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I can easily generalize the notion of a determinant to non-square matrices using a QR decomposition: $$A \in \mathbb{R}^{m \times n}$$ $$A = Q\cdot\begin{pmatrix} R \\ 0\end{pmatrix} \text{for } m\ge n$$ Now let $$\mathrm{Det}~A = \det R.$$

This gives me the signed volume of an n-dimensional parallelepiped in m dimensions.

We may also define $$A = Q\cdot\begin{pmatrix} R & B \end{pmatrix} \text{for } m<n$$ $$\mathrm{Det}~A = \det R,$$ but I have no geometric interpretation for this one.

The absolute value is the same as the square root of the Gramian determinant ($(\mathrm{Det}~A)^2 = \det(A^T\cdot A)$), but the QR decomposition has the advantage that we also obtain a sign. It should give the same result as multiplying the singular values of $A$ if we take the sign of the determinant of the orthogonal transformation matrices into account.

Does this kind of determinant already has an established name?

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    $\begingroup$ Your Det is not well defined. E.g. when $m>n\ge2$, if you multiply the first and the last columns of $Q$ by $-1$, the sign of $\det(Q)$ remains unchanged, but the sign of $\det(R)$ is flipped. $\endgroup$ – user1551 Mar 9 '18 at 9:21
  • $\begingroup$ Too bad. That must be the reason, why this definition has no name. $\endgroup$ – Lemming Mar 9 '18 at 10:27

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