= Solution
The first stage is
$$
(I-\mu A_y)\mathbf u^{n+1/2}=\mathbf u^n.
$$
Because $A_y=I\otimes T$, this consists of $m$ independent tridiagonal systems of dimension $m$. The <Thomas algorithm> solves each in $O(m)$ operations, for a total of $O(m^2)$.
The second stage,
$$
\mathbf u^{n+1}=(I+\mu A_x)\mathbf u^{n+1/2},
$$
is multiplication by a matrix with only $O(m^2)$ nonzero entries, so it also costs $O(m^2)$. Hence one complete step uses at most
$$
\boxed{O(m^2)}
$$
arithmetic operations.
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