$\det(I+A)=1+\operatorname{Tr}(A)$ 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.

Assume $A$ is diagonal. As its rank is $1$, it has only one non-zero eigenvalue $\lambda$. Then
$$ \det(I+A) = 1+\lambda = 1 + \mathrm{tr}\, A. $$

Assume that $A$ is diagonalisable, so that $A = PDP^{-1}$. Then
$$\det(I + A) = \det(I + PDP^{-1}) = \det(P(I + D)P^{-1} ) = \det(I + D)$$
Similarly
$$1 + \mathrm{tr}\,A = 1 + \mathrm{tr}\, D,$$
so we reduced this case to the previous one.

Now assume $A$ is an arbitrary complex matrix. Both sides of the equation are continuous and $A$ can be approximated by diagonalisable matrices. This finishes the proof.
