Find $\det(mI + U)$, where $U$ is a rational symmetric matrix of rank $1$ 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!
 A: If $U$ has rank one, then there is a basis consisting of an eigenvector corresponding to a non-zero eigenvalue (any non-zero element if the image of $U$ will do) and a basis of the kernel of $U$. There is then an invertible matrix $C$ such that $CUC^{-1}$ is diagonal, with exactly one non-zero element in the upper left corner.
Now $C(mI+U)C^{-1}$ and $mI+U$ have the same determinant. Can you compute the determinant of the former?
(That $U$ is symmetric or rational has nothing to do with this, by the way)
A: Your problem can be solved by Sylvester's determinant theorem which states that $\det(I_n + AB)=\det(I_m + BA)$ where $A$ is an $n\times m$ matrix and $B$ is $m\times n$. In your case you first have to factor out $m$ which yields $m^d\det(I+d^{-1}U)=m^d \det(1 + m^{-1}tr(U)) = m^{d-1}(m+tr(U))$. 
The reason why the $tr(U)$ shows up is that $U$ is is rank one. Therefore you can write it as the outer product between a vector $u$ and itself. Applying Sylvester's determinant theorem would turn $uu^\top$ into $u^\top u = tr(u^\top u) = tr(uu^\top) = tr(U)$.
