My math professor has asked me to do the following:
Let $V = \{(x,y,z) \in \mathbb{R}^3 : x + y + z = 0\}$. Let $\alpha = \{(1, 0, -1), (0, 1, -1)\}$ and $\beta = \{(1, 1, -2), (1, -1, 0)\}$ be ordered bases for $V$. Let $T: V\rightarrow V$ be the linear transformation defined by $T(x,y,z) = (y, z, x)$. Calculate $[T]_\alpha^\beta$.
I'm having a hard time understanding why it is correct to write $T(x,y,z) = (y,z,x)$ if $T$ maps to and from a two-dimensional space. In particular, the very definition of $T$ seems to imply to me that we are thinking of $T$ as an operator on $\mathbb{R}^3$, using the standard basis of $\mathbb{R}^3$. My suspicion is that this is a slight abuse of notation, i.e. that by writing basis $\alpha$ in coordinate form, i.e. $\alpha = \{(1, 0, -1), (0, 1, -1)\}$, we are actually implying that there is a "hidden" transformation $A: V \rightarrow \mathbb{R}^3$ that first "prepares" vectors in $V$ by sending them to $\mathbb{R}^3$ so that $T$, which is actually an operator on $\mathbb{R}^3$, knows what to do with these vectors. In fact, if $B: V \rightarrow \mathbb{R}^3$ similarly writes the $\beta$ vectors in coordinate form, then I believe I can get the desired result essentially by taking the product $[T]_\alpha^\beta = [B]^{-1} [T] [A]$, where I'm using the left-inverse of $[B]$.
I guess in short, is writing $T(x,y,z) = (y,z,x)$ technically a light abuse of notation?
Hint: $T$ is the linear transformation of $\Bbb R^3$ whose matrix is $\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}$ restricted to $V$.
Since $V$ is invariant under $T$, that is, $T(V)\subseteq V$, this makes sense.