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Write $a=(2,1{)}^T$, so the data map is $a^{T}u$. The Gaussian likelihood and prior distribution giveCompleting the square, or using Gaussian conjugacy for a normal linear model, identifies a multivariate normal distribution with precision matrix $C_{\delta }^{-1}=I_2+{\delta }^{-2}a{a}^T$. Since $a^{T}a=5$, its posterior mean and covariance matrix areThe parameters areFor example, $(I_2+{\delta }^{-2}a{a}^T)C_{\delta }=I_2$, which checks the square completion.

The orthogonal vectorsare eigenvectors of the covariance matrix. Their respective variances areThey are independent under the posterior distribution, by independence of uncorrelated jointly normal variables. Thus uncertainty is different in the two directions even for positive noise: the data constrain $e_{\parallel }$, whereas $e_{\perp }$ spans the null space of the data map.

As $\delta \to 0$, the posterior distribution converges to the law ofThis is supported on $2u_1+u_2=m$, and its posterior mean is the minimum-norm least-squares solution. The prior distribution still determines the distribution along the unobserved null space; it has not disappeared. This example of a Gaussian posterior in the zero-noise limit distinguishes exact recovery of an observed component from recovery of the whole unknown.