$\require{mathtools}
\newcommand{\nc}{\newcommand}
%
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%
% Environments
\newcommand{\al}[1]{\begin{align}#1\end{align}} % need this for \tag{} to work
\renewcommand{\r}{\mathrm}
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%
% Delimiters
% (I needed to create my own because the MathJax version of \DeclarePairedDelimiter doesn't have \mathopen{} and that messes up the spacing)
% .. one-part
\newcommand{\p}[1]{\mathopen{}\left( #1 \right)}
\renewcommand{\P}[1]{^{\p{#1}}}
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%% .. two-part
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\newcommand{\co}[2]{ {\left.\inco{#1}{#2}\right.}}
\newcommand{\cond}{\co} % deprecated
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%
% Greek
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% the following cause issues with real LaTeX tho :/ maybe consider naming it \fhi instead?
\let\fi\phi % because it looks like an f
\let\phi\varphi % because it looks like a p
\renewcommand{\th}{\theta}
\newcommand{\Th}{\Theta}
\newcommand{\om}{\omega}
\newcommand{\Om}{\Omega}
%
% Miscellaneous
\newcommand{\LHS}{\mathrm{LHS}}
\newcommand{\RHS}{\mathrm{RHS}}
\DeclareMathOperator{\cst}{const}
% .. operators
\DeclareMathOperator{\poly}{poly}
\DeclareMathOperator{\polylog}{polylog}
\DeclareMathOperator{\quasipoly}{quasipoly}
\DeclareMathOperator{\negl}{negl}
\DeclareMathOperator*{\argmin}{arg\thinspace min}
\DeclareMathOperator*{\argmax}{arg\thinspace max}
% .. functions
\DeclareMathOperator{\id}{id}
\DeclareMathOperator{\sign}{sign}
\DeclareMathOperator{\err}{err}
\DeclareMathOperator{\ReLU}{ReLU}
% .. analysis
\let\d\undefined
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\newcommand{\df}[2]{ {\f{\d #1}{\d #2}}}
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\newcommand{\partf}[2]{\f{\part #1}{\part #2}}
\newcommand{\parts}[2]{\sl{\part #1}{\part #2}}
\newcommand{\grad}[1]{\mathop{\nabla\!_{#1}}}
% .. sets
\newcommand{\es}{\emptyset}
\newcommand{\N}{\mathbb{N}}
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\newcommand{\zpmo}{\set{0,\pm 1}}
% .... set operations
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\newcommand{\inc}[1]{\union \set{#1}} % "including"
\newcommand{\exc}[1]{\setminus \set{#1}} % "except"
% .. over and under
\renewcommand{\ss}[1]{_{\substack{#1}}}
\newcommand{\OB}{\overbrace}
\newcommand{\ob}[2]{\OB{#1}^\t{#2}}
\newcommand{\UB}{\underbrace}
\newcommand{\ub}[2]{\UB{#1}_\t{#2}}
\newcommand{\ol}{\overline}
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\newcommand{\HAT}{\widehat} % deprecated
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\newcommand{\rt}[1]{ {\sqrt{#1}}}
\newcommand{\for}[2]{_{#1=1}^{#2}}
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% .... two-part
\newcommand{\f}{\frac}
\renewcommand{\sl}[2]{#1 /\mathopen{}#2}
\newcommand{\ff}[2]{\mathchoice{\begin{smallmatrix}\displaystyle\vphantom{\p{#1}}#1\\[-0.05em]\hline\\[-0.05em]\hline\displaystyle\vphantom{\p{#2}}#2\end{smallmatrix}}{\begin{smallmatrix}\vphantom{\p{#1}}#1\\[-0.1em]\hline\\[-0.1em]\hline\vphantom{\p{#2}}#2\end{smallmatrix}}{\begin{smallmatrix}\vphantom{\p{#1}}#1\\[-0.1em]\hline\\[-0.1em]\hline\vphantom{\p{#2}}#2\end{smallmatrix}}{\begin{smallmatrix}\vphantom{\p{#1}}#1\\[-0.1em]\hline\\[-0.1em]\hline\vphantom{\p{#2}}#2\end{smallmatrix}}}
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}
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% .. named tensors
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% ... information theory
\let\H\undefined
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% .. other divergences
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%
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% .. classical
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\newcommand{\TZ}{\mathtt{Z}}$
Personal summary of Exact solutions to the nonlinear dynamics of learning in deep linear neural networks.
