Are these two transformations different? How?

Question

Let $T_0$ be a transformation from $\mathbb{R}^3$ to $\mathbb{R}^3$.

$$ T_0 = \left[ {\begin{array}{ccc} a & b & c\\ d & e & f\\ 0 & 0 & 0\\ \end{array} } \right] $$

How is this really different from this transformation from $\mathbb{R}^3$ to $\mathbb{R}^2$ ?

$$ T = \left[ {\begin{array}{ccc} a & b & c \\ d & e & f \\ \end{array} } \right] $$

Answer 1

It is crucial to realize the following statement:

$\mathbb R^2$ is not a subset of $\mathbb R^3$. Instead, there is a way of realizing $\mathbb R^2$ as a subspace of $\mathbb R^3$, via the map $i(x,y) = (x,y,0)$.

Therefore, $T_0$ and $T$ are different, simply because they take the same input to different vectors! I mean, two functions are equal if their value at every point is equal, but here some function takes values in $\mathbb R^2$, while another takes value in $\mathbb R^3$, so obviously $T_0$ and $T$ are "different" as functions.

Now, how you would like to express the "similarity" between $T_0$ and $T$ is as follows : what $T_0$ is doing, is essentially $T$, except that it is adding this one more row of zeros. That is, for the same $(x,y,z) \in \mathbb R^3$, if $T(x,y,z) = (u,v)$, then $T_0(x,y,z) = (u,v,0)$.

So, then $i \circ T$, where I defined $i$ in the yellow box, does go from $\mathbb R^3 \to \mathbb R^3$, just like $T_0$. Furthermore, $i \circ T(x,y,z) = i(u,v) = (u,v,0) = T_0(x,y,z)$ for all $(x,y,z) \in \Bbb R^3$.

Hence, the accurate statement is : $i \circ T = T_0$ as functions. The "canonical" nature of $i$ as a submersion allows everybody reading this statement to realize that $T$ and $T_0$, while not being equal, are "essentially" equal.