I have the following recursive sequence:


$Z\_k = Z\_{k-1} - AA^TZ\_{k-1}xx^T$ where $Z\_k \in \mathbb R^{n \times d}, A \in \mathbb R^{n \times d}, d > n, rank(A) = n, x \in \mathbb R^{d \times 1}$


$A$ is a constant matrix, $x$ is a constant vector. Theoretically, this sequence of matrices $Z\_k$ is entirely determined by $Z\_0$ the initial element of the sequence.


If I give you $Z\_0$ you are able to find $Z\_k$ for any $k$.


Suppose I want to find $Z\_{100}$. Is it possible to find an expression for $Z\_k$ as a function of $Z\_0$ so that I don't actually have to find $Z\_1, Z\_2, ..., Z\_{99}$?



Recurrence relation in matrices
Let $B=AA^T$ and $X=xx^T$. Note that
$X^k=\|x\|^{2k-1}X$.


\begin{align\*}
 Z\_1& = Z\_{0} - BZ\_{0}X \\
 %
 Z\_2& = Z\_{1} - BZ\_{1}X 
 %
= Z\_{0} - BZ\_{0}X -B\big[Z\_{0} - BZ\_{0}X\big]X
 %
= Z\_{0} - 2BZ\_{0}X + B^2Z\_{0}X^2 \\
 %
 Z\_3& = Z\_{2} - BZ\_{2}X 
 %
= Z\_{1} - BZ\_{1}X -B\big[Z\_{1} - BZ\_{1}X\big]X
 %
= Z\_{1} - 2BZ\_{1}X + B^2Z\_{1}X^2 \\
 %
& = Z\_{0} - 3BZ\_{0}X + 3B^2Z\_{0}X^2-B^3Z\_{0}X^3 .
\end{align\*}


So that by induction,


\begin{align\*}
 %
 Z\_n& =\sum\_{k=0}^n-1^k\binom nk B^kZ\_0X^k
 %
= Z\_0 + \sum\_{k=1}^n-1^k\binom nk \|x\|^{2k-1}B^kZ\_0X\\
 %
 & = Z\_0-\frac1{\|x\|^2}Z\_0X + \frac1{\|x\|^2}
 %
\bigg[\sum\_{k=0}^n-1^k\binom nk \|x\|^{2k}B^k\bigg]Z\_0X \\
 %
& =Z\_0-\frac1{\|x\|^2}Z\_0X + \frac1{\|x\|^2}
 %
[I-\|x\|^2B]^nZ\_0X \\
 %
& =Z\_0-\frac1{\|x\|^2}Z\_0xx^T + \frac1{\|x\|^2}
 %
[I-\|x\|^2AA^T]^nZ\_0xx^T
 %
\end{align\*}


