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Let $M(\mathbb{R})$ be the set of all matrices (of any size) over $\mathbb{R}$. Let $v : M(\mathbb{R}) \rightarrow \mathbb{R}$ be a function which satisfies the following 5 properties:

  1. If $\mathbf{a}$ is an $n \times 1$ matrix, then $v(\mathbf{a}) = \max \mathbf{a}$ where $\max \mathbf{a}$ denotes the maximum entry in $\mathbf{a}$.

  2. $v(A^T) = -v(-A)$.

  3. If $\mathbf{a}$ is a convex combination of $\mathbf{a}_1,\dots,\mathbf{a}_k$, then $$v([\mathbf{a} ~ \mathbf{a}_1 ~ \cdots ~ \mathbf{a}_k]) = v([\mathbf{a}_1 ~ \cdots ~ \mathbf{a}_k])$$ where $[\mathbf{a}_1 ~ \cdots ~ \mathbf{a}_k]$ denotes the matrix with columns $\mathbf{a}_1,\dots,\mathbf{a}_k$.

  4. If $a \geq 0$ and $B$ is the matrix with the same dimensions as a $A$ but with every entry equal to $b \in \mathbb{R}$, then $$v(a A + B) = a v(A) + b.$$

  5. $v$ is continuous. That is, if $A, B$ are two matrices of the same dimensions, then for every $\epsilon > 0$ there is a $\delta > 0$ such that $$| v(A + \delta B) - v(A) | < \epsilon.$$

The value of a two person zero sum matrix game is one example of such a function. The question is this: can you provide a second example of such a function $v$?

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    $\begingroup$ Interesting question. Do you know of any other properties of the value which follow from these axioms? For example, do these axioms imply that $v$ is invariant under permutations of the rows and columns? $\endgroup$
    – Noah Stein
    Apr 22, 2013 at 20:33
  • $\begingroup$ Yes, I was thinking about adding the property concerning column order to the list. Maybe I should have. The "meta" question: The value of a matrix game can be defined as the maximum constant $v$ for which there is a probability vector $\mathbf{x}$ such that $\mathbf{x}^T A \geq v \mathbf{1}^T$. So the definition of value is connected to the definition of an optimal row or column strategy. However, I am looking for an alternative definition of the value of a matrix game which does not use mixed strategies. Can this be done with a sequence of axioms like those described above? $\endgroup$ Apr 22, 2013 at 22:26

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