Find the image of the plane $y = x$ (in $\mathbb{R}^3$) under the transformation $T$.

Question

I've been stuck on the following question:

The transformation $T$ is defined by $$T(x, y, z) \;=\; \left(\frac13(2x-2y+z),\; \frac13(x+2y+2x),\; \frac13(2x+y-2z)\right).$$ Find the image of the plane $y = x$ (in $\mathbb{R}^3$) under the transformation $T$.

I have tried the following:

$$TX = X'$$

$$T^{-1}TX = X'$$

$$X = T^{-1}X'$$

and express $X'$ accordingly:

$\begin{bmatrix}x\\y\\z\end{bmatrix} = \frac13 \begin{bmatrix}2&1&2\\-2&2&1\\1&2&2\end{bmatrix}$$\begin{bmatrix}x\\x\\z\end{bmatrix}$

Is this a good approach? Or I have started on the wrong footing?

The answer is: $4x - y + z =0$.

Answer 1

Since the plane $y=x$ is spanned by $(1,1,0)$ and by $(0,0,1)$, its image under $T$ is the subspace of $\mathbb R^3$ spanned by $T(1,1,0)\bigl(=(0,1,1)\bigr)$ and by $T(0,0,1)\left(=\left(\frac13,\frac23,-\frac23\right)\right)$.

Answer 2

Your idea seems good, but the details leave a bit to be desired.

If $T(X)=MX$ for some matrix $M$, then we have $X'=MX$. Writing the equation of the plane as $N^TX=0$ and assuming that $M$ is invertible, you can substitute: $$N^T(M^{-1}X') = (N^TM^{-1})X' = (M^{-T}N)^TX = 0,$$ so the vector of coefficients of the transformed equation is $M^{-T}N$. Here, the equation of the plane is $y=x$, which we rewrite as $x-y=0$, therefore $N=(1,-1,0)^T$. I leave the rest of the calculations to you.

Answer 3

I know I am 3 years late but for anyone who is still having difficulties solving this question, all you have to do is randomly choose 3 points on the plane y=x (as hinted y=x coordinates need to be in the form (x,x,y) )and find the transformation under T of the three points (T(x) = X'). Then we use the vector product of the three points AB X AC to find the normal and form the equation of the plane which results in 4x-y+z=0