Determine if the standard matrix $A$ representing the linear transformation is one-to-one.
$A = \begin{bmatrix}1&-1\\2&0\\3&-4\end{bmatrix}$
Answer: Yes
My question: When determining if multiplication by A is a one-to-one linear transformation, is it equivalent to determine if $ker(A)=\vec 0$ ? In other words, is the following method always a valid way to determine if $T_A$ is a one-to-one linear transformation?
$T_A$ is one-to-one if $ker$($T_A$)$=\vec 0$ , i.e, $A\vec x=\vec 0$ only has the trivial solution $\vec x=\vec 0$
$\begin{bmatrix}1&-1&0\\2&0&0\\3&-4&0\end{bmatrix}$ $\rightarrow$ $x=0,y=0$ $\rightarrow$ $\vec x=\vec 0$
So $T_A$ is one-to-one.
I am confused about this method because if $ker$($T_A$)$=\vec 0$ then that means the only vector that is mapped to $\vec 0$ after the linear transformation is $\vec 0$. So this only shows that the there is one input vector for the output vector $\vec 0$. However, a one-to-one linear transformation means that each unique vector in the domain maps to only one vector in the range. So why don't I have to show all vectors have this property and just show that zero vector has this property?
So my question is why is showing that the kernel of the transformation is $\vec 0$ enough to show that the transformation is one-to-one?
Best Answer
Like you have mentioned in your comment, the span of the column vectors of A (column space) represent the range of the linear transformation. This is because when you multiply the given standard matrix by column vector of the right number of rows,
$$A\vec x = \vec b$$
$$\begin{bmatrix}1&-1\\2&0\\3&-4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x-y\\2x\\3x-4y\end{bmatrix}$$
your resultant vector $\vec b$ is a linear system representing all possible linear combinations, i.e., all possible output vectors (range) reached by multiplying any input vector $\vec x$ (domain) by the standard matrix $A$ (transformation).
So since the column space corresponds to the range, if you end up with the trivial solution for your homogeneous augmented matrix that means your column vectors are linearly independent, in other words all output vectors in the range are also linearly independent. This then implies that your linear transformation is one-to-one because your output vectors cannot be expressed in terms of other output vectors, meaning that the preimage of each output vector was unique.