# Gradient descent of $f(w)=\frac12w^TAw-b^Tw$ viewed in the space of Eigenvectors of $A$

I'm reading Why Momentum Really Works, a post from the new distill journal. I'll paraphrase the main equations leading to the part which confuses me, the post describes the intuition in more detail.

The gradient descent algorithm is given by the following iterative process $$w^{k+1} = w^k-\alpha \nabla f(w^k)$$ where $w^k$ is the value of iteration $k$, the learning rate is $\alpha$ and $\nabla f(w)$ is the gradient of the function $f$ evaluated at $w$. The function $f$ you wish to minimize.

Gradient descent with momentum is given by adding "memory" to the descent, this is described by the pair of equations:

\begin{align} z^{k+1} &= \beta z^k + \nabla f(w^k) \\ w^{k+1} &= w^k - \alpha z^{k+1} \end{align}

In the next section "First Steps: Gradient Descent" the author considers a convex quadratic function $$f(w) = \frac12w^TAw-b^Tw, \quad w \in \mathbb{R}^n, A \in \mathbb{R}^{n,n}$$ which has gradient $$\nabla f(w) = Aw-b$$ If we assume $A$ is symmetric and invertable then $f$ has optimal solution $w^\star = A^{-1}b$.

If we were to use gradient descent then we would iterate towards this optimal solution in the following way \begin{align} w^{k+1} &= w^k - \alpha \nabla f(w) \\ &= w^k - \alpha (Aw^k -b) \end{align}

Then the article goes on to say "There is a very natural space to view gradient descent where all the dimensions act independently — the eigenvectors of $A$". I think this makes sense, although my intuition is kind of fuzzy.

Every symmetric matrix $A$ has an eigenvalue decomposition where $$A = Q\text{diag}(\lambda_1,\ldots,\lambda_n)Q^T.$$

Where $\lambda_1 > \ldots > \lambda_n$ and $Q$ is the vector with the corresponding eigenvectors as columns (right?).

This next part is where I don't understand what is going on:

If we perform a change of basis, $x^k = Q^T(w^k - w^\star)$, the iterations break apart, becoming:

\begin{align} x_i^{k+1} &= x_i^k - \alpha \lambda_i x_i^k \\ &=(1-\alpha\lambda_i)x_i^k &= (1- \alpha\lambda_i)^{k+1}x_i^0 \end{align}

Moving back to our original space $w$, we can see that

$$w^k - w^\star = Qx^k = \sum\limits_{i}^n = x_i^0(1-\alpha\lambda_i)^kq_i$$

What is going on here? Where is the motivation of taking $w^k - w^\star$ into the eigendomain? What is $x^k$? Why are we now looking at invidual elements of the vector? I tried to follow the caculations through, but $x^{k+1}$ depends on $w^{k+1}$ which depends on $z^k$, which I thought we were trying to eliminate. My question is can someone expand on these few steps with some intuition and calculations? Thanks.

In many mathematical applications, the motivation becomes clearer after deriving the result. So let's start off with the algebra.

Suppose we were to run GD for $T$ iterations. This will give us the set ${(w_k)}_{k=1}^T$.

Let's do a change of basis:

$w^k = Qx^k + w^*$ $\iff$ $x^k = Q^T(w^k-w^*)$

Now we have ${(x_k)}_{k=1}^T$. What can we say about them? Let's look at each coordinate separately. By substituting the above and using the update step of GD,

$x_i^{k+1}= (Q^T(w^{k+1}-w^*))_i = (Q^T(w^k-\alpha (Aw^k-b)-w^*))_i$

Arranging,

$x_i^{k+1}=(Q^T(w^k-w^*))_i-\alpha \cdot (Q^T(Aw^k-b))_i$

The first term is exactly $x_i^k$. For the second term, we substitute $A=Qdiag(\lambda _1 \dots \lambda _n)Q^T$. This yields,

$x_i^{k+1}=x_i^k-\alpha \lambda _i x_i^k=(1-\alpha \lambda _i)x_i^k$

Which was a single step. Repeating until we get all the way to $x_0$, we get

$x_i^{k+1}=(1-\alpha \lambda _i)^{k+1}x_i^0$

All this seems really useless at this point. Let's go back to our initial concern, the ${w}$s. From our original change of basis, we know that $w^k-w^*=Qx^k$. Another way of writing the multiplication of the matrix $Q$ by the vector $x^k$ is as $\sum_i x_i^kq_i$. But we've shown above that $x_i^{k}=(1-\alpha \lambda _i)^{k}x_i^0$. Plugging everything together, we have obtained the desired "closed form" formula for the GD update step:

$w^k-w^*=\sum_i x_i^0(1-\alpha \lambda _i)^{k} q_i$

This is essentially an expression for the "error" at iteration $k$ of GD (how far we are from the optimal solution, $w^*$). Since we're interested in evaluating the performance of GD, this is the expression we want to analyze. There are two immediate observations. The first is that this term goes to 0 as $k$ goes to infinity, which is of course good news. The second is that the error decomposes very nicely into the separate elements of $x_0$, which is even nicer for the sake of our analysis. Here I quote from the original post, since I think they explain it nicely:

Each element of $x^0$ is the component of the error in the initial guess in the $Q$-basis. There are $n$ such errors, and each of these errors follows its own, solitary path to the minimum, decreasing exponentially with a compounding rate of $1-\alpha \lambda_i$. The closer that number is to 1, the slower it converges.

I hope this clears things up for you enough that you can go on to continue reading the post. It's a really good one!

• Wow thank you so much, this is such an excellent answer! Perhaps I should have read a bit further to what the point of all this was. Easy to become discouraged when you're lost at the first bit of math in a journal devoted to clear explanations :P Apr 17, 2017 at 11:31
• I am not OP, but I want to thank you for this excellent answer. It's fantastic and I finally really understood what was going on in these steps. Mar 14, 2021 at 17:14

I have read the same paper, got stuck at the exact same place and worked through with the help of galoosh33's answer.

I just didn't find obvious the step:

$$$$\begin{split} x_{i}^{k+1} & = (Q^{T}(w^{k} - w^{*}))_{i} - \alpha (Q^{T}(Aw^{k} - b))_{i} \\ & = x_{i} - \alpha \lambda_{i} x_{i}^{k} \end{split}$$$$

So for those, who do not want to work through the algebra and do not immediately see, how we got rid of $$b$$, it is from substitution $$w^{k} = Qx^{k} + w^{*}$$ and $$w^{*} = A^{-1}b$$ and the fact that eigenvectors are orthogonal $$Q^{-1} = Q^{T}$$.

$$$$\begin{split} (Q^{T} A w_{k} - Q^{T}b)_{i} & = (Q^{T} A Q x^{k} + Q^{T} A \overbrace{w^{*}}^{A^{-1}b} - Q^{T}b)_{i} \\ & = (\underbrace{Q^{T} Q}_{I} \text{diag}(\lambda_1, \ldots, \lambda_n) \underbrace{Q^T Q}_{I} x^{k} \underbrace{+ Q^{T} \underbrace{A A^{-1}}_{I} b - Q^{T} b}_{0})_{i} \\ & = \lambda_{i} x_{i}^{k} \end{split}$$$$

• Thanks for this, I went through this derivation too but I never get to the last step, how do we simplify to just the ith component and eigenvalue? Jul 18, 2021 at 7:49

I'll provide a few comments in the language of machine learning that will hopefully lead you to a helpful logical conclusion.

First, minimizing that quadratic objective is like solving a least squares problem (if this is not obvious, try proving it as an exercise). Second, for any least squares problem, if the features are orthogonal, then estimating the coefficients seperately or sequentially (like doing exactly one round of coordinate descent) is equivalent to estimating them jointly. (If this isn't obvious, then suppose the features are orthogonal. Do you see this means $A$ must be diagonal? That means each entry of the solution does not depend on the others).

So now the question is: How can we solve the same problem, but with a diagonal matrix in place of $A$? Third, the $\ell_2$ norm is orthogonally-invariant, so if you left or right multiply whatever is inside the norm by an orthogonal matrix (which is interpreted as a rotation), you can just solve that problem then back out that orthogonal transformation at the end. Since $A$ is symmetric positive semi-definite, we can get those orthogonal matrices from the eigenvalue decomposition of $A$ (aka by "diagonalizing" $A$).

Back to statistics: This process is sometimes referred to as whitening or pre-whitening though I believe that there is a lack of concesus as to the usage of this term.

Put simply and loosely, in the eigenspace of $A$, the columns/rows of $A$ can be viewed as totally separate and unrelated pieces of information.