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For a bipartite pure state, entanglement entropy is the Von Neumann entropy of either reduced density matrix. When Alice makes a local measurement with pure conditional branches, Bob's reductions average to his original reduction. Concavity of Von Neumann entropy therefore gives ${\sum }_r{p}_{r}S({\rho }_{B,r})\le S({\rho }_B)$. When Bob measures, apply concavity to Alice's reductions. In each pure branch the two reduced entropies coincide by the Schmidt decomposition. Refining measurement outcomes to individual Kraus operators and iterating proves the inequality for arbitrary finite-round LOCC. Discarded local systems can be measured and their outcomes refined as well, so discarding does not evade the bound.

Give Alice the first qubit $A$ and Bob the second $B$. Compare the Bell state $|{\Phi }^{+}⟩$ with a preparation that tosses a fair classical bit and gives both parties its value. Their density operators, written in the computational basis ordered $00,01,10,11$, areThey have identical diagonal quantum state populations. The off-diagonal entries of the Bell state retain quantum coherence between $00$ and $11$; the classical mixture has none in this basis.

If both parties perform a Pauli measurement of $Z$, the Born rule gives $P(00)=P(11)=1/2$ and $P(01)=P(10)=0$ for both preparations. Perfect agreement here therefore cannot establish entanglement: an ordinary shared random bit already explains it.

Now both measure $X$, whose normalized eigenstates are $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Changing basis givesThe Bell state again has matching outcomes, each with probability $1/2$. For either product state $|00⟩$ or $|11⟩$, each of the four $X$ outcome pairs has probability $1/4$, so their classical mixture does too. Equivalently, the Pauli correlator $⟨X\otimes X⟩$ is one for the Bell state and zero for this mixture, while $⟨Z\otimes Z⟩$ is one for both.

Taking a partial trace gives ${\rho }_A={\rho }_B=I_2/2$ for both preparations. Thus every measurement confined to Alice's system has the same statistics in the two cases, and so does every measurement confined to Bob's. They reveal the difference by comparing records through classical communication. Conditional on Alice's result, Bob can predict a matching $X$ result for the Bell state; without that result, his outcomes remain evenly split. An uncommunicated local quantum channel, including a measurement with its outcome discarded, leaves his reduced density matrix unchanged. This is quantum no-signalling, even when conditional states change. ^[ref]

The Von Neumann entropy makes the joint-versus-local distinction numerical. With base-two logarithms, the Bell state has eigenvalues $1,0,0,0$, giving $S(AB)=0$, whereas the mixture has $1/2,1/2,0,0$, giving $S(AB)=1$. Both have $S(A)=S(B)=1$. Their quantum mutual information $I(A:B)=S(A)+S(B)-S(AB)$ is consequently two and one bits, respectively. The value two is not two readable classical bits in either local qubit: the matching $Z$ records have ordinary mutual information one bit in both cases. Random records can come from a globally pure state. ^[ref]

These two matching-axis measurements distinguish the specified preparations; they alone are not a test of a Bell inequality. A local hidden-variable theory could assign shared random answers separately to these two axes. Bell theorem concerns incompatible correlations across a suitable larger collection of measurement settings. Bell's 1964 paper established the obstruction for local hidden-variable descriptions of the predicted correlations. ^[ref]

For density operators ${\rho }_1,{\rho }_2$ and $0\le p\le 1$, Von Neumann entropy is concave:

A dephasing channel suppresses off-diagonal entries of a density matrix in a preferred basis while preserving quantum state populations. Complete dephasing is identical to a projective measurement in that basis with its outcome discarded.

For the Bell correlations versus a classical mixture example, completely dephasing Alice's qubit in the computational basis maps ${\rho }_{\mathrm{Bell}}$ to ${\rho }_{\mathrm{mix}}$. It preserves their perfect $Z$ agreement and each reduced density matrix, yet destroys the $X$ correlation. This is why testing only populations can miss damage to a quantum memory: the stored quantum coherence may have disappeared while every preferred-basis measurement still looks correct. The joint Von Neumann entropy rises from zero to one bit, although each local entropy stays fixed.