This paper shows you can decompose learning dynamics of a fully linear network into some (mostly independent) singular modes that learn at different rates (depending on the eigenvalues of the input-output correlation matrix). In fact, it’s as if the network is “simulating” another network for which the correlation between input and output coordinates is diagonal.
General setup
The model is $\ndef{\input}{input}\ndef{\hidden}{hidden}\ndef{\output}{output}\ndef{\data}{data}\newcommand{\We}{W^\mathrm{e}}\newcommand{\Wd}{W^\mathrm{d}}x\ndot\input\We\ndot\hidden \Wd$ (using named tensor notation), where
- $\We \in \R^{\input \times \hidden}$ “encodes” the input $x$ into a hidden layer,
- $\Wd \in \R^{\hidden\times \output}$ “decodes” the hidden layer into the output,
and both $\We$ and $\Wd$ are initialized as independent Gaussians.
It is trained on $N$ data points:
- inputs: $X \ce (x^{(1)}, \ldots, x^{(n)}) \in \R^{\data \times \input}$,
- outputs: $Y \ce (y^{(1)}, \cdots, y^{(n)}) \in \R^{\data \times \output}$.
Which means the square loss is
\[
\CL
%\ce \sum_{d=1}^D \norm{y^{(d)}-\Wd \ndot\hidden \We\ndot\input x^{(d)}}_\output^2
\ce \sum_\data\norm{Y - X \ndot\input \We\ndot\hidden \Wd}_\output^2.
\]
Taking the differential, we get
\[
\al{
\partial \CL
&= \sum_\data 2 \p{Y - X \ndot\input \We\ndot\hidden \Wd}\ndot\output\partial\p{Y - X \ndot\input \We\ndot\hidden \Wd}\\
&= -2 \sum_\data \p{Y - X \ndot\input \We\ndot\hidden \Wd}\ndot\output\b{X\ndot\input\p{\partial\We \ndot\hidden \Wd + \We \ndot\hidden \partial\Wd}}\\
&= -2 \b{X\ndot\data\p{Y - X \ndot\input \We\ndot\hidden \Wd} } \ndot{\input\\\output}\p{\partial\We \ndot\hidden \Wd + \We \ndot\hidden \partial\Wd}.
}
\]
Let’s assume that $X \ndot\data X = I_\input$ (i.e. the input coordinates have been standardized to have mean $0$, variance $1$ and be uncorrelated). Then we have
\[
%\CE =
X\ndot\data\p{Y - X \ndot\input \We\ndot\hidden \Wd} = \UB{X\ndot\data Y}_{\ec\ \Sigma} - \We \ndot\hidden \Wd,
\]
where the “input-output correlations” matrix $\Sigma \ce X \ndot\data Y \in \R^{\input \times \output}$ completely captures the problem. The matrix product $\We \ndot\hidden \Wd$ is “trying” to match these correlations $\Sigma$, and the learning dynamics are given by
\[
\al{
\frac{\d\We}{\d t} = -\frac{\partial\CL}{\partial\We} &= 2\p{\Sigma - \We\ndot\hidden\Wd}\ndot\output\Wd\\
\frac{\d\Wd}{\d t} = -\frac{\partial\CL}{\partial\Wd} &= 2\p{\Sigma - \We\ndot\hidden\Wd}\ndot\input\We.
}
\]
We’ll drop those factors $2$ for simplicity.
Diagonal case
Suppose that $|\input| = |\output|\ec d$, and that $\Sigma$ is diagonal with elements $s_1, \ldots, s_n$. Then we can decompose the dynamics into (somewhat) independent parts if we group the weights by which input/output coordinate they connect to:
\[
\al{
\frac{\d\We_{\input(i)}}{\d t} &= \p{s_i - \We_{\input(i)}\ndot\hidden\Wd_{\output(i)}}\Wd_{\output(i)} - \sum_{j \ne i} \p{\We_{\input(i)}\ndot\hidden\Wd_{\output(j)}}\Wd_{\output(j)}\\
\frac{\d\Wd_{\output(i)}}{\d t} &= \UB{\p{s_i - \We_{\input(i)}\ndot\hidden\Wd_{\output(i)}}\We_{\input(i)}}_\text{``feature benefit''} - \UB{\sum_{j \ne i} \p{\We_{\input(j)}\ndot\hidden\Wd_{\output(i)}}\We_{\input(j)}}_\text{``interference''}.
