$\det(I+A)=1+\operatorname{Tr}(A)$ if $\operatorname{rank}(A)=1$ 
Let $A$ be a complex matrix of rank $1$. Show that $$\det (I+A) = 1 + \operatorname{Tr}(A)$$ where $\det(X)$ denotes the determinant of $X$ and $\operatorname{Tr}(X)$ denotes the trace of $X$.

Any hint, please. I do not get how to combine the ideas of rank, determinant and trace. Thank you.

Suppose the eigenvalues of $A$ are $\lambda_j,j=1,\cdots,n$, since the rank of $A$ is 1, there is only one $\lambda_j$ is nonzero, then
$$\det(\lambda-A)=\prod_{j}(\lambda-\lambda_j)=\lambda^n-(\sum_{j}\lambda_j)\lambda^{n-1},$$
Now let $\lambda=-1$, one readily gets
$$
\det(I+A)=(-1)^n((-1)^n-(\sum_{j}\lambda_j)(-1)^{n-1})=1+\text{Tr}(A).
$$
The above procedure can be smoothly extended to $Rank\geq 1$ matrices. Moreover, if the size goes to infinite dimension, Fredholm Determinant comes.
