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Topic: Linear Algebra: Subspaces of Vector Spaces  (Read 10331 times)

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Offline HighTek

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Linear Algebra: Subspaces of Vector Spaces
« on: October 15, 2008, 04:41:57 PM »
I'm having a hard time determining if the following subset is a subspace or not:

To be a subspace, it has to be closed under addition AND scalar multiplication.

The set of all even functions: f(-x) = f(x). The book says that it is indeed a subspace, but I'm not convinced. It IS closed under addition, but the scalar multiplication I came up with looks like this:

c = -1, f(x) = x^2, x = 1

c * f(-x) = (-1) * (-1)^2 = (-1) * 1 = -1, therefore, c * f(-x) = - f (x). Is that right?

Or does the negative in front of the function NOT matter? I hope someone can follow what I'm trying to type. Thanks.
I'm experiencing both Alzheimer's and Deja Vu right now, I think I forgot this before.

Offline Yggdrasil

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Re: Linear Algebra: Subspaces of Vector Spaces
« Reply #1 on: October 16, 2008, 02:14:31 AM »
Given f(x) is even, we want to show that a scalar multiple of f(x) ([specifically (-1)*f(x)] is also even.

For simplicity, let's define g(x) = (-1)f(x)

To show that g(x) is even, we want to show that g(x) = g(-x)

Let's consider g(-x):
g(-x) = (-1)f(-x)

Since we know f(x) is even, f(-x) = f(x).  Therefore,
g(-x) = (-1)f(x) = g(x)

Offline HighTek

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Re: Linear Algebra: Subspaces of Vector Spaces
« Reply #2 on: October 16, 2008, 01:14:22 PM »
Thanks for the response. This class is way more abstract than I like. I'm trying to wrap my head around it, but I think I got. I do better with numbers than variables.

Thanks again...
I'm experiencing both Alzheimer's and Deja Vu right now, I think I forgot this before.

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