# Linear operator's linearity

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

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 $a$ –  Andrew Jan 20 '12 at 8:39
Are you looking for the largest integer $n$ such that $f\,$ is a linear operator? –  Henry Shearman Jan 20 '12 at 8:48
@Giuseppe ok, let's say $a$ is $0$, how do I write the matrix of $f$? –  Andrew Jan 20 '12 at 8:50
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 $\mathbb{R}^n$. –  Henry Shearman Jan 20 '12 at 9:12
@Henry: the variables $x$ and $y$ are supposed to be independent, no? –  anon Jan 20 '12 at 9:24

The fact that $n=3$ comes from inspection.

In order for $f:\mathbb{R}^3\to\mathbb{R}^3:(x,y,z)\mapsto(x-z,y,az^2)$ to be a linear operator you need

$$f(\vec{x}+\vec{u})=f(\vec{x})+f(\vec{u}), \quad\text{or}$$

$$\forall \vec{x},\vec{u}\in\mathbb{R}^3:\quad\begin{cases}(x+u)-(z+w)=(x-z)+(u-w) \\ (y+v)=(y)+(v) \\ a(z+w)^2=az^2+aw^2.\end{cases}$$

(Note: $\vec{x}=(x,y,z),\vec{u}=(u,v,w)$ here.) The first two check out but the last one implies $2azw=0$ for all $z,w\in\mathbb{R}$, which is obviously false unless $a=0$.

Now we have that the matrix associated to $f$ as a linear map is given by

$$f:\begin{pmatrix}x\\y\\z\end{pmatrix}\mapsto x\begin{pmatrix}1\\0\\0\end{pmatrix}+y\begin{pmatrix}0\\1\\0\end{pmatrix}+z\begin{pmatrix}-1\\0\\0\end{pmatrix}\quad\text{hence}\quad f(\vec{x})=\begin{pmatrix}1&0&-1\\0&1&0\\0&0&0\end{pmatrix}\vec{x}.$$

Finally, to find $\mathrm{Ker} f$, solve $x-z,y,0=0$, which is parametrized by $(t,0,t)$ and hence $\mathrm{dim}=1$.

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So, $z^2$ is nothing special, it's like we write it for $z$? –  Andrew Jan 20 '12 at 9:28
@Andrew: I don't understand your comment. $f$ is a function from $\mathbb{R}^3$ to $\mathbb{R}^3$; its three components are each functions from $\mathbb{R}^3$ to $\mathbb{R}$; the third component of $f$ is the map $(x,y,z)\mapsto az^2$ (and so is therefore independent of $x$ and $y$). Does that answer your question? –  anon Jan 20 '12 at 9:33
What I meant is that the matrix associated to $f$ doesn't really reflect that there is $z^2$. –  Andrew Jan 20 '12 at 9:36
@Andrew: Of course it doesn't; there is no $z^2:\;\;$ $0\cdot z^2 = 0$. –  anon Jan 20 '12 at 9:42
The $z^2$ vanishes with the choice of $a$. –  Henry Shearman Jan 20 '12 at 9:42