Find determinant of reversal matrix using permutation similarity

Question

This question is Problem 7.7 from Matrix Mathematics: A Second Course in Linear Algebra (2 ed) by Garcia and Horn:

Let $K_n$ be the reversal matrix \begin{equation*} K_n = \begin{bmatrix} & & & 1 \\ & & \unicode{x22F0} & \\ & 1 & & \\ 1 & & & \end{bmatrix}. \end{equation*} If $n = 2m$ or $n = 2m + 1$, show in two ways that $\det K_n = (-1)^m$:

(a) Use induction and a Laplace expansion.

(b) Show that $K_n$ is permutation similar to the direct sum of some copies of $K_2$ and $K_1$.

I was able to show part (a) without much issue. However, I am at a complete roadblock with part (b). I started by trying to work with the case $n = 4$ and manually testing some "simple" permutation similarities that have $K_2$ or $I$ as blocks. For example, \begin{equation*} \begin{bmatrix} 0 & K_2 \\ K_2 & 0 \end{bmatrix}\begin{bmatrix} K_2 & 0 \\ 0 & K_2 \end{bmatrix} \begin{bmatrix} 0 & K_2 \\ K_2 & 0 \end{bmatrix} = \begin{bmatrix} K_2 & 0 \\ 0 & K_2 \end{bmatrix} \end{equation*} is one of the several (non-fruitful) permutation similarities I tested.

I am hoping someone can find the permutation matrix that shows $K_2 \oplus K_2$ and $K_4$ are permutation similar, and possibly proceed to the general case. It might also be possible these are not permutation similar; in which case, are $K_2 \oplus K_1 \oplus K_1 = K_2 \oplus I_2$ and $K_4$ permutation similar instead?

The question here is related but does not use the method I am interested in. One of the answers is well-written solution giving the idea for part (a).

Answer 1

The goal here is to find a basis of $\mathbf{R}^n$ such that $K_n$ with respect to this matrix is block-diagonal with blocks consisting of $K_1$ and $K_2$.

To this end, consider that $K_n$ acts on the standard basis vectors $S = \{e_1, \dotsc, e_n\}$ according to $$ K_n e_j = e_{n - j + 1}. $$

Hence, we consider a reordered version of this basis $B = \{v_1, \dotsc, v_n\}$ according to \begin{align*} v_{2k - 1} &= e_k, \\ v_{2k} &= e_{n - k + 1} \end{align*} so that $$ K_n v_{2k-1} = K_n e_k = e_{n-k+1} = v_{2k}. $$

That is, $K_n$ permutes each pair $(v_1, v_2), (v_3, v_4), \dotsc$ (i.e., acts on the subspaces spanned by each pair according to $K_2$) and, if $n$ is off with $n = 2m+1$, $K_n$ fixes $v_{m+1}$ (i.e., acts on it according to $K_1$).

Thus, the matrix representation of $K_n$ according to the basis $B$ is $$ [K_{2m}]_B = \overbrace{K_2 \oplus \dotsb \oplus K_2}^{m\text{ terms}} $$ if $n = 2m$ is even, and $$ [K_{2m + 1}]_B = \overbrace{K_2 \oplus \dotsb \oplus K_2}^{m\text{ terms}} \oplus K_1 $$ if $n = 2m+1$ is odd.

If you want to write out the matrix similarity explicitly, it's a matter of finding the change-of-basis matrix between $S$ and $B$. But this is just the orthogonal matrix $P$ whose columns are $\{v_1, \dotsc, v_n\}$, which yields $K = P^{-1} [K]_B P$.