Introduction
In one of their smaller papers - “Superposition, Memorization, and Double Descent”, Anthropic investigated how model features were learnt over time under different dataset size regimes. They trained a toy ReLU autoencoder (a model that compresses an input vector from higher dimensionality to a lower dimensional latent space via weight matrix $W$, and then tries to reconstruct it by applying $W^\top$, adding a bias $b$, and finally, $\text{ReLU}$-ing it). They demonstrated that at lower dataset sizes, the model prefers to represent training examples (“memorization”), while at higher dataset sizes, the model prefers to represent features (“generalization”), and the representation of examples is very similar to the representation of features. In particular, we see the beautiful geometric arrangement of examples, where each example is assigned a direction, and the directions are as spaced out as possible in the latent space:
Each vector corresponds to a training example in latent space, over a range of dataset sizes (Anthropic)
Intuitively, this all makes sense - the model wants to represent what it learns with as little interference as possible, so it tries to space everything out as much as possible in its latent space, but this is way too anthropomorphic and doesn’t give us a good enough picture of why this works, how this works, and the limitations of such models, which are great toy models for the linear layers that exist in large models.
In particular, there are some good questions and spin-off work that were mentioned at the end of the Anthropic post, and this one caught my eye:
Screenshot of an interesting follow-up work cited by Anthropic
This is particularly interesting because I smell bullshit. Not to shade the researchers (I consider them, Chris Olah in particular, to be some of the best, field-defining interpretability researchers), but because they only offer intuitive explanations without any justification, and this is a question that has many intuitively plausible explanations.
The Questions
Based on their very short write-up above, these questions seem natural to investigate:
- When do points not organize themselves into a circle? By circle, I mean with equal magnitude and equally spaced apart.
- What are key features of the “Optimization Failure” solution? Is it having multiple circles of different radii? Is it being able to linearly separate each point from the rest of the points?
This post will turn into a bit of a rabbit-hole deep-dive, but I hope to put in my most lucid insights that I have managed to hallucinate over the past months, that I believe will be crucial
The Problem Set-Up
I’ll do a very quick summary of Anthropic’s Toy Model set up, and only for the variant I’m interested in. As mentioned, they define a toy ReLU autoencoder that tries to reconstruct $x$, call the reconstruction $x’$, by doing the following:
\[x' = \text{ReLU}(W^\top W x + b)\]Where $x \in \mathbb{R}^n, W \in \mathbb{R}^{m \times n}$. $n$ refers to the data’s dimensionality (input dimensionality) and is much higher than $m$, which is the latent dimensionality (aka “hidden dimensionality”). We focus on the case of $n = 10,000, m = 2$.
There is also a sparsity level of $S = 0.999$, which means that every entry of each data sample has a $0.999$ chance of being $0$, and being uniformly sampled from $(0, 1]$ otherwise. After sampling, these data vectors are normalized to have magnitude 1.
Lastly, they define their dataset to have $T$ examples, which is varied from very small ($3$) to very large (million, or infinity, since they generate new data on the fly).
A More Intuitive Description
Let’s break down $x’ = \text{ReLU}(W^\top W x + b)$ and get a more intuitive understanding of each of these parts. After all, all linear layers basically follow this form, so an understanding of this will be very powerful.
- $Wx$, also known as $h$, is the latent vector. It is a smaller-dimensional (2D) representation of the data.
Purpose: The point of compressing dimensionality is important in learning because it forces the model to learn only the most generalizable features in its training data. If it had unlimited representational dimensions, it would simply memorize the training data. This is not very relevant to our discussion here.
