Linear Algebra: find the Kernel and Image

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SUMMARY

The discussion focuses on finding the Kernel and Image of the linear transformation \(\varphi: V \rightarrow V\) defined by \(\varphi(f) = f' + f\), where \(V\) is the subspace spanned by \(\sin\) and \(\cos\). The Kernel is determined to be \(\text{Ker}(\varphi) = \{0\}\), as the only solution to \(\varphi(f) = 0\) occurs when both coefficients \(\lambda_1\) and \(\lambda_2\) are zero. The Image is characterized by all functions \(g \in V\) of the form \(g(x) = (\lambda_1 + \lambda_2)\cos x + (\lambda_1 - \lambda_2)\sin x\), leading to a basis for the Image being \(\{\cos x + \sin x, \cos x - \sin x\}\.

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  • Understanding of linear transformations in vector spaces
  • Familiarity with the concepts of Kernel and Image in linear algebra
  • Basic knowledge of derivatives and smooth functions
  • Proficiency in manipulating trigonometric functions
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Homework Statement


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

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 isn't 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?
 
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forgive me for my poor latex skills : (
 

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