I got to this by experimentation, but can someone maybe provide an explanation for this?
The
CBC
mode processed as $c_i = E_k(p_i \oplus c_{i-1})$ with $c_0 = IV$The
ECB
mode processed as $c_i = E_k(p_i)$
If you set the first block with $$c_o = IV = 0$$ in the CBC
mode, than it is calculated as $$c_1 = E_k(p_1).$$ This is exactly as ECB
mode.
The next blocks, however, will not be equal;
- in
ECB
mode $$c_i = E_k(p_i)$$ - whereas in
CBC
mode $$c_i = E_k(p_i \oplus c_{i-1}) \neq E_k(p_i) \text{ for } 1 < i \leq m$$ where $m$ is the number of blocks.
Therefore the equality is only valid for the first block.
-
$\begingroup$ Exactly as I suspected, thank you for a more formal explanation <3 $\endgroup$ – bbozo Oct 15 '18 at 10:51