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

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    $\begingroup$ It’s a bijection iff the kernel is just the zero vector. $\endgroup$ – Louis Feb 8 '18 at 8:13
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    $\begingroup$ So, can we say that it is a direct consequence of the rank nullity theorem? $\endgroup$ – MasaJuno Feb 8 '18 at 8:16
  • $\begingroup$ Yes. That’s correct. $\endgroup$ – Louis Feb 8 '18 at 8:17
  • $\begingroup$ @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 $\endgroup$ – gimusi Feb 9 '18 at 23:46
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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 $A^{-1}= E_1 E_2...E_k I.$

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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.

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