$\mbox{rank}(A)=1$ implies $\det(A+I)=\mbox{trace}(A)+1$ This problem comes from exercise from my lecturer
Let $A$ be a square matrix such that $\mbox{rank}(A)=1$. Prove that $\det(A+I)=\mbox{trace}(A)+1$.
($I$ is the identy matrix)
I usually haven't any idea in solving linear algebra problem.
 A: Edited in response to julien's comment
Recall that:
1) the rank is greater than or equal to the number of non-zero eigenvalues
2) the determinant is equal to the product of all eigenvalues
3) the trace is equal to the sum of all eigenvalues
You should also be able to show that if $\lambda$ is an eigenvalue of $A$ then $\lambda+1$ is an eigenvalue of $A+I$ 
I think you should be able to do the proof now. Let me know if you need more help.
A: Hint: If the rank of the matrix is $1$, it means that all columns can be written as multiples of one of them (say, wlog, the first one); since the determinant is invariant by linear combination of the columns, you can write $\det(A+I)$ as $\det(A'+I)$, where $A'$ and $A$ have same rank, but only the first column of $A'$ is non-zero.
A: In a suitable base, $A$ is a matrix $a e_{ij}$, where $a \ne 0$, and as usual $e_{ij}$ is the matrix that has all entries zero, except the $(i,j)$ one, which is $1$. 
(Whereas, as noted by @EricAuld, the identity map has matrix $I$ with respect to any base.)
There are two cases to distinguish.


*

*When $i \ne j$, then $1 + \operatorname{trace}(A) = 1 = \det(I + A)$.

*When $i = j$, then $1 + \operatorname{trace}(A) = 1 + a = \det(I + A)$.



Proof of the first statement. The image of $A$ has dimension $1$. Choose a non-zero vector $v$ in the image. 
Now if $A v \ne 0$, complete $v$ to a base of the whole space by adding a base of $\ker(A)$, which has dimension $n-1$, if the matrix is $n \times n$. This is case (2).
If $A v = 0$, choose an element $u$ such that $A u \ne 0$, and take as a base $u, v$ and $n-2$ further vectors that together with $v$ form a base of $\ker(A)$. This is case (1).
A: An example.
Let $A$ be the $2\times 2$ matrix $A=(A_1,A_2)$, with $A_1=A_2=(1,1)^t$.
So $tr(A)=2$ and $\det(A+I)=3=tr(A)+1$.
EDIT: the original formula for the determinant contained a mistake .Should I remove the answer? Without a more general proof it does not support the OP. Thanks
