It is well known that a determinant of a matrix is the product of eigenvalues:

$$\det({\bf A}) = \prod_{\forall i}\lambda_i({\bf A})$$ and product of singular values equal to the norm of determinant as @Batominovski pointed out: $$|\det({\bf A})| = \prod_{\forall i}\sigma_i({\bf A})$$

This has useful interpretations for many kinds of matrices in many applications. Now if we look at the $\bf M_2$ matrices from this answer, we see that we can have a matrix that performs "summing" of independent sub-spaces but will always have determinant 1. Could we generalize our notion of determinant to somehow measure this summing ability? If we are aware of the $2\times 2$ block structure we could "pretend" that each such block along the diagonal was an eigenvalue, and some kind of generalized determinant would then be the product of them which would yield a sum for those off-diagonal position, like this:

$$\text{bdet}({\bf M_2}(6)) = \left[\begin{array}{cc}1&0\\1&1\end{array}\right]\left[\begin{array}{cc}1&0\\1&1\end{array}\right]\left[\begin{array}{cc}1&0\\0&1\end{array}\right]=\left[\begin{array}{cc}1&0\\2&1\end{array}\right]$$ Where the two is the sum of prime exponents. b could stand for block, or something.

Does this seem useful or make any sense or am I just rambling about?

  • 1
    $\begingroup$ One thing to look into are permanents. Might not be exactly what you are looking for. en.m.wikipedia.org/wiki/Permanent $\endgroup$
    – user223391
    Aug 27 '16 at 2:07
  • $\begingroup$ But the determinant is not the product if singular values. Its norm is. $\endgroup$ Aug 27 '16 at 2:27
  • $\begingroup$ @Batominovski The second statement in your comment is in general incorrect. $\endgroup$
    – user1551
    Aug 27 '16 at 2:41
  • $\begingroup$ @user1551 What are you talking about? See math.stackexchange.com/questions/244743/singular-value-proofs. $\endgroup$ Aug 27 '16 at 2:47
  • $\begingroup$ Ohh, I see. You may have thought I meant a matrix norm. By "its norm," I meant "the complex norm of the determinant," or the absolute value. $\endgroup$ Aug 27 '16 at 2:55

If we require, that the generalized determinant $gdet$ is equivalent to the standard determinant for square matrices, then we cannot expect, that $gdet(AB) = gdet(A)\times gdet(B)$ for any $A$, $B$, unless $gdet(A) \equiv 0$ for all non square matrices $A$.

To see this it is enough to consider a diagonal square matrix $A$ and a matrix $B$ being $i$-th column of the identity matrix.

Without this property I don't think, that such generalization of the determinant could be useful.

  • $\begingroup$ -1 The OP seems to have a very specific motivation in mind, and in light of this motivation, there's no reason to assume that the determinant is multiplicative. $\endgroup$ Aug 27 '16 at 4:12
  • $\begingroup$ Thank you for the contribution but it is not exactly what I am searching for. $\endgroup$ Aug 28 '16 at 11:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.