# find a basis for the subspace and prove it is a basis, state the dimension

Each of the following sets spans a subspace of the space of all functions from $\bf R$ to $\bf R$.

$\exp x, \sin(x), \cos(x)$

$\exp x, \cosh(x), \sinh(x)$

this is just an extra credit problem, but I have no clue how to go about it. If someone please could help me:)

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Doesn't any set of vectors span a subspace of the space containing it...? – tomasz Nov 28 '12 at 20:27

I will provide some hints.

For the first family, show that it forms a linearly independent set. So it's a basis of the generating subspace of functions from $\Bbb R$ to itself.

For the second one, recall that $\cosh x+\sinh x=e^x$, so the family is not linearly independent. But it is one you remove $\exp x$ for example.

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an additional hint: To prove that $f_1 , f_2, f_3$ are linearly independent, assume $\lambda_1 f_1 + \lambda_2 f_2 + \lambda_3 f_3 = 0$ for some $\lambda_1, \lambda_2, \lambda_3 \in \mathbb{R}$ , that is for ALL $x \in \mathbb{R}$: $$\lambda_1 f_1(x) + \lambda_2 f_2(x) + \lambda_3 f_3(x) = 0$$. Then prove that $\lambda_1, \lambda_2, \lambda_3$ must all be $0$. To do this, it is usually enough to plug in some different values for $x$ and solve the system of linear equations you get from that. – Brusko651 Nov 28 '12 at 20:14
Thank you!!! i will try to do that:)) – Andriy Lysak Nov 28 '12 at 20:20
for the second one i dont understand the part about removing exp x, could you please explain a bit more? – Andriy Lysak Nov 28 '12 at 22:26
The generated family will be the same. – Davide Giraudo Nov 29 '12 at 9:40

The set of vectors $\{\sinh(x),\cosh(x),e^{x}\}$ is a linearly dependent set because the last vector (function) can be written as a linear combination of the two previous ones, more specifically as the sum of the previous two.

Hence if we remove it and consider only the functions spanned by $\{\sinh(x),\cosh(x)\}$ we get the same subspace!

This is true in general. If you have a linear dependent set of vectors, you can always remove a vector that is in the span of the others, and the new (smaller) set of vectors will generate the same subspace.

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Oh I got it now, Thank You!!! – Andriy Lysak Nov 28 '12 at 22:51