Is it possible for a matrix to not be onto or 1-1? Is it possible for a matrix to be neither onto nor 1-1?
 A: A matrix isn’t a function, so the question doesn’t actually make sense. However, I understand what you mean: you’re asking whether the function $T(x)=Ax$, where $A$ is some matrix, can be neither onto nor $1$-$1$. The answer is yes: in fact, if you’re talking about square matrices of real numbers, the transformation is onto if and only if it is $1$-$1$. This is immediate from the rank-nullity theorem.
A: Consider a $m \times n$ matrix with rank $r < \min(m,n)$. This is neither onto nor one to one.
EDIT
To make the above point clear, in case the OP is not aware of rank, consider matrices $m \times n$ matrices of the form $$A = U V^T$$ where $U \in \mathbb{C}^{m \times r}$ and $V \in \mathbb{C}^{n \times r}$, where $r < \min(m,n)$.
To see this is not one-to-one, choose a non-zero vector $z$ orthogonal to the columns of $V$, i.e., we have $V^Tz = 0$. (Why does a non-zero $z$ exists?) Now if you consider the vector $x$ and $x+z$, we have $$V^Tx = V^T(x+z) \implies UV^Tx = UV^T(x+z),$$which clearly shows that the linear mapping is not one-to-one.
To see this is not onto, choose a vector $z$ orthogonal to the columns of $U$, i.e., we have $U^Tz = 0$. Since $z$ is linearly independent (Why?) with respect to the columns of $U$, we can never write $z = Uy$ for some $y \in \mathbb{C}^{r \times 1}$. Hence, $$z \neq U(V^Tx)$$ for all $x \in \mathbb{C}^{n \times 1}$, which clearly shows that the linear mapping is not onto.
A: Take the null matrix: $A_{ij}=0$ for any $i,j$. Of course, when you say "matrix", you really mean the associated linear transformation, $(Ax^t)^t$, yes?
