Is ( mathbb{R}^3 ) a Vector Space Over ( mathbb{C} )?

In this article, we will delve into the fundamental question of whether ( mathbb{R}^3 ) qualifies as a vector space over the field of complex numbers ( mathbb{C} ). This exploration will focus on the essential properties of a vector space and the stringent requirements that must be met for ( mathbb{R}^3 ) to be considered a vector space over ( mathbb{C} ).

Basic Properties of a Vector Space

A vector space ( V ) over a field ( F ) must satisfy several key properties:

Closure under Addition: For any two vectors ( mathbf{u}, mathbf{v} in V ), their sum ( mathbf{u} mathbf{v} ) must also reside within ( V ).

Closure under Scalar Multiplication: For any vector ( mathbf{u} in V ) and any scalar ( c in F ), the product ( cmathbf{u} ) must also reside within ( V ).

Other properties including: associativity, commutativity, additive identity, additive inverse, distributive properties over vector addition, and distributive properties over scalar multiplication.

Examining ( mathbb{R}^3 ) as a Vector Space Over ( mathbb{C} )

To determine if ( mathbb{R}^3 ) is a vector space over ( mathbb{C} ), we first need to check if it meets the closure property under scalar multiplication with complex scalars.

Closure Under Addition

It is straightforward to verify that ( mathbb{R}^3 ) is closed under addition. For any two vectors ( mathbf{u}, mathbf{v} in mathbb{R}^3 ), their sum ( mathbf{u} mathbf{v} ) remains a vector in ( mathbb{R}^3 ) since the sum of two real vectors is another real vector.

Closure Under Scalar Multiplication

However, the second property - closure under scalar multiplication - is where ( mathbb{R}^3 ) fails to meet the requirements for being a vector space over ( mathbb{C} ).

Consider a vector ( mathbf{u} (x, y, z) in mathbb{R}^3 ) and a complex scalar ( c a bi ), where ( a, b in mathbb{R} ). The product ( cmathbf{u} ) is calculated as follows:

[cmathbf{u} (a bi)(x, y, z) (ax - by, ay bx, az - bz)]

If we examine the resulting vector ( (ax - by, ay bx, az - bz) ), we see that any non-zero scalar ( c ) (specifically, the purely imaginary components) results in components that might not be real. This failure to maintain real entries after scalar multiplication shows that ( mathbb{R}^3 ) is not closed under scalar multiplication with complex scalars.

Counterexample

Even if one were to attempt to define an operation where scalar multiplication was kept 'real', such an operation would still fail to meet the requirements for a vector space. For instance, if we define ( cr ) such that it only retains the real part of the complex product, nonzero purely imaginary scalars ( c ) would produce the zero vector when multiplied with any nonzero real vector ( r ).

Here is a concrete example to illustrate this:

Let ( r (1, 0, 0) ) and ( c i ) (a purely imaginary complex number).

[ cr i(1, 0, 0) (0, 0, 0) ]

This confirms that ( mathbb{R}^3 ) fails in satisfying the closure property under scalar multiplication with complex scalars.

Conclusion

In summary, ( mathbb{R}^3 ) does not qualify as a vector space over ( mathbb{C} ) because it fails to meet the closure property under scalar multiplication with complex scalars. However, it does qualify as a vector space over the field of real numbers ( mathbb{R} ).

References and Further Reading

If you wish to delve deeper into the theory of vector spaces and fields, you may find the following resources helpful:

Halmos, Paul R. Fundamental Concepts of Higher Algebra. Springer Verlag, 1958.

Halmos, Paul R. Fundamental Concepts of Linear Algebra. Springer Verlag, 1958. [Note: The specific reference provided is outdated, and the updated edition is Finite-Dimensional Vector Spaces].

These books provide rigorous and detailed explanations of the fundamental concepts in vector spaces over different fields, including complex and real numbers.