Find the determinant of the linear transformation $L(A)=A^T$ from $\mathbb{R}^{n\times n}$ to $\mathbb{R}^{n\times n}$ Find the determinant of the linear transformation $L(A)=A^T$ from $\mathbb{R}^{n\times n}$ to $\mathbb{R}^{n\times n}$. The solution is $(-1)^{\frac{n(n-1)}{2}}$. I've found some resources to do this when for the case from $\mathbb{R}^{2\times 2}$ to $\mathbb{R}^{2\times 2}$ (in fact from stackexchange), but in the $n$ case, I'm really confused about the formula.
Using the same logic as the 2x2 case, I get that in an equivalent mapping of $L:\mathbb{R}^{n^2} \rightarrow \mathbb{R}^{n^2}$ is: $L(e_{11},e_{12},e_{nn})=\begin{bmatrix}
    1 & 0 & 0 & \dots  & 0& 0 \\
    0 & 0 & 0 & \dots  & 1& 0 \\
    \vdots & \vdots & \vdots & \ddots & \vdots& \vdots \\
0 & 1 & 0 & \dots  & 0& 0 \\
    0 & 0 & 0 & \dots  & 0& 1
\end{bmatrix} \begin{bmatrix}
    e_{11} \\
    e_{21} \\
    \vdots\\
\vdots \\
    e_{nn}
\end{bmatrix}$ 
Where we can find the determinant of the large $0$ and $1$ matrix above (let's call it matrix B) to find the determinant of the linear transformation L. I know that $det(B)=(-1)^s$ where $s=$the number of swaps made to get B into rref form. However, I just can't see how we need $\frac{n(n-1)}{2}$ swaps and not $n$ swaps?
 A: The map $A\mapsto A^T$ is an involution: its square is the identity.
Thus it splits the space $\mathbb{R}^{n\times n}$ into two eigenspaces:
one with eigenvalue $+1$ and the other with eigenvalue $-1$. The determinant of the operator is then $(-1)^d$ where $d$ is the dimension of the $(-1)$-eigenspace.
The $(+1)$-eigenspace consists of the matrices with $A^T=A$, the symmetric
matrices.
The $(-1)$-eigenspace consists of the matrices with $A^T=-A$, the skew-symmetric
matrices.
So, what's the dimension of the space of skew-symmetric matrices?
A: The solution is correct, but you're right that it doesn't come naturally from counting the number of row swaps:
Your matrix is size $n^2\times n^2$. The first and last row are fine and do not need to swapped. Each of the other $n^2-2$ rows need to be moved; in fact you need to swap the second and second-to-last row, the third and third-to-last row, etc. 
When $n$ is even, since each swap fixes two rows, you need $\frac{n^2-2}{2}$ swaps total. When $n$ is odd, you do not need to swap the middle row, so you need $\frac{n^2-3}{2}$ swaps. Both cases can then be encoded by the formula $\frac{n^2-n}{2}$, since it has the same parity as the answer we calculated for each of the two cases.
A: By taking transpose, you swap every symmetric pair of off-diagonal elements $\{a_{ij},a_{ji}\}$ (with $i\ne j$) in the matrix $A$. Now there are $\frac{n(n-1)}2$ such pairs. Hence the answer.
