Linear Algebra - Suppose $CA=I_n$. Show that the equation $Ax = 0$ has only the trivial solution. Suppose $CA=I_n$.  Show that the equation $Ax = 0$ has only the trivial solution.  Explain why $A$ cannot have more columns than rows.
I really don't even know where to begin with this one.
 A: Hint: Suppose $Ax = 0$.  Now multiply both sides on the left by $C$.
A: As roundabout alternatives, you could think of the problem in terms of linear independence (or one-to-oneness) if it is more intuitive for you:
First, note that the columns of $I_n$ are linearly independent ($I_n$ has a pivot position in every column).
Since, $CA = I_n$, then the columns of $A$ must be linearly independent.
To see why this is true, let $B = [\mathbf{b}_1 \dotsb \mathbf{b}_n]$ be a matrix whose columns are linearly dependent.
Since the columns of $B$ are linearly dependent, there exists a vector $\mathbf{b}_j$ in $\{\mathbf{b}_1, \dotsc, \mathbf{b}_n\}$ that can be written as a linear combination of the other vectors in $\{\mathbf{b}_1, \dotsc, \mathbf{b}_n\}$. Because $CB = [C \mathbf{b}_1 \dotsb C\mathbf{b}_j \dotsb C \mathbf{b}_n]$, if $\mathbf{b}_j$ can be written as a linear combination of the other vectors in $\{\mathbf{b}_1, \dotsc , \mathbf{b}_n\}$, then $C\mathbf{b}_j$ can be written as a linear combination of the other vectors in $CB$. So the columns of $CA$ are linearly independent only if the columns of $A$ are linearly independent.
Finally, since the columns of $A$ are linearly independent if and only if $A\mathbf{x} = \mathbf{0}$ has only the trivial solution, $CA = I_n$ only if $A\mathbf{x} = \mathbf{0}$ has only the trivial solution.
Equivalently, you could explain it in the context of matrix multiplication as compositions of mappings and one-to-oneness.
Let $T(\mathbf{x})= A\mathbf{x}$, $S(\mathbf{y}) = C\mathbf{y}$, and $R(\mathbf{x}) = S(T(\mathbf{x})) = CA\mathbf{x}$.
If there exists  $\mathbf{p} \ne \mathbf{q}$ such that $T(\mathbf{p}) = T(\mathbf{q})$ ($T$ is not one-to-one), then since $R(\mathbf{x}) = S(T(\mathbf{x}))$, $R(\mathbf{p}) = R(\mathbf{q})$ ($R$ is not one-to-one). In other words, $R$ is one-to-one only if $T$ is one-to-one.
In this case, because $R(\mathbf{x}) = CA\mathbf{x} = I_n \mathbf{x} = \mathbf{x}$, $R(\mathbf{u}) = R(\mathbf{v}) \Leftrightarrow \mathbf{u} = \mathbf{v}$, so $R$ is one-to-one and, thus, $T$ is one-to-one.
Then since $T$ is one-to-one if and only if $A\mathbf{x} = \mathbf{0}$ has only the trivial solution, $CA = I_n$ only if $A\mathbf{x} = \mathbf{0}$ has only the trivial solution.
A: I think another useful way to look at this problem is to imagine $C $ and $A$ as functions.
You might have seen that if $ g\circ f $ is invertible, then it forces $ f $ to be injective(or one to one).
In our case, $ C $ is $g $ and $f $ is $ A $, and so $ A $ is injective, hence: $ Ker(A)=\{\vec{0}\} $
