Is the field of this vector space understood to be the reals?

"Let V denote the set of all differentiable real-valued functions defined on the real line."


Does this automatically mean that this vector space is over the field of reals?

Why or why not?

I ask because I need to prove this is a vector space. But, if I pick some element a from F (the field), then the scalar multiplication of a and an element of V is only real-valued if a is a real. This would make this scalar multiplication not an element of V, making it not a vector space, if F were a field that contained non-real elements. So, I must assume that this V is over the field of reals in order for it to prove it is a vector space, but why am I warranted to make that claim?

Sorry if this is a dumb question, I am just starting LA independently.

Thanks

Yes, your argument is correct. If a is a "scalar" and v is a "vector" then av must be a vector. If you multiply a "differentiable real-valued function defined on the real line" by a complex number, the result would no longer be "differentiable real-valued function defined on the real line". Now, it would be possible to have the space of "differentiable real-valued functions defined on the real line" over the field of rational numbers since the product of a rational and a real number is a real number- but that would be very unusual.

Quote:

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Originally Posted by HallsofIvy

Yes, your argument is correct. If a is a "scalar" and v is a "vector" then av must be a vector. If you multiply a "differentiable real-valued function defined on the real line" by a complex number, the result would no longer be "differentiable real-valued function defined on the real line". Now, it would be possible to have the space of "differentiable real-valued functions defined on the real line" over the field of rational numbers since the product of a rational and a real number is a real number- but that would be very unusual.

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Hey HoI,

So from your response, when being asked to prove that V is or is not a vector space, if the field is not mentioned, I should assume the field is one that would not make it immediately impossible for V to be a vector space.

So there is nothing about "the set of all differentiable real-valued functions defined on the real line" that inherently makes the field R.

And, this is a matter of taking it for granted that the field is an appropriate one that wouldn't trivialize the exercise.

Is that correct?

Thank you.

Yes, that seems like a reasonable interpretation.