This is a cross-post from this other question that I asked ~1 month ago in the mathematics forum, with no reaction. I am still stuck on this, looking for references or approaches to proofs. I hope I have a bit more luck in this forum! let me know if any clarifications would be needed as I am not a professional mathematician.


Consider the set of orthogonal matrices in the compact Stiefel manifold:

$\mathcal{M}_{N, k} := \{o \in \mathbb{R}^{N \times k} : \quad o^\top o = I_k \}$

We also define the projector $P_i$ of a matrix $o_i \in \mathcal{M}_{N, k}$ as:

$P_i = o_i o_i^\top$

And the principal angles $\theta_{i\rightarrow j}$ between 2 matrices $o_i, o_j$ (see e.g. Absil, Mahony, Sepulchre, 2003 or Qiu, Zhang, Li, 2005, s.2) as:

$SVD[o_i^\top o_j] = U \quad cos(\theta_{i\rightarrow j}) \quad V^T$

Finally, consider the following (squared and normalized) Frobenius projective metric between any 2 matrices $(o_i, o_j)$ in $\mathcal{M}_{N, k}$ (see e.g. Edelman, Arias, Smith, 1998, section 4.3):

$d_{pF}^2(o_i, o_j) = \frac{1}{k} \lVert P_i - P_j \rVert_F^2 = \frac{1}{k} \lVert sin^2(\theta_{i\rightarrow j}) \rVert_F^2 = 1 - \frac{1}{k} \lVert cos ^2(\theta_{i\rightarrow j}) \rVert_F^2 = 1 - \frac{1}{k} \lVert o_i^\top o_j \rVert_F^2 \in [0, 1]$


  1. Draw a random pair $(o_1, o_2) \sim (\mathcal{M}_{N, k} \times \mathcal{M}_{N, k})$ uniformly from the Haar measure (e.g. using the QR method)
  2. Compute $d_{pF}^2(o_1, o_2)$
  3. Repeat steps 1 and 2 for different random pairs, and average the results. We observe that $\frac{1}{k} \lVert o_i^\top o_j \rVert_F^2$ converges to $\frac{k}{N}$. Or equivalently, $\mathbb{E}[d_{pF}^2] = 1 - \frac{k}{N}$


I imagine this is a well known result in the literature. Where could I find it? Alternatively, how could I prove it?

The approach I thought of so far seems rather cumbersome, and I am not sure if it would be correct:

  1. Take the joint distribution $\mathcal{X}$ of principal angles for a Haar-uniform random orthogonal matrix in $\mathbb{R}^{N \times N}$ (presented e.g. in Rummler, 2002).
  2. Truncate $\mathcal{X}$ to $k$ entries, and calculate the normalized squared cosine, i.e. $\mathcal{Y} = \frac{1}{k} cos^2(\mathcal{X}_k)$
  3. Then, $\mathbb{E}[\mathcal{Y}]$ should equal $\frac{k}{N}$ (?)

Again, this is probably not necessary since I assume this result is known and I'm just missing the reference. Thanks in advance!


1 Answer 1


Fortunately, there's nothing deep going on here.

We'll use slightly different notation. Let $\mathbf{Q} \in \mathbb{R}^{N \times k}$ be a random matrix drawn uniformly from the Stiefel manifold of $N \times k$ orthonormal frames. In particular,

  • The columns of $\mathbf{Q}$ are orthonormal.
  • The marginal distribution of each column $\mathbf{q}_i$ is the uniform distribution over the Euclidean unit sphere, hence is isotropic: $\mathbb{E} [\mathbf{q}_i \mathbf{q}_i^* ] = N^{-1} \mathbf{I}_N$, where $\mathbf{I}_N$ is the $N \times N$ identity matrix.

The orthogonal projector onto the range of $\mathbf{Q}$ takes the form $\mathbf{P} = \mathbf{QQ}^*$, where $*$ is the transpose. The key observation here is that $$ \mathbb{E}[ \mathbf{P} ] = \sum_{i=1}^k \mathbf{E}[ \mathbf{q}_i \mathbf{q}_i^* ] = \sum_{i=1}^k \frac{1}{N} \mathbf{I}_N = \frac{k}{N} \mathbf{I}_N. $$

Given two independent realizations $\mathbf{Q}_1, \mathbf{Q}_2$, we form the associated orthogonal projectors $\mathbf{P}_1, \mathbf{P}_2$. Write $\mathbb{E}_1, \mathbb{E}_2$ for the expectations with respect to the randomness in $\mathbf{Q}_1, \mathbf{Q}_2$ respectively. Using the definition of the Frobenius norm in terms of the trace, linearity, and independence, we find that \begin{multline*} \mathbb{E} \Vert \mathbf{P}_1 \mathbf{P}_2 \Vert_{\mathrm{F}}^2 = \mathbb{E} \operatorname{trace}[ \mathbf{P}_2 \mathbf{P}_1^2 \mathbf{P}_2 ] = \mathbb{E} \operatorname{trace}[ \mathbf{P}_2 \mathbf{P}_1 \mathbf{P}_2 ] \\ = \mathbb{E}_2 \operatorname{trace}[ \mathbf{P}_2 \mathbb{E}_1[ \mathbf{P}_1] \mathbf{P}_2 ] = \frac{k}{N} \mathbb{E}_2 \operatorname{trace}[ \mathbf{P}_2^2 ] = \frac{k^2}{N^2} \operatorname{trace}[ \mathbf{I}_N ] = \frac{k^2}{N}. \end{multline*} This is the core part of the computation.

Finally, we compute the expected Frobenius projective distance between the pair: \begin{multline*} \mathbb{E} \mathrm{dist}_{\mathrm{F}}^2(\mathbf{Q}_1, \mathbf{Q}_2) = \mathbb{E}\big[ 1 - k^{-1} \Vert \mathbf{Q}_1^* \mathbf{Q}_2 \Vert_{\mathrm{F}}^2 \big] \\ = \mathbb{E}\big[ 1 - k^{-1} \Vert \mathbf{P}_1 \mathbf{P}_2 \Vert_{\mathrm{F}}^2 \big] = 1 - k^{-1} (k^2 / N) = 1 - k / N. \end{multline*} That's it.

  • $\begingroup$ Thanks a lot! crystal clear $\endgroup$
    – fr_andres
    Aug 14 at 19:06

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