Finding the largest rank of a general matrix

Question

Let's say $ A $ is a 71 x 61 matrix. What is the largest rank of A? I am wondering if my reasoning is correct: Since there 71 rows but only 61 columns, then there will be 10 zero rows in the reduced row echelon form, therefore the largest rank would be 71 - 10 = 61.

Answer 1

Yes, the answer is correct.

There will be at least $10$ rows of zeros in RREF and the rank is the largest when we have fewest number of rows.

We know that if $A \in \mathbb{R}^{m \times n}$, then $\operatorname{rank}(A) \le \min(m,n)$ and for our problem $$A=\begin{bmatrix} I_{61} \\ O_{10 \times 61}\end{bmatrix}$$ would attain $\min(m,n)$.

Answer 2

Another way you could word it: the rank of a matrix is the dimension of the columnspace, and this is the same as the dimension of the rowspace. The rowspace is the number of linearly independent rows of the matrix, and since we have at most $61$ rows, the rank is at most $61$. Note that even though there are $71>61$ columns, the number of linearly independent columns cannot exceed $61$ since then the dimension of columnspace $=$ dimension of rowspace rule would be broken.