TITLE: Given $W=ULV^T$ and a vector $\mathbf{x}$, can we compute $UL^kV^T\mathbf{x}$ without doing the SVD, for any integer k?
QUESTION [5 upvotes]: Consider a matrix $W \in \mathbb{R}^{n\times m}$ with corresponding singular value decomposition, $W = ULV^T$,  and a vector $\mathbf{x} \in \mathbb{R}^{m}$. Is it possible to compute matrix-vector-products of the form $U L^k V^T \mathbf{x}$ without explicitly computing the SVD?
If $k$ is odd, this is easy. For example, $W W^{T} W\mathbf{x} = UL^3V^T\mathbf{x}$. But I can't see an obvious way to solve this for even powers. 
For context, I want to avoid the SVD as computing the matrix vector products will typically be much cheaper (e.g. for $k \ll n)$. Ideally, a working algorithm would have complexity $O(kmn)$ instead of $O(mn^2 + nm^2)$.

REPLY [2 votes]: I think an important observation to make here, especially since the SVD is derived from the spectral theorem, is that the matrices $WW^T$ and $W^TW$ are symmetric positive semi-definite matrices.  As mentioned by iljusch in the comments, we will stick to the case that $n = m$.  In particular we find that
$$
WW^T \;\; =\;\; UL^2U^T \hspace{3pc} W^TW \;\; =\;\; VL^2V^T.
$$
Because they are positive semi-definite we can in fact find their square-roots since all of the singular values will be nonnegative:
$$
\sqrt{WW^T} \;\; =\;\; ULU^T \hspace{3pc} \sqrt{W^TW} \;\; =\;\; VLV^T.
$$
Therefore in computing even powers we find that square-roots can serve as a key to performing your computations without actually performing SVD:
\begin{eqnarray*}
\sqrt{WW^T}W & = & UL^2V^T \\
\sqrt{WW^T}(WW^TW) & = & \left (WW^T\right )^{3/2}W \\
&  = & UL^4V^T. \\
\end{eqnarray*}
Looks as if your general formula for positive integers is given by:
$$
UL^kV^T \;\; =\;\; \left (WW^T\right )^{\frac{k-1}{2}}W.
$$
This can just as well be formulated as
$$
UL^kV^T \;\; =\;\; W \left (W^TW\right )^{\frac{k-1}{2}}.
$$
The one remaining problem I see is whether or not this ultimately lowers your computational complexity.  This might be an interesting read for you.