If a linear transformation is represented by an non-invertible matrix $P$, then it might happen that two different vectors (points in $\mathbb{R^n}$) will be mapped to the same point.
However, if the matrix is invertible, then it allegedly cannot happen. One of the explanation I've seen is that:
Assume the mappings of two points $y_i=Px_i (i=1,2)$. Then, they can
be mapped to the same point $y_1=y_2$ if $P(x_2−x_1)=0$. If $P$ is
invertible then $x_1=x_2$ else $x_2=x_1+av$ for all $a\in R$ with
vector $v$ the eigenvector that corresponds to the zero eigenvalue. If
$P$ is not invertible infinite number of points can be mapped to the
same point.
Why $P(x_2−x_1)=0$ implies that they can be mapped to the same point?
How do we know that if matrix is invertible, then it has to be that $x_1=x_2$, if it's not then $x_2=x_1+av$, why is $v$ an eigenvector corresponding to zero eigenvalue? From this equation if follows the solution is unique, but the next sentence says that infinite number of points can be mapped to the same point.
Why, if $P$ is invertible, $x_1$ has to be equal $x_2$ to be mapped to the same point, and if $P$ is invertible, then $x_2$ is equal to $x_1$ plus $av$?
Could you please explain it to me in as simple terms as possible, so I can see the connection between these facts?
Best Answer
a) $P$ invertible:
a1) Then $$x_1=x_2\Longleftrightarrow y_1=Px_1=Px_2=y_2$$ a2) $$ x_1\neq x_2 \Longleftrightarrow y_1=Px_1\neq Px_2=y_2 $$
b) $P$ singular:
$$x_1=x_2\Longrightarrow y_1=Px_1=Px_2=y_2$$ but the inverse does not hold because an infinite number of different $x_i$'s can be mapped to the same $y$. If $v$ is the eigenvector that corresponds to the zero eigenvalue of $P$ then $Pv=0$. Consider now the maps of $x_1$ and $x_1+av$ for all $a\in\mathbb{R}$. Then $$y_2=Px_2=P(x_1+av)=Px_1+aPv=Px_1=y_1$$ even though $x_1\neq x_2$ for $a\neq 0$.