![PCA-like 2D]()
How one might represent a 3D dataset in 2D space by projecting it down onto a plane $W^\top W x$, aka $W^\top h$, is the decoded vector. The decoding operation (applying $W^\top$) simply takes the compressed representation ($h$) and projects it out to the higher dimensional space (same as the input space). Note that even though $W^\top h$ lives in $\mathbb{R}^n$ (a high dimensional space), its effective dimensionality is that of $h$ (2), since no matter how you linearly transform (stretch & scale) a 2-dimensional set of points, you can never make it more than 2-dimensional. Note: In real neural networks, the decoding function is not simply a linear function like $W$, but some more expressive non-linear function. The decoding function represents the construction of output features (or in this case, the reconstruction of input features, because the model is trained to reconstruct its inputs) from a more compressed representation ($h$).
$\text{ReLU}$, the most exciting part, is the non-linear function that is magically able to extract more dimensions of information than exists in the latent space (2). We will see how this works in a while. Mechanically, $\text{ReLU}$ simply takes the max of $0$ and its input value. It’s useful to think of this as clamping values to $> 0$:
![Graph of ReLU]()
Graph of the ReLU function$b$, the bias. The bias is, in my opinion, the most deceptive part of the linear layer. Intuitively, it’s simple: adding a bias simply shifts everything by an offset. $(W^\top W x + b)$ is basically $W^\top W x$, but $b$ distance away from the origin. But, this bias is incredibly powerful and has the potential to disrupt many assumptions of feature geometry. For example, if a model learns to represent features linearly from the origin, such as the days of the week in this following image I got from here:
![Days of the week features]()
Features of each day of the week, Engels et. alAdding a bias would suddenly shift this circle of features to be, say, in the positive quadrant, rather than centered about the origin. Suddenly, the “Sunday” feature will be in the same line of sight as the “Wednesday” feature, which would make the features NOT linear relative to the origin! Purpose: we will discuss the purpose(s) of the bias in this post.
Why is this toy model worth our time? It captures basically all of the properties of actual linear layers in deep models. I’ll just list 2 key reasons that this toy model forms a good surrogate for large model components:
- This toy model is basically a “Linear Layer” (although the name has “Linear” in it, in deep-learning parlance, it just means one linear transformation followed by a non-linear activation function like ReLU, hence it’s not linear), which are the basic building blocks of ALL deep models
- This toy model implements compression and decompression, which are key behaviors of ALL deep models
All of the local behaviors of deep models can basically be studied with this toy model.
ReLU: Free Dimensions, Somewhat
So, in the world of linear transformations, one can never extract more dimensions of information than the minimum that was available at any point. For example, in the above illustration that I’ll paste below, we transformed a 3D dataset into a 2D one. But, given the 2D points (green), you can never reconstruct the original 3D points, because you don’t know how far along the dotted lines they originally were away from the plane.
How one might represent a 3D dataset in 2D space by projecting it down onto a plane
It follows that when you’ve compressed a $10,000$-dimensional vector to $2$ dimensions, like Anthropic did in the problem set-up, you would not be able to construct the original vector. However, throw in a non-linearity and some sparsity in your data points and you can actually reconstruct the original training data fairly well, sometimes without any training loss! How does this happen? Let’s take a look.
