How to proceed for the following linear algebra problem Let $f:M_{2}(\mathbb{C})\rightarrow M_{3}(\mathbb{C})$, be a function such that $$f(X.Y)=f(X).f(Y),$$ for all $X,Y\in M_{2}(\mathbb{C})$. Show that $f$ is not a bijection.
I have tried like this:
First I put $X=Y=I$. obtained $f(I)^2=f(I)$. From this how to show?
 A: Here's a nice approach: first, show that for $f$ to be bijective and multiplicative, we must have $f(0) = 0$.  
In particular, we know that $f(0)$ must satisfy
$$
f(0) = f(0 X) = f(0) f(X).
$$
That is, $f(0)$ must be such that $f(0) f(X) = f(0)$ for every matrix $X$.  If we select $X$ such that $f(X) = 0$ (such an $X$ exists since $f$ is surjective), we have
$$
f(0) = f(0) f(X) = f(0)\cdot 0 = 0
$$
Now, note that the matrix 
$$
M = \pmatrix{0&1&0\\0&0&1\\0&0&0}
$$
satisfies $M^2 \neq 0$ and $M^3 = 0$.  Suppose (for the purpose of contradiction) that $f$ is a bijective, mulitplicative function and that $X \in M_2(\Bbb C)$ satisfies $f(X) = M$ (such an $X$ exists if $f$ is surjective).  Then we have
$$
0 = M^3 = f(X)^3 = f(X^3) = f(0).
$$
Since $f$ is injective, this implies that $X^3 = 0$.
On the other hand, we have
$$
f(X^2) = f(X)^2 = M^2 \neq 0
$$
from which we conclude that $X^2 \neq 0$.  Thus, $X$ is a matrix in $M_2(\Bbb C)$ for which $X^2 \neq 0$ but $X^3 = 0$.  We have reached a contradiction since (by the Cayley-Hamilton theorem, for instance) there is no such matrix.
