$\det(I+A)=1+\operatorname{Tr}(A)$ if $\operatorname{rank}(A)=1$

Solution 1:

The minimal polynomial of $A$ splits in $\Bbb C$. So, there is $P\in \text{GL}(n,\Bbb C)$ such that $PAP^{-1}$ is upper diagonal. Now, $\operatorname{rank}(A)=1$, so at most one diagonal entry of $P^{-1}AP$ is non-zero and all other diagonal entries of $P^{-1}AP$ are zero.

Hence, $\det(I+A)=\det\left(I+P^{-1}AP\right)=(1+\lambda)$, where $\lambda$ is the only non-zero diagonal entry of $P^{-1}AP$. Now, $\operatorname{tr}(A)=\operatorname{tr}(P^{-1}AP)=\lambda$. So, we are done.

Another case is also possible, all diagonal entries of $P^{-1}AP$ are zero, that is $A$ is nilpotent. In this case the equality $\det(I+A)=1+\operatorname{tr}(A)$, holds similarly.

Solution 2:

Since matrix $\rm A$ is rank-$1$, it can be written in the form $\rm A = u v^*$. Using the matrix determinant lemma and the cyclic property of the trace operator,

$$\det \left( {\rm I} + {\rm A} \right) = \det \left( {\rm I} + {\rm u v^*} \right) = 1 + {\rm v^* u} = 1 + \mbox{tr} \left( {\rm v^* u }\right) = 1 + \mbox{tr} \left( {\rm u v^*}\right) = 1 + \mbox{tr} \left( {\rm A}\right)$$

linear-algebra matrices rank-1-matrices determinant trace