# Inequality w.r.t determinant of a non-negative definite matrix.

I am reading a paper where the author mentioned the following property without proof. Neither can I prove it nor can I find the proof in various textbooks.

For any non-negative definite (i.e. positive semidefinite) matrix $$A$$, $$\text{det}(A) \leq \Pi_iA_{ii}$$

How to prove it?

First, there is the fact that for any $$n\times n$$ matrix $$P$$, if one writes $$P=[v_1, v_2, ..., v_n]$$, where $$v_k$$ is the $$k$$-th column of $$P$$, then $$|\text{det}P|\leq |v_1|\cdot|v_2|\cdot ...\cdot|v_n|;$$ here $$|v_k|$$ is the length of the vector $$v_k$$. This actually reaches the equal sign if $$v_1, ...,v_n$$ are mutually perpendicular; in that case $$[v_1/|v_1|, ..., v_n/|v_n|]$$ is an orthogonal matrix, thus has determinant $$\pm 1$$, on the other hand its determinant is also $$\text{det} P/(|v_1|\cdot ...\cdot |v_n|)$$. For the general case, we do a procedure similar to Gram-Schmidt: replace $$v_2$$ by $$v_2'$$, which is $$v_2$$ minus its projection to $$v_1$$; that will reduce $$|v_2|$$ but not changing $$\text{det}P$$. Replace $$v_3$$ by $$v_3'$$, which is $$v_3$$ minus its projection to the $$v_1, v_2$$ plane, ... and in the end get $$[v_1, v_2', v_3'...]$$ with orthogonal columns. Thus $$|\text{det} P|= |v_1|\cdot |v_2'|\cdot..\cdot|v_n'|\leq |v_1|\cdot...\cdot|v_n|$$.
Now $$A$$ is nonnegative, it is "well-known" that we can find orthogonal matrix $$Q$$ so that $$A=Q^T\Lambda Q$$, where $$\Lambda$$ is diagonal with $$\geq 0$$ entries. Write $$\sqrt \Lambda$$ be the diagonal matrix with $$(\sqrt\Lambda)^2=\Lambda$$, thus $$A=Q^T\sqrt\Lambda\sqrt\Lambda Q=(Q^T\sqrt\Lambda Q)(Q^T\sqrt\Lambda Q)$$. in conclusion, we can write $$A=P^TP$$, with $$P=Q^T\sqrt\Lambda Q$$.
Let $$P=[v_1, ...,v_n]$$. Then from $$A=P^TP$$ we see $$a_{11}=|v_1|^2$$, ... , $$a_{nn}=|v_n|^2$$. Thus $$\text{det}A= (\text{det} P)^2\leq (|v_1|\cdot...\cdot|v_n|)^2=|v_1|^2\cdot...\cdot|v_n|^2=a_{11}\cdot ...\cdot a_{nn}.$$
• Thanks, your proof is amazing. The first fact is also known as Hadamard's inequality. If you assign $P = \sqrt{\Lambda}Q$, then the expression $A = P^TP$ still holds true, and it's simpler. – Ly Minh Hoang Apr 22 at 4:56