This discussion will pertain solely to the $\text{ReLU}$ non-linearity. The intuition will transfer somewhat to other activation functions, but activation functions have other kinds of inductive biases that will likely cause the representation of features that models end up learning to be different. This is out of the scope of this discussion
We start by considering the dataset. The dataset is incredibly sparse ($S = 0.999$), meaning that on average, about $10$ entries out of the $10,000$ in a single data vector are non-zero. Allowing multiple entries to be non-zero actually imposes several preferential optimization pressures on the model’s learning process that may be the cause of “Optimization Failure,” so to keep things simple, let’s look at the ideal case where on average, only about $1$ entry is active. That’s what Anthropic effectively did anyway, in the paper where they originally discovered the beautiful geometry (I still remember Zac Hatfield-Dodds going “whaaaaaat why are there pentagons here!!”) - Toy Models of Superpositions. This means that instead of having vectors that point in random direction in space, you essentially have the standard basis vectors (axial lines):
The three axial vectors in 3D space
I know that we are operating in $10,000$-dimensional space, but we can only focus on the dimensions where there is data with non-zero entries in that dimension. Suppose we have only 3 data points, then we can basically only look at the 3 dimensions where the data points are active in. Let’s look at how these 3 ‘data archetypes’ can be represented in a 2D latent space. There are 2 explanations:
Explanation 1: The effective dimensionality of the points is low (2D)
In this case, this is correct. You can observe that the 3 data points (the axial vectors) form a plane (a “2D simplex”). Since the effective dimensionality of the data is 2, then of course we can represent it without loss of information in 2D. However, it only works in the case of trying to represent $n$-dimensional data in a $(n-1)$-dimensional latent space. In general, $n$ points in $n$-dimensional space will form an $(n-1)$-dimensional simplex. If your latent space is coincidentally also $(n-1)$-dimensional or bigger, this is fine and dandy, but if your latent space is smaller than $(n-1)$ dimensions, then it will not be able to capture your $(n-1)$-dimensional object perfectly, and this explanation does not suffice.
Explanation 2: The ReLU Hypercube
I’m not quite sure how to phrase this as an explanation, but I’ll explain the intuition that makes it super clear. I mentioned that a helpful way of thinking about what ReLU does is that it clamps values to be $>= 0$, i.e. that it snaps all points to the edges of the positive half-space (1D) / quadrant (2D) / octant (3D), and so on; I’ll call the high-dimensional generalization of this the “ReLU hypercube,” because why not. Let’s see how this clamping visually looks in 1 to 3 dimensions:
Before (left) and after applying ReLU (right)
Notice how, because ReLU effectively zeros out all negative values, many points are snapped against the positive edges of the ReLU hypercube. This means that a lot of the post-ReLU points are scalar multiples of the axial vectors. Visually, you can see this happening by observing that the axis lines are particularly packed with data points in the right column. This also means that ReLU is particularly good at extracting axial vectors, which is what breaks the rotational invariance (“symmetry”) that is otherwise present in purely linear systems - the axial vectors are hence “privileged,” to borrow the term from Anthropic.
Also, because we are working in the ideal case where the data points are axis-aligned, this is particularly convenient for us, because the type of points that ReLU loves to output (axis-aligned), is exactly the type of points that we are looking to reconstruct. This may look like a beautiful coincidence, and you may think that in real models, you can’t expect axis-aligned vectors or features like this, but the reality is the these things are actually super common! By the end of this section, hopefully the intuition behind how ReLU gives us free dimensions is clear, and it’ll become obvious why learning such axis-aligned feature vectors is beneficial to the model, and is hence what happens in real models.
Because ReLU sort of “wraps” your data points around the ReLU hypercube at the origin, this mechanism has the effect of giving you “free dimensions.” In particular, if you wanted to be able to construct a dataset that comprises axial vectors, you could represent them in your latent space (that is, pre-ReLU) in such a way where the first axial vector’s ($e_1$) latent representation ($h(e_1)$) will be snapped onto $e_1$ by ReLU, the second axial vector’s ($e_2$) latent representation ($h(e_2)$) will be snapped onto $e_2$ by ReLU, and so on. To achieve this, you simply have to achieve the following criteria, which I’ll call the “unique single positive value” criteria:
The latent representation of $e_i$ must have a positive $i$-th element and all other elements be non-positive.
In 2D, you can achieve such criteria by putting all your datapoints’ latent representations on a line that extends into the $(\leq 0, +)$ and $(+, \leq 0)$ region. The purple dashed line illustrates such a line; the green points are the ReLU-ed versions of each of the datapoints’ latent representations.
Data latents (purple) being ReLU-ed onto the ReLU hypercube
If you looked from the under-side of the line (from the bottom-left to the top-right), you would see the axial vectors as the feature vectors (entries of $W$, since $W \in \mathbb{R}^{1 \times 2}$).
