Is V Isomorphic to R^2 Under the Given Mapping?

Mitch_C
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Homework Statement



Let V={a cosx + b sinx | a,b \in R}

(a) Show that V is a subspace of the R-vector space of all maps from R to R.

(b) Show that V is isomorphic to R^2, under the map
f: V\rightarrowR^2
a cosx + b sinx \rightleftharpoons [ a over b ] (this is supposed to be a matrix with a above and b below, couldn't find it in the Latex reference)



The Attempt at a Solution



I have done part (a) okay so it's just part (b) I need a hand with. So I know V is isomorphic to R^2 if the map f is linear and the dimV = dim R^2.

I think I sort of showed that the dimensions are equal by taking a basis of {0 , cosx + sinx}. Is that basis okay? And am I right in thinking that because that has two elements dimV=2 and obviously dimR^2= 2 yeah?

So assuming that all that's ok so far I'm kind of stuck showing that it's linear. The properties of a linear map are f(u+v)=f(u)+f(v) and f(av)= af(v) where u,v\inV and a\in R. I'm trying to show those properties are true at the minute but having some difficulty. Am I on the right track at least?

thanks in advance!
 
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How do you construct sinx from the set {0, cosx + sinx}?

Also any set containing zero is not linearly independent since 1*0=0 so there is a non-trivial linear combination which gives zero.
 
Mitch_C said:

Homework Statement



Let V={a cosx + b sinx | a,b \in R}

(a) Show that V is a subspace of the R-vector space of all maps from R to R.

(b) Show that V is isomorphic to R^2, under the map
f: V\rightarrowR^2
a cosx + b sinx \rightleftharpoons [ a over b ] (this is supposed to be a matrix with a above and b below, couldn't find it in the Latex reference)



The Attempt at a Solution



I have done part (a) okay so it's just part (b) I need a hand with. So I know V is isomorphic to R^2 if the map f is linear and the dimV = dim R^2.

I think I sort of showed that the dimensions are equal by taking a basis of {0 , cosx + sinx}. Is that basis okay?
No, it is NOT okay. A basis never includes the 0 vector. Try {cos x, sin x} instead.
And am I right in thinking that because that has two elements dimV=2 and obviously dimR^2= 2 yeah?

So assuming that all that's ok so far I'm kind of stuck showing that it's linear. The properties of a linear map are f(u+v)=f(u)+f(v) and f(av)= af(v) where u,v\inV and a\in R. I'm trying to show those properties are true at the minute but having some difficulty. Am I on the right track at least?

thanks in advance!
"\begin{bmatrix} a \\ b\end{bmatrix}" gives
\begin{bmatrix} a \\ b\end{bmatrix}

If u= a cos(x)+ b sin(x) and v= c cos(x)+ d sin(x), what is u+ v?

f(u)= \begin{bmatrix}a \\ b\end{bmatrix}
and
f(v)= \begin{bmatrix}c \\ d \end{bmatrix}

what is f(u+ v)?
 
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Thanks for that. Looking back on it the basis I picked is obviously not a basis I just wasn't thinking. Spelling out for me what f(u) equals really helped. When I came back to it I solved it in about 10mins!

Thanks again! :)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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