I am trying to figuring out what is meant by Q and F in the 4th equation mentioned in this paper: Diachronic Word Embeddings Reveal Statistical Laws of Semantic Change

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They haven't define in the paper what is meant by Q and F. So, I am assuming that they are some fixed mathematical notations.

  • 5
    $\begingroup$ The $Q$ is defined right there in the notation. You're minimizing over all the possible orthogonal matrices $Q$. This probably means that $W(t)$ is a matrix in some arbitrary coordinate system, which may be different than the coordinate system of $W(t+1)$, so to compare them, you need to "align" the coordinate systems somehow. Looks like the authors opted to measure the distance between the W's by finding the minimal possible distance over all possible coordinate systems. The $F$, I don't know, it's a decoration on the norm they are using to compare matrices. There must be some context. $\endgroup$ Jul 10, 2019 at 5:38
  • 5
    $\begingroup$ F is probably the Frobenius norm (square root of sum of squares of the entries). $\endgroup$
    – Flounderer
    Jul 10, 2019 at 5:40
  • 1
    $\begingroup$ See en.wikipedia.org/wiki/Orthogonal_Procrustes_problem $\endgroup$
    – Henry
    Jul 10, 2019 at 5:42
  • $\begingroup$ There are 4 votes to close this Q but it seems perfectly clear to me. It should stay open. $\endgroup$
    – amoeba
    Jul 13, 2019 at 19:40

2 Answers 2


The answers have already been provided in the comments. Just so this question has an answer attached...

$Q$ is defined in the optimization problem itself as the variable we're minimizing with respect to. The expression

$$\underset{Q^T Q = I}{\text{argmin }} f(Q)$$

could also be written as:

$$\underset{Q}{\text{argmin }} f(Q) \quad \text{s.t. } Q^T Q = I$$

That is: find the value of $Q$ that minimizes the objective function $f$, subject to the constraint that $Q^T Q =I$. This means that $Q$ is an orthogonal matrix.

$\| \cdot \|_F$ denotes the Frobenius norm. For a matrix $A$:

$$\| A \|_F = \left ( \sum_i \sum_j A_{ij}^2 \right)^\frac{1}{2}$$

  • $\begingroup$ you beat me to it by a few seconds ;-) $\endgroup$
    – Gregg H
    Jul 10, 2019 at 14:00
  • $\begingroup$ @GreggH Now we have the answer doubly memorialized =) $\endgroup$
    – user20160
    Jul 10, 2019 at 14:03

The commenters essentially answered this question, but I will memorialize it here.

The argmin (or argmax) notation can be a bit confusing, because it often introduces a dummy variable (much like the dx or dt in an integral). As Matthew Drury's comment indicates, the $\mathbf{Q}$ is the dummy variable here (so it won't be introduced elsewhere in the paper, as it only serves a place holder function).

Next, the argmin operator asks you to figure out which value of $\mathbf{Q}$ gives the smallest value. However, instead of returning the smallest value for the expression in the argmin, you instead want the value generating this smallest number. With this in mind, your $\mathbf{Q}$ is essentially your $\mathbf{R}^{(t)}$...so, $\mathbf{Q}$ is defined however $\mathbf{R}^{(t)}$ is defined.

Lastly the $F$ subscript on the norm function $||·||$ most likely indicates the Frobenius norm (https://en.wikipedia.org/wiki/Matrix_norm#Frobenius_norm; as suggested by @Flounderer). This is just the square-root of the sum of the squares of all of the entries in the matrix inside the norm function.


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