In 3D, there is such a similar way of arranging your latent representations, fulfilling the same criteria:
Data latents (on the plane) being ReLU-ed onto the ReLU hypercube in 3D
Note that points in the bluish purple regions of the plane (the latent space) fulfill this criteria and will get snapped onto the axial vectors (edges of ReLU hypercube), while points in the red regions of the plane do not, and will get snapped onto the faces of the ReLU hypercube, as opposed to the edges. Note too how the 3 axial vectors form that trigonal planar arrangement when viewed from the underside of the plane (right diagram), and these are precisely the columns of $W \in \mathbb{R}^{2 \times 3}$ in Anthropic’s Toy ReLU autoencoder that was trained to reconstruct 3D points with a 2D latent space:
Anthropic’s ReLU autoencoder features (rows of W), image from their paper. Note that m in this image represents n in this post.
Even though we (and Anthropic) achieve the geometrically perfect solution (the 3 vectors are equally spaced apart), in reality, the 3 feature vectors do not have to be equally spaced apart, because the plane has some wiggleroom to pivot at various different angles about the origin, such that the cube appears to be rotated at various angles when viewed from the underside of the plane. Any mildly rotated solution will work, so long as the axial vectors still fall inside their respective purple zones on the plane. If you’re trying to be able to extract more axial vectors from a low-dimensional latent space, the amount of wiggle room decreases, just because the regions of space that satisfy the single unique positive value criteria decreases. For example, when we try to represent the four $4$-dimensional axial lines on a 2D plane such that $\text{ReLU}$ can snap them back onto their respective axes, we find that the regions of space that satisfies the criteria has no area (no wiggleroom at all!) - it is just a set of 4 lines, that correspond to the origin-bound edges of the 4D hyper-cube shown below (yes, this is what 4D hyper-cube looks like when projected onto a 2D surface):
Four 4D axial vectors embedded into 2D space, + some rotation
This is precisely the job of $W$: their columns act as feature vectors, while their rows act as the span of the $m$-dimensional hyperplane in $n$-dimensional space.
Note that this description of $W$ is specific to this context where the compression weights $W$ are the same as the decoding weights ($W^\top$). In reality, they are commonly different, and so you have something like $x’ = \text{ReLU} (MWx + b)$, hence it would be that the rows of $M$ act as feature vectors.
The role of $W$
If you don’t have a very intuitive understanding of linear transformations, then what I wrote above about the “job of $W$” probably sounds very unobvious, so I’ll quickly illustrate the connection between the plane, the ReLU hypercube, and $W$.
We’ve established that to get the free dimensions via the ReLU Hypercube explanation requires you to rotate your plane (an $m$-dimensional object, in this case, 2) in $n$-dimensional space properly such that when you view the axis lines from the underside of the plane, they fall into their respective sectors of the plane that fulfill the criteria of having 1 positive entry and all else being non-positive. In linear algebra, applying a matrix to a vector (e.g. $Mx$) simply re-interprets $x$ to be not in the standard axes, but in the axes denoted by the columns of $W$. There will be a section on this from my linear algebra primer I’ll release in the future, but for now, here is a quick image of what I mean:
Left: an example vector x (purple). Right: Mx, where the maroon and orange lines are the columns of M
If $M$ were a tall matrix (more rows than columns, i.e. each column is tall), then the changed basis simply just lives in higher-dimensional space. If $M$ were a wide matrix (more columns than rows, i.e. each column is short), then in addition to changing your basis, you’re also reducing the dimensionality of $x$, because $x$ is going from a num_columns-dimensional vector to a $Mx$, a num_rows-dimensional vector.
So if we have $x \rightarrow Wx$, this means that the 2-dimensional columns of $W$ represent the basis vectors of the latent space (a 2D plane) in a 2D world. If we further have $Wx \rightarrow W^\top W x$, this means that the 3-dimensional columns of $W^\top$ represent the basis vectors of the latent space (same 2D plane) in a 3D world. This is why the rows of $W$ act as the span of the $m$-dimensional hyperplane in $n$-dimensional space.
