I have studied that for orthogonality to exist between two binary sequences:
[Number of bit agreements – Number of bit disagreements]/sequence length=0
Eg, for an orthogonal matrix X given by:
\begin{bmatrix}0&0&0&0\\0&1&0&1\\0&0&1&1\\0&1&1&0\end{bmatrix}
Any two rows satisfy the rule above, showing that any two binary sequences defined by two different rows of X are orthogonal.
However, if we consider the case of Generator(G) and Parity Check(H) Matrix, the rows of G are orthogonal to rows of H. But the rule stated above, doesn't prove it. Why so?
1) How do we verify that two binary sequences are orthogonal?
2) Can we extend the same testing rule to check orthogonality between the rows of two matrices?
Example:
G =\begin{bmatrix}1&1&0&1&0&0\\0&1&1&0&1&0\\1&0&1&0&0&1\end{bmatrix}
and H = \begin{bmatrix}1&0&0&1&0&1\\0&1&0&1&1&0\\0&0&1&0&1&1\end{bmatrix}
These two matrices claim to have orthogonal rows . So, if we take (let's say) first row of G and H and apply the formula I mentioned, why don't they give 0?
They do give 0 by modulo 2 addition of the product of two rows, but why the formula that I mentioned, is not applicable?
Best Answer
For binary vectors to be orthogonal, you are taking the normal dot product, and adding $\bmod{2}$. So they have to agree in "1" an even number of times to be orthogonal.
Each nonzero vector represents a 1-dimensional subspace of $\mathbb{F}_{2}^{k}$. If $v \neq 0$, then $v^{\perp}$ will be a subspace of $\mathbb{F}_{2}^{k}$ with dimension $k-1$. Think of the vector $[0,0,1]$ in $\mathbb{F}_{2}^{3}$: it is orthogonal to $\{[0,0,0], [1,0,0], [0,1,0], [1,1,0]\}$. This is the hyperplane satisfying $x_{3} = 0$. This is similar to standard orthogonality, where $v^{\perp}$ in $\mathbb{R}^{k}$ will have dimension $k-1$.
I think your formula involves representing binary vectors as 1/-1 vectors, and orthogonality being over $\mathbb{R}$ in the usual sense.