Problem
Show $\vert \mathrm{det}(A) \vert \le \prod_{j=1}^n \Vert a_j \Vert_2$, where $a_j$ denotes the $j$th column of $A$, which is $n \times n$ matrix.
Try
When $A$ is singular, the result is trivial.
Otherwise, let us consider the matrix $U$, whose $j$th column, $u_j$ has unit length, i.e. $\Vert u_j \Vert_2 = 1, \forall j$. Then
$$ \det(U) \le 1 $$
where equality holds iff $u_j$'s are orthogonal.
I think I need to show
$$ \mathrm{det}(A) = \left( \prod_{j=1}^n \Vert a_j \Vert_2 \right) \det(U) \ \ \ (\ast) $$
because if it is true then
$$ \left( \prod_{j=1}^n \Vert a_j \Vert_2 \right) \det(U) \le \prod_{j=1}^n \Vert a_j \Vert_2 $$
But I'm stuck at showing the above $(\ast)$.
Any help will be appreciated.
