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?
Best Answer
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 $$\text{det}A= (\text{det} P)^2\leq (|v_1|\cdot...\cdot|v_n|)^2=|v_1|^2\cdot...\cdot|v_n|^2=a_{11}\cdot ...\cdot a_{nn}. $$