= 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 $\mathbf1^Tv=0$ costs zero. Its terminal <financial payoff>, in the exact one-factor model, is
$$
v^T(\mathbf1+r)=v^Ta+(v^Tb)f.
$$
If also $v^Tb=0$, the <financial payoff> is the constant $v^Ta$. A nonzero value would be an <arbitrage> after choosing the sign of $v$. Absence of <arbitrage> thus says that $a$ annihilates $\ker[\mathbf1\ b]^T$. By finite-dimensional <linear algebra>,
$$
a\in(\ker[\mathbf1\ b]^T)^\perp=\operatorname{span}\{\mathbf1,b\}.
$$
Write $a=\lambda_0\mathbf1+\kappa b$. Taking <expectations>, assuming the factor has finite mean, proves
$$
\boxed{\bar r_i=\lambda_0+b_i\lambda_1,\qquad\lambda_1=\kappa+\mathbb Ef.}
$$
This 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.
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