Is it possible to use non square matrices to solve systems of simultaneous linear equations?

Question

Today I learnt non square matrices generally cannot have determinants because the matrix cannot fulfill all properties of determinants as seen in square matrices. So I want to know if given a system of linear equations whose matrix is non square can be solved if it can't have determinants.

Answer 1

Totally. Search Gauss elimination, and you would find something related. Also for overdetermined systems, we can find the best approximation in some sense. Check Linear Algebra textbooks of you like.

Answer 2

A fruitful way of thinking about an $n \times m$ matrix $A$ is as a function $A:\mathbb{R}^m \rightarrow \mathbb{R}^n$. So $A$ acts on every vector $v \in \mathbb{R}^m$ by left multiplication, giving a new vector $Av\in \mathbb{R}^n$. If $n=m=2$ you can think of $A$ as a way of moving around the plane, keeping straight lines straight, and keeping the origin fixed. Because of this, $A$ sends every $1 \times 1$ square in the plane to a parallelogram, and all of those parallelograms have the same area. The (absolute value of the) determinant of $A$ tells you that area.

If general for $m=n$, the matrix (though of as a function) $A:\mathbb{R}^n \rightarrow \mathbb{R}^n$ sends any $\underbrace{1 \times \dots \times 1}_{n \textrm{ times}}$ hypercube to a parallelotope and all such parallelotopes have the same volume. $\det(A)$ tells you the signed $n$ -dimensional volume of such a parallelotope.

Answer 3

They don't use Gaussian elimination for non-square matrices. This is a series of charts from the matlab site for the backslash operator which is used to solve systems of equations i.e. $ Ax= b$

ml-divide full, for non-sparse matrices

ml-divide for sparse matrices

as you can see it when it isn't square it uses the $QR$ decomposition. That is

$$ Ax =b \\ QRx = b \\ Rx = Q^{-1}b $$ then it uses back substitution to solve this

It is also possible to use the SVD decomp

$$ Ax=b \\ U \Sigma V^{T} x = b \\ x = V \Sigma^{-1}U^{T}b $$