Let A and B be $n \times n$ real symmetric matrix. Suppose A is positive definite and denote its smallest eigenvalue as $\lambda_{\min}(A)>0$. All elements of B are positive and bounded, i.e. $0\le a <B_{ij}<b, B_{ij}=B_{ji}$. Denote the hadamard product of two matrices as $C=A\circ B$.

Here are two questions:

  • Can we know whether $B$ is positive definite or not? (constant a and b can be chosen arbitrarily to satisfy the requirements)
  • Can we lower bound the smallest eigenvalue of C?

P.S. I'm encountered with this question in an engineering computation problem. The motivation is to control $-v^{\top} C v$ or $\|v^{\top} C\|$ and we can make some restrictions on B, for example, bounding the entries of $B$ or set $n$ as large as we need. I have seen some results in On majorization and Schur products while $B$ has to be positive definite, which may not be guaranteed in my question.

  • $\begingroup$ I don't think that there is any difference between saying "the entries of $B$ are positive" and "the entries of $B$ are positive and bounded" unless $B$ has infinitely many entries. $\endgroup$ Commented Jan 11, 2022 at 15:53
  • $\begingroup$ The answer to your first question is no. For example, if we have $a < c < d < b$, then the matrix $$ B = \pmatrix{c&d\\d&c} $$ will never be positive definite. $\endgroup$ Commented Jan 11, 2022 at 15:54
  • $\begingroup$ One answer to your second question is that $\lambda_{\min}(C) \geq \lambda_{\min}(A)\lambda_\min(B)$. $\endgroup$ Commented Jan 11, 2022 at 15:56
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    $\begingroup$ Please note, however, that your question does not meet the standards for this site in its current form. Askers are expected to provide context for their questions, as is explained here. For example, it would be helpful if you could edit your post to address the following. Where did you encounter this problem? What are your thoughts on the problem? What have you tried so far? $\endgroup$ Commented Jan 11, 2022 at 15:59
  • $\begingroup$ It would also help to know a bit about your mathematical background; in particular, do you know what the Kronecker product is? Are you familiar the Cauchy interlacing theorem? $\endgroup$ Commented Jan 11, 2022 at 16:00

1 Answer 1


As discussed in the comments, the positivity of the entries of $B$ does not ensure that $B$ is positive definite. I further suspect that there's no good way to take advantage of the positivity of the entries in checking whether $B$ is positive definite.

Here's a proof of a corrected version of the result that I referred to

Claim: If $A$ is positive definite and $B$ is symmetric with at least one negative eigenvalue, then $\lambda_{\min}(A \circ B) \geq \lambda_\max(A)\lambda_\min(B)$.

Proof: We note that $A \circ B$ is a principal submatrix of the Kronecker product $A \otimes B$. The eigenvalues of $A \otimes B$ are of the form $\lambda = \lambda_i(A) \lambda_j(B)$ for $1 \leq i,j \leq n$. By the Cauchy interlacing inequality, it follows that $$ \lambda_{\max}(A)\lambda_\min(B) = \lambda_{\min}(A \otimes B) \leq \lambda_{\max}(A \circ B). \qquad \square $$

Alternative proof: Let $\mu = \lambda_{\min}(B)$, so that $B - \mu I$ is positive semidefinite. It follows that $A \circ (B - \mu I)$ is positive semidefinite, so that its minimal eigenvalue is at least $0$. Thus, $$ \lambda_\min(A \circ B) = \lambda_\min(A \circ (B - \mu I) + \mu A) \\ \geq \lambda_\min(A \circ (B - \mu I) + \lambda_\min(\mu A) \\ = \lambda_\min(A \circ (B - \mu I) - |\mu|\lambda_\max(A) \\ \geq 0- |\mu|\lambda_\min(A) = \lambda_\max(A)\lambda_\min(B). $$

  • $\begingroup$ Thank you! So there's no way to know whether $\mu$ is positive or the absolute value under current conditions, right? Is there any possibility to upper bound $v^{\top} C v$ or $\|v^{\top} C\|$ without knowing these? $\endgroup$
    – Woody
    Commented Jan 12, 2022 at 3:59
  • $\begingroup$ Well, in general we can say that $$ \lambda_\min(A\circ B) \geq \min\{\lambda_\min(A)\lambda_\min(B),\lambda_\max(A)\lambda_\min(B)\}. $$ Perhaps that's enough for your purposes. $\endgroup$ Commented Jan 12, 2022 at 4:00

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