0
$\begingroup$

Let $V = \{(a_1, a_2, ..., a_n): a_i \in \mathbb{C}$ for $ i = 1, 2, .., n\}$; so $V$ is a vector space over $\mathbb{R}$. Is V a vector space over the field of real numbers with the operations of coordinatewise addition and multiplication?

The solution says,

Yes. All the conditions are preserved when the field is the real numbers.

I don't understand how $V$ can be a vector space over $\mathbb{R}$? $V$ itself is specified as being over the field of complex numbers ($V = \{(a_1, a_2, ..., a_n): a_i \in \mathbb{C}$), so how can the question then claim that $V$ is a vector space over the real numbers?

I would greatly appreciate it if someone could please take the time to clarify my misunderstanding.

  • 0
    So, colloquially speaking, you take your complex vector space, and then "forget" all the non-real scalar multiplications. In category theory we have the "forgetful functor" for situations like this. – 2017-01-27

1 Answers 1

2

Check the definition of a vector space over a field, for example on Wikipedia: https://en.wikipedia.org/wiki/Vector_space#Definition

Being over $\mathbb{R}$ means that vectors can be multiplied by real numbers and you get again a vector in $V$. This is certainly true in your example since complex numbers multiplied by real numbers are again complex numbers.

  • 0
    Ahh, this makes sense. The set itself contains complex elements, but the field that the set is over is itself the real numbers. Am I understanding this correctly? – 2017-01-27
  • 1
    I'm not sure what you mean by "the field that the set is over is itself the real numbers". The point is that a vector space being defined over some field $F$ means that there is an operation for multiplying vectors with elements in $F$. In the example you gave, you can define such operation for the field of real numbers, but also for complex or rational numbers since $\mathbb{C}$ is closed under multiplication by such numbers. – 2017-01-27
  • 0
    I understand. Thank you for the assistance. – 2017-01-27