Determinant of a linear transformation defining matrix transpose So if I define a linear transformation $ T: M_{n\times n}(R) \rightarrow M_{n\times n}(R) $ and $ T(A)=A^t $ what would be its determinant? 
 A: The determinant is most easily determined in a basis of eigenvectors. There are $n$ matrices with a single $1$ on the diagonal, $n(n-1)/2$ matrices with two $1$s in symmetric positions and $n(n-1)/2$ matrices with a $1$ and a $-1$ in symmetric positions. These form a basis together. The diagonal and symmetric matrices have eigenvalue $1$ under transposition, the antisymmetric matrices have eigenvalue $-1$, so the determinant is $(-1)^{n(n-1)/2}$, which is $+1$ if $n\bmod4$ is $0$ or $1$ and $-1$ if it's $2$ or $3$.
A: Let us construct the matrix representation with respect to the standard basis vectors ordered in a special way. The first $n$ are $\mathbb{e}_{ii}$ for $1\le i \le n$. Then pair the remaining matrices in transpose pairs. If you construct the resulting matrix, you will have a block diagonal matrix beginning with an identity block along with $\frac{n^2 - n}{2}$ anti-diagonal blocks of size $2$. It follows that the determinant is given by $(-1)^{\frac{n^2 - n}{2}}$.
A: EDIT---
 I think (now that I have seen other answers) this argument also works. We want the determinant of linear transformation say $T$ such that $vec(A^{T})=Tvec(A)$ (the $vec$ operator which writes a matrix as a long vector column wise). It is easy to see that $vec(A^{T})$ can be obtained by rearranging the rows of the vector $vec(A^{T})$. Each one of the row swappings makes your determinant multplied by (-1). Now how many swappings you have to make is the question. You have to swap every element beneath the main diagonal with another from above the diagonal. So there are $c=n(n-1)/2$ swappings. So your determinant is $(-1)^c$. 
PREVIOUS ATTEMPT--(BEFORE I UNDERSTOOD THE QUESTION PROPERLY)
First of all, note that $det(AB)=det(A)det(B)$, now appy this recursively. You will get $det(A^{t})=(det(A)det(A^{t-1}))=(det(A))^t$
