Defining Linear Transformations I am currently stuck on a problem (this is not a homework problem) mainly because I am weak at DEFINING functions.
The problem states: Suppose $U$ is a subspace of $\Bbb R^3$ defined as $$U=\{(x,y,z) \ : \ x,y,z \in \Bbb R \ ; \  x+y+z=0\}.$$ Find a linear map other than the identity map defined from $U$ to $U$ such that $U$ is mapped to itself.
I tried to start with a basis (without knowing if it will work) and could not proceed any further. It would be helpful if you also please give the idea on how to define maps.
Thanks.
 A: Notice that $U$ is an hyperplane of $\Bbb R^3$ as it's a kernel of a linear form so $\dim U=2$ and we see that with $v_1=(1,-1,0)$ and $v_2=(1,0,-1)$ we have $(v_1,v_2)$ is a basis for $U$. Now we know that a linear transformation is entirely defined by giving the image of the vectors of a  basis so we may define the desired linear transformation $T$ by
$$T(v_1)=v_2\quad;\quad T(v_2)=v_1$$
and there are infinitely many other choices.
A: Every linear transformation from $\mathbb R^3$ to $\mathbb R^3$ looks like this:
$$
f(x,y,z) = (a_{11}x + a_{12}y + a_{13}z,\,a_{21}x + a_{22}y + a_{23}z,\,
a_{31}x + a_{32}y + a_{33}z).
$$
In other words, you can define a linear transformation by choosing values for $a_{11}$, $a_{12}$, $\dots$, $a_{33}$. For example, all of the following functions are linear transformations:


*

*$f(x,y,z) = (x,y,z)$ (the identity transformation)

*$f(x,y,z) = (0,0,0)$ (the zero transformation)

*$f(x,y,z) = (z,x,y)$

*$f(x,z,y) = (x + 2y -z,-\pi x + \pi y + \pi z, 0)$


For your particular problem, you need to find a linear transformation that changes the $x$, $y$, and $z$ coordinates but does not change their sum.
A: What about the negative identity, i.e. $\mathrm{f}(x,y,z) = (-x,-y,-z)$? 
This is such that $\mathrm{f}(U) = U$.
You can find the set of all linear maps as follows:
Your subspace has the equation $x+y+z=0$, i.e. $z=-x-y$. Every point is of the form
$$\left(\begin{array}{c} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} x \\ y \\ -x-y \end{array}\right)=x\left(\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right)+y\left(\begin{array}{c} 0 \\ 1 \\ -1 \end{array}\right)$$
It follows that $u=(1,0,-1)^{\top}$ and $v=(0,1,-1)^{\top}$ span $U$, in fact they form a basis for $U$. A general linear map $\mathrm{f} : \mathbb{R}^3 \to \mathbb{R}^3$, with respect to the usual basis, has a matrix representation
$${\bf F} := \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array}\right)$$
If you calculate ${\bf F}u$ and ${\bf F}v$ you can find conditions for ${\bf F}u \in U$ and ${\bf F}v \in U$. For example
$${\bf F}u = \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array}\right)\!\left(\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right)=\left(\begin{array}{c} a-c \\ d-f \\ g-i \end{array}\right)$$
For ${\bf F}u \in U$ you need $x+y+z=0$, i.e. $(a-c)+(d-f)+(g-i)=0$.
$${\bf F}v = \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array}\right)\!\left(\begin{array}{c} 0 \\ 1 \\ -1 \end{array}\right)=\left(\begin{array}{c} b-c \\ e-f \\ h-i \end{array}\right)$$
For ${\bf F}v \in U$ you need $x+y+z=0$, i.e. $(b-c)+(e-f)+(h-i)=0$.
Solving both equations simultaneously gives, 
$$\begin{eqnarray*}
h &=& a-b+d-e+g \\ \\
i &=& a-c+d-f+g
\end{eqnarray*}$$
Let ${\bf G}$ be the matrix given by substituting these solutions into ${\bf F}$. We have
$$\det({\bf G}) = (a+d+g)(ae-af+cd+bf-bd-ce)$$
If we allow $ae-af+cd+bf-bd-ce=0$, then ${\bf G}$ does not preserve $U$ as a plane. The kernel of ${\bf G}$ is contained within $U$. If $a+d+g=0$ then 
$$\ker{\bf G} \subset U \iff ae-af+cd+bf-bd-ce=0$$
Finally, we conclude that the set of linear maps with $\mathrm{f}(U) = U$ have matrices
$$\left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array}\right)$$
for which the following conditions are met:
$$\begin{eqnarray*}
a-c+d-f+g-i &=& 0 \\ \\
b-c+e-f+h-i &=& 0 \\ \\
ae-af+cd+bf-bd-ce &\neq& 0
\end{eqnarray*}$$
In the case $a+d+g=0$ we have $\mathrm{f}(\mathbb{R}^3) = U$.
