Question

Show that if A_(n×n) is invertible then the inverse is unique

Answer 1

Since A is a $n\times n$ matrix, the linear transformation $T:x\rightarrow Ax$ is one-to-one

$\Rightarrow$  linear transformation $T:x\rightarrow Ax$ is invertible

$\Rightarrow$  A is invertible.

Suppose that $Ax=b$ and $Av=Au=0,v≠u.$

Then $A(x+v)=A(x+u)=b$ , whereby $x+v≠x+u.$

$\therefore$ $Ax=b$ for some unique $x$  

$\Rightarrow$ either there is no $v$ such that $Av=0$ or there is a unique $v$ such that $Av=0$ .

Since $A0=0$ , we conclude that there is a unique $v$ such that $Av=0$ , and thus $T$ is one-to-one.

Therefore, A is invertible.