Checking if a linear code exists - singleton , hamming and gilbert-varshamov bounds do not help. Suppose I want to check if a (11, 6, 4) code exists.
I cannot prove non-existence using the singleton and the hamming bound. I also cannot prove existence using the gilbert-varshamov bound. I'm not sure what is the best way to go then -  shall I just try to create a parity-check matrix?
 A: An extended Hamming code has parameters $(16,11,4)$. By shortening it in five positions you get the desired $(11,6,4)$-code. The process can also be described as follows. Let $C$ be that $(16,11,4)$-code. Select a set of five bit positions. Let $C'$ be the subspace consisting of those words of $C$ that have a zero at all those five positions. This amounts to adding five extra parity checks, so $C'$ is a linear code. Because all the words of $C'$ are also words of $C$, its minimum weight is still $\ge4$. Because we place 5 extra parity checks, the dimension of $C'$ is at least $11-5=6$. Finally, if we throw away the five shared zeros of all the words of $C'$, we get a code of length $11$. 

Equivalently you can just throw away any five columns from the parity check matrix 
$$
H=\left(\begin{array}{l}
1111 1111 1111 1111\\
0000 0000 1111 1111\\
0000 1111 0000 1111\\
0011 0011 0011 0011\\
0101 0101 0101 0101
\end{array}\right)
$$
of the $(16,11,4)$ extended Hamming code.
A: I tried to simply create a parity check matrix for the code and it turned out to be quite easier than I thought.
Assuming $r = n - k = 5$
$$
H=\left(\begin{array}{l}
11100\\
01110\\
00111\\
10011\\
11001\\
11111\\
10000\\
01000\\
00100\\
00010\\
00001
\end{array}\right)
$$
Any 3 $(d-1)$ rows are linearly independent and there is at least one set of of 4 $(d)$ linearly dependent rows.
