Determining whether an $m \times n$ matrix is one-to-one $$ A = 
\begin{bmatrix} 
1 & 2 & 1\\
0 & 1 & 1\\
1 & 1 & 0\\
1&0&-1
\end{bmatrix}
$$
Given this matrix, I am supposed to determine whether multiplication by $A$ is a one-to-one matrix transformation. I know that if this was an $n \times n$ matrix I could discover if it was one-to-one by checking its determinant but I don't understand how to do it for an $m \times n$.
I read that it was something about if the matrix had the property $m \ge n$ but I don't see how this applies. Thank you for any help.
 A: Let $$
T = \begin{bmatrix} 
1 & 2 & 1\\
0 & 1 & 1\\
1 & 1 & 0\\
1&0&-1
\end{bmatrix}
$$
Show that in the equation $$T \bf{x} = 0 \iff \bf{x}=0$$
i.e., the trivial solution is the only one that solves this.
A: The condition is that all $3\times3$ subdeterminants are not all $0$. If you know about exterior powers, this means the exterior power $\stackrel{3}{\bigwedge}A\ne 0$.
More generally, if $A$ is an $m\times n$ matrix, representing a linear map $\varphi\colon\mathbf R^n\longrightarrow\mathbf R^m$, the condition is that the $n$-th exterior power of this map:
$$\stackrel{n}{\bigwedge}\varphi\colon\stackrel{n}{\bigwedge}\mathbf R^n\simeq\mathbf R\longrightarrow\stackrel{n}{\bigwedge}\mathbf R^m\simeq\mathbf R^{\binom mn}$$
is injective, i.e. non-zero.
In terms of matrices,, this means the minors of order $n$ are not all $0$.
Note : This is valid for vector spaces over any field.
There is also an extension for finitely generated free modules over a commutative ring: the condition being that the ideal generated by the minors of order $n$ is faithful, i.e. its annihilator is $0$.
A: Recall that for a linear map $T:V\to W$, the rank-nullity theorem states that
$$
\dim\DeclareMathOperator{image}{image}\image(T)+\dim\ker(T)=\dim(V)
$$
Also recall that $T$ is one-to-one if and only if $\dim\ker(T)=0$. 
Now, let $A$ be an $m\times n$ matrix and suppose our $T$ is the linear map $T:\Bbb R^n\to \Bbb R^m$ given by $T(x)=Ax$. Then $\dim\image(T)=\DeclareMathOperator{rank}{rank}\rank(A)$ so the rank-nullity theorem takes the form
$$
\rank(A)+\dim\ker(T)=\dim(\Bbb R^n)
$$
which can be rearranged as
$$
\dim\ker(T)=n-\rank(A)
$$
This gives a satisfying answer to your question:

Let $A$ be an $m\times n$ matrix. Then the linear map $T:\Bbb R^n\to \Bbb R^m$ given by $T(x)=Ax$ is one-to-one if and only if $\rank(A)=n$. 

Equivalently:

Let $A$ be a matrix. Then the linear map given by $T(x)=Ax$ is one-to-one if and only if $\rank(A)=\#\DeclareMathOperator{columns}{columns}\columns(A)$.

Let's test this with your example. Let
$$
A=
\left[\begin{array}{rrr}
1 & 2 & 1 \\
0 & 1 & 1 \\
1 & 1 & 0 \\
1 & 0 & -1
\end{array}\right]
$$
Row-reducing gives
$$
\DeclareMathOperator{rref}{rref}\rref(A)=
\left[\begin{array}{rrr}
1 & 0 & -1 \\
0 & 1 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]
$$
Hence $\rank(A)=2<3=\#\columns(A)$. This proves that the linear map $T:\Bbb R^3\to\Bbb R^4$ given by $T(x)=Ax$ is not one-to-one.
Note that in the special case that $A$ is square our result can be restated in terms of determinants, using the fact that $\rank(A)=\#\columns(A)$ if and only if $\det(A)\neq0$.
A: You matrix (let's name it) $A$ represents a linear map $\phi: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$. For $\phi$ to be one-to-one in particular it has to be onto. But since we have $m \geq n$ and the image of a n-dimensional vector space is itself a vector space of dimension $\leq n$ the map cannot be onto.
Here I'm supposing you are working with real vector spaces (but there is no difference in this respect using other fiels).
