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This is an exercise in linear algebra:

Let $A=(a_{ij})$ and $B=(b_{ij})$ both be positive semidefinite $n \times n$ (real) matrices. Prove that $$ \sum_{i=1}^n \sum_{j=1}^n a_{ij}b_{ij} \geq 0\,. $$

This is a known fact when $A$ and $B$ are symmetric. For example, here is an analogous fact for positive definite symmetric matrices.

darij grinberg
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Dr-MS
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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Sep 02 '19 at 16:59
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    Hint: the quantity you are interested in is the trace of $AB$. Show that $AB$ is a sum pf $M_{x,z}=xx^Tzz^T$ for vectors $x,z$. Note that the trace of $M_{x,z}$ is the trace of $(x^Tz)(z^Tx)$ where the two matrix products between parentheses are $1 \times 1$ matrices, the transpose one of the other. So they are equal and the trace is the square of $x \cdot z$. – Aphelli Sep 02 '19 at 17:02

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Here is a related (counter)example if we remove the "symmetric" assumptions on $A$ and $B$: \begin{align} A &= \left[\begin{array}{cc} 1 & -1 \\ 3 & 1\end{array}\right]\\ B &= \left[\begin{array}{cc} 1 & 3 \\ -1 & 1\end{array}\right] \end{align} It can be shown that $x^TAx \geq 0$ and $x^TBx \geq 0$ for all $x \in \mathbb{R}^2$, but $$ \sum_{ij} a_{ij}b_{ij} = (1)(1) + (-1)(3) + (3)(-1) + (1)(1) = -4$$

On the other hand, if $A, B$ are real-valued $n \times n$ matrices that satisfy $x^TAx \geq 0$ and $x^TBx \geq 0$ for all $x \in \mathbb{R}^n$, and if $A$ is symmetric, then the situation is different: We can say $A = R^TR$ for some square matrix $R$ and so $$ \sum_{i=1}^n\sum_{j=1}^n a_{ij}b_{ij}=trace(A^TB) = trace(R^TR B)= ...$$ so you can use properties of the trace to eventually prove $trace(A^TB)\geq 0$. In particular, the desired non-negativity result holds as long as at least one of the matrices $A, B$ is also assumed to be symmetric.

Michael
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