}
\]
That is, there are two forces:
- feature benefit: the dot product between $\We_{\input(i)}$ and $\Wd_{\output(i)}$ is “trying” to be $s_i$ (they tend to align),
- interference: $\We_{\input(i)}$ and $\Wd_{\output(j)}$ are trying to be perpendicular when $i \ne j$ (they tend to repell each other).
This is very similar to the setup in Toy Models of Superposition, except that
- there are no ReLUs at the end,
- the encoding and decoding weights $\We$ and $\Wd$ are not tied together,
- the target values $s_i$ can be different from $1$.
Speed of learning
To simplify things, let’s assume that all $s_i$ are positive, $\We$ and $\Wd$ start out with
- $\We_{\input(i)}$ perpendicular to $\Wd_{\output(j)}$ when $i \ne j$,
- $\We_{\input(i)}=\Wd_{\output(i)}$.
Then interference disappears, and $\We_{\input(i)}$ will remain equal to $\Wd_{\output(i)}$ throughout training. Because of this, we’ll just refer to them both as $W_i$. Introducing a new variable $u_i \ce W_i\ndot\hidden W_i = \norm{W_i}_\hidden^2 \in \R$ to track their squared norm, we get
\[
\al{
\frac{\d u_i}{\d t}
&= 2\frac{\d W_i}{\d t}\ndot\hidden W_i\\
&= 2\b{\p{s_i - W_i\ndot\hidden W_i}W_i}\ndot\hidden W_i\\
&= 2\p{s_i - u_i}u_i.
}
\]
This is the differential equation of a logistic function: it solves to
\[
u_i(t) = \frac{s_i}{1+e^{-2s_i(t-t_0)}}.
\]
This means that $u_i$ takes time
- $\approx \log(s_i/\eps)/s_i$ to go from $\eps$ to $s_i/2$;
- $\approx \log(s_i/\eps)/s_i$ to go from $s_i/2$ to $s_i-\eps$.
In other words, the rate at which the values of the correlation matrix $\Sigma$ get learned is (roughly) proportional to those values!
Speed of lowering interference
This section is my own speculation.
Let’s keep the assumption that $\We_{\input(i)}=\Wd_{\output(i)}$ at the start (and therefore, throughout training), but relax the assumption that $W_i$ starts out perpendicular to $W_j$ when $i \ne j$. Then we can study at which rate the interference decreases.
It turns out that the right quantity to look at is the cosine similarity
\[
\eps_{ij} \ce \frac{W_i \ndot\hidden W_j}{\norm{W_i}_\hidden \norm{W_j}_\hidden}.
\]
Indeed, $\eps_{ij}$ doesn’t change when $W_i$ or $W_j$ gets scaled up by some factor (without changing direction), so the “feature benefit” part doesn’t affect $\eps_{ij}$: for the purposes of studying how fast $\eps_{ij}$ decreases at any point in time, the only relevant changes in $W_i$ are
\[
\frac{\d W_i}{\d t} = -\sum_{j \ne i}\p{W_i \ndot\hidden W_j}W_j.
\]
Then we have
\[
\al{
&\frac{\d \p{W_i \ndot\hidden W_j}}{\d t}\\
&\quad= \frac{\d W_i}{\d t}\ndot \hidden W_j + W_i \ndot\hidden \frac{\d W_j}{\d t}\\
&\quad= -\p{W_i \ndot \hidden W_j}\p{\norm{W_i}_\hidden^2 + \norm{W_j}_\hidden^2} - \UB{2\sum_{k \ne i,j}\p{W_i \ndot \hidden W_k}\p{W_j \ndot \hidden W_k}}_\text{second-order effects}
}
\]
Ignoring the second-order effects, we get
\[
\frac{\d \eps_{ij}}{\d t} \approx -\eps_{ij}\p{\norm{W_i}_\hidden^2 + \norm{W_j}_\hidden^2}.
