Ini akan menghapus kemajuan dan data obrolan Anda untuk semua bab dalam kursus ini, dan tidak dapat dibatalkan!

Glosarium

Pilih salah satu kata kunci di sebelah kiri…

Multivariable CalculusMatrix differentiation

Waktunya membaca: ~10 min

Just as elementary differentiation rules are helpful for optimizing single-variable functions, matrix differentiation rules are helpful for optimizing expressions written in matrix form. This technique is used often in statistics.

Suppose is a function from to .Writing , we define the Jacobian matrix (or derivative matrix) to be

Note that if m=1, then differentiating f with respect to \mathbf{x} is the same as taking the gradient of f.

With this definition, we obtain the following analogues to some basic single-variable differentiation results: if A is a constant matrix, then

\begin{align*}\frac{\partial}{\partial \mathbf{x}} (A \mathbf{x}) &= A \\\ \frac{\partial}{\partial \mathbf{x}} (\mathbf{x}' A) &= A' \\\ \frac{\partial}{\partial \mathbf{x}} (\mathbf{u}' \mathbf{v}) &= \mathbf{u}'\frac{\partial \mathbf{v}}{\partial \mathbf{x}} + \mathbf{v}'\frac{\partial \mathbf{u}}{\partial \mathbf{x}}\end{align*}

The third of these equations is the rule.

The Hessian of a function f:\mathbb{R}^n \to \mathbb{R} may be written in terms of the matrix differentiation operator as follows:

Some authors define \frac{\partial f}{\partial \mathbf{x}'} to be \left(\frac{\partial f}{\partial \mathbf{x}}\right)', in which case the Hessian operator can be written as \frac{\partial^2}{\partial \mathbf{x} \partial \mathbf{x}'}.

Exercise Let f: \mathbb{R}^n \to \mathbb{R} be defined by f(\mathbf{x}) = \mathbf{x}' A \mathbf{x} where A is a symmetric matrix. Find \frac{\partial f}{\partial \mathbf{x}}.

Solution.We can apply the product rule to find that

Exercise Suppose A is an m\times n matrix and \mathbf{b} \in \mathbb{R}^m.Use matrix differentiation to find the vector \mathbf{x} which minimizes |A \mathbf{x} - \mathbf{b}|^2.Hint: begin by writing |A \mathbf{x} - \mathbf{b}|^2 as (A \mathbf{x} - \mathbf{b})' (A \mathbf{x} - \mathbf{b}).You may assume that the rank of A is n.

Solution.We write

\begin{align*}|A \mathbf{x} - \mathbf{b}|^2 &= (A \mathbf{x} - \mathbf{b})' (A \mathbf{x} - \mathbf{b}) \\\ &= \mathbf{x}' A' A \mathbf{x} - \mathbf{b}' A \mathbf{x} + \mathbf{x}' A' \mathbf{b} + |\mathbf{b}|^2.\end{align*}

To minimize this function, we find its gradient

\begin{align*}\frac{\partial}{\partial \mathbf{x}}|A \mathbf{x} - \mathbf{b}|^2 = 2\,\mathbf{x}' A' A - \mathbf{b}' A + (A'\mathbf{b})' = 2\mathbf{x}' A' A- 2\mathbf{b}' A\end{align*}