One-to-one and onto I think this is a fairly easy question, but I want to ask anyway, since I don't know if I remember all correctly.
Suppose a linear transformation $T: \Bbb R^n \rightarrow \Bbb R^n$ has the property that $T(\mathbf{u}) = T(\mathbf{v})$ for some pair of distinct vectors $\bf{u}$ and $\bf{v}$ in $\Bbb R^n$. Can $T$ map $\Bbb R^n$ onto $\Bbb R^n$? Why or why not?
What I think, is that because of the distinct vectors in $T(\mathbf{u}) = T(\mathbf{v})$, $T$ is not one-to-one, and then because of the Invertible Matrix Theorem it cannot map $\Bbb R^n$ onto $\Bbb R^n$. Is this that simple? It's been a year since I took the previous course in linear algebra...

You can remember this theorem (rank-nullity): $$\dim \Bbb R^n = \dim \mathrm{ker}~ T + \dim \mathrm{Im}~T $$
where $\mathrm{ker}~T = \{ \mathbf{x} \in \Bbb R^n : T(\mathbf{x}) = \mathbf{0} \}$. Another theorem states that $T$ is one-to-one if and only if $\mathrm{ker} ~T = \{0\}$. 
If $T(\mathbf{u}) =$ $T(\mathbf{v})$, clearly $T$ isn't one-to-one, but to remember better, you can think that $T(\mathbf{u} - \mathbf{v}) = 0$, hence $\mathbf{u} - \mathbf{v} \in \mathrm{ker}~T \neq \{0\}$ and from this, $T$ can't be one-to-one. And $\dim \mathrm{Im}~T \geq 1$, so from the first expression, $\dim \mathrm{Im}~T < n$, so $T$ also can't be onto. 
