1. ## dimension

hello
what's the dimension of the subspace $\displaystyle F$ of $\displaystyle \mathfrak{F}(\mathbb{R},\mathbb{R})$ spanned by $\displaystyle f_1(x)=\sin^2(x),f_2(x)=\cos^2(x),f_3(x)=\sin(2x)$ and $\displaystyle f_4(x)=\cos(2x)$
thanks.

2. Originally Posted by Raoh
hello
what's the dimension of the subspace $\displaystyle F$ of $\displaystyle \mathfrak{F}(\mathbb{R},\mathbb{R})$ spanned by $\displaystyle f_1(x)=\sin^2(x),f_2(x)=\cos^2(x),f_3(x)=\sin(2x)$ and $\displaystyle f_4(x)=\cos(2x)$
thanks.
you should explain very clearly what your vector space is. just using a non-standard and weird notation such as $\displaystyle \mathfrak{F}(\mathbb{R},\mathbb{R})$ is not good enough. i guess your vector space is the $\displaystyle \mathbb{R}$-vector space of

all real-valued functions defined on $\displaystyle \mathbb{R}$. then $\displaystyle \dim_{\mathbb{R}}F=2$ and $\displaystyle \{\sin^2 x, \sin (2x) \}$ is a basis for $\displaystyle F.$

Edit: i meant $\displaystyle \dim_{\mathbb{R}} F=3$ and $\displaystyle \{1,\sin^2x, \sin(2x) \}$ is a basis for $\displaystyle F.$ sorry for the mistake.

3. Originally Posted by NonCommAlg
you should explain very clearly what your vector space is. just using a non-standard and weird notation such as $\displaystyle \mathfrak{F}(\mathbb{R},\mathbb{R})$ is not good enough. i guess your vector space is the $\displaystyle \mathbb{R}$-vector space of

all real-valued functions defined on $\displaystyle \mathbb{R}$. then $\displaystyle \dim_{\mathbb{R}}F=2$ and $\displaystyle \{\sin^2 x, \sin (2x) \}$ is a basis for $\displaystyle F.$
your guess was right,anyway would you show me how you got that basis.
thank you.

4. Originally Posted by NonCommAlg
you should explain very clearly what your vector space is. just using a non-standard and weird notation such as $\displaystyle \mathfrak{F}(\mathbb{R},\mathbb{R})$ is not good enough. i guess your vector space is the $\displaystyle \mathbb{R}$-vector space of

all real-valued functions defined on $\displaystyle \mathbb{R}$. then $\displaystyle \dim_{\mathbb{R}}F=2$ and $\displaystyle \{\sin^2 x, \sin (2x) \}$ is a basis for $\displaystyle F.$
I don't think this is right since assume there exist $\displaystyle a,b \in \mathbb{R}$ such that $\displaystyle \cos ^2 (x)= a\sin (2x) + b\sin ^2 (x)=2a \sin (x) \cos (x) + b\sin ^2 (x)$ then evaluation in $\displaystyle 0$ gives $\displaystyle 1=0$. Thus $\displaystyle A= \{ \sin ^2 (x), \cos ^2 (x) , \sin (2x) \}$ is l.i. and since $\displaystyle \cos ^2 (x) - \sin ^2 (x) = \cos (2x)$ we get that $\displaystyle A$ is a basis for $\displaystyle F$.

5. Originally Posted by Jose27
I don't think this is right since assume there exist $\displaystyle a,b \in \mathbb{R}$ such that $\displaystyle \cos ^2 (x)= a\sin (2x) + b\sin ^2 (x)=2a \sin (x) \cos (x) + b\sin ^2 (x)$ then evaluation in $\displaystyle 0$ gives $\displaystyle 1=0$. Thus $\displaystyle A= \{ \sin ^2 (x), \cos ^2 (x) , \sin (2x) \}$ is l.i. and since $\displaystyle \cos ^2 (x) - \sin ^2 (x) = \cos (2x)$ we get that $\displaystyle A$ is a basis for $\displaystyle F$.
hi "Jose27" would you please explain the method you used to find that basis.
thanks.

6. Assume $\displaystyle \sin ^2(x) = a \sin (2x)$ for some $\displaystyle a\in \mathbb{R}$ then evaluation in $\displaystyle \frac{\pi }{2}$ will give $\displaystyle 1=0$ so these two functions are l.i. Then by my previous post we get that $\displaystyle \cos ^2 (x) \notin span \{ \sin ^2 (x), \sin (2x) \}$ so it must be that these three are l.i. but again by my previous post $\displaystyle \cos (2x) \in span(A)$ so $\displaystyle A$ is a basis for $\displaystyle F$.

7. Originally Posted by Jose27
Assume $\displaystyle \sin ^2(x) = a \sin (2x)$ for some $\displaystyle a\in \mathbb{R}$ then evaluation in $\displaystyle \frac{\pi }{2}$ will give $\displaystyle 1=0$ so these two functions are l.i. Then by my previous post we get that $\displaystyle \cos ^2 (x) \notin span \{ \sin ^2 (x), \sin (2x) \}$ so it must be that these three are l.i. but again by my previous post $\displaystyle \cos (2x) \in span(A)$ so $\displaystyle A$ is a basis for $\displaystyle F$.
one final question
what is the first step to do to answer this kind of questions ?
thanks "Jose27"

8. I guess trying to determine if they're l.i. To do this pick one vector then add another, see if it's l.i then add another etc.

9. Originally Posted by Jose27
I guess trying to determine if they're l.i. To do this pick one vector then add another, see if it's l.i then add another etc.
Got that,thanks "Jose27"

10. Originally Posted by Jose27
I don't think this is right since assume there exist $\displaystyle a,b \in \mathbb{R}$ such that $\displaystyle \cos ^2 (x)= a\sin (2x) + b\sin ^2 (x)=2a \sin (x) \cos (x) + b\sin ^2 (x)$ then evaluation in $\displaystyle 0$ gives $\displaystyle 1=0$. Thus $\displaystyle A= \{ \sin ^2 (x), \cos ^2 (x) , \sin (2x) \}$ is l.i. and since $\displaystyle \cos ^2 (x) - \sin ^2 (x) = \cos (2x)$ we get that $\displaystyle A$ is a basis for $\displaystyle F$.
i actually had $\displaystyle \{1,\sin^2x, \sin(2x) \}$ in my mind but i wrote something else. thanks for noticing that.