Determinant of the identity matrix with columns in reverse order

Question

If $C \in M_n(\mathbb{R})$ such that $(C)_{ij} = (I)_{(i)(n+1-j)}$, how do I prove that $\det (C) = -1$.

What I tried: I know that $\det(C) = -1$ for $C\in M_{2}(\mathbb{R})$, but I don't know how to prove it for $M_n(\mathbb{R})$.

Answer 1

One of the definitions of determinant is:

$$\det(A) = \sum_{\sigma\in S_n}\text{sgn}(\sigma)\prod\limits_{i=1}^n a_{i,\sigma_i}$$

In hopefully simpler terms, we range over all possible "patterns" (ways of picking $n$ entries such that every row and every column is represented exactly once), computing the product of the entries in the pattern, applying a sign change depending on the number of pairs of entries in the pattern where one is above and to the right of other entries in the pattern (if odd, change sign. if even, keep sign the same), and then summing.

In the case of the anti-identity, there is exactly one pattern which contributes a nonzero sum.

Furthermore, as every entry in the pattern is up and to the right of every other previous entry of the pattern, the number of "up-and-right" pairs will be $\binom{n}{2}$, so the signature will be $(-1)^{\binom{n}{2}}$

The product will be one, as every nonzero entry is one, giving:

$\det(A)=(-1)^{\binom{n}{2}}$

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as an aside, to show this formula in action, consider the following matrix:

$\begin{bmatrix}0&2&1\\2&-1&1\\0&1&2\end{bmatrix}$. There are only two patterns which will contribute a nonzero total to the sum: $\begin{bmatrix}0&\color{red}{2}&1\\\color{red}{2}&-1&1\\0&1&\color{red}{2}\end{bmatrix}$ and $\begin{bmatrix}0&2&\color{red}{1}\\\color{red}{2}&-1&1\\0&\color{red}{1}&2\end{bmatrix}$. In the first case, the only up-right pair are the twos in the first and second column, so the signature is $-1$. In the second pattern, there are two occurrences, the two in the first column and the one in the third as well as the one in the second column with the one in the third, so its signature is $1$.

This gives the determinant as being $(-1)(2\cdot 2\cdot 2)+(1)(2\cdot 1\cdot 1) = -8+2 = -6$ for the determinant.

Answer 2

It’s not true when $n=4$, for instance:

$$\begin{align*} \left|\matrix{0&0&0&1\\ 0&0&1&0\\ 0&1&0&0\\ 1&0&0&0}\right|&=-\left|\matrix{0&0&1\\0&1&0\\1&0&0}\right|=-\left|\matrix{0&1\\1&0}\right|=1 \end{align*}$$

by repeated cofactor expansion along the top row.

In fact it’s true precisely when $\binom{n}2$ is odd: interchanging two columns of a matrix multiplies its determinant by $-1$, and it takes $\binom{n}2=\sum_{k=1}^{n-1}k$ interchanges to invert the order of the columns of an $n\times n$ matrix.

Answer 3

The permutation of columns which changes the identity matrix if dimension $n$ into the ‘anti-identity’ matrix is the product of all transpositions $(i,n+1-i)$, $\;1\le i\le \bigl\lfloor\frac n2\bigr\rfloor$. Hence the determinant is $(-1)^{\lfloor n/2\rfloor}$.