$A$ is positive semidefinite, $A^{(−1)} = [a^{−1}_{i j} ]$ and, $A^{(−1)}$ be positive semidefinite.Why does $rank A = 1$? Let $A = [a_{i j }]\in M_n$ be positive semidefinite and suppose that each of its entries is nonzero.
Consider  $A^{(−1)} = [a^{−1}_{i j} ]$, and $A^{(−1)}$ be
positive
semidefinite.
Why does $rank A = 1$?
 A: Let us prove by induction on $n$ that if $A \in M_n(\mathbb{R})$ satisfies the hypothesis of the question then $\mathrm{rank}(A) = 1$.
If $A \in M_1(\mathbb{R})$, there is nothing to prove. If $A \in M_{2}(\mathbb{R})$, one can prove the claim by direct calculation using Sylvester's criterion. Assume the hypothesis holds for matrices of dimension $n - 1$ with $n > 2$ and let $A \in M_n(\mathbb{R})$. Denote by $B \circ C$ the Hadamard Product of $B$ and $C$. Then, since $A$ and $A^{(-1)}$ are positive semidefinite, we have
$$ 0 = \det(\hat{I}_n) = \det(A \circ A^{(-1)}) \geq \det(A) \det(A^{(-1)}) \geq 0 $$
where $\hat{I}_n$ is the $n \times n$ matrix that has $1$ in all entries. Thus, either $\det(A) = 0$ or $\det(A^{(-1)}) = 0$.
If $\det(A) = 0$, consider an arbitrary principal square submatrix $M$ of $A$ of order $n-1$ (obtained by a row and the corresponding column). By Sylvester's criterion, $M$ is also positive semidefinite, with non-zero entries and the same goes for $M^{(-1)}$ (as it is a principal square submatrix of $A^{(-1)}$). By induction hypothesis, $\mathrm{rank}(M) = 1$ which implies in particular that all principal minors of $A$ of order $2 \leq j \leq n - 1$ are zero. Together with $\det(A) = 0$, we see that all principal minors of $A$ of order $j \geq 2$ are zero.
Now, if a diagonalizable matrix $A \in M_n(\mathbb{F})$ has all principal minors of order $j \geq k$ equal to zero then $\mathrm{rank}(A) < k$. Why? If $\chi_A(x) = \sum_{i=0}^n (-1)^i c_i x^{n-i}$ is the characteristic polynomial of $A$ then each $c_i$ is the sum of all principal minors of $A$ of size $i$. If all the minors of size equal or greater than $k$ vanish, then $c_n = \ldots = c_k = 0$ and so $x^{n-k+1}$ divides $\chi_A(x)$. Since $A$ is diagonalizable, we must have $\dim \ker A \geq n - k + 1$ or $\mathrm{rank}(A) < k$.
This shows that $\mathrm{rank}(A) < 2$ and clearly $\mathrm{rank}(A) \geq 1$ so $\mathrm{rank}(A) = 1$.
If, instead we have $\det(A^{(-1)}) = 0$, the same argument shows that $\mathrm{rank}(A^{(-1)}) = 1$ but this immediately implies that also $\mathrm{rank}(A) = 1$.