Further, notice that because applying $W$ to $x$ is a reinterpretation of $x$’s entries as values in each of the columns of $W$, this is a remapping of $x$’s features from the standard basis world (e.g. first entry $=$ first feature $= [1,0,0]$), to the $W$ basis world (e.g. first feature is now represented via $W$’s first column - a 2D vector, instead of $[1,0,0]$). This is why the columns of $W$ act as feature vectors.
Now then, it becomes obvious that $W$’s role over training is just to adjust it values properly such that it defines a good $m$-dimensional hyperplane in $n$-dimensional space, such that ReLU can recover all the free dimensions that it needs to perfectly reconstruct the $n$-dimensional vectors with only an $m$-dimensional intermediate representation.
Limit of number of Free Dimensions
I find this intuition very clear: if you pivot a plane about the $n$-dimensional hypercube at the origin and look at the hypercube from the underside of the plane, you can always find a way to rotate the plane about that pivot point until you can see all the axial vectors (edges of the hypercube). But does this mean that you can always partition the plane into at least $n$ sectors that will correctly snap to their corresponding axial vectors after $\text{ReLU}$ (i.e. that fulfill the above criteria)? NO!
In general, the combined powers of $W$ and $\text{ReLU}$ can only perfectly reconstruct up to $2m$-dimensional axial vectors from an $m$-dimensional latent space. This is because you cannot select a set of $(n > 2m)$ $n$-dimensional vectors, where they satisfy the requirement of 1 positive element in a unique position and all else being non-positive, while having them all be on the same $m$-dimensional object. This means that, for example, a 2-dimensional latent space is expressive enough to reconstruct up to 4 dimensions (if we only care about the axial vectors), a 3-dimensional latent space can reconstruct up to 6 dimensions, and so on. This is great, but it’s definitely not infinite.
Repeat: four 4D axial vectors embedded into 2D space, + some rotation
While I am very confident that this upper bound is correct, and can be achieved with vector sets similar to $[1, -1, 0, 0], [-1, 1, 0, 0], [0, 0, 1, -1], [0, 0, -1, 1]$ (in 4D), I have no proof. It would be satisfying to have a proof, and I would greatly appreciate one.
So, $W$ and $\text{ReLU}$ are definitely not sufficient to account for what Anthropic found - “in fact, the ReLU output model can memorize any number of orthonormal training examples using $T$-gons if given infinite floating-point precision,” where $T$ represents the size of the dataset. The memorized training examples have these latent representations:
Latent vectors of datasets as T increases (under a certain threshold limit), image from their paper
So how on earth does this happen? This is where the bias $b$ comes in. But before that, let’s take a quick look at what happens when we try to memorize more than $2m$ $m$-dimensional axial vectors with only an $m$-dimensional latent space.
Imperfect Reconstruction: $n > 2m$
When $n > 2m$, the model will still try to learn a $W$ matrix whose columns are geometrically regular. In this case, we’ll examine $n = 5, m = 2$. Aside: if you tried to project an $n$-dimensional hypercube down to the $m$-dimensional hyperplane in a way that mapped the $n$-dimensional axial lines to the columns of $W$, the $n$-dimensional hypercube would look VERY regular (which is special; any randomly sampled orientation of a hypercube is exceedingly likely to have a 2D projection that looks very asymmetrical). I trained such an autoencoder and got the following columns of $W$ (and plotted the corresponding projection of the implied 5D hypercube):
Five 5D axial vectors embedded into 2D space, + some rotation
These five 2D feature vectors, when projected back out to 5D space via $W^\top$, (i.e. the columns of $W^\top W$), do not fulfill the unique positive value criteria; each of them has more than 1 positive entry. As such, $\text{ReLU}$ is unable to snap these 5D feature vectors cleanly back to the axial lines. The reconstruction is not perfect and has some loss. In particular, the extraneous / “wrong” positive entries of each column of $W^\top W$ will contribute some error.