A mixture of two orthogonal pure qubit states has Von Neumann entropy equal to the binary entropy of its weights. Its entropy difference from the first pure state is exactly $h_2(u)$ and its trace distance from that state is $u$. The elementary bound $-(1-u)\ln (1-u)\le u$ gives $h_2(u)\le u{\log }_2(e/u)$; entropy also never exceeds one bit. For small $u$, the leading term is $u{\log }_2(1/u)$.

The classical-quantum state is block diagonal. If ${\lambda }_{xj}$ are the eigenvalues of ${\rho }_x$, its eigenvalues are $p_x{\lambda }_{xj}$. Inserting these into the definition of Von Neumann entropy gives the displayed sum of classical Shannon entropy and mean conditional quantum entropy. Its quantum mutual information is therefore the Holevo quantity $S({\sum }_x{p}_x{\rho }_x)-{\sum }_x{p}_{x}S({\rho }_x)$.

For density operators supported on mutually orthogonal subspaces, the eigenvalues of their mixture are $p_i{\lambda }_{ij}$, where ${\lambda }_{ij}$ are the eigenvalues of ${\rho }_i$. Substitution into the definition of Von Neumann entropy gives the displayed decomposition into Shannon entropy of the label and the mean entropy within each block. Zero eigenvalues use $0\log 0=0$.

For the retention parameter $p$, the Holevo capacity isEvery output has Von Neumann entropy at least the value on a pure input, $h_2((1+p)/2)$. Two equiprobable orthogonal pure inputs attain this minimum entropy individually and have maximally mixed average output, attaining the bound. Additivity of Holevo capacity makes the unassisted classical capacity of a quantum channel equal to this value. King's depolarizing-channel capacity paper proves additivity with an arbitrary other channel.

A linear map $V:\mathcal{H}\to \mathcal{K}$ is an isometry when it preserves inner products, equivalently $V^{†}V=I$. It need not be surjective. It maps an orthonormal basis to an orthonormal family, and $\rho \mapsto V\rho V^{†}$ preserves the nonzero eigenvalues and the Von Neumann entropy of a density operator.

On a $d$-dimensional Hilbert space, the Von Neumann entropy is maximized uniquely by $I/d$. Indeed, nonnegativity of quantum relative entropy giveswith equality exactly when $\rho =I/d$.

A pure-state source emits independent states from a fixed quantum state ensemble. Its average block density operator is the tensor power of the one-use average. Unassisted reliable quantum source compression preserves the emitted quantum states, rather than only their classical labels. Schumacher compression gives the optimum asymptotic qubit rate $S(\rho )$ for such a source, with every larger rate achievable. Nonorthogonal equally likely pure states need not have entropy one: their classical labels have entropy one but their average density operator can have lower Von Neumann entropy.

The transmitted branch is the Kraus operatorFor nonzero loss it is not trace preserving: $\mathrm{tr}(E\rho E^{†})$ is the transmission probability. A complete channel includes an orthogonal vacuum or erasure flag, with a loss operator $L=\sqrt{R}|\mathrm{vac}⟩⟨0|$ and $E^{†}E+L^{†}L=I$. Postselecting only the first outcome gives the single-operator description in the question.

For $|\pm ⟩$, the unnormalized transmitted vector is $(\sqrt{1-R}|0⟩\pm |1⟩)/\sqrt{2}$. Its norm squared and normalized conditional state areThe reduced density matrix of either member of the initial pair has eigenvalues $\alpha ,1-\alpha$. Since the joint state is pure, its Von Neumann entropy is zero. With logarithms to base two and $h_2$ the binary entropy function,The two nonzero Schmidt coefficients show that this pure pair is entangled exactly when $0<\alpha <1$. Its mutual information rises from zero to two bits at $\alpha =1/2$ and returns to zero.

For this known pure family, every positive mutual information signals entanglement. For an otherwise unspecified two-qubit state, positive mutual information can also be classical. The mutual information bound for separable states gives the stronger general witness $I(A:B)>1$ bit. On this curve it holds for approximately $0.110028<\alpha <0.889972$; the rest of the open interval is still entangled, but that generic one-bit witness alone would not certify it.

