# 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!

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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)

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

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Thanks, seems like an interesting fact! –  badatmath Nov 23 '11 at 7:54