Prove or disprove: There exists $T \colon \mathrm M_2(\mathbb R) \to\mathbb{R}^3$ which is 1-1. Prove or disprove the claim: 
  There exists $T \colon \mathrm M_2(\mathbb R) \to\mathbb{R}^3$ which is 1-1.


I have no clue how to start this problem. We only learned in class about determining whether a defined linear transformation is 1-1 or onto.
Since you tagged your question with linear-algebra, I am assuming you want $T$ to be linear.

You get from the linearity of $T$ that $\mathrm {Im}(T)$ is a vector space of $\mathbb R^3$, and since $T$ is 1-1, you have

$$4=\dim \mathrm M_2(\mathbb R)=\dim \mathrm{Im} (T).$$

Which is a contradiction since $\dim \mathbb R^3=3$.

Thus, such a $T$ can not exist.
The dimension of $\mathcal M_2(\Bbb R)$ is 4 so by the rank nullity theorem

$$\dim \ker T=\dim \mathcal M_2(\Bbb R)- rank T\ge 4-3=1$$
so $T$ can't be 1-1.