Lastly, let’s address the intuition for why $W$ is still pressured to space these feature vectors out and achieve that regular geometry. There are some big caveats here that could be the cause(s) of optimization failure; take everything in the next 2 paragraphs at face-value. Remember that the Toy ReLU Autoencoder was trained using the Mean Squared Error (MSE) loss function, which is quadratic:
\[\mathcal{L}(W, X) = \frac{1}{|X|} \sum_{x \in X} (x' - x)^2\]This means that the greater the extraneous positive entries, the quadratically greater the loss. This makes it prefer evenly spaced out columns of $W$, and here’s why. Because angular space is finite, preferentially spacing out any pair of feature vectors will necessarily mean that other pairs of feature vectors will be squeezed closer together. The additional loss incurred by the squeezed vectors will weigh quadratically as much as the loss “saved” by the spaced out vectors, resulting in greater overall loss. This is why $W$ still faces optimization pressure to distribute the error evenly to all feature vectors and achieve that geometric regularity. This is standard behavior whenever you use some sort of convex loss function, which MSE is, but I thought I would just point it out.
Caveat 1: Loss is not actually quadratic, only convex
Actually, loss here is not quadratic with respect to the angle between the vectors, but a composition of the quadratic MSE Function and the cosine function, which ends up being a sinusoidal function:
\[\begin{align*} \mathcal{L}\left(\theta, x^{(1)}, x^{(2)}\right) &= \left(\max(0, \cos(\theta))\right)^2 \\ &= \frac{1}{2}\left(\cos(2\theta) + 1 \right) \end{align*}\]This is not quadratic, but it is still convex near $\theta = \pm \frac{\pi}{2}$.
Loss over angle between 2 vectors
Caveat 2: Loss is only convex near $\theta = \pm \frac{\pi}{2}$
This means that the system faces different optimization preferences when the feature vectors (columns of $W$) are in the range $\frac{-\pi}{2} < \theta < \frac{\pi}{2}$. We will investigate more later.
The role of $b$
By now, we can observe 2 things:
- Because of the optimization pressure to achieve geometric regularity, the representations of these axial vectors in the 2D latent space mimic the spokes of a wheel (generalizable to higher-dimensional hyperspheres).
- Trying to represent more than $2m$ axial vectors in an $m$-dimensional latent space will violate the unique positive value criteria.
Previously, I said that the combined powers of $W$ and $\text{ReLU}$ can only allow us to hydrate $2m$ $n$-dimensional axial lines from an $m$-dimensional latent space. The intuition for that is basically that in an $m$-dimensional latent space, you can only fit $2m$ orthogonal pairs of antipodal lines. If your vectors are forced to be less than $90^\circ$ apart from each other, then you get some interference. But, it turns out, that $b$ can basically absorb all that interference by trading off some “sensitivity”. Let’s see how that works. Much of the proof was worked out by Henighan in this Google Colab Notebook.
The intuition thus far was for the latent space (a plane in all our examples so far) to be pivoted at the origin, but it doesn’t have to be pivoted at the origin. In particular, you could add a negative bias in all dimensions such that you can shift the latent representations away from the origin, into the region of space where all dimensions are negative:
Feature representations in latent space, with bias
This isn’t super useful, because if all your features were represented in the negative regions of space, $\text{ReLU}$ would come along and zero everything; you have no information. However, you can shift the feature wheel in a way such that 1 feature vector is in a region of space with 1 positive dimension, while the rest of the feature vectors are in a region of space of non-negatives:
With just 3 vectors, you may think that the shifting is unnecessary, because even with the wheel of features pivoted at the origin ($0$ bias), the 3 feature vectors already fulfilled the unique positive value criteria. However, if we had more than 3 vectors, e.g. we have 8 $8$-dimensional axial vectors projected down onto a 2D wheel similarly, then the shifting is necessary to ensure that each one of those unique positive value criteria-fulfilling regions only has 1 feature vector.
