Proof - Square Matrix has maximal rank if and only if it is invertible

Could someone help me with the proof that a square matrix has maximal rank if and only if it is invertible?

Thanks to everybody

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What is a quadratic matrix? – Qiaochu Yuan Feb 14 '11 at 12:43
@Qiaochu Yuan he obviously means square matrix – Listing Feb 14 '11 at 12:46
Yeah sorry for my english :) – markzzz Feb 14 '11 at 13:27
@user3123: I asked because it sounded like the OP could have been referring to a quadratic form rather than a matrix. – Qiaochu Yuan Feb 14 '11 at 14:01
please make your posts self-contained. Don't rely on the subject: put the entire information on the body. – Arturo Magidin Feb 14 '11 at 14:18

1 Answer

Suppose $A\in F^{n \times n}$. If A is invertible then there is a matrix B such that $AB=I$ so the standard basis $e_i$ (the columns of I) is in the image of A (these vectors are just the image Av where v are the columns of B) - this shows that $\dim (Im(A)) = n$.

On the other hand, if $\dim (Im (A))=n$ then for every i there is $v_i$ such that $A v_i = e_i$. Let B be the matrix with columns $v_i$ then $AB=I$ and A is invertible.

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Nice elementary proof. – Listing Feb 14 '11 at 12:36
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