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I realize this is a simple problem, but I'm working through a book and trying to practice the concepts. I have posted a solution in the answers.

Problem: Working over the field with two elements, find the rank of the following matrix: \begin{bmatrix} 1 & 1 & 0 \\[0.3em] 0 & 1 & 1 \\[0.3em] 1 & 0 & 1 \end{bmatrix}

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Because the rank is equal to the dimension of the row space, we consider the span of the following three vectors:

$\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}.$

Notice that adding the first and second produces the third. It is easy to see that the first and second are linearly independent (one would have the be a multiple of the other, but the only thing we can multiply by in this field is 1 or 0, both of which don't work) so the rank is $2$.

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    +1 This is one of the better ways of looking at the problem. A coding theorist might say that all the columns have an even number of 1s. IOW they are "orthogonal" to $(1,1,1)^T$, and hence cannot span the entire space. That doesn't tell nearly the whole story, though. It is just a quick trick specific to $GF(2)$. Your approach works more generally.2012-06-03