Let $V$ be the set of real-valued continuous functions on the interval $[-3, 3]$. $S$ is set of real-valued functions with condition $f(-1) = f(1)$.
Is $S$ a subspace of $V$? Prove, and if not, why?
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3$\begingroup$ What's causing you problems? Do you know what conditions you need to check for $S$ to be a subspace? $\endgroup$– ah11950Mar 25, 2014 at 23:29
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1$\begingroup$ What do you need to show to show something is a subspace of a vector space? $\endgroup$– Thomas AndrewsMar 25, 2014 at 23:33
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$\begingroup$ I need to show that it contains the zero vector, that it contains the sum of two integers, and the scalar multiple… I just can't figure it out. @ThomasAndrews $\endgroup$– user136438Mar 25, 2014 at 23:38
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$\begingroup$ @user136438: You mean the zero function? If $z$ is the zero function, then $z(-1) = z(1) = 0$. $\endgroup$– copper.hatMar 25, 2014 at 23:40
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2 Answers
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Define $\phi:V \to \mathbb{R}$ by $\phi(f) = f(1)-f(-1)$. Note that $\phi$ is linear, and $S=\ker \phi$.
answered Mar 25, 2014 at 23:33
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$\begingroup$ +1 Nice simple solution, which leaves a lot left to solve. But this sort of problem usually comes before learning about linear functions and kernels, and what you have left to prove is really just the definition of "subspace." $\endgroup$ Mar 25, 2014 at 23:35
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$\begingroup$ @ThomasAndrews: Which just is an argument for introducing linear functions right from the start in a linear algebra course, before even introducing subspaces. Recognising linear maps at sight is quite easy, and most of the time can be justified without going back to the definition of linear maps, once a few fundamental examples are done, and combination rules are known. Here the fundamental example of polynomial evaluation, and the fact that linear combinations of linear maps are linear suffices. And as for subspaces, 99% of all examples are easily kernels or images of linear maps, as here. $\endgroup$ Mar 28, 2014 at 10:09
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Suppose that $f$ and $g$ are two real-valued functions with domain $\left[ -3, 3 \right] $ satisfying $f(-1) = f(1)$ and $g(-1) = g(1)$. Then $(f + g)(-1) = f(-1) + g(+1) = f(1) + g(1) =(f+ g)(1)$. Pick $r \in \mathbb{R}$. Then $(rf)(-1) = rf(-1) = rf(1) = (rf)(1)$.
