Can you help me show that $\det(mI + U) = m^{v-1}(m + \mathrm{tr}(U))$? ($U$ has rank $1$) They said to use the spectral theorem, and using that I got that the determinant had to be the product of the eigenvalues of $mI + U$. However, I have no idea how to find the eigenvalues. It seems that the eigenvalues are at the points $\lambda$ such that $\det((m-\lambda)I + U) = 0$, but calculating that determinant seems about as hard as calculating the original determinant.
Thanks!