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.

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  • $\begingroup$ Exactly as I suspected, thank you for a more formal explanation <3 $\endgroup$ – bbozo Oct 15 '18 at 10:51

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