Wednesday, June 12, 2013

bonus question: can you show that the determinant of a skew-symmetric matrix $A$, $A^t=-A$, is equal to zero if the dimension is odd?
1. bonus question: the determinant of $A$ is equal to the product of its eigenvalues (by diagonalizing). The eigenvalues of $A$ are minus the eigenvalues of $A^t$ (as it is equal to $-A$). As the number of (possibly degenerate) eigenvalues is equal to the dimension of the matrix, if this dimension is odd there must be a zero eigenvalue, making the determinant zero.
1. yes, quite so! another way of seeing that would be $\mathrm{det}(A)=\mathrm{det}(A^t)=\mathrm{det}(-A)=(-1)^n\mathrm{det}(A)$ using multilinearity. of course $\mathrm{det}(A)=(-1)^n\mathrm{det}(A)$ can only be true if $\mathrm{det}(A)=0$ in the case of odd $n$.