# Linear Algebra: find the Kernel and Image

## Homework Statement

(a)Find the Kernel and Image of each of the following linear transformations.
...
(iv)$$\varphi : V\rightarrow V$$ given $$\varphi(f)=f'+f$$ where V is the subspace of the space of smooth functions $$\Re\rightarrow\Re$$ spanned by sin and cos, and f' denotes the derivative.
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## The Attempt at a Solution

first I consider an arbitrary vector $$f_1 \in V$$. then $$f_1$$ has the following form:
$$f_1=\lambda_1 sinx + \lambda_2 cosx$$ with$$\lambda_1\lambda_2\in \Re$$
Then
$$\varphi(f_1)=\lambda_1 cosx-\lambda_2 sinx + \lambda_1 sinx+\lambda_2 cosx$$
$$=(\lambda_1+\lambda_2)cosx+(lambda_1-lambda_2)sinx$$ call (*)

By definition,
$$Ker(\varphi)={f\in V: \varphi (f)=0}$$

but from the above, $$\varphi (f)=0$$ iff $$(\lambda_1+\lambda_2)=0, (\lambda_1-\lambda_2)=0$$ since this only occurs when both lambdas are zero, then
$$Ker(\varphi)={0}$$

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 $$g\in V$$ such that $$g(x)=(\lambda_1+\lambda_2)cosx+(\lambda_1-\lambda_2)sinx, \lambda_1,\lambda_2\in \Re$$

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

accordingly a basis for the image would be $${cosx+sinx, cosx-sinx}$$

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