Eigenvalues of a quantum state after partial tracing

There are no effective bounds on how small or how large the smallest non-zero eigenvalue can become upon taking the partial trace, other than the obvious constraint that it must be in $(0, 1]$.

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Let $lambda_{min}(rho)$ denote the smallest non-zero eigenvalue of operator $rho$.

Fix a real $p in (0, frac{1}{2}]$ and consider the pure two-qubit state

$$ |psirangle = sqrt{p}|00rangle + sqrt{1 - p}|11rangle. $$

Note that $rho_{AB} = |psiranglelanglepsi|$ has eigenvalues $0, 0, 0, 1$ so $lambda_{min}(rho_{AB}) = 1$.

On the other hand,

$$ rho_A = mathrm{tr}_B(rho_{AB}) = p|0ranglelangle 0| + (1 - p)|1ranglelangle 1| $$

has eigenvalues $p, 1 - p$ and so $lambda_{min}(rho_A) = p$.

Consequently, not only can partial trace decrease the value of $lambda_{min}$ (as already noted in the comments), but there is no positive lower bound on $lambda_{min}(mathrm{tr}_Brho)$. More precisely, for any $epsilon > 0$ we can find a state $rho_{AB}^epsilon$ such that $lambda_{min}(rho_{AB}^epsilon) = 1$ yet $lambda_{min}(mathrm{tr}_B rho_{AB}^epsilon) = epsilon$.

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In order to find a bound on how much larger the smallest non-zero eigenvalue can become after taking the partial trace, fix a real $p in (0, frac{1}{2}]$ and consider the state

$$ rho_{AB} = p|00ranglelangle 00| + (1 - p)|01ranglelangle 01| = |0ranglelangle 0| otimes (p|0ranglelangle 0| + (1 - p)|1ranglelangle 1|). $$

Note that its eigenvalues are $0, 0, p, 1-p$ and so $lambda_{min}(rho_{AB}) = p$.

On the other hand,

$$ rho_A = mathrm{tr}_B(rho_{AB}) = |0ranglelangle 0| $$

has eigenvalues $0, 1$ and so $lambda_{min}(rho_A) = 1$.

Consequently, upon taking the partial trace $lambda_{min}$ can increase all the way to the maximum $1$ regardless of $lambda_{min}$ of the original state.