# trace and determinant of backward diagonal matrix

I am just wondering whether there is any general formula or relation between trace of backward / anti diagonal matrix just like we have in a diagonal matrix of order $$n*n$$ where trace is the sum of diagonal elements and determinant is the product of diagonal elements.

• This is interesting Dec 21, 2018 at 17:16
• The trace is not the product of the diagonal elements... Dec 21, 2018 at 17:20
• Here is another post about antidiagonal matrices that you might find interesting Dec 21, 2018 at 17:26

The trace of an anti diagonal matrix is still the sum of the [main] diagonal elements. If $$n$$ is even, then the trace is zero since all diagonal elements are zero. If $$n$$ is odd, then the $$\frac{n+1}{2}, \frac{n+1}{2}$$ element at the very center of the matrix is the only element on both the main diagonal and anti diagonal, so the trace is equal to this element.
The determinant of an anti diagonal matrix can be shown to be the product of the anti diagonal elements possibly multiplied by $$-1$$ depending on the value of $$n$$ (specifically, if $$n \equiv 2 \mod 4$$ or $$n \equiv 3 \mod 4$$ then you need to multiply by $$-1$$). A quick way to verify this is to note that you can perform column swap operations to obtain a diagonal matrix (whose determinant is the product of the diagonal entries), and then use property 13 here.
Take $$M = \pmatrix{&&a_1\\&\cdots\\a_n}$$ to be your anti-diagonal matrix. We have $$\operatorname{trace}(M) = \begin{cases} a_{(n+1)/2} & n \text{ is odd}\\ 0 & n \text{ is even} \end{cases}$$ and $$\det(M) = (-1)^{\lfloor n/2\rfloor} a_1 \cdots a_n$$, where $$\lfloor x \rfloor$$ is the greatest integer $$m$$ such that $$m \leq x$$.