I was watching 3blue1brown's video on inverse matrices, and I didn't understand what he said about the case where a matrix $A$ has no inverse when $\det(A)=0$.
Considering $Ax=v$ where $x$and $v$ are vectors, he made two statements:
1) If a transformation squishes all of plane onto a single line for a $2\times 2$ matrix then no transformation can unsquish that line back to a plane. (I didn't understand his explanation based on functions)
2) It is still possible that solution exists
even if inverse doesn't exist when $\det(A)=0$ if vector $v$ "lives" somewhere on that line.
I don't understand why no such transformation exists in the first statement.
I also don't understand what vector $v$ "living" on the line means.
Here's the link to the video:
Video by 3blue1brown
Best Answer
It might help to look at an example. For instance, consider the matrix $$ A = \pmatrix{1&0\\0&0} $$ The transformation associated with this matrix is $$ f(x) = Ax = \pmatrix{1&0\\0&0}\pmatrix{x_1\\x_2} = \pmatrix{x_1\\0} $$ Geometrically, this transformation "squishes" $\Bbb R^2$ onto the $x_1$-axis (or the $x$-axis, if you prefer thinking in terms of $x$ and $y$).
The point of an inverse transformation is to undo the effect of $f$. That is, we're looking for a transformation $g(x)$ such that $g(f(x)) = x$. That is, we would like to make a function that satisfies $$ g(f(x_1,x_2)) = g(x_1,0) = (x_1,x_2) $$
Intuitively, we see that this is impossible since, after applying $f$, we have lost the information of what $x_2$ was.
More formally and specifically: we know that every point of the form $(1,x_2)$ (i.e. every point of the form $x_1 = 1$) will get "squished" to the point $(1,0)$. So, for $g$ to undo $f$, we would need to have the following be true: $$ g(f(1,3)) = g(1,0) = (1,3)\\ g(f(1,-7)) = g(1,0) = (1,-7) $$ The problem here is that since $g$ is a function, we can only send the point $(1,0)$ to one output. So, there is no one function $g$ that can give back the coordinates of every point we put in.
Regarding your second point: even though $\det(A) = 0$, it is certainly possible for $Ax = v$ to have a solution if $v$ lives in the line (in our case, the $x_1$-axis). For instance, if we have $v = (-4,0)$, then $f(x) = v$ will be true whenever $x = (-4,x_2)$, i.e. whenever $x$ is a point that gets squished to $v$. On the other hand, if we have $v = (-4,2)$, then no $x$ will satisfy $f(x) = v$: $f(x)$ is necessarily on the $x_1$-axis, but $v$ doesn't "live" there.