Prove that if $B$ is an antisymmetric matrix with real entries, then $\det(B+I) \neq 0$.
2026-04-07 22:59:03.1775602743
Prove that if $B$ is an antisymmetric matrix, then $\det(B+I) \neq 0$
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The matrix $B$ satisfies $B^T=-B$; suppose $\det(B+I)=0$; then there exists $x\ne0$ such that $(B+I)x=0$. It follows that $Bx=-x$ and so $$ x^TBx=-x^Tx $$ Transpose both sides, getting $$ x^TB^Tx=-x^Tx $$ and use the hypothesis $B^T=-B$ to get a contradiction.
(It can be made without contradiction, of course.)