Let $A$ be a $n \times n$ real matrix such that $$ A + A^t = 2I, $$ where $I$ is the $n \times n $ identity matrix.
Prove that $\det(A) \geq 1$.
It is obvious that tr$(A) = n$. Furthermore, we also have that $$A – 2I = -A^t, $$ so we get that $$\det(A-2I) = (-1)^n \cdot \det(A).$$
Now, we let $\lambda_1, \lambda_2, \cdots, \lambda_n \in \mathbb{C}$ be the eigenvalues of $A$, so we get that $$(\lambda_1 – 2)(\lambda_2 – 2) \cdots (\lambda_n – 2) = (-1)^n\lambda_1\lambda_2 \cdots \lambda_n, $$ but I couldn't derive anything about the product $\lambda_1\lambda_2 \cdots \lambda_n = \det(A)$ (the only known thing is $\lambda_1 + \cdots + \lambda_n = n$).
Also, I tried the same approach for $2 \times 2$ and $3 \times 3 $ matrices, but it didn't lead to anything (especially since the eigenvalues can be complex numbers).
Best Answer
Note that $A - I$ is a real skew-symmetric matrix, so all of its eigenvalues are pure imaginary (or zero) and appear in complex conjugate pairs. Therefore the eigenvalues of $A$ are either $1$ or of the form $1 + c i, 1-c i$ for some $c\in\mathbb{R}$. Note that $(1+ci)(1-ci) = 1+c^2 \geq 1$. Therefore
$$\det(A) = \prod (1 + c^2) \geq 1$$