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S is the vector space consisting of the set of all linear combinations of the functions $f_1(x)=e^x$, $f_2(x)=e^{-x}$, $f_3=sinh(x)$. Find a basis for S and find the dim[S].

First, I let

$c_1f_1$+$c_2f_2$+$c_3f_3$=$f$

Then, since $sinh(x) =\frac{e^x-e^{-x}}{2}$,

$f_3=\frac{f_1-f_2}{2}$

Substituting back in, I can get

$c_1f_1$+$c_2f_2$+$c_3\frac{f_1-f_2}{2}$ =$f$

$(c_1+\frac{1}{2}c_3)f_1$+$(c_2-\frac{1}{2}c_3)f_2$ =$f$

Am I on to something?

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  • 1
    $\begingroup$ Nice name, by the way. :) $\endgroup$ – FundThmCalculus Nov 29 '14 at 17:58
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Yup! :) You are correct. You have shown that $f_3$ is a linear combination of the other two basis vectors, and thus not needed to form the basis of the vector space (although you could have just as easily proved any one of the three given functions wasn't needed). As a result, since you have two basis functions, your subspace dimension is 2.

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$f_1$ and $f_2$ are linearly independent and, as you point out, $f_3$ is dependent on these two, so a basis is $B=\{f_1,f_2\}$ and the dimension is 2.

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  • $\begingroup$ I hate deciding who to "accept an answer" from..... You answered first, but FundThmCalculus provided more insight.... I feel bad haha $\endgroup$ – Math StackExchange Nov 29 '14 at 18:13
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$f_1$ and $f_2$ are linearly independent (check using the definition), while $f_3$ is a linear combination of the first two. So the basis (i.e. linearly independent set which spans the space) is just $\{f_1,f_2\}$. Since this set contains two elements, the space is two dimensional.

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