Let $\vec{b}=\langle b_1,\dots,b_n\rangle ^T$ be an n-dimensional vector of coefficients. Let $\vec{x}_1,\dots,\vec{x}_n$ be $n$ $p$-dimensional vectors. Let $G(\vec{b})=\log\det\left( \sum_{i=1}^n b_i \vec{x}_i\vec{x}_i^T\right)$.

Let $A=\sum_{i=1}^n b_i \vec{x}_i\vec{x}_i^T$. If one wants to compute the $i$-th component of the gradient, we get

\begin{eqnarray} \nabla_i G(\vec{b}) &=& \text{Tr}\left(\partial_i A \right) \\ &=& \text{Tr}\left( A^{-1} \vec{x}_i\vec{x}_i^T \right) \\ &=& \text{Tr}\left(\vec{x}_i^T A^{-1} \vec{x}_i \right) \\ &=& x_i^T A^{-1} x_i \end{eqnarray}

I am filling in the details so far of this paper (page 19, before equation (33)). So far I agree with their calculation. However, I do not understand their calculation of the line (33) and (34) in which they calculate the Hessian.

They claim that $$ \nabla^2_{ij} (G(\vec{b})) = -(\vec{x}_i^T A^{-1}\vec{x}_j)^2. \tag1 $$

I get something different. Using the Matrix Cookbook (equation (61)), I see that \begin{eqnarray} \partial_j(\vec{x}_i^T A^{-1} \vec{x}_i) &=& -A^{-1}\vec{x}_i\vec{x}_i^T A^{-1}\cdot\partial_i(A) \tag2\\ &=& -A^{-1}\vec{x}_i\vec{x}_i^T A^{-1} \vec{x}_j\vec{x}_j^T, \end{eqnarray} which is a matrix and not a scalar!

I know I must be making a mistake somewhere. I am still not quite comfortable with matrix calculus.

Can someone help me figure out where I'm going wrong?


The derivation in (2) is wrong. You apply the formula (61) from the Matrix Cookbook, but this formula is for the derivative of the quadratic form with respect to the whole matrix. That's why it is a matrix. You need to calculate the derivative of $G_i'$ with respect of the variable $b_j$ only, so you are rather to use the equation (59) in the Cookbook. Here how it goes $$ G_{ij}''=[G_i']_j'=\frac{\partial x_i^TA^{-1}x_i}{\partial b_j}= x_i^T\frac{\partial A^{-1}}{\partial b_j}x_i\stackrel{\text{Eq.(59)}}{=} x_i^T\Bigl(-A^{-1}\underbrace{\frac{\partial A}{\partial b_j}}_{x_jx_j^T}A^{-1}\Bigr)x_i=\\=-x_i^TA^{-1}x_j\cdot x_j^TA^{-1}x_i $$ which is exactly their expression (1).

UPDATE explaining the formula (61) in the Matrix Cookbook.

In one dimensional calculus we have $f\colon\mathbb{R}\to\mathbb{R}$ and differentiating according to $$ f(x+h)-f(x)=\underbrace{f'(x)\cdot h}_{\partial_x f}+o(|h|).\tag{*} $$ When $x$ becomes a vector and $f\colon\mathbb{R}^n\to\mathbb{R}$, the derivative becomes a vector of partial derivatives (gradient) which normally has the same size as the vector $x$ so that on the place for $x_k$ one gets the partial derivative $\frac{\partial f}{\partial x_k}$ $$ x=\left[\matrix{x_1\\x_2\\\vdots\\x_n}\right]\qquad\Rightarrow\qquad\frac{\partial f}{\partial x}= \left[\matrix{\frac{\partial f}{\partial x_1}\\\frac{\partial f}{\partial x_2}\\\vdots\\\frac{\partial f}{\partial x_n}}\right]. $$ The differential in (*) is then no longer a multiplication, it becomes the scalar product, i.e. the sum of pointwise multiplications between the gradient and the variable $h$ $$ f(x+h)-f(h)=\langle\frac{\partial f}{\partial x},h\rangle+o(\|h\|). $$ When one has a function $f\colon\mathbb{R}^{m\times n}\to \mathbb{R}$, for example, $f(X)=a^TX^{-1}b$ as in (61), it is also convenient to organize the matrix of partial derivatives in the same way that on the position for $x_{ij}$ we get $\frac{\partial f}{\partial x_{ij}}$, and this matrix is denoted by $\frac{\partial f}{\partial X}$, however, the differential is not $\partial f=\frac{\partial f}{\partial X}\partial X$ as you wrote in (2), but as in the vector case the scalar product, i.e. $$ \partial f=\langle\frac{\partial f}{\partial x},\partial X\rangle= \text{tr}\Bigl(\biggl(\frac{\partial f}{\partial x}\biggr)^T\partial X\Bigr). $$ So the corrected version of (2) would be $$ \text{tr}(-A^{-1}x_ix_i^TA^{-1}x_jx_j^T)=-(x_j^TA^{-1}x_i)^2. $$

