 Solution

Use unrestricted long and short portfolios, as required by the usual arbitrage pricing theory argument. A vector $v\in {\mathbb{R}}^n$ of initial monetary positions with $1^{T}v=0$ costs zero. Its terminal financial payoff, in the exact one-factor model, isIf also $v^{T}b=0$, the financial payoff is the constant $v^{T}a$. A nonzero value would be an arbitrage after choosing the sign of $v$. Absence of arbitrage thus says that $a$ annihilates $\ker [1b{]}^T$. By finite-dimensional linear algebra,Write $a={\lambda }_{0}1+\kappa b$. Taking expectations, assuming the factor has finite mean, provesThis is exact factor pricing without a traded risk-free asset. The PDF prints an asset-indexed ${\lambda }_i$; the stronger valid result has the same ${\lambda }_1$ for every asset, so it also satisfies the printed relation by taking all its ${\lambda }_i$ equal. If every loading is equal, the spanning vectors are dependent and the coefficients need not be unique. For a nondegenerate factor and a zero-exposure unit-cost portfolio, ${\lambda }_0$ is that portfolio's certain return; if a risk-free asset is explicitly traded, law of one price makes it the risk-free asset's return. No equilibrium preferences are needed for this argument.