Does $\det(I+A) = 1 + \mbox{tr}(A)$ hold if $A$ is a rank-$1$ complex matrix? 
If $A$ is a complex $n \times n$ matrix of rank $1$, then $$\det(I+A) = 1 + \mbox{tr}(A)$$

How to approach this problem?
Rank-$1$ matrices have special properties. Also, thinking about the determinant of a matrix as the product of its eigenvalues and the trace of the matrix as the sum of its eigenvalues, 
$$\prod_{k=1}^{n}(1+\lambda_{k}) = 1  +\sum_{k=1}^{n} \lambda_{k}$$
I could not proceed.
 A: Let $\rm A = u v^*$ be a rank-$1$ matrix. Using the Weinstein-Aronszajn determinant identity,
$$\det \left( \mathrm I_n + \mathrm A \right) = \det \left( \mathrm I_n + \mathrm u \mathrm v^* \right) = 1 + \mathrm v^* \mathrm u = 1 + \mbox{tr} \left( \mathrm v^* \mathrm u \right) = 1 + \mbox{tr} \left( \,\mathrm u \mathrm v^* \right) = 1 + \mbox{tr} \left( \mathrm A \right)$$

linear-algebra matrices rank-1-matrices determinant trace
A: There is a much simpler solution:

*

*$A$ has only one nonzero eigenvalue, say $\lambda$.


*The eigenvalues of $A+I$ are the eigenvalues of $A$ plus $1$ (so $n-1$ eigenvalues of $A+I$ are 1, and 1 eigenvalue is $\lambda+1$).


*$\det(A+I)$ is the product of the eigenvalues: $\lambda+1$.


*The trace is the sum of the eigenvalues, so $\text{tr}(A)=\lambda$.
A: The Jordan normal form of a rank $1$ matrix either has the form$$\begin{pmatrix}\lambda&0&0&\ldots&0\\0&0&0&\ldots&0\\&\vdots&&\ddots&\vdots\\0&0&0&\ldots&0\end{pmatrix}$$for some $\lambda\neq0$, or the form$$\begin{pmatrix}0&1&0&\ldots&0\\0&0&0&\ldots&0\\&\vdots&&\ddots&\vdots\\0&0&0&\ldots&0\end{pmatrix}.$$and the formula that you want to prove holds in both cases.
