How to compute the hessian of the log barrier function According to note, the log barrier function is given by (page 10):
$f(x) = - \sum\limits_{i = 1}^m \log(b_i - a_i^Tx)$ where $b_i$ is a scalar, $a_i, x$ are $n$ dimensional vectors
I have successfully computed the gradient to be 
$\nabla f(x) = A^Td$, where $d_i = \dfrac{1}{b_i-a_i^Tx}$ (pg 11)
I am trying to compute the Hessian but I cannot figure out why the final solution is $A^T diag(d)^2 A$, where did the diag came from?
Recall the hessian is given by $\nabla^2 f(x) = [\dfrac{\partial^2 f(x)}{\partial x_i \partial x_j}]$
I am certain that the coefficients are not zero for off-diagonal entries...since I am literally taking the derivative of the inner affine function $b_i-a_i^Tx$
Can someone explain why the final solution has a diagonal form?
 A: There are a couple of identities involving the Hadamard ($\circ$) product that you'll need in order to see how the Diag() operation arises. 
Let 
  $\,\,\,\,x,y$ = arbitrary vectors 
  $\,\,\,\,1$ = vector of all ones 
  $\,\,\,\,{\rm Diag}(x)$ = function which returns a diagonal matrix whose diagonal equals the vector $x$  
Then
$$\eqalign{
  {\rm Diag}(x)\,y &= x\circ y \cr
  &= y\circ x \cr
}$$
The last relation is due to the fact that the Hadamard product is commutative. 
Also, since we'll be talking about differentials, let's denote your vector $d$, by the vector $w$ instead. 
Since you already know that the gradient of $f$ is
$$
  g = A^Tw
$$
we'll start from there. 
For convenience, define the variables
$$\eqalign{
  z &= Ax \cr
  y &= b - z \cr
  w &= 1/y \cr
  W &= {\rm Diag}(w) \cr
}$$
The expression for $w$ uses Hadamard division, so we could express the relationship between $y$ and $w$ as
$$\eqalign{
  1 &= w\circ y \cr
  0 &= d(w\circ y) = w\circ dy + y\circ dw \cr
  dw &= -w\circ w\circ dy \cr
}$$
Now expand the differential of the gradient, substituting variables step-by-step
$$\eqalign{
 dg &= A^T dw \cr
    &= A^T (-w\circ w\circ dy) \cr
    &= A^T (w\circ w\circ dz) \cr
    &= A^T (W W dz) \cr
    &= A^T (W W A\,dx) \cr
    &= A^T W^2 A\,dx \cr
}$$
And so we arrive at an expression for the Hessian as
$$
  A^T W^2 A
$$
