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## Homework Statement

(a)Find the Kernel and Image of each of the following linear transformations.

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(iv)[tex]\varphi : V\rightarrow V[/tex] given [tex] \varphi(f)=f'+f[/tex] where V is the subspace of the space of smooth functions [tex]\Re\rightarrow\Re[/tex] spanned by sin and cos, and f' denotes the derivative.

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## Homework Equations

## The Attempt at a Solution

first I consider an arbitrary vector [tex]f_1 \in V[/tex]. then [tex]f_1[/tex] has the following form:

[tex] f_1=\lambda_1 sinx + \lambda_2 cosx [/tex] with[tex] \lambda_1\lambda_2\in \Re[/tex]

Then

[tex]\varphi(f_1)=\lambda_1 cosx-\lambda_2 sinx + \lambda_1 sinx+\lambda_2 cosx[/tex]

[tex]=(\lambda_1+\lambda_2)cosx+(lambda_1-lambda_2)sinx[/tex] call (*)

By definition,

[tex]Ker(\varphi)={f\in V: \varphi (f)=0}[/tex]

but from the above, [tex]\varphi (f)=0 [/tex] iff [tex] (\lambda_1+\lambda_2)=0, (\lambda_1-\lambda_2)=0[/tex] since this only occurs when both lambdas are zero, then

[tex]Ker(\varphi)={0}[/tex]

Originally, by (*) i wanted to say that the image of phi was the whole subpsace V, but this isnt true since the coefficients depend on each other (lambda1-lambda 2 and lambda1+lambda2). Rather the image would be all [tex]g\in V[/tex] such that [tex]g(x)=(\lambda_1+\lambda_2)cosx+(\lambda_1-\lambda_2)sinx, \lambda_1,\lambda_2\in \Re[/tex]

Does this seem right? I am skeptical for some reason.

accordingly a basis for the image would be [tex]{cosx+sinx, cosx-sinx}[/tex]

Can someone check this for me and let me know if if my argument seems right?