# Invertibility and rank

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

• It’s a bijection iff the kernel is just the zero vector. – Louis Feb 8 '18 at 8:13
• So, can we say that it is a direct consequence of the rank nullity theorem? – MasaJuno Feb 8 '18 at 8:16
• Yes. That’s correct. – Louis Feb 8 '18 at 8:17
• @MasaJuno Please, if you are ok, you can accept the answer and set it as solved. Thanks! cdn.sstatic.net/img/faq/faq-accept-answer.png – gimusi Feb 9 '18 at 23:46

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 $A^{-1}= E_1 E_2...E_k I.$
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.