TITLE: Dimensions of self-affine sets
QUESTION [13 upvotes]: Let $A$ be a $2\times 2$ matrix which we assume to be contracting, i.e., the exists $\alpha\in(0,1)$ such that
$$
\|A {\mathbf x}\|_2\le \alpha\|{\mathbf x}\|_2,\quad \forall {\mathbf x}\in\mathbb R^2.
$$
Let for $i=1,2$,
$$
T_i(\mathbf x)=A\mathbf x+\mathbf v_i
$$
for some vectors $\mathbf v_1, \mathbf v_2$. Let now $\Lambda$ denote the attractor for the iterated function system $\{T_1, T_2\}$, i.e., the unique set such that $\Lambda=T_1(\Lambda)\cup T_2(\Lambda)$ (a self-affine set). 
Question. Assuming that $T_1(\Lambda)\cap T_2(\Lambda)=\varnothing$, are there closed formulae for the Hausdorff and box-counting (Minkowski) dimensions of $\Lambda$? 
I know that there are plenty of papers on this subject but it looks like every time I follow yet another link, I only fall further down the rabbit hole...

REPLY [6 votes]: As the previous two answers correctly indicate, there is no simple, closed form expression for the dimension of a self-affine set.  Here is a concrete example illustrating why we might expect difficulty in finding such a formula.  While not a proof, it at least shows that dimension is discontinuous with respect to the IFS.
This is not my example.  I believe I learned it out of one of Falconer's books but I don't have the reference immediately handy.
Let 
$$A=\left(
\begin{matrix}
 1/2 & 0 \\
 0 & 1/4
\end{matrix}
\right)
$$
and consider the iterated function system
\begin{align}
T_1({\mathbf x}) &= A{\mathbf x}+\langle0,1\rangle \\
T_2({\mathbf x}) &= A{\mathbf x}+\langle\alpha,0\rangle,
\end{align}
where $\alpha$ is a positive parameter.  Here is an illustration of the effect of this IFS for $\alpha=0.2$ on the unit square.

We see the image of the large square under the two functions as rectangles in the upper left and lower middle.  Inside those are smaller rectangles formed by the second iterate of the IFS.  Inside those is an approximation of the attractor.  The key observation is that there is overlap between the projections onto the $x$-axis of each rectangle in every approximation.  As a result the projection onto the $x$-axis of the fractal attractor is an interval and, therefore has dimension at least one.  This picture is for $\alpha=0.2$ but is indicative of the behavior for all $\alpha>0$.
Let's now consider the case $\alpha=0$, as shown below.  The dimension is now $\log(2)/\log(4)=1/2<1$.  Thus, the dimension is not continuous as a function of $\alpha$ at $\alpha=0$.