Find a linear transformation $T: \mathbb{R^3} \to \mathbb{R^3}$ which takes an $XZ$ plane (i.e $X=0$) to a plane parallel to it (i.e $X=1$).

I thought along these lines:

Let $T$ be such a linear transformation. Then I assigned $T(0,0,1)=(1,0,1)$, $T(0,1,0)=(1,1,0)$,$T(1,0,0)=(1,0,-1)$

But the problem with it is that for any linear combination of $(0,1,0)$ and $(0,0,1)$, the transformation should yield something like $(1,y_1,z_1)$.

Thanks for the help!!


There is no linear map like this. What you want is an affine map. Many people say "linear" when they mean "affine," but it can get confusing when the difference is important.


Linear maps preserve scalar multiplication and vector addition. $L$ is linear if $$L(x+u,y+v,z+w)=L(x,y,z)+L(u,v,w)$$ and $$L(ax,ay,az)=aL(x,y,z).$$

An affine map is linear plus a constant. $T$ is affine if $$T(x,y,z)=L(x,y,z)+v$$ where $L$ is some linear map and $v$ is a constant vector.

I would normally call the plane where $x=0$ the $YZ$- rather than $XZ$-plane.

An Affine Solution

Let $T(x,y,z)=(1,y,z)$. In matrix form, $$T(x,y,z)= \begin{bmatrix} 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} + \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} $$

It is very easy to show that $T$ is affine, and the plane where $x=0$ (and in fact all of $\mathbb{R}^3$) is mapped to the plane where $x=1$.

There is no such linear map

You can easily prove there is no such linear map: suppose $T$ is linear and there are two points in the $YZ$-plane--call them $(0,y_1,z_1)$ and $(0,y_2,z_2)$--that get mapped to the $x=1$ plane--let's say $T(0,y_1,z_1)=(1,u_1,v_1)$ and $T(0,y_2,z_2)=(1,u_2,v_2)$. Then $$\begin{align}T(0,y_1+y_2,z_1+z_2) =&T(0,y_1,z_1)+T(0,y_2,z_2)\\ =&(1,u_1,v_1)+(1,u_2,v_2)\\ =&(2,u_1+u_2,v_1+v_2) \end{align}$$ which is not in the $x=1$ plane. So the entire $YZ$-plane does not get mapped to the $x=1$ plane by any linear map.


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