  • $\begingroup$ Thanks @A.G., but I am not quite following how you justify integrating "through" the quadratic form to the inverse. I.e., can you explain how you get the third equality? I had thought I was applying the chain rule, essentially. This is why I integrated first with respect to the whole matrix $A^-1$. $\endgroup$ – Lepidopterist Sep 19 '15 at 14:42
  • $\begingroup$ @Lepidopterist The third equality is the derivative of the product (Eq. (37) in the Cookbook). For three matrices one gets $(\partial x_i^T)A^{-1}x_i+x_i^T(\partial A^{-1})x_i+x_i^TA^{-1}(\partial x_i)$, but the first and the last terms are zeros since $x_i$ does not depend on $b_j$. Chain rule for matrices should respect the order of multiplication, since the factors are not commutative, derivation is applied to matrices at their positions, you cannot just plug in $\partial_j A$ at the end. $\endgroup$ – A.Γ. Sep 19 '15 at 17:15
  • $\begingroup$ Thanks for your comment. I wonder if you might briefly explain where (61) comes from then. As I understand your reasoning, the derivative should look like $a^tA^{-2}b$, which of course it is not... $\endgroup$ – Lepidopterist Sep 19 '15 at 22:07
  • $\begingroup$ @Lepidopterist Ok, I think I understand now your question. It comes from the interpreting the derivative $\partial f/\partial X$ for a matrix variable $X$. I've added the explanation to the answer, hope it helps. $\endgroup$ – A.Γ. Sep 19 '15 at 23:40

You can recast this problem into a purely matrix form.

Let $X\in {\mathbb R}^{p\times n}$ denote the matrix whose columns are the {$x_k$} vectors, and let the matrix $\,\,B={\rm Diag}(b)$.

Then the function can be written $$\eqalign{ A &= X\,B\,X^T \cr G &= \log \det (A) = {\rm tr}\,\log(A) \cr }$$ Notice that $A$ and $B$ are symmetric.

Find the differential of the function, and successively substitute variables until we arive at an expression involving $db$ $$\eqalign{ dG &= A^{-T}:dA \cr &= A^{-1}:X\,dB\,X^T \cr &= X^TA^{-1}X:dB \cr &= X^TA^{-1}X:{\rm Diag}(db) \cr &= {\rm diag}(X^TA^{-1}X):db \cr }$$ Since $dG = \frac{\partial G}{\partial b}\!:\!db,\,$ the gradient is $$\eqalign{ G^\prime = \frac{\partial G}{\partial b} &= {\rm diag}(X^TA^{-1}X) \cr\cr }$$

To find the Hessian, start by taking the differential of the gradient $$\eqalign{ dG^\prime &= {\rm diag}(X^T\,dA^{-1}\,X) \cr &= -{\rm diag}(X^TA^{-1}\,dA\,A^{-1}X) \cr &= -{\rm diag}(X^TA^{-1}X\,dB\,X^TA^{-1}X) \cr &= -{\rm diag}(C\,dB\,C) \cr &= -{\rm diag}(C\,{\rm Diag}(db)\,C) \cr }$$ where I have collected some common matrix factors into $C=(X^TA^{-1}X),\,$ which is also a symmetric matrix.

Here is a handy formula from Thomas Minka which converts the diag-diag operation into a Hadamard (aka element-wise) product: $$ {\rm diag}(A\,{\rm Diag}(x)\,B) = (B^T\circ A)\,x $$ Applying Minka's formula yields $$\eqalign{ dG^\prime &= -(C\circ C)\,db \cr }$$ from which the Hessian is seen to be $$\eqalign{ G^{\prime\prime} &= -(C\circ C) \cr &= -(X^TA^{-1}X)\circ(X^TA^{-1}X) \cr }$$


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