So again, the task is
Consider map $T: \mathbb R^3 \rightarrow \mathbb R^2$ defined by $$ T\bigr(\begin{bmatrix} x & y & z \end{bmatrix}^T \bigr) = \begin{bmatrix} xy & xz \end{bmatrix}^T$$
Evaluate dim(ker(T))
The solution given by the textbook I'm reading:
He says
"The entries of the vector $\bf x$ such that $T(\mathbf x) = \bf O$ gives $x=0$, $y=r$, and $z=s$."
What about vector $\bf x$ such that
$$ \mathbf x = \begin{bmatrix} c\\ 0\\ 0 \end{bmatrix} $$
where $c ≠ 0$? If I am not missing anything, the author fails to mention that case.
So the correct kernel for $T$ must be
$$ker(T) = \{ \begin{pmatrix} 0 & y & z \end{pmatrix}^T \mid y,z \in \mathbb R \} \cup \{\begin{pmatrix} c & 0 & 0 \end{pmatrix}^T \mid c \in \mathbb R \}$$
And seems like in this case, $nullity(T) = 3$, instead of $2$.
So
Question
Is he correct? If so, what am I missing?
$T$ isn't a linear map so talking about its "kernel" and a basis for that kernel is weird. As you see, $T(1,0,0) = T(0,1,0) = T(0,0,1) = 0$ so if $T$ were linear then $T(a,b,c) = 0$ for any $a, b, c \in \mathbb{R}$.
The zero-set of $T$ (I don't want to call it a kernel) is the set of all $(x,y,z)$ such that $xz = 0$ and $xy = 0$. We can write this as
$$ [(x = 0) \text{ or } (z = 0)] \text{ and } [(x = 0) \text{ or } (y = 0)] = (x = 0) \text{ or } (y = 0 \text{ and } z = 0). $$
So we can see that the zero set is the union of the $yz$-plane and the $x$-axis. Namely: $$ \{(x,y,z) : T(x,y,z) = 0\} = \{(0,y,z) : y, z \in \mathbb R\} \cup \{(x,0,0) : x \in \mathbb R\}. $$
We can't expect this to be a linear subspace because $T$ isn't linear. Since it is the union of a plane (2-dimensional) and a line (1-dimensional) it doesn't have a well defined "dimension" either.