# Question about the relationship between Weak Axiom and Slutsky Matrix

We know that if a differentiable Walrasian demand function $$x(p,w)$$ satisfies Walras' law ($$p^Tx=w$$), homogeneity of degree zero ($$x(\alpha p,\alpha w)=x(p,w)$$), and the weak axiom of revealed preference, then at any $$(p,w)$$, the Slutsky matrix $$\begin{equation} S(p,w)=D_px(p,w)+D_wx(p,w)x(p,w)^T \end{equation}$$ is negative semidefinite.

My question is: if $$S(p,w)$$ is negative semidefinite, then what can we say about the demand function $$x(p,w)$$? Can we conclude that $$x(p,w)$$ satisfies weak axiom?

## 1 Answer

It is almost true.

There are examples of demand that have a negative definite Slutsky matrix but fails the Weak Axiom.

However, if we ask that $$v \cdot S(p,w) v <0$$ whenever $$v \not = \alpha p$$ for any scalar $$\alpha$$ (i.e. $$S$$ is negative definite for all vectors except those proportional to price), then the Weak Axiom holds.

• I see. Is that a stronger condition than negaive semidifinite? Because it could be the case that vS(p,w)v=0 for some v that is not porpotional to p when S(p,w) is negaive semidifinite. – Rui Nov 13 '19 at 17:33
• @Huaixin You're correct. If we weaken negative definite to negative semidefinite, again we can obtain an example that fails the Weak Axiom. It does however satisfy a weaker Weak Axiom. This result is by Kihlstrom, Mas-Colell and Sonnenschien (1976), they call it WWA (if you're interested). – Walrasian Auctioneer Nov 13 '19 at 18:15
• get it, thanks for the precise answer. – Rui Nov 16 '19 at 11:14