Determinant of a non-square matrix

I wrote an answer to this question based on determinants, but subsequently deleted it because the OP is interested in non-square matrices, which effectively blocks the use of determinants and thereby undermined the entire answer. However, it can be salvaged if there exists a function \det defined on all real-valued matrices (not just the square ones) having the following properties.

  1. \det is real-valued
  2. \det has its usual value for square matrices
  3. \det(AB) always equals \det(A)\det(B) whenever the product AB is defined.
  4. \det(A) \neq 0 iff \det(A^\top) \neq 0

Does such a function exist?

Answer

Such a function cannot exist. Let A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0\end{pmatrix} and B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}. Then, since both AB and BA are square, if there existed a function D with the properties 1-3 stated there would hold
\begin{align}
\begin{split}
1 &= \det \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \det(BA) = D(BA) = D(B)D(A) \\
&= D(A)D(B) = D(AB) = \det(AB) = \det \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} = 0.
\end{split}
\end{align}

Attribution
Source : Link , Question Author : goblin GONE , Answer Author : Dan Fox

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