Introduction

To show that the set S={(1,0,0),(1,1,0),(1,1,1)} is a basis of C³(C), we need to demonstrate two things: linear independence and spanning.

Linear Independence

To prove that the set S is linearly independent, we need to show that the only solution to the equation a(1,0,0) + b(1,1,0) + c(1,1,1) = (0,0,0) is a=b=c=0.

Let's solve the equation:

a(1,0,0) + b(1,1,0) + c(1,1,1) = (0,0,0)
(a+b+c, b+c, c) = (0,0,0)

This gives us the following system of equations:

a+b+c = 0 ...(1)
b+c = 0 ...(2)
c = 0 ...(3)

From equation (3), we can immediately see that c=0. Substituting this value into equation (2), we get b=0. Finally, substituting the values of b=0 and c=0 into equation (1), we find a=0.

Therefore, the only solution to the equation is a=b=c=0, which proves that the set S is linearly independent.

Spanning

To prove that the set S spans C³(C), we need to show that for any vector (x,y,z) in C³(C), there exist scalars a, b, and c such that a(1,0,0) + b(1,1,0) + c(1,1,1) = (x,y,z).

Let's solve this equation:

a(1,0,0) + b(1,1,0) + c(1,1,1) = (x,y,z)
(a+b+c, b+c, c) = (x,y,z)

This gives us the following system of equations:

a+b+c = x ...(4)
b+c = y ...(5)
c = z ...(6)

From equation (6), we can immediately see that c=z. Substituting this value into equation (5), we get b=y-z. Finally, substituting the values of b=y-z and c=z into equation (4), we find a=x-y.

Therefore, for any vector (x,y,z) in C³(C), we can choose a=x-y, b=y-z, and c=z such that a(1,0,0) + b(1,1,0) + c(1,1,1) = (x,y,z).

Conclusion

We have shown that the set S={(1,0,0),(1,1,0),(1,1,1)} is both linearly independent and spans C³(C). Therefore, it is a basis of C³(C).