1.) I think I can answer the case when the rows are linearly independent vectors:

Since the rows of the matrix $M$ are linearly independent, we cannot create an all $0$ row in the matrix therefore the $rank(M) = n$.

2.) How do you proceed if you want to deduce a similar result for the case when the columns are linearly independent?

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    $\begingroup$ Can you tell me the definition of the rank of the matrix ? What is the relation between rank of $A$ and $A^t$ $\endgroup$ – Chiranjeev_Kumar Jul 13 '15 at 10:26
  • $\begingroup$ Well I wasn't aware of this definition since there is a different one in my textbook but from Wikipedia: "The rank of a matrix $A$ is the dimension of the vector space generated by its columns." And since the columns are linearly independent, the dimensions of the vector space generated by them is equal to their number, that is $n$. Is that a correct line of reasoning? Also the rank of ($A$) = rank of ($A^t$), which could be used alternatively, to deduce it from the result of 1.). Thanks. $\endgroup$ – pseudomarvin Jul 13 '15 at 10:40

TIP: $col(A)=row(A^T)$ and $rank(A^T)=rank(A)$

You should study the Fundamental Theorem for invertible matrices for proofs.

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  • $\begingroup$ Thanks I was not aware of that. $\endgroup$ – pseudomarvin Jul 13 '15 at 10:42

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