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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.