I am trying to show that
$$\det \left(I_{n} + \begin{bmatrix} a_{1} \\ a_{2} \\ \vdots \\ a_{n} \end{bmatrix} \begin{bmatrix}b_{1}&b_{2}&\cdots&b_n \end{bmatrix} \right) = 1 + \sum_{m=1}^n a_{m}b_{m}$$
I know I have to use the fact that $$\ M=QDQ^{T}$$
I can see that with a diagonal matrix its determinant its equal to its trace, but I am having a hard time figuring out what to do with that identity and that 1?
Hint: Assuming the problem is to find the determinant of the matrix $\ I_n + ab^\top\ $, the determinant can be expressed as a product of that matrix's eigenvalues. One of its eigenvectors is $\ a\ $. What is the corresponding eigenvalue? There are $\ n-1\ $ eigenvectors that are perpendicular to $\ b\ $. What are the corresponding eigenvalues?