Inequality involving Hadamard's inequality

Question

Let $A$ be matrix in $\mathbb{R^{m \times n}}$. Let $A$ and $B$ be quadratic submatrices of $M$ such that $\det(A)< \det(B)$.

Does this imply $\prod_{i=1}^n \|A^i\| < \prod_{i=1}^n \|B^i\|$ (using Hadamard's inequality), where $A^i$ and $B^i$ denotes the columns of $A$ and $B$, respectively.

Answer 1

If I have understood the question correctly, this is a counterexample:

Let $$A=\left(\begin{matrix} 10 & 0 & 0\\ 0 & 10 & 10\\ 0 & 0 & 0 \end{matrix}\right) $$

$$B=\left(\begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{matrix}\right) $$

Then $0=\det(A)<\det(B)=1$, but $\prod_{i=1}^3 \|a_i\| \ge \prod_{i=1}^3 \|b_i\|$.

Answer 2

I don't think that being submatrices of a given matrix is actually necessary. Indeed, given any two $n\times n$ matrices you can always concatenate them to form a $n\times 2n$ matrix of witch (by construction) they both are square submatrices).

As a counterexample to your question, consider the matrices $$ M=\begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & \frac{1}{2} \end{pmatrix} $$ $$ A=\begin{pmatrix} 1 & 2 \\ 0 & \frac{1}{2} \end{pmatrix} $$

$$ B=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$ Then $\det(A)=\frac{1}{2}<1=\det(B)$ but $$ \prod_{i=1}^2 \|A^i\| = \sqrt{4+\frac{1}{4}} = \frac{\sqrt{17}}{2} > 1 = \prod_{i=1}^2 \|B^i\| $$