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Suppose I have two quadratic forms on $\mathbb R^n$, represented as symmetric matrices $A$ and $B$ on the usual basis. I am interested in approximating the function $x \mapsto \max(x^TAx, x^TBx)$ while remaining within the space of quadratic forms.

Is there a nice way to define a "maximum" operation on symmetric matrices, such that $C = \max(A,B)$ if $C$ is, in some sense, the "smallest" symmetric matrix satisfying $x^TCx \ge x^TAx$ and $x^TCx \ge x^TBx$ for all vectors $x$?

I've purposely left the notion of the "smallest" matrix $C$ undefined, as I'll accept any formalization that allows its solution to be elegantly expressed and/or easily computed. One possibility is minimizing the trace of $C$. Another, if we restrict ourselves to positive semidefinite matrices, is minimizing a convenient matrix norm.

In any case, $\max$ certainly must be commutative, and must satisfy $\max(A,A) = A$. Also, if $A$ and $B$ share the same eigenvectors, with eigenvalues $\lambda_i$ and $\mu_i$ respectively, then it seems natural that $\max(A,B)$ should also have eigenvectors the same, and eigenvalues $\max(\lambda_i,\mu_i)$. Beyond that, I can't really tell.

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    Rahul, your countere$x$ample is right. However, $y$ou have a great a$n$swer from Davide.2012-01-25

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Let $P$ an orthogonal matrix such that $P^T(B-A)P=\operatorname{diag}(\alpha_1,\ldots,\alpha_n)$. We define $|B-A|$ as the matrix such that $P^T|B-A|P=\operatorname{diag}(|\alpha_1|,\ldots,|\alpha_n|)$, namely $|B-A|=P\operatorname{diag}(|\alpha_1|,\ldots,|\alpha_n|)P^T$. We put $\max(A,B):=\frac 12\left(A+B+|A-B|\right)$. Then we have for a fixed $x\in\mathbb R^n$: \begin{align*} x^T\max(A,B)x-x^TAx&=\frac 12x^T(B-A+|A-B|)x\\ &=\frac 12x^T(P\operatorname{diag}(\alpha_1,\ldots,\alpha_n)P^T+P\operatorname{diag}(|\alpha_1|,\ldots,|\alpha_n|)P^T)x\\ &=\frac 12(P^Tx)^T(\operatorname{diag}(\alpha_1,\ldots,\alpha_n)+\operatorname{diag}(|\alpha_1|,\ldots,|\alpha_n|))P^Tx\\ &\geq 0 \end{align*} since $\operatorname{diag}(\alpha_1+|\alpha_1|,\ldots,\alpha_n+|\alpha_n|)$ is positive semidefinite. By the same way $x^T\max(A,B)x-x^TBx=\frac 12(P^Tx)^T(\operatorname{diag}(-\alpha_1,\ldots,-\alpha_n)+\operatorname{diag}(|\alpha_1|,\ldots,|\alpha_n|))P^Tx\geq 0.$

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    Any idea if this operation would be associative? It certainly doesn't look like it at first glance, but then neither does $\max(x,y) = \frac12(x + y + |x-y|)$ on the real numbers, yet it is nonetheless.2012-01-27