Suppose you have a linear transformation $T: M_{2\times 2}\to M_{2\times 2}$ given by
$ \begin{pmatrix} a & b \\ c & d\end{pmatrix}\mapsto \begin{pmatrix} a+b & a \\ c & c+d\end{pmatrix}$
How can I tell quickly, if it is invertible?
Suppose you have a linear transformation $T: M_{2\times 2}\to M_{2\times 2}$ given by
$ \begin{pmatrix} a & b \\ c & d\end{pmatrix}\mapsto \begin{pmatrix} a+b & a \\ c & c+d\end{pmatrix}$
How can I tell quickly, if it is invertible?
Hint: 1) You need to check the linear transformation part.
2) If someone gives you an arbitrary matrix, can you uniquely identify the matrix that it came from?
For example, let $B=\begin{pmatrix} 12 & -3 \\ 4 & \pi\end{pmatrix}.$ What would be the only $A$ such that $T(A)=B$? Note that "$a"$ must be $-3$, and $c$ must be $4$. Thus $b$ must be $12-(-3)$ and $d$ must be $\pi -4$.
See if it's possible to find two elements of $M_{2\times2}$ that map to the same thing. If you can, it's not invertible, if not then explain why not rigorously.
Alternatively, $M_{2 \times2} $is isomorphic to $\mathbb{R^4}$ so we could write the map as a $4\times4$ matrix and see if it's invertible or not.
Since the domain and codomain are of same dimension, it is enough if u can check that it has a trivial kernel. That will do.