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$W = \{1,\cosh x,\sinh x, \cosh 2x, \sinh 2x\}$ is a subspace of the vector space of continuous functions on $\mathbb{R}$ with real values, and

$\mathcal{B} = \{1,\cosh x,\sinh x, \cosh 2x, \sinh 2x\}$

$\mathcal{C} = \{\cosh^2x,\cosh x, \sinh x, \sinh 2x, \sinh^2x\}$

are bases for $W$. Find the change of basis matrixes $\underset{\mathcal{C}\leftarrow\mathcal{B}}{P}$ and $\underset{\mathcal{B}\leftarrow\mathcal{C}}{P}$.

Finding a change of basis matrix from a "traditional" $m\times n$ matrix with scalars, I have no problem with. But with this problem i have no idea how to even begin to write down the unit vectors of these subspaces. And I think that I have to do this, if not, I'd have some nasty expressions in the reduces augmented matrix, trying to turn i.e. $\cosh x$ into $1$.

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Note that $\mathcal{B}$ and $\mathcal{C}$ share three basis elements. For the other two basis elements, note that $\cosh^2 x - \sinh^2 x = 1$, and $\cosh 2x = \cosh^2 x + \sinh^2 x$ (And similarly, $\cosh^2 x = \frac{1 + \cosh 2x}{2}$ and $\sinh^2 x = \frac{-1 + \cosh 2x}{2}$). This shows that a transformation matrix from $\mathcal{B}$ to $\mathcal{C}$ is given by $$\left(\begin{matrix} 1 & 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 1\\ -1 & 0 & 0 & 1 & 0\\ \end{matrix}\right).$$

A matrix from $\mathcal{C}$ to $\mathcal{B}$ can be found the same way.

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  • $\begingroup$ Thank you! This is pretty obvious now that I see how it's done. My book says that I can just find the inverse of that matrix to find the one from C to B! $\endgroup$ – Kexmonster Oct 15 '19 at 18:16

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