If a binary linear code $C = \{0,1\}$, then it is $1$-dimensional and its generator matrix $G$ is $[1]$.
It's dual code will be $\{0\}$ and $0$-dimensional and has basis $\{\}$.
Then what is a check matrix $H$ for $C$??
As $H$ is defined to be a matrix whose rows form a basis for $C$'s dual, I gather from the above that H does not exist.
But we know that $HG^T=0$, which means that $H=[0]$.
We also know that if $G=[I|A]$, then $H=[-A^T|I]$, which means that $H=[1]$.
I'm contradicting myself all over; please help me sort this one out.
It doesn't come up much in elementary linear algebra classes, but a matrix can have dimension $0 \times n$, $n \times 0$, or even $0 \times 0$ (where $n$ is a positive integer).
It can be a little tricky to deal with if it's your first time dealing with vacuous things.
So by the definition you gave, $H$ has to be a $0 \times 1$ matrix.
The product $HG^T$ is the product of a $0 \times 1$ matrix with a $1 \times 1$ matrix, and thus has dimension $0 \times 1$. There is only one $0 \times 1$ matrix, so $HG^T$ is automatically the zero matrix of that dimension.
Some other things to note: if $A$ is defined by that matrix equation with $G$, then that means $A$ is a $1 \times 0$ matrix. This is consistent with the formula for $H$ which adjoins $-A^T$ -- a $0 \times 1$ matrix -- with the $0 \times 0$ identity matrix.
(note that the $0\times 0$ identity matrix is the same as the $0 \times 0$ zero matrix)
Since there is only one $0 \times 1$ matrix, it does indeed give the correct formula for $H$.