\]
So $\eps_{ij}$ decreases exponentially at a rate proportional to the larger of the two squared norms $\norm{W_i}_\hidden^2$ and $\norm{W_j}_\hidden^2$. This means that the angle between $W_i$ and $W_j$ will start to approach $90\degree$ faster when one of $W_i$ or $W_j$ gets big. And that, as we’ve seen, depends on how big $s_i$ and $s_j$ are.
Summary of the dynamics
So we can summarize the dynamics of learning this way:
- the input-output correlations get learned one by one, in order of their goal value $s_i$;
- when $\We_{\input(i)}$ and $\Wd_{\output(i)}$ get learned, they quickly align and acquire the same length ($\We_{\input(i)} \approx \Wd_{\output(i)}$ if $s_i>0$ and $\We_{\input(i)} \approx -\Wd_{\output(i)}$ if $s_i < 0$);
- they eventually reach length $\sqrt{|s_i|}$;
- when $\We_{\input(i)}$/$\Wd_{\output(i)}$ get learned, they simultaneously repel all the other vectors into their perpendicular space.
Non-diagonal case
The striking thing with fully linear networks is that they’re completely rotationally invariant. For us, this means that even if $\Sigma$ is not diagonal, things will still happen the exact same way, but over the singular value decomposition of $\Sigma$ instead.
Suppose that $|\input| = |\output|\ec d$. Let’s rewrite $S$ as a sum of some singular modes (some of which might be $0$ if $\Sigma$ is not full-rank):
\[
\ndef{\inmode}{inmode}
\ndef{\outmode}{outmode}
\Sigma = \sum_{i=1}^D s_i(U_{\inmode(i)} \odot V_{\outmode(i)}) = U \ndot\inmode S \ndot\outmode V
\]
where $U \in \R^{\inmode\times\input}$, $V \in \R^{\outmode \times \output}$ are orthonormal bases of $\R^{\input}$ and $\R^{\output}$, and $S \in \R^{\inmode \times \outmode}$ is diagonal with values $s_1, \ldots, s_n$. Let’s also represent $\We$ and $\Wd$ under the new bases:
\[
\We = U \ndot\inmode\ol\We \qquad \Wd = \ol\Wd \ndot\outmode V.
\]
Then we have
\[
\al{
\partial \CL
&= 2\p{\Sigma - \We \ndot\hidden \Wd}\ndot{\input\\\output}\p{\partial\We \ndot\hidden \Wd + \We \ndot\hidden \partial \Wd}\\
&= 2\b{U \ndot{\inmode}\p{S - \ol\We \ndot\hidden \ol\Wd}\ndot\outmode V}\\
&\qquad\ndot{\input\\\output}\b{U \ndot\inmode \p{\partial\ol\We \ndot\hidden \ol\Wd + \ol\We \ndot\hidden \partial \ol\Wd} \ndot\outmode V}\\
&=2\p{S - \ol\We \ndot\hidden \ol\Wd}\ndot{\inmode\\\outmode}\p{\partial\ol\We \ndot\hidden \ol\Wd + \ol\We \ndot\hidden \partial \ol\Wd},
}
\]
(where the last equality is by orthonormality of $U,V$) so the new learning dynamics are
\[
\al{
\frac{\d\ol\We}{\d t} = -\frac{\partial\CL}{\partial\ol\We} &= 2\p{S - \We\ndot\hidden\Wd}\ndot\outmode\ol\Wd\\
\frac{\d\ol\Wd}{\d t} = -\frac{\partial\CL}{\partial\ol\Wd} &= 2\p{S - \ol\We\ndot\hidden\ol\Wd}\ndot\inmode\ol\We.
}
\]
This is exactly the same as what we had earlier, except that $\Sigma, \We, \Wd$ changed to $S, \ol\We,\ol\Wd$ respectively, and $S$ is guaranteed to be diagonal. That is, by performing a simple change of basis, we’ve completely reduced to the diagonal case over a new “virtual” linear neural network.
Also note that by rotational symmetry, if the initializations of $\We$ and $\Wd$ are i.i.d. Gaussians, then so are the initializations of $\ol\We$ and $\ol\Wd$.