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.

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
