How to show that $\det(A+I)\ne 0$ How to show that for any skew symmetric real matrix $A$, we have $\det(A+I)\ne 0?$
Where to begin? I'm looking for some clue only.
 A: Hint: What are the eigenvalues of a real skew symmetric matrix like? If $v$ is an eigenvector of $A$, what is $(A+I)v$? 
A: When I learned how to do proofs (I am still learning :) ) one thing that felt true was to pay very close attentions to definitions... 
so my hints would be...

1). what is the definition of non-zero skew symmetric real martix A. 
2)  what does adding identity matrix do to A. 
3). what does $det(A + I) \neq 0$ really mean....think invertibility, which will bring us full circle back to Joseph G's help. 
Happy Hunting
A: let $\lambda$ be an eigen value of a skew symmetric matrix $A$ and $X\ne 0$ be an eigen vector correspomding to this $\lambda$ so $AX=\lambda X\Rightarrow X^*A^*=\bar{\lambda}X^*\Rightarrow X^*(-A)X=\bar{\lambda}X^*X\Rightarrow  -X^*\lambda X=\bar{\lambda}X^*X\Rightarrow (\lambda+\bar{\lambda})X^*X=0\Rightarrow \lambda=\bar{\lambda}$ as $X$ was non null, so Eigen values of real skew symmetric matrices arepurely imaginary  or $0$, The problem you have stated says that $-1$ is an eigen value of $A$ so contradiction. 
