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Can somebody please help me with this problem? I've got no clue how to prove them. Thanks so much!:)

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closed as off-topic by This is much healthier., Michael Albanese, Claude Leibovici, RecklessReckoner, drhab Jul 5 at 7:40

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There's nothing here. –  Christopher A. Wong Mar 26 '13 at 22:56
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Please read the definition of these terms, get an empty paper and start to verify the assertions using the definitions. –  Berci Mar 26 '13 at 23:04
    
miss-morning: Please use the LaTeX capabilities of this site rather than uploading the problems as images externally on imagehosters (which usually are ephemeral and often have connection problems. –  darij grinberg Mar 28 '13 at 0:43
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1 Answer

I'll show you the first, I think you'll get the reasoning behind this, so you take care of the others. The idea behind the proof is: let's take a linear combination of two vectors, plug into $T^\ast(f)$ and use the linearity properties to prove what we want. So let's go:

Let $v_1, v_2 \in V$ and $\lambda \in F$. By definition of $T^\ast(f)$ we have that:

$$T^\ast(f)(\lambda v_1 + v_2) = f(T(\lambda v_1+v_2))$$

However, by hyphotesis we have $T$ linear, so $T(\lambda v_1 + v_2) = \lambda T(v_1) + T(v_2)$, which leads us to

$$T^\ast(f)(\lambda v_1+v_2)=f(\lambda T(v_1)+T(v_2))$$

Our hypothesis again grants us that $f$ is linear, and is this case we get

$$T^\ast(f)(\lambda v_1+v_2)=\lambda f(T(v_1))+f(T(v_2))$$

Finally, the definition of $T^\ast(f)$ shows us that:

$$T^\ast(f)(\lambda v_1 + v_2) = \lambda T^\ast(f)(v_1) + T^\ast(f)(v_2)$$

Hence $T^\ast(f)$ is linear as proposed.

Look now at the flow of this proof: we want to show that some map is linear, so we get a linear combination of vectors and plug into the map and use properties that are granted by hypothesis find linearity. Try to use this to get yourself started on trying the others. Good luck!

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