##AB = I \implies BA = I##, for square matricies ##A,B##

[Buffu]

Let $(AB)_j$ be the jth column of $AB$, then $\displaystyle (AB)_j = \sum^n_{r= 1} B_{rj} \alpha_r$ where $\alpha_r$ is the rth column of $A$.

Also $(BA)_j = B \alpha_j \implies A(BA)_j = \alpha_j$ susbtituting this in the sum $\displaystyle (AB)_j = \sum^n_{r = 1} B_{rj}A(BA)_r$

Now if we multiply any matrix by some column of $I$ we get that column of the matrix as the result,

So $\displaystyle B_j = BA \sum^n_{r= 1} B_{rj} (BA)_r$

Also $\displaystyle [(BA)B]_j = \sum^n_{r= 1} B_{rj}(BA)_r$

therefore $\displaystyle B_j = (BA) [(BA)B]_j = (BA)[B(AB)]_j = (BA)B_j$ Therefore $BA = I$

Is this correct ? Is there a easy way to do this ?

 

[mathwonk]

this is a non trivial fact so i don't think
there is an easy way to do it from scratch. the key point is that a linear map from k^n to itself that is injective is also surjective, or that a set of n vectors that is independnt is also spanning. Thus if you assume one of those non trivial facts, which can be proved abstractly and perhaps in an easier looking way, then you can deduce your result. i.e. BA = I implies the map A is injective hence also surjective so it has a right inverse C such that AC = I. But then B = B.I = B(AC) = (BA).C = I.C = C, so AB = AC = I.

you can also use the theory of row operations to show that if BA = I then A can be row reduced to the identity by means of elementary row operations, hence by left multiplication by elementary matrices that can be checked to hace two sided inverses. Thus there is an invertible matrix D such that DA = I. But since D has a 2 sided inverse E, such that DE = ED = I, then A = I.A = (ED)A = E(DA) = E.I = E. So A also has a 2 sided inverse, which must equal B as above.

 

[Buffu]

this is a non trivial fact so i don't think

So my proof is incorrect ? :(

there is an easy way to do it from scratch. the key point is that a linear map from k^n to itself that is injective is also surjective, or that a set of n vectors that is independnt is also spanning. Thus if you assume one of those non trivial facts, which can be proved abstractly and perhaps in an easier looking way, then you can deduce your result. i.e. BA = I implies the map A is injective hence also surjective so it has a right inverse C such that AC = I. But then B = B.I = B(AC) = (BA).C = I.C = C, so AB = AC = I.

you can also use the theory of row operations to show that if BA = I then A can be row reduced to the identity by means of elementary row operations, hence by left multiplication by elementary matrices that can be checked to hace two sided inverses. Thus there is an invertible matrix D such that DA = I. But since D has a 2 sided inverse E, such that DE = ED = I, then A = I.A = (ED)A = E(DA) = E.I = E. So A also has a 2 sided inverse, which must equal B as above.

Thank you for the second proof I will try to fill the missing parts :).

 

[zwierz]

nonsense was written here :(

 

[mathwonk]

forgive me i did not read your proof in detail as i find such computational proofs hard to read. i prefer logical ones. but based on the length of your proof i would guess it could be right. you asked if it could be made easy, i said not to my knowledge.

 

[zwierz]

Let's multiply equality $AB=I$ by $A^{-1}$ from the left: $B=A^{-1}$. The last equality multiply by $A$ from the right: $BA=I$.
From $AB=I$ it follows that $\det A\det B=1$ thus $\det A\ne 0$ and the above argument is correct

 

[StoneTemplePython]

Here is a long form approach which borrows from a technique which may be worth re-visiting later on (i.e. in showing $AB$ and $BA$ have the same non-zero eigenvalues).

First, recognize the definition of the Identity Matrix is a matrix that maps every vector to itself and does not change length or orientation. To avoid confusion, note that the following $\mathbf x \neq \mathbf 0$ aka $\Vert \mathbf x \Vert_2 \gt 0$ and $\mathbf y \neq \mathbf 0$ aka $\Vert \mathbf y \Vert_2 \gt 0$-- i.e. the zero vector is dealt with separately.

Stated again, we can define the identity matrix as:
$I \mathbf x = \mathbf x$, for all $\mathbf x$

(and of course $I \mathbf 0 = \mathbf 0$)

Side note: to the extent we're interested in row vectors, we can also mention $\mathbf x^T I = \mathbf x^T$, for all $\mathbf x^T$

Because $(AB) = I$ we know $(AB)\mathbf x = \mathbf x$, for all $\mathbf x$

We know that $A$ and $B$ each must be full rank.
- - - -
Why? The gist is $(AB)$ cannot be the identity matrix if there is some $B\mathbf x = \mathbf 0$ or else you'd have $(AB)\mathbf x = A(B\mathbf x) = A\mathbf 0 = \mathbf 0 \neq I \mathbf x$. You can extend this logic for $A$, or alternatively, use the equivalence of size of left and right nullspace of square matrix $A$ and note there cannot be $\mathbf x^T A = \mathbf 0^T$ or else that would mean we have $\mathbf x^T A B = (\mathbf x^T A) B = \mathbf 0^T B = \mathbf 0^T \neq \mathbf x^T I$.
- - - -
From here left multiply by $B$ and we get

