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Let $A \in \mathcal{M}_n(\mathbb{C})$ such that $\text{rank} A = 1$. Prove that $$\det(X+A)=\det X+ \text{tr}(X^*A), \: \forall X \in \mathcal{M}_n(\mathbb{C})$$ where $X^*$ is the matrix such that $X \cdot X^*=\det X \cdot I_n$.

I know that for $2 \times 2$ matrices it is true that $\det(A+B)=\det A + \text{tr}(A^*B)+\det B$. Since in our problem the rank of $A$ is 1, this means that all its minors of order greater than 1 are zero and maybe we could get a formula similar to the one in the $2 \times 2$ case, but I don't know how to expand $\det(X+A)$.

AndrewC
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    The notation is super confusing. $X^\ast$ usually denotes the Hermitian adjoint (i.e. conjugate transpose, which is also denoted by $X^H$), but what you have here is the classical adjoint (i.e. the adjucate), which is conventionally denoted by $\operatorname{adj}(X)$. Anyway, the formula is a well-known formula for the determinant of a rank-$1$ update. See Wikipedia, for instance. – user1551 Mar 05 '18 at 18:10
  • Thank you! I wasn't aware of that formula. – AndrewC Mar 05 '18 at 19:36

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