# key size of public key in McEliece

The length of public-key matrix in McEliece cryptosystem is $n \times k$. Then the key size is $\frac{n \times k}{8}$ bytes. But I find that the key size is $\frac{(n-k) \times k}{8}$ bytes. why ?

Let $G$ be the public key matrix. With Gaussian elimination, you can find a Matrix $G'$ with $G = (E_k | G')$, where $E_k$ is the identity matrix with dimension $k$. Then, you only have to store $G'$. And $G'$ has the dimension $(n-k) \times k$.