Invertibility and rank

Question

How do you formally prove that a matrix A is invertible if and only if it has full rank, without using determinants?

Answer 1

If a matrix $A$ has full rank the row reduced echelon form of $A$ will be the identity matrix.

We can find the inverse of $A$, multiplying I by the elementary row operations.

Note that if $E_1 E_2...E_k A= I$, then $A^{-1}= E_1 E_2...E_k I.$

Answer 2

If A is not full rank let consider $x\in ker(A)$ then $Ax=0$ and $A(2x)=0$ thus it is not injective and therefore not invertible.

If A is full rank it is surjective (column space span $\mathbb{R^n}$) and injective ($x\neq y \implies Ax\neq Ay$) therefore it is invertible.

If A is invertible $ker(A)=\emptyset$ then A is full rank.