$\begingroup$
$\endgroup$
4
How do you formally prove that a matrix A is invertible if and only if it has full rank, without using determinants?
2 Answers
$\begingroup$
$\endgroup$
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 $A^{-1}= E_1 E_2...E_k I.$
$\begingroup$
$\endgroup$
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.
