# Proof that determinant rank equals row/column rank

Let $A$ be a $m \times n$ matrix with entries from some field $F$. Define the determinant rank of $A$ to be the largest possible size of a nonzero minor, i.e. the size of the largest invertible square submatrix of $A$. It is true that the determinant rank is equal to the rank of a matrix, which we define to be the dimension of the row/column space.

It's not difficult to see that $\text{rank} \geq \text{determinant rank}$. If some submatrix of $A$ is invertible, then its columns/rows are linearly indepedent, which implies that the corresponding rows/columns of $A$ are also linearly indepedent.

Is there a nice proof for the converse?

## 1 Answer

If the matrix $A$ has rank $k$, then it has $k$ linearly independent lines. Those form an $k\times n$ submatrix, which of course also has rank $k$. But if it has rank $k$, then it has $k$ linearly independent columns. Those form a $k\times k$ submatrix of $A$, which of course also has rank $k$. But a $k\times k$ submatrix with rank $k$ is a full-rank square matrix, therefore invertible, thus is has a non-zero determinant. And therefore the determinant rank has to be at least $k$.

• Nice! Now that I see this proof, I can't believe I missed it.. – spin Aug 27 '12 at 15:33
• Why do you say "And therefore the determinant rank has to be at least k"? Doesn't your answer show that it will be exactly equal? Why "at least"? – agdhruv May 6 '18 at 13:05
• @agdhruv: Because that is what is asked for in the question. – spin Jun 4 '18 at 5:57