# Prove that the rank of a matrix is the number of non-zero rows of its row-reduced form

As a reference to this and this and similarly many sources, it has been sait that the rank of a matrix, say $A$, is the number of non-zero rows of the row-reduced form of $A$.

However, why is this the case ? How can we prove it ?

I mean in the second question that I have linked, the answerer says the non-zero row form a basis etc. which I think does not connect to the rank of matrix.

## 1 Answer

If you agree that row operations don't affect the rank of the matrix, then this follows from one of the definitions of the rank of a matrix, which is the dimension of the row space of the matrix.

The rows of that matrix span a vector space, and the dimension of that vector space is equal to the number of non-zero vectors that span it.

The dimensionality is clearly not greater than the number of rows. But its also not less: there are no redundant vectors, because every row contains an entry where it is the only non-zero element in its column, so it is impossible to build that row from a linear combination of the other rows.

• Intuitively, I can see that the row operations should not affect the rank of a matrix, but mathematically I can not prove it.
– Our
Oct 30, 2017 at 7:49
• Oct 30, 2017 at 7:57