Syndrome is Zero iff Vector is Codeword

Theorem
Let $C$ be a linear $\tuple {n, k}$-code whose master code is $\map V {n, p}$

Let $G$ be a (standard) generator matrix for $C$.

Let $P$ be a standard parity check matrix for $C$.

Let $w \in \map V {n, p}$.

Then the syndrome of $w$ is zero $w$ is a codeword of $C$.

Proof
Let $G = \paren {\begin{array} {c|c} \mathbf I & \mathbf A \end{array} }$.

Let $c \in \map V {n, p}$.

Then, by definition of $G$, $c$ is a codeword of $C$ $c$ is of the form $u G$, where $u \in \map V {k, p}$.

Thus $c \in C$ :

where:
 * $v = u \mathbf A$


 * $\paren {\begin{array} {c|c} u & v \end{array} }$ denotes the $1 \times n$ matrix formed from the $k$ elements of $u$ and the $n - k$ elements of $v$.

Let $w \in \map V {n, p}$.

$w$ can be expressed in the form:
 * $w = \paren {\begin{array} {c|c} u_1 & v_1 \end{array} }$

where $u_1 \in \map V {k, p}$.

The syndrome of $v$ is then calculated as:

It follows that the syndrome of $w$ is zero $w$ is the concatenation of $u_1$ and $v_1$, where:
 * $v_1^\intercal = \mathbf A^\intercal u_1^\intercal = \paren {u_1 \mathbf A}^\intercal$

Thus the syndrome of $w$ is zero $w$ is a codeword of $C$.