How to Easily Compute Matrix Derivatives, Fréchet Style

The routine, traditional interpretation of a derivative is as a slope of the tangent line. This is great for a function of a single variable, but if you have a multivariate function $f: \mathbb{R}^m \to \mathbb{R}^n$, how can you talk about slopes? One approach is to make a conceptual leap: a derivative is a linear approximation to your function; the tangent line itself is the derivative, rather than the slope of the tangent line. This is called the Fréchet derivative.

Suppose you have a multivariate function $f: \mathbb{R}^m \to \mathbb{R}^n$. At some point in the domain $x_0 \in \mathbb{R}^m$, we want to say that if you make a small change $\epsilon \in \mathbb{R}^m$ around $x_0$, the change in $f$ is approximately a linear function in $\epsilon$:

$f(x_0 + \epsilon) \approx f(x_0) + Df_{x_0}(\epsilon)$

Here $Df_{x_0} : \mathbb{R}^m \to \mathbb{R}^n$ is our notation for the derivative function (which is linear, and can be represented as a matrix applied to $\epsilon$). $Df_{x_0}$ could depend highly on $x_0$; after all, $f$ will probably have different slopes at different points. More formally, we want a function $Df_{x_0}$ such that

$f(x_0 + \epsilon) = f(x_0) + Df_{x_0}(\epsilon) + o(\epsilon)$

where $o(\epsilon)$ is a function that “is tiny compared to $\epsilon$” e.g. every entry of the vector is the square of the largest entry of $\epsilon$.

Computing Fréchet Derivatives is Incredibly Elegant

To compute $Df_{x_0}(\epsilon)$, just plug in $x_0 + \epsilon$ to $f$, collect all the terms that are linear in $\epsilon$, and drop all the terms that “get small as $\epsilon \to 0$“. In practice, this means we drop all “higher-order terms” in which we multiply two coordinates of $\epsilon$. The best way to illustrate is with examples.

Example 1: f(x) = x^Tx

If we change $x$ a little, how does $x^\top x$ change? Here $f: \mathbb{R}^n \to \mathbb{R}$ is $f(x) = x^\top x$.

$f(x + \epsilon) = (x + \epsilon)^\top (x + \epsilon) = x^\top x + x^\top \epsilon + \epsilon^\top x + \epsilon^\top \epsilon$

$= f(x) +x^\top \epsilon + \epsilon^\top x + \epsilon^\top \epsilon$

The part linear in $\epsilon$ here is $x^\top \epsilon + \epsilon^\top x$, and $\epsilon^\top \epsilon$ is tiny compared to $\epsilon$. So

$Df_x(\epsilon) = x^\top \epsilon + \epsilon^\top x = 2x^\top \epsilon$

$f(x+\epsilon) \approx f(x) + 2x^\top \epsilon$

If we want to see where $x^\top x$ is minimized, the critical points are found by finding where the function $Df_x$ is 0; that is, where we can’t move a little bit to go up or down. In this case,

$\forall \epsilon.\; 2x^\top \epsilon = 0 \iff x = 0$

Example 2: Derivative of the Matrix Inverse

Let’s look at a function of a matrix,

$f: \mathbb{R}^{n \times n} \to \mathbb{R}^{n \times n}$

$f(M) = M^{-1}$

As we change $M$ by a matrix $X$ with small entries (in particular, small enough to ensure $M+X$ is still invertible), how does $(M+X)^{-1}$ change from $M^{-1}$? Just thinking about dimensions, the answer to that question should be by adding a small matrix to $M^{-1}$.

$f(M+X) = (M+X)^{-1} \approx M^{-1} + Df_M(X)$

The hard part is massaging $(M+X)^{-1}$ to look like $M^{-1}$ plus a linear function of $X$. To start,

$(M+X)^{-1} = M^{-1}(I + XM^{-1})^{-1}$

The matrix $I + XM^{-1}$ can be inverted using the Neumann series, which is just a fancy name for the geometric series summation formula

${1 \over I + XM^{-1}} = I - XM^{-1} + (XM^{-1})^2 - (XM^{-1})^3 + \cdots$

Note that the series converges for small enough $X$ because $XM^{-1}$ has all eigenvalues less than 1.

