Rank of a matrix and its submatrices

[Grothard]

Homework Statement

Let A be a nonzero matrix of size n. Let a k*k submatrix of A be defined as a matrix obtained by deleting any n-k rows and n-k columns of A. Let m denote the largest integer such that some m*m submatrix has a nonzero determinant. Prove that rank(A) = k.

Now conversely suppose that rank(A) = m. Prove that some m*m submatrix has a nonzero determinant.

Homework Equations

Determinant formulas

The Attempt at a Solution

Not quite sure if I should proceed by examining the solution space of A or rather just do something clever with the determinants. I feel like there's a property of determinants that I'm missing that'd make this much easier.

 

[lippyka]

For the first part, I could assume the opposite is true and then try to derive a contradiction. Suppose rank(A) > k. Then by the rank-nullity theorem, there must be at least n-k linearly independent columns of A (for if there were fewer, the rank would be less than n-k). When we delete these n-k columns from A, the new matrix must have a nonzero determinant, since it is not the zero matrix. But then m = n-k > k, which is a contradiction. Now for the second part, suppose rank(A) = m. Then by the rank-nullity theorem, there must be exactly m linearly independent columns of A. Let the columns of A be denoted by c1, ..., cm. I think that from here, I need to do something clever with the determinant formulas. Let det(A) denote the determinant of A. Since c1, ..., cm are linearly independent, they form a basis of R^n. Then det(A) = det(c1, ..., cm), where ci denotes the column vector of A corresponding to the ith column. From here, I'm really not sure what to do. Any help would be greatly appreciated.