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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.

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  • $\begingroup$ If $A$ has rank 1 it can be written as $u v^T$ for some vectors $u,v$. $\endgroup$ – copper.hat Nov 23 '17 at 17:09
  • $\begingroup$ How can $\det(A+I)$ imply anything? $\endgroup$ – copper.hat Nov 23 '17 at 17:10
  • $\begingroup$ It may be wrong. It is just what I was thinking. $\endgroup$ – user444042 Nov 23 '17 at 17:11
  • $\begingroup$ I want to solve this question according to the hints given in the question. $\endgroup$ – user444042 Nov 23 '17 at 17:15
  • $\begingroup$ Is A assumed to be diagonalisable? $\endgroup$ – klirk Nov 23 '17 at 18:04
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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.

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