I have a $k$-dimensional topological manifold $M$ embedded in $\mathbb{R}^n$ ($k<n$). And a linear transformation given by $T:\mathbb{R}^n \rightarrow \mathbb{R}^m$. Assuming that $m>n$, I want to prove that the resulting manifold: ($Tx$ $\forall x \in M$) it still $k$-dimensional.
For $n=m$ it is obvious because the transformation is a homeomorfism (if the transformation matrix is non-singular). But the problem to me is the fact that the linear transformation is non-square.
Edit: Assume dim Im $T$ = n
HINT: By a linear change of coordinates you can reduce yourself to the case $$ T=\begin{bmatrix} 1 &\ldots & 0 \\ \ldots & \ldots &\ldots \\ 0& \ldots & 1 \\ 0 & \ldots & 0 \\ \ldots & \ldots & \ldots \\ 0 & \ldots & 0\end{bmatrix}.$$