Basis and vector space

Question

Please I need help solving this problem: Find the basis and the dimension of the vector space V spanned by the given vectors: $$1, \cos^2x, \sin^2x, \cos 2x$$ and $$1, \cosh^2x, \sinh^2x, \cosh 2x, e^{2x}, e^{-2x}.$$ I know I will have to bring out a matrix from the vectors then transform the matrix to an echelon matrix, but how do I bring out the matrix?

Answer 1

To bring out a matrix, you need to have a basis, which you don't. Instead do some (hyperbolic) trigonometry to eliminate those vectors which linearly depend on the others.

• For $\langle 1,\cos^2x,\sin^2x,\cos 2x\rangle$, the linearisation formulæ: $$\cos^2x=\frac{1+\cos 2x}2,\quad\sin^2x=\frac{1-\cos 2x}2, $$ show that $\langle 1,\cos^2x,\sin^2x,\cos 2x\rangle=\langle 1,\cos 2x\rangle$.

On the other hand, it is easy to check that any linear relation: $$\forall x,\enspace\lambda \cdot 1+\mu\cos 2x=0$$ implies $\lambda=\mu=0$, so the constant function $1$ and $\cos 2x$ are linearly independent, and $$\dim(\langle 1,\cos^2x,\sin^2x,\cos 2x\rangle)=\dim(\langle 1, \cos 2x\rangle)=2.$$

• Similarly, $$\cosh^2x=\frac{\cosh 2x+1}2,\quad\sinh^2x=\frac{\cosh 2x-1}2, $$ so $\;\bigl\langle 1,\cosh^2x,\sinh^2x,\cosh 2x,\mathrm e^{2x},\mathrm e^{-2x}\bigr\rangle=\bigl\langle 1,\cosh 2x,\mathrm e^{2x},\mathrm e^{-2x}\bigr\rangle.$

On the other hand, $\;\cosh 2x=\dfrac{\mathrm e^{2x}+\mathrm e^{-2x}}2$, so $$\bigl\langle 1,\cosh 2x,\mathrm e^{2x},\mathrm e^{-2x}\bigr\rangle=\bigl\langle 1,\mathrm e^{2x},\mathrm e^{-2x}\bigr\rangle$$

and one checks the constant function $1$, $\mathrm e^{2x}$ and $\mathrm e^{-2x}$ are linearly independent, so that $$\dim\bigl(\bigl\langle 1,\cosh^2x,\sinh^2x,\cosh 2x,\mathrm e^{2x},\mathrm e^{-2x}\bigr\rangle\bigr)=\dim\bigl(\bigl\langle 1,\mathrm e^{2x},\mathrm e^{-2x}\bigr\rangle\bigr)=3.$$