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} $.
Solution
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.
Solution
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.
No comments:
Post a Comment