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I apologize in advance if this is a very easy exercise. Let $k$ be a field and let $A$ be an $n \times n$ matrix over $k$ such that $A$ is a row-reduced echelon matrix without zero rows. Prove that $A$ is the $n\times n$ identity matrix.

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  • $\begingroup$ Doesn't this just follow from the definition of row-reduced echelon form? What is your definition of RREF? $\endgroup$ – user137731 May 12 '15 at 15:12
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Every row contains a non-zero entry, that is first. Thus, there are at least $n$ entries, one in each row.

However, from the definition of row-reduced echelon matrix, the first non-zero entry in each row has no non-zero entries in that column. But there are only $n$ columns, so we have only $n$ entries, such that each row and each column contains one entry only. This makes it a diagonal matrix.

However, the first non-zero entry, and thus the only one has got to be $1$. This is a diagonal matrix with $1$ as entries. Thus, it is the identity matrix.

This is a non-rigorous proof, but I have shown you the idea so you can formulate it. Also, do excuse my English as it is not my first language.

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