Why don’t non-square matrices have determinants? The determinant is just the matrix’s scale factor (i.e. the “size” of the linear transformation), and I don’t see why a rectangular matrix wouldn’t have one.

My answer to a question in Quora: Why don’t non-square matrices have determinants? The determinant is just the matrix’s scale factor (i.e. the “size” of the linear transformation), and I don’t see why a rectangular matrix wouldn’t have one.

What follows is an answer that I would give to my students, if they asked it in the lecture. When you wish to generalise determinants to non-square matrices, but preserve their interpretation as “scale factors”, you have to preserve the multiplicativity of determinants: scale factors of consecutively executed transformations should multiply — otherwise why call them scale factors? Hence you perhaps wish to have, for this “extended” determinant, the property

\(\det AB = \det A \cdot \det B\) whenever the product \(AB\) exists.

Perhaps you would also wish this “new extended” determinant to coincide with the traditional determinant when applied to square matrices.

Alas, this is impossible: take

\(A = \begin{pmatrix} 1 & 1 \end{pmatrix} \)

and

\(B = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \)

then

\(AB = \begin{pmatrix} 2\end{pmatrix} \)

and

\(BA = \begin{pmatrix} 1 &1 \\ 1&1 \end{pmatrix} \)

therefore

\(\det A \cdot \det B = \det AB = \det(2) = 2 \)

and

\(\det B \cdot \det A = \det BA = \det\begin{pmatrix} 1 &1 \\ 1&1 \end{pmatrix} = 0, \)

hence

\(\det A \cdot \det B \ne \det B \cdot \det A\)

—an obvious contradiction.