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

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Homework Help Overview

The discussion revolves around the vector space V defined as V={a cosx + b sinx | a,b ∈ R} and the task of demonstrating its isomorphism to R^2 through a specified mapping. Participants are exploring the properties of linearity and dimensionality in the context of vector spaces.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the dimensionality of V and its basis, questioning the validity of the original basis proposed. There is an exploration of the linearity of the mapping f and its properties, with attempts to clarify how to demonstrate these properties.

Discussion Status

Some participants have provided guidance on the choice of basis and the implications of including the zero vector. There is an acknowledgment of the initial confusion regarding the mapping and its linearity, with one participant expressing progress after receiving clarification.

Contextual Notes

There is a noted constraint regarding the definition of a basis in vector spaces, specifically the requirement that it cannot include the zero vector. Participants are also navigating the properties of linear mappings in their attempts to establish the isomorphism.

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)?
 
Last edited by a moderator:
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! :)
 

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