Expectation of $QQ^T$ where $Q^TQ=I$ It's exercise 1.1 on p.2 of this book.
The goal is to is to show that, for some random matrix $Q \in \mathbb{R}^{n\times k}$ where $k<n$ and the columns of $Q$ are orthogonal (i.e. $Q^T Q = I$; and assuming $Q$ is uniformly distributed in whatever space it's a member of), then for $u \in \mathbb{R}^{n}$ and $v = \sqrt{n/k}Q^Tu$, the following holds:
$$E(\|v\|^2)=\|u\|^2$$
I tried a few monte carlo experiments, and it does work out. It seems that
$$E(QQ^T) = \textrm{diag}(k/n,\ldots,k/n)$$
If $Q$ is relaxed to be standard gaussian, I can work it out analytically. But I'm stumped how this works if $Q$ has an orthogonality constraint.
So how does $$E(\|v\|^2)=\|u\|^2$$ work?
 A: Your observation that $E(QQ^T) = \operatorname{diag}(k/n,\dotsc,k/n)$ points in the right direction. The expected value of the off-diagonal elements of $QQ^T$ vanishes by symmetry, since flipping the sign of a row of $Q$ preserves $Q^TQ=1$ but changes the sign of the off-diagonal elements of $QQ^T$. The diagonal elements must all be equal by symmetry, and $\operatorname{tr}QQ^T=\operatorname{tr}Q^TQ$ shows that they must be $k/n$.
Then by the cyclic invariance of the trace and the linearity of trace and expectation
$$
\begin{eqnarray}
E\left(\lVert v\rVert^2\right)
&=&
E\left(v^Tv\right)
\\
&=&
E\left(\operatorname{tr}\left(v^Tv\right)\right)
\\
&=&
\frac nkE\left(\operatorname{tr}\left(u^TQQ^Tu\right)\right)
\\
&=&
\frac nkE\left(\operatorname{tr}\left(uu^TQQ^T\right)\right)
\\
&=&
\frac nk\operatorname{tr}\left(E\left(uu^TQQ^T\right)\right)
\\
&=&
\frac nk\operatorname{tr}\left(uu^TE\left(QQ^T\right)\right)
\\
&=&
\operatorname{tr}\left(uu^T\right)
\\
&=&
\operatorname{tr}\left(u^Tu\right)
\\
&=&
u^Tu
\\
&=&
\lVert u\rVert^2\;.
\end{eqnarray}
$$
Alternatively, you can argue as follows. Since permuting the columns of $Q$ preserves $Q^TQ=1$, the expected values of the squares of the components of $v$ are all the same by symmetry, so $E(\lVert v\rVert^2)$ is just $k$ times one of these expected values. A component of $v$ is $\sqrt{n/k}$ times the scalar product of a random unit vector with $u$. If we rotate to a basis in which $u$ has only one non-zero component, this scalar product becomes $\lVert u\rVert$ times that component of the random unit vector. By symmetry, the expected value of the square of a component of an $n$-dimensional unit vector is $1/n$, and the result follows.
