1. **Problem statement:** We are given a linear code $C$ with a spanning set $\{11110111, 01100011, 00110010, 11001010, 01010001\}$. We need to find a basis for $C$.
2. **Recall:** A basis for a linear code is a minimal set of linearly independent vectors that span the code. Since the given set spans $C$, the basis is a linearly independent subset of these vectors.
3. **Step:** Write the vectors as rows of a matrix and perform row reduction to find the linearly independent vectors.
The matrix is:
$$\begin{bmatrix}
1 & 1 & 1 & 1 & 0 & 1 & 1 & 1 \\
0 & 1 & 1 & 0 & 0 & 0 & 1 & 1 \\
0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 \\
1 & 1 & 0 & 0 & 1 & 0 & 1 & 0 \\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 1
\end{bmatrix}$$
4. **Row reduce:**
- Use the first row as pivot.
- Subtract first row from fourth row:
$$\text{Row}_4 \to \text{Row}_4 - \text{Row}_1 = \cancel{1}-\cancel{1}, \cancel{1}-\cancel{1}, 0-1, 0-1, 1-0, 0-1, 1-1, 0-1 = 0,0,-1,-1,1,-1,0,-1$$
- Since we are in binary (mod 2), $-1 \equiv 1$, so:
$$0,0,1,1,1,1,0,1$$
5. **Update matrix:**
$$\begin{bmatrix}
1 & 1 & 1 & 1 & 0 & 1 & 1 & 1 \\
0 & 1 & 1 & 0 & 0 & 0 & 1 & 1 \\
0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 1 & 1 & 1 & 0 & 1 \\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 1
\end{bmatrix}$$
6. **Check linear dependence:**
- Rows 3 and 4 are similar except for the fifth and sixth entries.
- Subtract row 3 from row 4:
$$\text{Row}_4 \to \text{Row}_4 - \text{Row}_3 = 0,0,\cancel{1}-\cancel{1}, \cancel{1}-\cancel{1}, 1-0, 1-0, 0-1, 1-0 = 0,0,0,0,1,1,1,1$$
7. **Update matrix:**
$$\begin{bmatrix}
1 & 1 & 1 & 1 & 0 & 1 & 1 & 1 \\
0 & 1 & 1 & 0 & 0 & 0 & 1 & 1 \\
0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 1
\end{bmatrix}$$
8. **Check row 5:**
- Subtract row 2 from row 5:
$$\text{Row}_5 \to \text{Row}_5 - \text{Row}_2 = 0, \cancel{1}-\cancel{1}, 0-1, 1-0, 0-0, 0-0, 0-1, 1-1 = 0,0,-1,1,0,0,-1,0$$
- Mod 2:
$$0,0,1,1,0,0,1,0$$
9. **Update matrix:**
$$\begin{bmatrix}
1 & 1 & 1 & 1 & 0 & 1 & 1 & 1 \\
0 & 1 & 1 & 0 & 0 & 0 & 1 & 1 \\
0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\
0 & 0 & 1 & 1 & 0 & 0 & 1 & 0
\end{bmatrix}$$
10. **Rows 3 and 5 are identical, so row 5 is dependent and can be removed.**
11. **Final basis vectors correspond to rows 1, 2, 3, and 4:**
$$\{11110111, 01100011, 00110010, 00011111\}$$
**Answer:** A basis for $C$ is $\{11110111, 01100011, 00110010, 00011111\}$.
Basis Linear Code B39F9A
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