Conceptual question of coplanar matrix

Question

I know that the matrix rank has to be $2$ or less so that the matrix can be coplanar but what is the reason for that?And a coplanar matrix can't be invertible if the rank is higher than $2$?

Sorry if my question is too basic and for my bad english.

Answer 1

To be honest I find the term 'coplanar matrix' a bit confusing (this is not criticism of you, but of the person introducing this term). A set of points in some space (say $\mathbb{R}^n$) is said to be coplanar if they all lie in a single two-dimensional plane.

From your question I infer that the definition of coplanar matrix is this:

An $n \times n$-matrix is coplanar if the $n$ points in $\mathbb{R}^n$ described by its columns lie in the same plane through the origin

or equivalently

An $n \times n$-matrix is coplanar if the $n + 1$ points in $\mathbb{R}^n$ described by its columns and the zero vector are coplanar in the ordinary sense.

With this cleared up the answer to the question is to look at the column-space of the matrix, that is the space spanned by the columns of the matrix, i.e. the set of all linear combinations of the columns. This is a $k$-dimensional subspace of $\mathbb{R}^n$ for some number $k$ (with $0 \leq k \leq n$). This $k$ is by definition the rank of the matrix.

If all columns lie in the same plane through the origin, then all their linear combinations do too and so the entire column space lies in this plane. In other words, we have a $k$-dimensional space that is somehow contained in a plane, that is: in a 2-dimensional space. It is intuitively obvious that this can only be the case if $k \leq 2$.

This hopefully answers your first question.

For the second question: if $n > 2$ (so the matrix has more than 2 rows) then a coplanar matrix can indeed not be invertible. Conceptually this can be seen by interpreting the matrix as a linear map: this map maps the entire $n$-dimensional space into the plane spanned by its columns. Necessarily this means mapping various points in space to the same point in the plane. Now we can see why it is hard to write an inverse: if many points in space are mapped to one point $x$ in the plane, to which of the original points should the inverse map send $x$?

EDIT: here is a concrete example. Suppose you have the map $M$ that sends $(x, y, z) \in \mathbb{R}^3$ to $(x, y, 0)$. So after applying the map every point lies in the $x, y$-plane, making the matrix corresponding to this map coplanar. (I leave it to you to actually write down the matrix.) Now the problem with finding an inverse map $M^{-1}$ is clear. $M$ maps $(1, 2, 3)$ to $(1, 2, 0)$. It also maps $(1, 2, 4)$ to $(1, 2, 0)$. Now to which point should the map $M^{-1}$ send $(1, 2, 0)$? To $(1, 2, 3)$? To $(1, 2, 4)$? To something else? This is unanswerable so $M^{-1}$ cannot exist.