TITLE: Gram-Schmidt over GF$(2)$
QUESTION [7 upvotes]: I am reading the paper The Steganographic File System by Ross Anderson, Roger Needham, and Adi Shamir. On page 4, paragraph 2, the authors write:

Finally, we use the Gram-Schmidt method to orthonormalise all the
  vectors from $i$ onwards by subtracting from the candidate $K_i$ all
  its components along later $K_j$ which the user knows by the chaining
  property of the $p_j$’s.

This simply means that the authors use the Gram-Schmidt algorithm with the ground field GF$(2)$, and from the context, each of the original vectors the algorithm is applied to has norm $1$. However, in this case the algorithm produces a basis which is not necessarily orthonormal, not even orthogonal. Am I right? Is this a serious flaw?

REPLY [4 votes]: The answer below was posted before the piece of information about all the participating vectors get tagged with random bits to make sure their weights are odd.

IMHO a more serious problem with using Gram-Schmidt over $GF(2)$ is that Gram-Schmidt may call for division by zero! This is because in the projection formula
$$
y\mapsto y-\frac{(y,x)}{(x,x)}x
$$
the denominator $(x,x)$ vanishes when $x$ is orthogonal to itself.
For example consider the following subspace of $GF(2)^3$:
$$
V=\{(a,b,c)\in GF(2)^3\mid a+b+c=0\}.
$$
You will quickly realize that every vector of $V$ is orthogonal to itself.
Have fun trying to Gram-Schmidt orthogonalize e.g. the basis $v_1=(1,1,0)$,
$v_2=(1,0,1)$ of $V$ :-)