Verander de taal

English

Log in bij Mathigon

of

Nieuw account     Wachtwoord opnieuw instellen     

Multivariable Calculus
by Samuel S. Watson
Introduction
Taylor Series
Differentiation
Optimization
Matrix differentiation
The chain rule
Multivariable integration
Custom integration

Delen

Voortgang resetten

Hierdoor worden je voortgang en chatgegevens voor alle hoofdstukken in deze cursus verwijderd en kan dit niet ongedaan worden gemaakt!

Woordenlijst

Selecteer een van de zoekwoorden aan de linkerkant ...

Multivariable CalculusMatrix differentiation

Leestijd: ~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 \mathbf{f} is a function from \mathbb{R}^n to \mathbb{R}^m. Writing \mathbf{f}(\mathbf{x}) = \mathbf{f}(x_1, \ldots, x_n), we define the Jacobian matrix (or derivative matrix) to be

\begin{align*}\frac{\partial \mathbf{f}}{\partial \mathbf{x}} = \left[ \begin{array}{cccc} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \cdots & \frac{\partial f_1}{\partial x_n} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \cdots & \frac{\partial f_2}{\partial x_n} \\ \vdots & & \ddots & \vdots \\ \frac{\partial f_m}{\partial x_1} & \frac{\partial f_m}{\partial x_2} & \cdots & \frac{\partial f_m}{\partial x_n} \end{array}\right]\end{align*}

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:

\begin{align*}\mathbf{H}(\mathbf{x}) = \frac{\partial}{\partial \mathbf{x}} \left(\frac{\partial f}{\partial \mathbf{x}}\right)'.\end{align*}

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

\begin{align*}\frac{\partial f}{\partial \mathbf{x}} = \mathbf{x}' \frac{\partial}{\partial \mathbf{x}}(A\mathbf{x}) + (A\mathbf{x})' \frac{\partial \mathbf{x}}{\partial \mathbf{x}} = \mathbf{x}' A + \mathbf{x}' A = 2\mathbf{x}' A.\end{align*}

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*}

and set it equal to \boldsymbol{0} to get

\begin{align*}\mathbf{x}' = \mathbf{b}' A(A' A)^{-1} \implies \mathbf{x} = (A' A)^{-1} A' \mathbf{x}.\end{align*}

(We know that A' A has an inverse matrix because its rank is equal to that of A, which we assumed was m.)

Bruno
Bruno Bruno