Does the IV need to be known by AES (CBC mode)? I was thinking about this today and thought I should ask. I think I understand IV's enough to say that they are basically the same thing as Salts when talking about hashes. They are there to improve randomness between messages. If the IV is simple appended, or prepended to the plain text, why does AES need to know the IV to decrypt it? Can't it just decrypt it, drop the first x amount of bytes and then it has the original plaintext or does AES do some special jizz-jazz with the first bytes in the stream?
The initialization vector is XORed against the first plaintext block before encryption in CBC mode, as shown in the Wikipedia article on block cipher modes. After the first block is decrypted, you still have an intermediate value which has been XORed with the plaintext — without this, you have little hope of recovering the plaintext. However, you do not need the IV to decrypt subsequent blocks.

You could perform CBC in a way that would remove the need to know the initialization vector (note: this is not recommended or encouraged, just pointing it out for the novelty). If you use a null IV and use a random value for the first block of plaintext, you can discard this value and only transmit the ciphertext. Note that this actually gains you nothing, because now the ciphertext is one whole block longer!