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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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    please make your posts self-contained. Don't rely on the subject: put the entire information on the body.2011-02-14

1 Answers 1

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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.2011-02-14