<p>While there are $n$ players, a player's strategy is not single dimensional.  The total dimension of the strategies is $m$.  So the vector of strategies $s$ is $m$ dimensional.  $g(s,z)$ is also $m$ dimensional, which is why the Jacobian is $m \times m$</p>

CC BY-SA 3.0
2016-09-27T12:22:51.083Z
2016-09-27T12:22:51.083Z
user10521
<p>Rosen's paper (<a href="http://www-inst.eecs.berkeley.edu/~ee228a/fa03/228A03/papers/rosen.pdf" rel="nofollow noreferrer">J. B. Rosen. Existence and uniqueness of equilibrium points for concave n-person games. Econometrica, 33(3):520–534, 1965</a>) presents a condition for the uniqueness of the Nash Equilibrium in $n$ players game. The setup is described in simple details by <a href="https://economics.stackexchange.com/questions/11217/rosens-diagonal-strict-concavity-condition">this other economics exchange post</a>. My curiosity is why is the Jacobian of $g(\mathbf{s},\mathbf{z})$, where $g(\mathbf{s},\mathbf{z})$ is the pseudogradient of the function $\sigma$, a square matrix?  If $s \in \mathbb{R}^m$ and there are $n$ players, it seems to me that the matrix would be $n \times m$ instead of $m \times m$. I have left out many of the details of the problem set up as they are covered in the other post; however, I can restate the set up or provide additional information if necessary.</p>

<game-theory><nash-equilibrium>
Rosen's uniqueness theorem: Why is the Jacobian Square?
eddymac
# Rosen's uniqueness theorem: Why is the Jacobian Square?

(Asked by: username_0 on 2016-08-11)

Rosen's paper ([J. B. Rosen. Existence and uniqueness of equilibrium points for concave n-person games. Econometrica, 33(3):520–534, 1965](http://www-inst.eecs.berkeley.edu/~ee228a/fa03/228A03/papers/rosen.pdf)) presents a condition for the uniqueness of the Nash Equilibrium in $n$ players game. The setup is described in simple details by [this other economics exchange post](https://economics.stackexchange.com/questions/11217/rosens-diagonal-strict-concavity-condition). My curiosity is why is the Jacobian of $g(\mathbf{s},\mathbf{z})$, where $g(\mathbf{s},\mathbf{z})$ is the pseudogradient of the function $\sigma$, a square matrix? If $s \in \mathbb{R}^m$ and there are $n$ players, it seems to me that the matrix would be $n \times m$ instead of $m \times m$. I have left out many of the details of the problem set up as they are covered in the other post; however, I can restate the set up or provide additional information if necessary.

**Tags:** game-theory, nash-equilibrium

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## 1 answers

### Answer by username_1 on 2016-09-27. Score: 3

While there are $n$ players, a player's strategy is not single dimensional. The total dimension of the strategies is $m$. So the vector of strategies $s$ is $m$ dimensional. $g(s,z)$ is also $m$ dimensional, which is why the Jacobian is $m \times m$


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