= Solution
For one factor, write $\lambda\in\mathbb R^p$ for the loading vector and $v=\sigma^2>0$. Let $\lambda_t$ be its current value and $s=v+\|\lambda_t\|^2$. The E-step of the <expectation-maximization algorithm> uses the <conditional multivariate normal distribution> to obtain
$$
Z_i\mid Y_i,\lambda_t\sim N(m_i,u),\qquad
m_i=\frac{\lambda_t^TY_i}{s},\qquad u=\frac vs.
$$
Thus $\mathbb E[Z_i^2\mid Y_i]=u+m_i^2$. The terms of the expected complete-data <log-likelihood> that depend on a candidate $\lambda$ are
$$
Q(\lambda\mid\lambda_t)
=-\frac1{2v}\sum_{i=1}^n\left(\|Y_i\|^2-2m_i\lambda^TY_i+(u+m_i^2)\|\lambda\|^2\right)+\text{constant}.
$$
This is a strictly concave quadratic in $\lambda$, since $\sum_i(u+m_i^2)>0$. Its unique maximizer is
$$
\boxed{\lambda_{t+1}=\frac{\sum_iY_im_i}{\sum_i(u+m_i^2)}.}
$$
For an explicit expression involving only the current loading and data, put $S=n^{-1}\sum_iY_iY_i^T$. Substitution gives the <EM update for a single-factor Gaussian model>
$$
\boxed{\lambda_{t+1}=\frac{(v+\|\lambda_t\|^2)S\lambda_t}
{v(v+\|\lambda_t\|^2)+\lambda_t^TS\lambda_t}.}
$$
No empirical centering is applied because the model fixes the mean to zero. If initialized at zero, this update remains at zero; monotonicity alone does not guarantee that such a stationary point is a global maximum. This part is present in the PDF but missing from the TeX transcription.
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