I am having difficulty in proving the following question.
$\mbox{rank}(A)=1$ implies $\det(A+I)=\mbox{trace}(A)+1$
My thought is det$(A+I)$ implies that eigenvalue of $A$ is $-1$ and since rank of $A$ is $1$ determinate of all minor of order $n-1$ is zero so in characteristic polynomial expansion formula we get $det(A+I)=(-1)^n(\lambda^n+trace(A))=1+trace(A)$.
Is my approach correct? Could you please write the solution according to the hints given in this question? Thank you.
We know that for any square matrix, its trace is the sum of its eigenvalues and its determinant is the product of the eigenvalues. (The eigenvalues could be complex).
Further, if $t$ is an eigenvalue of a Matrix $M$, then $1+t$ is an eigenvalue of $M+I$:
$$\det [(A+I)-(1+t) I] = \det [A-tI]=0$$
Let $A$ be a $n \times n$-matrix of rank $1$.
Since $A$ has rank $1$, it has the eigenvalues $0$ with multiplicity $n-1$ and $\lambda \neq 0$ with mulitplicity $1$.
Thus we get $\det (A) = 1^{n-1} \cdot(\lambda+1)= \lambda+1$ and $\operatorname{tr}(A) = (n-1)\cdot 0+\lambda= \lambda $
This implies the claim.
Edit: (This is more like a comment, but too long)
I see, you want to use that for $\lambda$ an eigenvalue of $M$
$$0= \det(M-\lambda I) = \sum_{i=0}^n b_i (-\lambda)^{n-1} $$
with $b_i$ the sum of all principal minors of order $i$.
In particular $b_0:=1$, $b_1 = \operatorname{tr} M$, $a_n= \det M$.
So for $M=A+I$:
$$ 0= \det (A+I-\lambda I) = (-\lambda)^n + (-\lambda)^{n-1}\operatorname{tr}(A+I)+\det(A+I) + \dots ,$$ and thus for $\lambda=1: $
$$ 0=\det(A) = (-1)^n (1-\operatorname{tr}(A+I)+ \dots) +\det(A+I) + .$$ This implies $$\det(A+I)= (-1)^{n+1} (1-\operatorname{tr}(A) -n+\dots) $$
But I don't see how we can use $\operatorname{rk} A=1$ in order to calculate the remaining summands.
