Why is it important for a matrix to be square?
Solution 1:
Remember that an $n$-by-$m$ matrix with real-number entries represents a linear map from $\mathbb{R}^m$ to $\mathbb{R}^n$ (or more generally, an $n$-by-$m$ matrix with entries from some field $k$ represents a linear map from $k^m$ to $k^n$). When $m=n$ - that is, when the matrix is square - we're talking about a map from a space to itself.
So really your question amounts to:
Why are maps from a space to itself - as opposed to maps from a space to something else - particularly interesting?
Well, the point is that when I'm looking at a map from a space to itself inputs to and outputs from that map are the same "type" of thing, and so I can meaningfully compare them. So, for example, if $f:\mathbb{R}^4\rightarrow\mathbb{R}^4$ it makes sense to ask when $f(v)$ is parallel to $v$, since $f(v)$ and $v$ lie in the same space; but asking when $g(v)$ is parallel to $v$ for $g:\mathbb{R}^4\rightarrow\mathbb{R}^3$ doesn't make any sense, since $g(v)$ and $v$ are just different types of objects. (This example, by the way, is just saying that eigenvectors/values make sense when the matrix is square, but not when it's not square.)
As another example, let's consider the determinant. The geometric meaning of the determinant is that it measures how much a linear map "expands/shrinks" a unit of (signed) volume - e.g. the map $(x,y,z)\mapsto(-2x,2y,2z)$ takes a unit of volume to $-8$ units of volume, so has determinant $-8$. What's interesting is that this applies to every blob of volume: it doesn't matter whether we look at how the map distorts the usual 1-1-1 cube, or some other random cube.
But what if we try to go from $3$D to $2$D (so we're considering a $2$-by-$3$ matrix) or vice versa? Well, we can try to use the same idea: (proportionally) how much area does a given volume wind up producing? However, we now run into problems:
If we go from $3$ to $2$, the "stretching factor" is no longer invariant. Consider the projection map $(x,y,z)\mapsto (x,y)$, and think about what happens when I stretch a bit of volume vertically ...
If we go from $2$ to $3$, we're never going to get any volume at all - the starting dimension is just too small! So regardless of what map we're looking at, our "stretching factor" seems to be $0$.
The point is, in the non-square case the "determinant" as naively construed either is ill-defined or is $0$ for stupid reasons.
Solution 2:
Lots of good answers already as to why square matrices are so important. But just so you don't think that other matrices are not interesting, they have analogues of the inverse (e.g., the Moore-Penrose inverse) and non-square matrices have a singular-value decompition, where the singular values play a role loosely analogous to the eigenvalues of a square matrix. These topics are often left out of linear algebra courses, but they can be important in numerical methods for statistics and machine learning. But learn the square matrix results before the fancy non-square matrix results, since the former provide a context for the latter.
Solution 3:
In linear algebra matrices usually represent linear transformations between vector spaces. An $m \times n$ matrix $M$ represents a linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$
When the matrix is square ($m=n$) it can be thought of as representing a transformation from $\mathbb{R}^n$ to itself. That's when the concepts of eigenvalues and the characteristic polynomial make sense.
Solution 4:
To add to @Ethan's answer what is fundamental when a linear map goes from a vector space to itself is that you can compare a vector with its image:
- was it rotated?
- was its length scaled?
- was it turned into a multiple of itself?
Or a set of vectors with its image:
- by how much was the volume of this parallelepiped stretched?
All these questions make sense (and carry an awful lot of information on the linear map) only when the domain and codomain coincide. The determinant, characteristic polynomial, etc. are there to answer these questions.