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

Let $$A$$ be a complex matrix of rank $$1$$. Show that $$\det (I+A) = 1 + \operatorname{Tr}(A)$$ where $$\det(X)$$ denotes the determinant of $$X$$ and $$\operatorname{Tr}(X)$$ denotes the trace of $$X$$.

Any hint, please. I do not get how to combine the ideas of rank, determinant and trace. Thank you.

• Not sure if this helps but $A$ can be written as $uv^T$ for some nonzero column vectors $u, v \in \Bbb C^n$. – Aryaman Maithani Jun 27 '20 at 15:15
• – Rodrigo de Azevedo Jul 6 '20 at 12:23

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.

• Sorry, bad notation. The actual terminology is upper triangular. – Mathlover Jun 27 '20 at 15:26
• Brilliant, thanks – user598858 Jun 27 '20 at 15:27

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,

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

• beautiful proof. – DuFong Jun 28 '20 at 22:14

Assume $$A$$ is diagonal. As its rank is $$1$$, it has only one non-zero eigenvalue $$\lambda$$. Then $$\det(I+A) = 1+\lambda = 1 + \mathrm{tr}\, A.$$

Assume that $$A$$ is diagonalisable, so that $$A = PDP^{-1}$$. Then $$\det(I + A) = \det(I + PDP^{-1}) = \det(P(I + D)P^{-1} ) = \det(I + D)$$ Similarly $$1 + \mathrm{tr}\,A = 1 + \mathrm{tr}\, D,$$ so we reduced this case to the previous one.

Now assume $$A$$ is an arbitrary complex matrix. Both sides of the equation are continuous and $$A$$ can be approximated by diagonalisable matrices. This finishes the proof.

Suppose the eigenvalues of $$A$$ are $$\lambda_j,j=1,\cdots,n$$, since the rank of $$A$$ is 1, there is only one $$\lambda_j$$ is nonzero, then $$\det(\lambda-A)=\prod_{j}(\lambda-\lambda_j)=\lambda^n-(\sum_{j}\lambda_j)\lambda^{n-1},$$

Now let $$\lambda=-1$$, one readily gets $$\det(I+A)=(-1)^n((-1)^n-(\sum_{j}\lambda_j)(-1)^{n-1})=1+\text{Tr}(A).$$

The above procedure can be smoothly extended to $$Rank\geq 1$$ matrices. Moreover, if the size goes to infinite dimension, Fredholm Determinant comes.