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Zero matrix

Let A\in \mathcal{M}_n\left ( \mathbb{C} \right ) with n\geq 2. If \det \left ( A+X \right )=\det A+\det X for every matrix X \in \mathcal{M}_n\left ( \mathbb{C} \right ) then prove that A=\mathbb{O}_{n}.


Suppose that A \neq 0, say A_{ij} \neq 0 for some i,j. Let P be any permutation matrix with P_{ij}=1 and let Q be the matrix obtained from P by changing its ij-entry to 0. Finally let X = xQ where x \in \mathbb{C}.

We have that \det X = 0 and that \det X = \det(A+X) - \det A is a polynomial in x. Furthermore, the coefficient of x^{n-1} of this polynomial is \pm A_{ij} depending on the sign of the corresponding permutation. So the polynomial is not identically zero, a contradiction.

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