solve det(I + uv^t)

prove det(I+uvt)=1+utv

First, let us list several basic formulations:

Assume Aa=λa, then we have A−1a=1λa, meanwhile, we can also get (A+I)a=(λ+1)a

Afterwards, I would like to quote a conclude and two exercises from math guidebook for graduate entrance examination.









Therefore, the conclusion can be easily attained:

one eigenvalue of uvt is tr(uvt) which equals to utv and all the others would be zero. Given by the equation (A+I)a=(λ+1)a results in the fact that one eigenvalue of I+uvt is 1+utv and all the others would be one.

Remember the determinant of a matrix equals to the multiply of all its eigenvalues, hence, det(I+uvt)=(1+utv)∗1=(1+utv), proved.