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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.

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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.

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    $\begingroup$ Sorry, bad notation. The actual terminology is upper triangular. $\endgroup$ – 0-th User Sumanta Jun 27 at 15:26
  • $\begingroup$ Brilliant, thanks $\endgroup$ – user598858 Jun 27 at 15:27
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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,

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


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    $\begingroup$ beautiful proof. $\endgroup$ – DuFong Jun 28 at 22:14
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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.

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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.

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