Under the assumption that the data is generated from a linear function where each observation exhibts i.i.d. Gaussian noise (i.e., each noise variable \(\nu_i\) is sampled independently from the same/identical Gaussian distribution), the negative log-likelihood is proportional to the squared sum of the residuals. Thus, minimizing the negative log-likelihood (maximizing the log-likelihood) leads to the same solution as the least squares method.
Given a dataset \(\mathcal{D} = \{(\textbf{x}_1, y_1), \dots, (\textbf{x}_N, y_N)\}\) and the model assumption:
\[ y_i = \textbf{w}^{\text{T}} \textbf{x}_i + \beta + \nu_i \quad \quad \nu \sim \mathcal{N}(0, \sigma^2) \]
the parameters of the model are \(\boldsymbol{\theta} = (\textbf{w}, \beta, \sigma^{2})\). In terms of maximum likelihood, we want to maximize:
\[ \begin{align} p(\mathcal{D} | \boldsymbol{\theta}) &= p (\textbf{x}_1) \cdot p(y_1| \textbf{x}_1; \boldsymbol{\theta}) \cdot ... \cdot p(\textbf{x}_N| \textbf{x}_1, \dots, \textbf{x}_{N-1}) \cdot p(y_N| \textbf{x}_N; \boldsymbol{\theta})\\&= p_X (\textbf{x}_1, \dots, \textbf{x}_N) \prod_{i=1}^N p(y_i | \textbf{x}_i; \boldsymbol{\theta}) \end{align} \]
where the density \(p_X\) is unknown and independent of \(\boldsymbol{\theta}\). Given our model assumption, we can write
\[ p(y_i|\textbf{x}_i, \boldsymbol{\theta}) \sim \mathcal{N} \left(\textbf{w}^{\text{T}} \textbf{x}_i + \beta, \sigma^2 \right) = \frac {1}{\sqrt{2\pi \sigma^2}} \exp\left( - \frac {(y_i - \textbf{w}^{\text{T}} \textbf{x}_i - \beta)^2} {\sigma^2} \right) \]
Thus, we get for the negative log-likelihood
\[ -\ln p(\mathcal{D}|\theta) = \sum_{i=1}^N \frac {1}{2} \log \big(2 \pi \sigma^2 \big) + \frac {1}{2\sigma^2} \big(y_i - \textbf{w}^{\text{T}} \textbf{x}_i - \beta \big)^2 \propto \sum_{i=1}^N \big(y_i - \textbf{w}^{\text{T}} \textbf{x}_i - \beta \big)^2 \]
As stated above, minimizing the negative log-likelihood (left side) leads to the same solution as minimizing the sum of the squares of the residuals (right side), i.e., the method of least squares. For a more detailed derivation, see d2l.ai. Note that in linear regression, we do not intent to estimate \(\sigma^2\). However, it may be helpful since a large variance indicates that our linear regression model will produce highly biased estimates.