I want to know the name of the following property of determinant $$\det \left(C-X^T\right)=\det\left(C-X\right)$$ where $C$ and $X$ are same size matrices. Further, $C$ is a diagonal matrix with every element in the diagonal being positive. I do not know if there is some name of this property (I am interested in its proof). I will be very thankful if somebody could provide its proof or the name of this property.

  • $\begingroup$ @mvw I think this condition is true for diagonal matrices. I mean who has only nonzero elements in their diagonal. BTW what is the name of this property. Or pleas provide the proof of it $\endgroup$ – Frank Moses Nov 1 '16 at 9:25

This is simply the property that, for any matrix $A$, $\det(A) = \det(A^T)$. Here, take $A = C - X$. Since $C$ is diagonal, as you point out above, $C = C^T$. Hope this helps! :)

Regarding the proof of this fact, you might find it helpful to consider the following SE questions. Determinant of transpose? and How do I prove that $\det A= \det A^t$?.

  • $\begingroup$ thanks I got it. So nice of you $\endgroup$ – Frank Moses Nov 1 '16 at 9:30

This works only because $C=C^T$ as well, is a symmetric matrix, so $$ (C - X)^T = C^T - X^T = C - X^T $$ and boils down to $\DeclareMathOperator{det}{det}\det A = \det A^T$.

So $$ \det(C - X^T) = \det((C-X)^T) = \det(C - X) $$ Here is a proof for the invariance of the determinant regarding matrix transposition: \begin{align} \DeclareMathOperator{sgn}{sgn} \det A^T &= \sum_{\pi\in S_n} \sgn(\pi) \, a^T_{1\pi(1)} \dotsb a^T_{n\pi(n)} \\ &= \sum_{\pi\in S_n} \sgn(\pi) \, a_{\pi(1)1} \dotsb a_{\pi(n)n} \\ \end{align} where $S_n$ is the set of all permutations of $n$ elements (also known as symmetric group of order $n$).

Now $\pi(n)=k_n \iff n = \pi^{-1}(k_n)$ so $(\pi(n),n) = (k_n, \pi^{-1}(k_n))$ and we can rearrange each product, sorting by increasing $k_i$, to get \begin{align} \det A^T &= \sum_{\pi\in S_n} \sgn(\pi) \, a_{k_1\pi^{-1}(k_1)} \dotsb a_{k_n\pi^{-1}(k_n)} \\ &= \sum_{\pi\in S_n} \sgn(\pi) \, a_{1\pi^{-1}(1)} \dotsb a_{n\pi^{-1}(n)} \\ \end{align} The sign of a permutation $\pi$ is $$ \sgn(\pi) = (-1)^k $$ where $k$ is the number of tanspositions (exchanges of two elements) it can be decomposed into. From this it follows $\sgn(\pi) = \sgn(\pi^{-1})$ and we have \begin{align} \det A^T &= \sum_{\pi\in S_n} \sgn(\pi^{-1}) \, a_{1\pi^{-1}(1)} \dotsb a_{n\pi^{-1}(n)} \\ &= \sum_{\pi^{-1}\in S_n} \sgn(\pi^{-1}) \, a_{1\pi^{-1}(1)} \dotsb a_{n\pi^{-1}(n)} \\ &= \sum_{\sigma\in S_n} \sgn(\sigma) \, a_{1\sigma(1)} \dotsb a_{n\sigma(n)} \\ &= \det A \end{align} where we used that the summation over all the permutations of $S_n$ will feature the same permutations, albeit in different order, as the summation over all inverse permutations.

  • $\begingroup$ thank you I understand it now. $\endgroup$ – Frank Moses Nov 1 '16 at 9:30
  • $\begingroup$ Nice clear explanation of why $\det A = \det A^T$ :) +1 $\endgroup$ – Sam T Nov 1 '16 at 15:11

There is no name for this particular property, which is a little specialized.

As it boils down to $\det(A)=\det(A^T)$, because $(C-X^T)=(C-X)^T$, you can just say that the determinant is transposition invariant.


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