TITLE: $A \in M_{n}(\mathbb{R})$ satisfies $A+A^{t}=I$, then does it imply $\text{det}(A)>0$
QUESTION [8 upvotes]: I am stuck on this simple question for a long time. 

If $A \in M_{n}(\mathbb{R})$ satisfies $A+A^{t}=I$, then does it imply $\text{det}(A)>0$?

I tried finding a counter-example as well as tried proving it. But couldn't succeed. One question, which I would like to ask the experts is: How does one Judge the "Truth/False ness" of the question by seeing it. Is it something which comes by experience.

REPLY [2 votes]: It is enough to show that $\det(\lambda I+B)>0$ for any skew symmetric matrix $B$ and any $\lambda>0$. I will argue by induction on the dimesion $n$ of the matrix.  For $n=1$
$$
\det((\lambda))=\lambda>0.
$$
To reduce the $n$-dimensional case to the $(n-1)$-dimensional, use Chio's pivotal condensation.