$B(AB)\mathbf x = B \mathbf x$ for all $\mathbf x$

using associativity, we can re-write this as

$(BA)(B\mathbf x) = (B \mathbf x)$ for all $\mathbf x$

define $B \mathbf x := \mathbf y$ (noting that this means $\mathbf x = A \mathbf y$)

$(BA)(\mathbf y) = \mathbf y$ for all $\mathbf y$

(and of course $(BA)\mathbf 0 = \mathbf 0$)

So we see that $(BA) = I$

 

[Buffu]

$\Vert \mathbf x \Vert_2 \gt 0$ and $\mathbf y \neq \mathbf 0$ aka $\Vert \mathbf y \Vert_2 \gt 0$

What does $\Vert\mathbf x \Vert_2$ mean ? I guess $\Vert$ means magnitude but what about $2$ subscript ?

 

[StoneTemplePython]

What does $\Vert\mathbf x \Vert_2$ mean ? I guess $\Vert$ means magnitude but what about $2$ subscript ?

yea -- length 2 norm aka euclidean length (aka generalization of Pythagorean theorem) but feel free to disregard the comments on length.

The reality is it's probably better to just write $\mathbf x \neq \mathbf 0$ and $\mathbf y \neq \mathbf 0$ and not mention length. Technically you lose some generality and need to be in an inner product space for length to be well defined. I mostly think in terms of inner product spaces so I sometimes forget that. The flip side is -- I suspect--- knowing that the zero vector is the only vector with zero length can be intuitive and helpful for some people.

 

[zwierz]

Funny point is that this assertion is not valid for infinite dimensional case. Indeed, consider a space of infinite sequences $X=\{(x_1,x_2,\ldots)\},\quad x_i\in\mathbb{R}$ and consider operators $A,B:X\to X$ defined as follows $$B(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots),\quad A(x_1,x_2,\ldots)=(x_2,x_3,\ldots).$$
It is clear $AB=I,\quad BA\ne I$

 

[WWGD]

Funny point is that this assertion is not valid for infinite dimensional case. Indeed, consider a space of infinite sequences $X=\{(x_1,x_2,\ldots)\},\quad x_i\in\mathbb{R}$ and consider operators $A,B:X\to X$ defined as follows $$B(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots),\quad A(x_1,x_2,\ldots)=(x_2,x_3,\ldots).$$
It is clear $AB=I,\quad BA\ne I$

It is not true in the f.d case either if the matrices are not square. You may have just a one-sided inverse.

 

[StoneTemplePython]

It is not true in the f.d case either if the matrices are not square. You may have just a one-sided inverse.

indeed where $A$ is $m$ x $n$ and $B$ is $n$ x $m$, we can always use the cyclic property of the trace, i.e. recognize that $trace(AB) = trace(BA)$

if $AB = I_m$ and $BA = I_n$, then

$m = trace(I_m) = trace(AB) = trace(BA) = trace(I_n) = n$

which tells us we can only have this relationship if $m = n$

 

[WWGD]

If the matrices are not square then the corresponding operators have different domains and anything you want can be in such a situation

There is a general trivial fact. Let $X$ be an arbitrary set and $A:X\to X$ be a bijection, it is clear $A^{-1]A=AA^{-1}$. Then what is the initial problem about ?
It is about that if $X$ is a vector space and $A$ is a linear operator then $A^{-1}$ is also linear. t is also very simple thing I guess

I was just making a simple point as a refresher; I do forget these at times.Yes; surjections and injections have (at least) 1-sided inverses.

 

[zwierz]

O sorry I deleted my post, now I restor it:

It is not true in the f.d case either if the matrices are not square. You may have just a one-sided inverse.

If the matrices are not square then the corresponding operators have different domains and anything you want can be in such a situation

There is a general trivial fact. Let $X$ be an arbitrary set and $A:X\to X$ be a bijection, it is clear $A^{-1}A=AA^{-1}$. Then what is the initial problem about ?
It is about that if $X$ is a vector space and $A$ is a linear operator then $A^{-1}$ is also linear. It is also very simple thing I guess

 

[Mark44]

It is not true in the f.d case either if the matrices are not square. You may have just a one-sided inverse.

In the thread title, both matrices are assumed to be square. I couldn't tell if you were making a general comment or one specific to this question.

 

[WWGD]

In the thread title, both matrices are assumed to be square. I couldn't tell if you were making a general comment or one specific to this question.

Sorry, it was a general comment in reply to Zwierz' post