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

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?

• Consider to include the reference in the question. To give you a hint: Think about what the determinant and the rank means in terms of the spectral decomposition (i.e. eigenvalues, eigenvectors etc.) of a hermitian operator. More concretely: What dimension has the space $\rho_{AB}$ acts on? How many possible eigenvalues can $\rho_{AB}$ thus have? What if at least one is zero? Jul 14, 2022 at 14:08
• I actually don't know what either of those mean for this, but I'll try to figure it out. Jul 14, 2022 at 14:11
• So according to Wikipedia: "The rank of [a matrix] equals the number of non-zero singular values". Since the density matrices are 4x4, a rank of 2 would mean that two of the eigenvalues would be zero, making the determinant zero. Right? Jul 14, 2022 at 17:50

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 $$4$$.