Math Problem Statement

If C is the field of complex numbers, which vectors in C 3 are linear combinations of (1, 0, −1), (0, 1, 1), and (1, 1, 1)?

Solution

We are asked to find which vectors in C3\mathbb{C}^3 (the space of 3-dimensional complex vectors) can be written as linear combinations of the three given vectors: (1,0,1)(1, 0, -1), (0,1,1)(0, 1, 1), and (1,1,1)(1, 1, 1).

A vector v=(x,y,z)C3\mathbf{v} = (x, y, z) \in \mathbb{C}^3 is a linear combination of (1,0,1)(1, 0, -1), (0,1,1)(0, 1, 1), and (1,1,1)(1, 1, 1) if there exist complex scalars aa, bb, and cc such that:

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

This results in the following system of equations for the components of the vectors:

a+c=x(from the first component)a + c = x \quad \text{(from the first component)} b+c=y(from the second component)b + c = y \quad \text{(from the second component)} a+b+c=z(from the third component)-a + b + c = z \quad \text{(from the third component)}

Now, let's solve this system for aa, bb, and cc in terms of xx, yy, and zz.

Step 1: Solve the system

  1. From the first equation a+c=xa + c = x, we can solve for aa:

a=xca = x - c

  1. From the second equation b+c=yb + c = y, solve for bb:

b=ycb = y - c

  1. Substitute a=xca = x - c and b=ycb = y - c into the third equation a+b+c=z-a + b + c = z:

(xc)+(yc)+c=z-(x - c) + (y - c) + c = z x+c+yc+c=z-x + c + y - c + c = z x+y+c=z-x + y + c = z

  1. Solve for cc:

c=z+xyc = z + x - y

Step 2: Find aa and bb

Now that we know c=z+xyc = z + x - y, substitute this back into the expressions for aa and bb:

a=x(z+xy)=yza = x - (z + x - y) = y - z b=y(z+xy)=2yxzb = y - (z + x - y) = 2y - x - z

Step 3: Conclusion

Thus, for any vector v=(x,y,z)C3\mathbf{v} = (x, y, z) \in \mathbb{C}^3, we can write it as a linear combination of the given vectors if and only if a=yza = y - z, b=2yxzb = 2y - x - z, and c=z+xyc = z + x - y, where aa, bb, and cc are complex scalars.

This means that every vector in C3\mathbb{C}^3 can be expressed as a linear combination of the given three vectors, and they span C3\mathbb{C}^3. Therefore, any vector in C3\mathbb{C}^3 is a linear combination of (1,0,1)(1, 0, -1), (0,1,1)(0, 1, 1), and (1,1,1)(1, 1, 1).


Would you like further details on this process or clarification on any step? Here are some related questions that might help:

  1. How do we check if a set of vectors is linearly independent?
  2. What is the concept of a vector space basis, and how does it relate to this problem?
  3. How can we extend this idea to higher-dimensional spaces like Cn\mathbb{C}^n?
  4. What is the rank of a matrix, and how does it relate to the span of vectors?
  5. How does the solution change if we work with real numbers instead of complex numbers?

Tip: When working with linear combinations, it's helpful to start by writing out the equations clearly, then use substitution or elimination to solve for the coefficients of the linear combination.

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Math Problem Analysis

Mathematical Concepts

Linear Combinations
Vector Spaces
Complex Numbers
System of Linear Equations
Spanning Sets

Formulas

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

Theorems

Linear Combination Theorem
Spanning Set Theorem

Suitable Grade Level

Grades 11-12