Because $X$ is small, $X^2$ is negligibly tiny. For our linear approximation, we can drop the terms that are too tiny to be noticed by the linear approximation, which is all but the first two terms of this series.

${1 \over I + XM^{-1}} \approx I - XM^{-1}$

$M^{-1}(I + XM^{-1})^{-1} \approx M^{-1}(I - XM^{-1}) = M^{-1} - M^{-1}XM^{-1}$

Remembering $f(M) = M^{-1}$, altogether we just computed

$f(M+X) \approx f(M) - M^{-1}XM^{-1}$

Noting that this second term is linear in $X$, we have the derivative

$Df_{M}(X) = -M^{-1}XM^{-1}$

As a remark: whenever we wrote $\approx$, we could have written $+ O(X^2)$ to be a little more formal about carrying the higher-order terms through the computation.

Example 3: Derivative of the Log of the Determinant

Now let’s do another example of a matrix function, the log of the determinant

$f: \mathbb{R}^{n \times n} \to \mathbb{R}$

$f(M) = \log |M|$

This function arises frequently in machine learning and statistics when computing log likelihoods.

Change $M$ by a matrix $X$ which has small entries

$f(M + X) = \log |M + X| = \log |M| |I + M^{-1}X| = \log |M| + \log |I + M^{-1}X|$

Now let’s look at $|I + M^{-1}X|$. Each of the entries of $M^{-1}X$ is “small”, though they’re essentially the same size as elements of $X$ (just differing by a few constants depending on $M^{-1}$). The determinant of a matrix $A$ is a sum over all permutations

$|A| = \displaystyle\sum_{\sigma \in S_n} \prod_{i = 1}^n A_{i, \sigma(i)}$

When performing this sum for $I + M^{-1}X$, if a product contains two off-diagonal terms, those will be two “small” terms from $M^{-1}X$ multiplied together, and multiplying two small terms makes a negligible term that we drop. So we only need to sum up $\sigma$ that include at least all but 1 diagonal elements. The only permutation that does this is the identity permutation.

$|I + M^{_1}X| \approx \prod_{i = 1}^n (I + M^{-1}X)_{i,i} =\prod_{i = 1}^n (1 + M^{-1}X_{i,i})$

Same trick: this sum has $2^n$ terms, but any term that multiples an $M^{-1}X_{i,i}$ with a $M^{-1}X_{j,j}$ is negligibly small. To get a term with at most one of the $M^{-1}X_{j,j}$, either we (1) take 1 from every factor, or (2) take 1 from all but one factor

$\prod_{i = 1}^n (1 + M^{-1}X_{i,i}) \approx 1^n + 1^{n-1}M^{-1}X_{1,1} +1^{n-1}M^{-1}X_{2,2} + \cdots + 1^{n-1}M^{-1}X_{n,n}$

$= 1 + \text{tr}(M^{-1}X)$

This is our linear approximation $|I + M^{-1}X| \approx 1 +\text{tr}(M^{-1}X)$. To incorporate the log, note that we’re taking the log of something that is pretty much 1. The normal derivative of log at 1 is 1, and in our Fréchet derivative formulation

$\log (1 + \epsilon) = \epsilon + O(\epsilon^2)$

$\log |I + M^{-1}X| \approx \log(1 + \text{tr}(M^{-1}X)) \approx\text{tr}(M^{-1}X)$

And so the derivative is $Df_{M}(X) = \text{tr}(M^{-1}X)$.

As a remark: incorporating log illustrates the composability of Fréchet derivatives. This makes sense when spoken aloud: if we have a linear approximation $Df$ to $f$ and a linear approximation $Dg$ to $g$, a linear approximation of $f \circ g$ is $Df \circ Dg$.

Other Viewpoints

Actually, $Df_{x_0}(\epsilon) = J\epsilon$ where $J$ is the Jacobian matrix of partial derivatives ${\partial f_i \over \partial x_j}$. But it’s often easier to not compute the Jacobian entirely, and keep $Df_{x_0}$ in an “implicit” form like in Example 2.