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7 the dimension of C plus the dimension of C ⊥ equals the length n, that is, C ⊥ has dimension n−k. Calculating dimensions twice, we learn that C ⊥⊥ has dimension k. As this space contains C and has the same dimension as C, it is equal to C. 7) Lemma. If C is an [n, k] linear code over F , then its dual C ⊥ is an [n, n − k] linear code over F and C ⊥⊥ = C. 2 The linear code C is self-orthogonal if C ⊥ ≥ C and is self-dual if C ⊥ = C. So, for instance, a binary repetition code of even length is self-orthogonal, as is the [7, 3] binary dual Hamming code.

We shall meet many M DS codes later when we discuss the generalized Reed-Solomon codes. 16 ) Problem. Prove that the dual of an M DS codes is also an M DS code. 17 ) Problem. Prove that a binary M DS code of length n is one of {0}, the repetition code, the parity check code, or all Fn 2. 2 Encoding and information If we are transmitting with an [n, k] linear code over the field F , then we think of our message as being provided as k-tuples from F , members of the space F k . We can encode using the generator matrix G by mapping the message k-tuple x to the codeword xG.

Indeed each occurrence of ? may filled arbitrarily by an alphabet letter (typically 0 for a linear code) and then flagged as unreliable. , the erasure symbol. Which point of view is the best will depend upon the particular situation. Since C contains codewords of weight d = dmin (C) as well as 0 of weight 0, we could never hope to correct d erasures; but we can decode up to d − 1 erasures correctly. 3) Proposition. Let C be an [n, k, d] linear code over F with check matrix H whose rows are hi , for i = 1, .

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