# How to prove that determinant of identity matrix with columns in reverse order is -1

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 $M_n(\mathbb{R})$.

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@AhmedHussein Thx and done. – Raton Feb 20 at 23:16
what happens to the value of the determinant when you interchange columns? – vnd Feb 20 at 23:17
The action of the matrix is to swap the first entry of a vector with the last entry, the second with the second to last, third with third to last, and so forth. So, you can see what the eigenvectors and eigenvalues must be by inspection. Vectors like $(1,0,\dots,0,1)$ are eigenvectors with eigenvalue $1$, whereas vectors like $(1,0,\dots,0,-1)$ are eigenvectors with eigenvalue -1. Just count the number of negative eigenvalues and see if it is even or odd. – Nick Alger Feb 21 at 4:40

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:

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

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.

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So basically the only way for $det(A)$ to equal $-1$ is if $\dfrac{n}{2}$ is odd? – Raton Feb 20 at 23:32
@Raton $\frac{n}{2}$ might not be an integer if $n$ is odd in the first place. You will find that it will be $-1$ when $n\equiv 2\pmod{4}$ (which is an even integer such that $\frac{n}{2}$ is odd) or $n\equiv 3\pmod{4}$ is one way of looking at it. Another way of looking at it is when $\binom{n}{2}$ is odd. – JMoravitz Feb 20 at 23:39
Good point and great solution. – Raton Feb 20 at 23:40

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.

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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 $(-1)^{\lfloor n/2\rfloor}$.

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