I'm exploring preconditioned gradient descent using a similar toy problem described in the first part of Lecture 8: Accelerating SGD with preconditioning and adaptive learning rates.

I have the function $f(x,y) = x^2 + 10\,y^2$ which has a gradient of $[2x, 20y]$.

I know the ideal form of the function is $f(x,y) = x^2 + y^2$ which has a gradient of $[2\,x, 2\,y]$.

How do I solve for the precondition matrix in the equation, $w_{t+1} = w_t - \alpha\,P\,\nabla f(w_t)$, like the first activity box asks? In this case $P$ would just be $[1, \frac{1}{10}]$?

At the bottom of the second page it says: enter image description here

I'm having trouble understanding how to formally solve for it in variance context described in the answer here Preconditioning gradient descent. I also see in that answer, reference of the third approach w/ $P = [H f(x^*)]^{-1}$.

I am unable to find any other examples walked through online.

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    $\begingroup$ @LmnICE Thanks for the edit, wasn't aware that you can just type latex in directly! $\endgroup$ Sep 8, 2020 at 15:30
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    $\begingroup$ you're welcome! There are other formatting tricks; you can find them here. With LaTeX you can even use a few statements (e.g. \begin{align}). $\endgroup$
    – LmnICE
    Sep 8, 2020 at 15:52
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    $\begingroup$ I'm really interested in this question; I'll be checking on it when I have more time. One thing to note - $P = RR^T$ where $R$ is a fixed matrix, so $P$ must be symmetric. This paper might help too - arxiv.org/pdf/1512.04202.pdf $\endgroup$
    – Elenchus
    Sep 10, 2020 at 13:42
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    $\begingroup$ @Elenchus I actually didn't see the 3 approaches listed at the bottom of the second page which I added into the question. So the 2 approaches in that other answer are definitely legit, but I just need to know how to apply them to this example problem! $\endgroup$ Sep 11, 2020 at 12:17
  • $\begingroup$ @Quantoisseur I missed that too; nice edit to the question, it's really clear now. I've brushed up a bit on optimisation (I swear I forget how matrices work every time I look away from them) so I'll have a go at an answer $\endgroup$
    – Elenchus
    Sep 11, 2020 at 13:10

3 Answers 3


Since the hessian is constant $P = H^{-1}$ is optimal (you fully recover Newton's method): $P = H^{-1} = \text{diag}(1/2, 1/20)$. This is equivalent to making the change of variables $x = \frac{1}{\sqrt{2}} x'$ and $y = \frac{1}{\sqrt{20}} y'$ which gives you the desired $\frac{1}{2} (x'^2 + y'^2)$ form.

As for approach 2, this is just a systematic way of estimating the scale of your variables. Get an empirical estimate for the covariance matrix $\Sigma \approx \text{Cov}(x)$. The answer you referenced shows how $P = \Sigma$ follows from a change of variable and chain-rule.

What distribution do you sample if you want to do approach 2? If you treat your objective function as a negative log-likelihood, then you should be sampling from a normal distribution with $\frac{1}{2} \Sigma^{-1} = \operatorname{diag}(1, 10)$. So you should get $\Sigma = \operatorname{diag}(1/2, 1/20)$. If you were to take many random samples from this distribution and calculate their covariance, you would get the same answer since $\Sigma_{i,j} := \operatorname{E} [(x_i - \mu_i)( x_j - \mu_j)] = \operatorname{Cov}(x)$ (see wikipedia).

These approaches are equivalent when your data is normally-distributed and you have a lot of it. In practice, approach 2 is more generally applicable to situations where the Hessian is not constant, but can be somewhat crude.

Automatically computing a dynamic preconditioning matrix is a very well studied topic. Optimization methods that do this are called Quasi-Newton methods.

  • $\begingroup$ Appreciate the answer! I have a question on approach 2. To test, I am sampling along the x and y axes which gives me a covariance matrix (between the two vectors) scaled roughly from $\Sigma = \begin{bmatrix} 1 & 10 \\ 10 & 100 \end{bmatrix}$. Note: Can't be exact as that's not positive definite. I then calculate $L^T = \begin{bmatrix} 1 & 10 \\ 0 & 0 \end{bmatrix}$. How do I extract P from this? $\endgroup$ Sep 11, 2020 at 20:15
  • $\begingroup$ I think you must have been sampling from the wrong distribution. I added a detailed explanation of how to use #2 in this case. $\endgroup$
    – Reid Hayes
    Sep 13, 2020 at 13:05
  • $\begingroup$ Ya, I was not using the correct sampling distributions. Could you please explain or link to an explanation of the transformation from the objective function to normal distributions: "If you treat your objective function as a negative log-likelihood, then you should be sampling from a normal distribution"? I see how once you get the covariance matrix it works out to P, but in this case shouldn't we find P to be diag(1, 1/10) like the first method? $\endgroup$ Sep 13, 2020 at 18:45
  • $\begingroup$ Or are they not equivalent because our data is not normally-distributed? $\endgroup$ Sep 13, 2020 at 18:57
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    $\begingroup$ Ah, yes. I made a mistake. There should be another L factor in there since dx and dx’ are different. I’ll fix up the answer in a bit $\endgroup$
    – Reid Hayes
    Sep 13, 2020 at 19:06

I'm fairly sure you can just use the Hessian, which in this case is $$ 2\begin{bmatrix} 1 & 0\\ 0 & 10 \end{bmatrix} $$

This comes from method 3 (using second-partial derivative information), and possibly also method 1.

