A question on invertible matrixes

[kostas230]

I'm making notes for linear algebra, and I'm using the weaker definition for the invertible matrix:

"A matrix A is called invertible it there exists a matrix B such that BA = I"

How do we prove that AB = I using only matrix theory?

 

[Square]

You mean how one would prove that $AB = I$ provided that $BA = I$?

If you assume that $BA = I$ and thus that $A^{-1} = B$ and $A = B^{-1}$ you could try multiplying from the left with $B^{-1}$ and from the right with $A^{-1}$, which gives you

$B^{-1}BAA^{-1} = B^{-1}IA^{-1} \ .$​

 

[Erland]

Square, you are assuming what should be proved: You assume that there is a matrix $A^{-1}$ such that $AA^{-1}=A^{-1}A=I$, but this is what should be proved.

To prove that $BA=I$ implies $AB=I$ is not entirely easy. First of all, we must assume that $A$ and/or $B$ are square matrices. Otherwise, this is false. For example, if

$B=[1\quad 1]$ and $A=[2\quad -1]^{T}$, then $BA=[1]=I_1$ and $AB\neq I_2$.

So, assume that $A$ and $B$ are $n\times n$-matrices, for some $n\ge 1$, such that $BA=I$.

Now, consider the homogeneous system

$A\bf x= 0$ (1).

If $\bf x$ is a solution of (1), then

${\bf x}=I{\bf x}=(BA){\bf x}=B(A{\bf x})=B{\bf 0}={\bf 0}$.

This means that the homogeneous system (1) has only the trivial solution ${\bf x} ={\bf 0}$. Thus, if we solve (1) by elimination, transforming $A$ to reduced echelon form by a sequence of elementary row operations, the resulting reduced echelon form must be $I$ (otherwise, there would be non-pivot columns which correspond to free variables).

Now, consider the inhomogeneous system

$A{\bf x} ={\bf b}$, (2),

where $\bf b$ is an arbitrary vector in $R^n$.

If we perform the same sequence of row operations as in the solution of (1), the resulting coefficient matrix is again $I$, since we started with the same coefficient matrix $A$.
This means that (2) has a unique solution.

This holds for all vectors ${\bf b}\in R^n$. In particular, it holds for the standard basis vectors ${\bf e_1}, {\bf e_2}\dots, {\bf e_n}$. Let us denote the solutions for these vectors by ${\bf x_1}, {\bf x_2},\dots,{\bf x_n}$, respectively.

Let $X$ be the $n\times n$-matrix with the vectors ${\bf x_1}, {\bf x_2},\dots,{\bf x_n}$ as columms, that is $X=[{\bf x_1}\, {\bf x_2}\dots{\bf x_n}]$. Also, notice that $[{\bf e_1}\,{\bf e_2}\dots{\bf e_n}]=I$.

It follows that $AX=I$.

Next, $B=BI=B(AX)=(BA)X=IX=X$, that is, $B=X$.
Hence, $AB=I$.

 

[mathwonk]

it is not true unless the matrices are square.