MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there an inequality of the form $|\det(A)| \geq F(v_1, \ldots, v_n)$ for a real $n\times n$-matrix $A$ with columns $v_i$, $F \geq 0$?

share|cite|improve this question
for example, $\detA)\geq -\prod \|v_i\|$ (yes, it is Hadamard inequality itself, not reverse). Or do you mean $|\det A|$ and non-negative $F$? – Fedor Petrov Feb 6 '11 at 17:04
Yes, I stated it incorrectly. – Timo Keller Feb 6 '11 at 17:07
The question is still vague. You could just define F to be the function that is identically 0. Obviously that's not what you are looking for, so please be more careful in telling us what properties or shape you are thinking about for your lower bound function F. – KConrad Feb 6 '11 at 17:40
I even found the best inequality of this type: $F=|det A|$! – GH from MO Feb 6 '11 at 18:03

Agreeing with commenters that the question is a bit vague, I will still go ahead and try to anticipate the intent.

A feature of Hadamard's inequality is that the columns are considered independently, another one is that the columns only enter by their norm. If one wants to preserve this, there is no way to obtain a nontrivial lower bound.

For one thing, without considering 'interactions' of the columns, there is no way to guarantee that they are linearly independent and thus the determinant nonzero.

Moreover, the fact that replacing, say, $v_1$ by $v_1 + \lambda v_2$ for any $\lambda$, leaves the determinant unchanged (allowing arbitrarily 'large' columns in a matrix with a fixed determinant), further illustrates that 'interactions' between the $v_i$ need to be considered.

share|cite|improve this answer

Not sure how helpful this is, but in Tao and Vu give the formula: $$ |\det(A)| = \prod_{j=2}^{n} dist(v_j, W_{j-1}) $$ where $v_j$ is the j-th column of $A$ and $W_{j}$ the subspace of $\mathbb{R}^n$ spanned by $v_1, \dots, v_j$.

Estimating these distances, they then conclude lower bounds on the determinant. Of course, this is an equality! But the advantage is that it contains geometric quantities, which one can estimate.

share|cite|improve this answer
Your formula doesn't look right; all the $W_j$ are equal. – dvitek Feb 6 '11 at 21:25
$W_j$ is meant to be the subspace spanned by $v_1,\dots,v_j$. – Gerry Myerson Feb 6 '11 at 23:31

Reverse inequality occur in particular for so-called homogeneous hyperbolic polynomials, when restricting to the forward cone $\Gamma$. For instance (G{\aa}rding), if $P$ is hyperbolic of degree $d$, then $P^{1/d}$ is concave over $\Gamma$. Actually, $$(P(v_1)\cdots P(v_d))^{1/d}\ge\phi(v_1,\ldots,v_d)\qquad \forall v_1,\ldots,v_d\in\Gamma$$ where $\phi$ is the multilinear form associated with $P$. Here are examples. If $Q$ is quadratic of signature $(1,n)$, then it satisfies the reverse Cauchy-Schwarz inequality in the forward cone: $$\sqrt{Q(v)Q(w)}\ge\phi(x,y), \qquad \forall v_1,v_2\in\Gamma$$ The determinant, restricted to Hermitian matrices, is hyperbolic and its forward cone is that of positive (semi-)definite matrices. You therefore have the concavity mentionned above (related to Brunn-Minkowski inequality) and the inequality $$(\det H_1\cdots\det H_n)^{1/n}\ge\Delta(H_1,\ldots,H_n),$$ where $\Delta$ is the so-called mixed determinant.

But this might not be what you are looking for ...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.