In deriving the parameter estimate in OLS, we differentiate the following (in matrix form)

$$y^T y - 2\beta^T X^T y + \beta^T X^T X \beta$$

The part of the differentiation I don't understand is why $$\beta^T X^T X \beta$$ differentiates to

$$2X^T X \beta. $$

The way I thought it would work, was:

$$\triangledown(\beta^T)X^TX\beta + \beta^TX^TX\triangledown (\beta) = X^TX \beta + \beta^TX^TX $$


In your way, the dimensions of the summands don't match. If you write the term as $\beta^TA\beta$ openly, we'll have the following expression ($A$ is symmetric): $$\beta^TA\beta=\sum_{i,j}A_{ij}\beta_i\beta_j=\sum_i A_{ii}\beta_i^2+2\sum_{i<j}A_{ij}\beta_i\beta_j$$ Derivative with respect to $\beta_i$ is $$\frac{\partial \beta^TA\beta}{\partial \beta_i}=2\beta_iA_{ii}+2A_{ij}\beta_j$$ This expression is basically the dot product of $i$-th row of $A$ and $\beta$ multiplied by $2$, i.e. $2A_i^T\beta$, where $A_i$ denotes the $i$-th row.

In numerator layout notation, which is commonly used in matrix calculus cheatsheets, scalar differentiated by a vector produces a horizontal vector, so we'll concatenate the derivative for each $\beta_i$ horizontally:

$$\frac{\partial\beta^TA\beta}{\partial\beta}=[2A_1^T\beta\dots2A_n^T\beta]=2[A_1^T\dots A_n^T]\beta$$

If we transpose each row of A and concatenate them horizontally in their respective order, we obtain $A^T$, which is $A$ since $A$ was symmetric. Therefore, the middle matrix in the above expression is $A$. Letting $A=X^TX$, we have:


which matches your formula.

  • $\begingroup$ How can I look at it from simply a matrix point of view? $\endgroup$ – Bill Jan 23 at 17:37
  • 1
    $\begingroup$ For a rigorous account of matrix differentiation, see stats.stackexchange.com/a/257616/919. But as a practical matter, dimensional considerations like those used in the present answer are a powerful way to avoid mistakes. $\endgroup$ – whuber Jan 23 at 18:22

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