Hausdorff measure and linear transformation in R^n

Question

Is there a formula of the type $$H^s(TA)=c H^t(A)$$,

where $T:R^n\mapsto R^m$ and $s$ is the Hausdorff dimension of $TA=\{Ax\ |\ x\in A\}$ and $t$ is the Hausdorff dimension of $A$?

Answer 1

HINT:

Unless $T$ is a multiple of an isometry, there exist $v_1$, $v_2$ two unit vectors such that $0<|T v_1|<|T v_2|$. So we get two sets of dimension $1$ and measure $1$ with images of dimension $1$ and different measure.

Answer 2

Usually, no.

Maybe the simplest case: $n=2, m=1, s=2, t=1$ and $T(x_1, x_2) = x_1$ the projection of the plane onto the $x$-axis. If you know the area of a set $A$ in the plane, does that determine the length of the projection $T(A)$ of the set?

Not even if $A$ is an ellipse (with its interior): rotating the ellipse will give you projections with different lengths.

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Also, the formula is not correct for dilations: $A$ expanded to $2A$. Then the formula would tell us

$$ H^s(T(2A)) = H^s(2 TA) = 2^s H^s(TA) = 2^s c H^t(A) = 2^{s-t} c H^t(2A) $$ So even if constant $c$ works for $A$, it doesn't work for $2A$.