In the above image, we can only see that 3 feature vectors are in the correct regions that make them fulfill the unique positive value criteria, but that is merely because we are unable to visualize an 8-dimensional space. If we could plot the above wheel in an 8D space as opposed to a 3D one, we would see that all 8 feature vectors are in their correct respective regions.
Can this really generalize to higher dimensions?
The last paragraph above is not intuively obvious because we can’t visualize more than 3 dimensions, so we can fall back on analysis to see that this really works. Here is where I will liberally quote the work of Henighan, in that Colab Notebook. This section is more mathematical analysis and less intuition, so feel free to skip it.
Henighan starts of by declaring the dataset $X$ the identity matrix, without loss of generality. This works, because thus far we’ve been assuming that the dataset is just points that align with the axis lines. This means that the data points (the rows of $X$) are one-hot, with their activated element being in unique positions. The identity matrix is just a neat way of arranging these rows, with the $i$-th row being active in the $i$-th position. This makes the objective of the ReLU toy model to reconstruct the identity matrix:
\[\begin{align*} X' & = \text{ReLU} \left( X W^\top W + b \right)\\ \Rightarrow I &= \text{ReLU} \left( W^\top W + b \right) \end{align*}\]This means that for this to work, the diagonal entries of $(W^\top W + b)$ need to be the ONLY entries $> 0$; hence the goal becomes to find $W$ and $b$ such that this is true.
He starts off with an a guess for $W$, based on the assumption that the columns (2D vectors) of $W$ will represent vectors that are spaced out evenly over a circle:
\[\begin{align*} W = \begin{bmatrix} \cos(0) & \cos(\phi) & \cdots & \cos(-\phi) \\ \sin(0) & \sin(\phi) & \cdots & \sin(-\phi) \end{bmatrix} \text{, where } \phi = \frac{2 \pi}{\text{num examples / features}} \end{align*}\]This guess turns out to work in most cases, because of my above explanation on how the partially convex loss function causes $W$ to distribute the total $2 \pi$ radians in a circle over each consecutive pair of vectors as evenly as possible. (Foreshadow possible optimization failure)
And so, $W^\top W$ looks like this, after a bunch of massaging with the trigonometric identities:
\[\begin{align*} W^\top W = \begin{bmatrix} \cos(0) & \cos(\phi) & \cos(2\phi) & \cdots & \cos(-\phi) \\ \cos(-\phi) & \cos(0) & \cos(\phi) & \cdots & \cos(-2\phi) \\ \cos(-2\phi) & \cos(-\phi) & \cos(0) & \cdots & \cos(-3\phi) \\ \cos(-3\phi) & \cos(-2\phi) & \cos(-\phi) & \cdots & \cos(-4\phi) \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \cos(2\phi) & \cos(3\phi) & \cos(4\phi) & \cdots & \cos(\phi) \\ \cos(\phi) & \cos(2\phi) & \cos(3\phi) & \cdots & \cos(0) \\ \end{bmatrix} \end{align*}\]But since that doesn’t look like much, let’s look at the matrix in visual form (if you had $10$ 10D vectors now):
A visual depiction of W.T @ W
I’ve highlighted the maximum value of each row (column too) in the above image. Remember that because $X’ = \text{ReLU}(W^\top W X + b)$, and $X = I$, each row here represent the pre-ReLU, pre-bias representations of each data point. From here, it’s obvious that you can just subtract the second-highest value of each row (i.e. $\cos(\phi)$) from every single value, and every row will just have ONE value that is $> 0$ - the blue entries, a.k.a. the diagonal. This explanation works for any $n$, so it will indeed generalize to any $n$.