# Expected value of the projective metric between random orthogonal Stiefel matrices in $\mathbb{R}^{N \times k}$ equals $1 - \frac{k}{N}$

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.

### Definitions:

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):

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

### Setup:

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, $$\mathbb{E}[d_{pF}^2] = 1 - \frac{k}{N}$$

### Question:

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!

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.

• Thanks a lot! crystal clear Aug 14 at 19:06