The reason we're using the exact matrix here instead of an approximation is because this is a simple case; as noted elsewhere efficiency can be an issue with larger dimensions. how to make preconditioning efficient

We've already got a diagonal matrix, so even if we did approximate as suggested it would be the same. Note this matrix also satisfies the criteria of being symmetric and positive semi-definite.

  • $\begingroup$ Oh, I forgot about the variance context part. I'm also not certain that it should be the Hessian of the overall function - it could be that it's supposed to be recalculated at each time step using the selected values for the function. The notation in the lecture notes made me think not, but the other answer made me think so. Happy to keep thinking about this $\endgroup$
    – Elenchus
    Sep 11, 2020 at 13:31
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    $\begingroup$ The hessian of $f(x,y) = x^2 + 10y^2$ is $H = 2 \begin{bmatrix} 1 & 0 \\ 0 & 10 \end{bmatrix}$ $\endgroup$ Sep 11, 2020 at 13:48
  • $\begingroup$ The inverse is what you wrote which I see is written in the update equation so it still works out. $\endgroup$ Sep 11, 2020 at 13:55
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    $\begingroup$ Ah, the hessian is constant in this case so even if you recalculated at all places, it'll be the same! Okay, I'm happy with how to use the hessian now, thank you. I'd still be interested in the variance context as well, as it seems easily generalized to approximations. $\endgroup$ Sep 11, 2020 at 14:18
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    $\begingroup$ @kevin the Hessian is the matrix of second-partial derivatives. See khanacademy.org/math/multivariable-calculus/… $\endgroup$
    – Elenchus
    May 29, 2022 at 4:22

This is the problem that preconditioning tries to solve:

enter image description here

From this blog (also discussed here on stackexchange: Fitting SIR model with 2019-nCoV data doesn't conververge)


  • With gradient descent methods you follow a path down-hill. But, your algorithm is often not as smooth as water and doesn't flow straight down in the most efficient way. Instead, often the algorithm will 'over-shoot' and follow an (inefficient) zigzag path.
  • Another problem is that the algorithm might have some basic rules to decide when to stop. For instance, deciding to stop when the step sizes are getting very small. However, for that stopping rule, when you are not looking into the direction of the optimum (because you are looking along the slow zigzag path), it might be that your step sizes are small while the optimum is still far away.

(the first problem is about the speed of getting a solution/convergence, the second problem is about the accuracy of the result)

Methods to 'pre-condition':

So it would be nice if you could change the scale of the different variables somehow (make them more equal instead of having this valley). You only need to know how.

  • method 1: use the formula for the problem or some intuition about the problem.

    In your case, the function $f(x,y) = x^2 + 10y^2$, you can compute the scaling/pre-conditioning based on the formula. In the formulation of your reference... $P$ should be a matrix. That is $P$ is not $\left[1,1/10\right]$ but instead $$P = \begin{bmatrix} 1 & 0 \\ 0 & 1/10 \end{bmatrix}$$ It is the scaling of the variables that is a vector, namely $R = \left[1,\sqrt{1/10}\right]$.

    Without formula, you can use 'intuition'. For instance, you might be having features with widely varying scales. I often just do this. Scale the variables such that the range of the 'expected' solution will be similar for each variable. (In the image below you see this as well. The range of the $R_0$ and $K$ parameter is a lot different)

  • method 2: Use the data/variance. I believe that this refers to using a sample distribution of the features (or some other information about the scale of spread). The variance of the features relates to the curvature of the likelihood function (observed information matrix). Maybe this method is also much the same as 'use intuition'.

    Aside from just looking at the scale of the parameters, it is also important to look at the 'variation'. If some parameter is around $1 \, kilometer \pm 0.001 \, kilometer$, then the loss function has more curvature than a parameter around $10 \, meter \pm 2 \, meter$. (you can see this in the image below, the parameter $R_0 \approx 1$ has a larger scale than the $K \approx 0.4$ parameter, but the variation is in a much smaller range)

  • method 3: You may not have a formula or intuition like in method 1, but you could compute the Hessian at some point during the descent and adapt the scaling accordingly.

    This method 3 is particularly nice if the problem is not continuous. For instance in the image below you get that the hessian and the scale is varying. Intuitively, the line that the "half-pipe" follows is bended (in the beginning the $K$ parameter has more curvature, but later the $R_0$ parameter). In these cases you can not easily solve problems of convergence by just setting a (single) pre-conditioning from the start, and you need to change it in multiple steps (along the path).


  • $\begingroup$ Appreciate the answer! $\endgroup$ Sep 11, 2020 at 20:22
  • $\begingroup$ I'm trying to solve via approach 2 in the comment on Reid Hayes answer. Could you possibly help? $\endgroup$ Sep 12, 2020 at 15:16

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