If $\rho_{AB}$ is at most rank 2, why is $\operatorname{det}(\rho_{AB})=0$?

Question

The book I'm reading (https://arxiv.org/abs/1508.02595 pages 108-109) is trying to demonstrate why a two qubit state $\rho_{AB}$ has a symmetric extension iff $$\operatorname{Tr}\left(\rho_{B}^{2}\right) \geq \operatorname{Tr}\left(\rho_{A B}^{2}\right)-4 \sqrt{\operatorname{det}\left(\rho_{A B}\right)}$$ where $\rho_B = \operatorname{Tr}_A(\rho_{AB})$

They use as example where $\rho_{AB}$ is "pure symmetric extendable" to $\rho_{ABC}=|\psi_{ABC}\rangle \langle \psi_{ABC} |$.

They write, using Schmidt decomposition $$|\psi_{ABC}\rangle = \sum_{\alpha}\lambda_{\alpha} |\alpha_{AB}\rangle |\alpha_C \rangle$$ and then say

This means that the nonzero eigenvalues of $\rho_{A B}=\operatorname{Tr}_{C}\left|\psi_{A B C}\right\rangle\left\langle\psi_{A B C}\right|$ is the same as those of $\rho_{B}=\rho_{C}=\operatorname{Tr}_{A B}\left|\psi_{A B C}\right\rangle\left\langle\psi_{A B C}\right|$, where $\rho_{B}=\rho_{C}$ comes from the symmetry assumption between qubits $B$ and $C$. Therefore we have $\operatorname{Tr}\left(\rho_{B}^{2}\right)=\operatorname{Tr}\left(\rho_{A B}^{2}\right)$. And because $\rho_{A B}$ is at most rank 2 , so $\operatorname{det}\left(\rho_{A B}\right)=0$.

Meaning that the equality of the first equation above holds. What I don't understand is

because $\rho_{A B}$ is at most rank 2 , so $\operatorname{det}\left(\rho_{A B}\right)=0$

Why is this true? Is this just some simple property of determinants that I'm not aware of?

Answer 1

For an $n \times n$ matrix with non-zero determinant, the rank is by definition $n$. In your case $n = 4$ ($2$-qubit state), so for the determinant to be nonzero, rank must be exactly $4$.