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$$f:R^n \to R^3 \ \ \ \ \ \ \ \ \ f(x,y,z)=(x-z,y,az^2)$$

I have to find $n$ and $a$ such that $f$ is a linear operator.

$$x-z=0$$ $$y=0$$ $$az^2=0$$

I found $n$ to be 3.

For $az^2$ to be equal to $0$, either $z$ is $0$ or $a$ is $0$, right? The $z^2$ is confusing me, I don't know from what $R^n \to R^3$ it is. Any idea please? After findng $a$ and $n$, I have to write the matrix of $f$ and find the $dim(KerF)$ Thank you.

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    @anon I updated the question, sorry. Yes, I found $n=3$. Also I should find $a$ such that $f$ is linear. I also thought it is not linear since there is $z^2$ but however, I'm asked to find $a$. Assuming my teacher knows what he is doing, maybe there is an $a$2012-01-20
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    Are you looking for the largest integer $n$ such that $f\,$ is a linear operator?2012-01-20
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    @Giuseppe ok, let's say $a$ is $0$, how do I write the matrix of $f$?2012-01-20
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    Once you have found your $n$ and $a$, you can write the matrix representation of $f \,$ by considering $f$'s actions on the standard basis for $\mathbb{R}^n$.2012-01-20
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    @HenryShearman I'd know how to write it if I had it without the $z^2$.2012-01-20
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    @Henry: the variables $x$ and $y$ are supposed to be independent, no?2012-01-20
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    Yes, don't mind me. I confused myself for a second.2012-01-20

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