# Determinant of outer product matrices

1. Sep 15, 2011

### seanbow

1. The problem statement, all variables and given/known data
Given $u, v \in \mathbb{R}^{n}$, and $A \in \mathbb{R}^{n \times n}$, $\mathrm{det}\left(A\right) \neq 0$, find $\mathrm{det}\left( A + uv^{T} \right)$

2. Relevant equations
Generic determinant and eigenvalue equations, I suppose.

3. The attempt at a solution
Hoping to gain some insight, I found the determinant for the special case $A = I$. In that case we can see that the eigenvalues of $I + uv^T$ satisfy
$$\mathrm{det}\left(I + uv^T + \lambda I\right) = \mathrm{det}\left( uv^T + \left( \lambda + 1\right) I\right) = 0$$
so we see that the eigenvalues of $uv^T + I$ are those of $uv^T$ plus one.

Outer product matrices have rank 1, so $uv^T$ has only one nonzero eigenvalue (which is pretty easy to see is equal to $\mathrm{tr}\left( uv^T \right) = u^{T}v$).

The determinant of $uv^T + I$ is equal to the product of its eigenvalues. Since it only has one eigenvalue not equal to 1, the determinant will equal that eigenvalue, or $\mathrm{tr}\left(uv^T\right) + 1 = 1 + u^Tv$

I tried similar methods replacing $I$ with the more generic $A$ without any luck. I also tried starting from the equation
$$A\vec{\eta} = \lambda\vec{\eta}$$
again without any luck.

I got to the equation
$$\mathrm{det}\left(A + uv^T\right) = \mathrm{det}(A) \mathrm{det}\left( I + A^{-1} u v^T \right)$$
but I don't think it's very useful.