Problem 1620

1. Given the matrix equation $A + A^{\text{T}} = I$, we start by analyzing the properties of $A$. Let $A = [a_{ij}]$. From the given equation, we have:
$$A + A^{\text{T}} = I$$
This implies that for the diagonal elements $a_{ii}$:
$$a_{ii} + a_{ii} = 1 \implies 2a_{ii} = 1 \implies a_{ii} = \frac{1}{2}$$
For the off-diagonal elements $a_{ij}$ where $i \neq j$:
$$a_{ij} + a_{ji} = 0 \implies a_{ij} = -a_{ji}$$
Therefore, $A$ can be written as:
$$A = \frac{1}{2}I + B$$
where $B$ is a real skew-symmetric matrix (i.e., $B^{\text{T}} = -B$).

2. Next, we consider the characteristic polynomial of $A$. Let $p_M(x) = \det(xI - M)$ be the characteristic polynomial of a matrix $M$. For $A$, we have:
$$p_A(x) = \det(xI - A) = \det\left(xI - \left(\frac{1}{2}I + B\right)\right) = \det\left(\left(x - \frac{1}{2}\right)I - B\right) = p_B\left(x - \frac{1}{2}\right)$$
This relationship shows that the eigenvalues of $A$ are shifted by $\frac{1}{2}$ from the eigenvalues of $B$.

3. Since $B$ is a real skew-symmetric matrix, its eigenvalues are either zero or purely imaginary numbers. Specifically, the eigenvalues of $B$ are of the form $ci$ for some real number $c$. If $c \neq 0$, the eigenvalues appear in conjugate pairs $ci$ and $-ci$.

4. Therefore, the eigenvalues of $A$ are either:
$$\frac{1}{2} \quad \text{(if the eigenvalue of } B \text{ is } 0\text{)}$$
or
$$\frac{1}{2} + ci \quad \text{and} \quad \frac{1}{2} - ci \quad \text{(if the eigenvalues of } B \text{ are } ci \text{ and } -ci\text{)}$$

5. To find the determinant of $A$, we multiply all its eigenvalues. Let $k$ be the number of eigenvalues equal to $\frac{1}{2}$ and let $m$ be the number of pairs of purely imaginary eigenvalues. Then:
$$\det A = \left(\frac{1}{2}\right)^k \prod_{j=1}^m \left(\frac{1}{2} + c_ji\right)\left(\frac{1}{2} - c_ji\right)$$
Simplifying the product of conjugate pairs:
$$\left(\frac{1}{2} + c_ji\right)\left(\frac{1}{2} - c_ji\right) = \left(\frac{1}{2}\right)^2 - (c_ji)^2 = \frac{1}{4} + c_j^2$$
Therefore:
$$\det A = \left(\frac{1}{2}\right)^k \prod_{j=1}^m \left(\frac{1}{4} + c_j^2\right)$$
Since $\frac{1}{4} + c_j^2 > 0$ for all real $c_j$, and $\left(\frac{1}{2}\right)^k > 0$, it follows that:
$$\det A > 0$$

The final answer is $\boxed{ \det A > 0 }$