# Does a non-invertible matrix transformation "really" not have an inverse?

I understand that, for example, if we have a 2 by 2 matrix, when its determinant is zero, then it does not have an inverse matrix. Hence, the linear transformation is non-invertible.

If we have a 1 by 1 square on the 2D plane, then, geometrically, this non-invertible 2D matrix transforms it into a 1D line segment.

However, on the other hand, don't a line segment and a square have the same amount of points? If so, there has to be a mapping that maps each point on this segment to the square, right? So does this mean even though the matrix is non-invertible, the transformation itself is still invertible?

• Welcome to Mathematics Stack Exchange. Not all transformations are linear; also, a function that is not one-to-one does not have an inverse function Apr 17, 2020 at 6:13

There are no inverse linear transformations. And in linear algebra we rarely, if ever, bother with functions that aren't linear transformations. So we say it's not invertible for that reason.

However, your function isn't injective. So regardless of whether we limit ourself to linear transformations or to general functions, there is no inverse to your function. Most points on your line segment is the image of an entire line segment from your square, and you can't reverse that mapping with our modern conventional understanding of what a function is.

There is no linear transformation that maps a segment to a square. (By linearity, the images of three aligned points must be three aligned points and the image of a segment is a segment - or a single point.)

There is no logical implication from

There exists a map from $$A$$ to $$B$$ that has an inverse.

to

Every map from $$A$$ to $$B$$ has an inverse.