Square matrices have a diagonal. They also have an anti-diagonal.

There are upper-triangular and lower-triangular matrices. We could easily imagine an "anti-upper-triangular" matrix, such as $\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]$, which is nonzero above its anti-diagonal. We can also imagine "anti-lower-triangular" matrices, which are nonzero under their anti-diagonal.

We can imagine anti-circulant and anti-Toeplitz matrices, which are constant along the direction of the anti-diagonal rather than along the direction of the diagonal.

So, the question: Is there a symmetry here? Can theorems involving diagonal "stuff" (and there are many of those) generally be transformed into similar theorems involving anti-diagonal stuff? I suspect the answer may be "no", but if so, I'd be interested in knowing why not. Why is the diagonal direction so much more interesting than the anti-diagonal direction?

  • $\begingroup$ Part of the reason is that matrices that are diagonal corespond to just a number of copies of whatever ring you are working over. This is not at all the case for anti-diagonal ones (which are not even closed under multiplication). $\endgroup$ – Tobias Kildetoft Apr 7 '16 at 10:52
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    $\begingroup$ I think, this is a consequence of the definition of the matrix-product.If we consider diagonal matrices : They probably do not have an analogue meaning if we choose the other diagonal to be the "main"-diagonal. The idendity matrix also shows that the diagonals do not have a similar meaning. $\endgroup$ – Peter Apr 7 '16 at 10:52
  • $\begingroup$ Further, if we consider square matrices as endomorphisms of a vector space with a fixed basis, then the diagonal has a nice description in terms of coefficients of vectors in the image of themselves, whereas the anti-diagonal has no such (and in fact depends on how we order the basis, which the diagonal does not). $\endgroup$ – Tobias Kildetoft Apr 7 '16 at 10:54
  • $\begingroup$ At least, the determinant (possibly upto the sign) does not change, if we mirror the matrix. $\endgroup$ – Peter Apr 7 '16 at 10:57

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