. As there are infinitely many choices of $\alpha$, $A$ has infinitely many right inverses.

Finally, statement 4 is correct. You may apply the proof of statement 3 to $A^T$. It is obvious that $B^T$ is a right inverse of $A^T$ if and only if $B$ is a left inverse of $A$.

user1551
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  • Why is $f$ necessary in the second last paragraph? – Coriolanus Mar 12 '20 at 02:14
  • @Coriolanus We want to exhibit a right inverse that is different from $B$. – user1551 Mar 12 '20 at 02:15
  • In particular if we remove $f$ then $w \mapsto Bw + \alpha v$ is not linear – Coriolanus Mar 12 '20 at 02:20
  • @Coriolanus Yes. And $w\mapsto f(w)v$ is just a rank-one linear transformation. In terms of matrices, if the matrix representation of $f$ is a row vector $u^T$, the linear map $w\mapsto f(w)v$ will be represented by the rank-one matrix $vu^T$ (so that each $w$ is mapped to $vu^Tw$). – user1551 Mar 12 '20 at 02:25
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Here's an answer to one of your questions (it's generally a good idea to stick to one question per question here). Let $V,W$ be finite-dimensional vector spaces of equal dimension. Let $T:V\to W$ and $S:W\to V$ be linear. Then $TS$ is the identity on $W$ if and only if $ST$ is the identity on $V$.

Do you want to see a proof, or would you rather have the feeling of accomplishment that comes with finding one on your own?

Gerry Myerson
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