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The distance ''d'' of a linear code ''C'' also equals the minimum number of linearly dependent columns of the check matrix ''H''.
''Proof:'' Because <math>\boldsymbol{H} \cdot \boldsymbol{c}^T = \boldsymbol{0}</math>, which is equivalent to <math>\sum_{i=1}^n (c_i \cdot \boldsymbol{H_i}) = \boldsymbol{0}</math>, where <math>\boldsymbol{H_i}</math> is the <math>i^{th}</math> column of <math>\boldsymbol{H}</math>. Remove those items with <math>c_i=0</math>, those <math>\boldsymbol{H_i}</math> with <math>c_i \neq 0</math> are linearly dependent. Therefore, <math>d</math> is at least the minimum number of linearly dependent columns. On another hand, consider the minimum set of linearly dependent columns <math>\{ \boldsymbol{H_j} | j \in S \}</math> where <math>S</math> is the column index set. <math>\sum_{i=1}^n (c_i \cdot \boldsymbol{H_i}) = \sum_{j \in S} (c_j \cdot \boldsymbol{H_j}) + \sum_{j \notin S} (c_j \cdot \boldsymbol{H_j}) = \boldsymbol{0}</math>. Now consider the vector <math>\boldsymbol{c'}</math> such that <math>c_j^{'}=0</math> if <math>j \notin S</math>. Note <math>\boldsymbol{c'} \in C</math> because <math>\boldsymbol{H} \cdot \boldsymbol{c'}^T = \boldsymbol{0}</math> . Therefore, we have <math>d \le wt(\boldsymbol{c'}) </math>, which is
== Example: Hamming codes ==
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