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#1 streek405   User is offline

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Linear Algebra: Determine if the set is a vector space

Posted 28 March 2015 - 02:05 AM

Am I doing this correctly?

Prove or disprove that the following is a vector space or not:

A.
the set {(x. y): x >= 0, y is a real number} with standard operations in R2.

let r = -1 and (x,y) = (1, 2)
then r(x, y) = (-1, -2), but x must be >= 0 (not sure about this part).
thus this is not a vector space.

B.
the set {(x,x): x is a real number} with standard operations in R2.

let (x1, x1) and (x2, x2) be an element of R2
then (x1, x1)+ x2, x2) = (x1 + x2, x1 + x2), which is an element of R2
thus closed under addition

let r be any real number and (x1, x1) be an element of R2
the, r(x1, x1) = (rx1, rx1) which is an element of R2
thus closed under scaler multiplication

thus this is a vector space

C. the set of all continuous functions defined on the interval [0, 1] with standard operations.

I thought this was false, but the book says its true.

My reasoning for this is because it is possible that if two functions are both on the interval [0, 1] and you add them up, then it is also possible that it may exceed the interval of [0, 1], ie [0,1] + [0,1] = [0,2]. Or am I thinking about this totally wrong?

This post has been edited by streek405: 28 March 2015 - 01:10 PM


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Replies To: Linear Algebra: Determine if the set is a vector space

#2 macosxnerd101   User is offline

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Re: Linear Algebra: Determine if the set is a vector space

Posted 13 April 2015 - 09:34 PM

A) More concisely, it fails to have additive inverses, so not a vector space.

B ) You must show (0, 0) \in V to show that the set is a vector space by the subspace test.

C) If you take two continuous functions and add them, they are still continuous. Same for multiplication. A function and its additive inverse are continuous. Similarly, multiplying a continuous function by a constant returns a continuous function. The proofs of these rely on delta-epsilon arguments. You can check the other axioms similarly. Note that they are all defined over the interval [0, 1], which is the domain. Nothing is said about their images.
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