This question rises from
For a linear transformation $T \colon V \to W$,
When $\dim V = m$ and $\dim W = n$ then $T$ is invertible iff $m$ equals $n$.
, which is false.
When I asked about it in previous question, I was told its counter example would be
'a matrix which maps every element in $\mathbb{R}^m$ to zero vector ($\in \mathbb{R}^m$)' .
For me, it does not make sense.
V is domain, W is range(or image) and codomain would be entire $\mathbb{R}^m$ space.
and I've known that dimension of vector space depends on the number of linearly independent basis.
In this example, W is zero vector so it has zero basis (Is it correct?) thus dimW is 0 (?).
So how can it be counter example?
In this case $\mathcal T$ is clearly not invertible, since it is not injective, i.e. "if $\mathcal Tv_1 = \mathcal Tv_2$ then $v_1=v_2$" does not hold. Indeed $v_1,v_2$ could be any pair of vectors in $V$. Invertible mappings are injective at least.
Also, $W$ is the codomain instead of the range.