Dear Linear Algebra experts,
According to Schur Product Theorem, if both $A \succ 0$ and $B \succ 0$ are positive definite, then the Hadamard product of $(A \circ B) \succ 0$ is also positive definite.
However, can Hadamard product $(A \circ B) \succ 0$ be positive definite,
if $A \succeq 0$ (positive semi-definite) and $B \succ 0$ (positive definite)?
If yes, can you prove it? Thank you so much.
Best Answer
When $A\succeq0$ and $B\succ0$, the Hadamard product $A\circ B$ is positive definite if and only if $A$ has not any zero rows.
Proof 1. $A\circ B$ is positive semidefinite. It is positive definite if and only if it is non-singular. By Oppenheim's inequality, whenever $A$ and $B$ are positive semidefinite matrices, $\det(B)\prod_ia_{ii}\le\det(A\circ B)$. So, in your case, if $A\circ B$ is singular, some $a_{ii}$ must be zero and hence the $i$-th row of $A$ is zero. Conversely, if $A$ has a zero row, clearly $A\circ B$ is singular.
Proof 2. One of the usual proofs of Schur product theorem (such as the one on Wikipedia or the one in Horn and Johnson's Matrix Analysis) can be modified to prove our necessary and sufficient condition for the positive definiteness of $A\circ B$.
Suppose the positive semidefinite matrix $A$ has rank $k$. By spectral decomposition, we may write $A=\sum_{i=1}^kv_iv_i^\ast$ where $\{v_1,\ldots,v_k\}$ is a set of mutually orthogonal eigenvectors corresponding to the positive eigenvalues of $A$. Note that $$ x^\ast(A\circ B)x =\sum_{i=1}^k x^\ast\left((v_iv_i^\ast)\circ B\right)x =\sum_{i=1}^k x^\ast\left(\operatorname{diag}(v_i)B\operatorname{diag}(v_i^\ast)\right)x =\sum_{i=1}^k (x\circ\bar{v_i})^\ast B(x\circ\bar{v_i}). $$ Since $B$ is positive definite, $x^\ast(A\circ B)x=0$ if and only if $x\circ\bar{v}_i=0$ for every $i$. The latter statement is in turn equivalent to $\operatorname{diag}(x)\overline{V}=0$, or equivalently, $\operatorname{diag}(\bar{x})V=0$, where $V$ is the augmented matrix $[v_1|v_2|\cdots|v_k]$. It follows that $A\circ B$ is positive definite if and only if $V$ has not any zero row, i.e. if and only if $A$ has not any zero row.