Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question
up vote 2 down vote accepted

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)

share|cite|improve this answer
Thank you, I understand how to do it now. :) – badatmath Nov 23 '11 at 7:53
Wait sorry, how did you get the first statement? I think there's some theorem I'm missing. Or an entire linear algebra course. – badatmath Nov 23 '11 at 16:00
FInd a non zero element of the image and call it $v_1$; find a basis of the kernel, and call its elements $v_2$, $\dots$, $v_n$; show that the set $\{v_1,\dots,v_n\}$ is a basis. – Mariano Suárez-Alvarez Nov 23 '11 at 16:15

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 $u^\top u = tr(u^\top u) = tr(uu^\top) = tr(U)$.

share|cite|improve this answer
Thanks, seems like an interesting fact! – badatmath Nov 23 '11 at 7:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.