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


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*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. 
 A: Write $A$ as the form $\frac{1}{2}I+B$, where $B$ is an anti-symmetric matrix. We need to show that $\det(\frac{1}{2}I+B)>0$ for any such $B$. For that we'll look at the polynomial 
$P(\lambda)=\det(I-\lambda B)$.  Clearly $P(0)=1$. To show it only take positive value we just need to show it has no real root.  The roots of $P(\lambda)$ are exactly the reciprocal of the nonzero eigenvalues of $B$.  Let's write $\langle\cdot, \cdot \rangle$ for the usual inner product on $\mathbb{R}^n$. Assume that $\mu$ is a real eigenvalue of $B$ with eigenvector $v$, then 
$$ \mu\langle v, v\rangle=\langle Bv, v\rangle = \langle v, B^T v\rangle=\langle v, -Bv\rangle= -\mu \langle v, v\rangle.$$  
Since $\langle v, v\rangle\neq 0$, it follows that $\mu=0$. So  $B$ has no nonzero real eigenvalues, and hence $P(\lambda)$ has no real roots. 
A: Yes. Since $A = A^2 + A^T A = A^2 + A A^T$, it follows that $A$ is normal, hence by the spectral theorem has an orthonormal basis of eigenvectors $v_1, ... v_n$. Moreover, the given condition says precisely that the corresponding eigenvalues $\lambda_i$ satisfy $A v_i + A^T v_i = \lambda v_i + \overline{\lambda} v_i = v_i$, hence all have real part exactly $\frac{1}{2}$. Since $A$ is real, the set of its eigenvalues is closed under complex conjugation, so every eigenvalue $\frac{1}{2} + iy$ is paired with its complex conjugate $\frac{1}{2} - iy$ (and the eigenvalue $\frac{1}{2}$ can occur with any multiplicity). The product of these is positive, so $\det A > 0$.
Here is an alternate argument which does not rely on the spectral theorem. The condition gives
$$\langle v, v \rangle = \langle (A + A^T) v, v \rangle = \langle Av, v \rangle + \langle A^T v, v \rangle = 2 \langle Av, v \rangle$$
for $v \in \mathbb{R}^n$. It follows that any such matrix $A$ is invertible, since $Av = 0$ implies $\langle v, v \rangle = 0$. On the other hand, the set of all matrices $A$ satisfying this condition is an affine subspace - in particular, it is path-connected. A path-connected set of invertible matrices must lie in a path component of $\text{GL}_n(\mathbb{R})$, hence the determinant must be the same sign for every element of the set. And $\frac{1}{2} I$ is in this set. 
Note that the set of matrices satisfying this condition is precisely a translate of the set of skew-symmetric matrices by $\frac{1}{2} I$. 
A: 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.