If Bob's qubit passes through the channel, the unnormalized retained state is $\sqrt{\alpha }|11⟩+\sqrt{(1-\alpha )(1-R)}|00⟩$. ThusFor $0\le R<1$, equal Schmidt coefficients require $\alpha =(1-\alpha )(1-R)$, givingThe retained state is then $(|00⟩+|11⟩)/\sqrt{2}$. At $R=1$, this limiting compensation has zero transmission probability and yields no maximally entangled pairs. Conditioning can increase the entanglement of retained pairs, while the discarded fraction accounts for the cost.

Use logarithms to base two and set $0{\log }_{2}0=0$. For a density matrix with eigenvalues $({\lambda }_j)$, the Von Neumann entropy isThus it is the Shannon entropy of the spectrum. In dimension $d$, each term is nonnegative, and entropy vanishes exactly when one eigenvalue is one, namely for a pure state. The Jensen inequality applied to the strictly concave function $-x{\log }_{2}x$ gives $S(\rho )\le {\log }_{2}d$, with equality precisely at the maximally mixed state $I/d$. Unitary conjugation preserves eigenvalues, so $S(U\rho U^{†})=S(\rho )$. Product eigenvalues giveThe finite-dimensional entropy is continuous because the eigenvalues depend continuously on the matrix and $-x\log x$ is continuous also at zero.

Define the partial trace over ${\mathcal{K}}_2$ by choosing an orthonormal basis $(e_b)$ and settingEquivalently it is uniquely characterized by $\mathrm{Tr}({\rho }_{1}A)=\mathrm{Tr}(\rho (A\otimes I))$ for every operator $A$ on ${\mathcal{K}}_1$. This identity proves basis independence. Define ${\rho }_2$ similarly, with $\mathrm{Tr}({\rho }_{2}B)=\mathrm{Tr}(\rho (I\otimes B))$. Hermiticity is preserved by the displayed sum. For each $u\in {\mathcal{K}}_1$,Taking the trace in a product orthonormal basis gives $\mathrm{Tr}{\rho }_1=\mathrm{Tr}\rho =1$, and the same proof works for ${\rho }_2$. Thus both reductions are density matrices. This is the partial trace positivity from product vectors argument.

For a bipartite pure state $\rho =|\psi ⟩⟨\psi |$, expand $\psi$ in a product orthonormal basis with coefficient matrix $C$. A singular value decomposition gives the Schmidt decompositionwith orthonormal families on both sides. Taking the two partial traces makes all cross terms vanish:The spectra may have different numbers of zeros but have identical nonzero eigenvalues. Consequently the reduced entropies coincide:This common value is the entanglement entropy. In infinite-dimensional trace-class systems the same partial-trace and Schmidt arguments apply, and the equality remains valid with the value $+\infty$ allowed; the bound ${\log }_{2}d$ is specifically finite-dimensional.

For completeness, the basic mixing and correlation properties also admit short proofs. Define quantum relative entropy by $D(\tau \Vert \sigma )=\mathrm{Tr}\tau ({\log }_2\tau -{\log }_2\sigma )$ when supports are included, and by $+\infty$ otherwise. Here is its nonnegativity proof without merely citing an operator inequality. Diagonalize $\tau ,\sigma$ with eigenvalues $r_i,s_j$, and let $q_{ij}$ be the squared overlaps of their eigenbases. Both row and column sums of $q$ are one. The scalar inequality $r\ln (r/s)\ge r-s$, with zero cases taken by limits, givesEquality forces $r_i=s_j$ on every nonzero overlap, hence $\tau =\sigma$. For $\overline{\rho }={\sum }_i{p}_i{\rho }_i$, expanding the trace definition yieldswhich proves Concavity of Von Neumann entropy.

The complementary entropy bounds for a quantum mixture areTo prove the upper bound, write the spectral pure-state decompositions of the ${\rho }_i$ as one ensemble with weights $w_{ij}=p_i{\lambda }_{ij}$. Form the column matrix with columns $\sqrt{w_{ij}}|{\psi }_{ij}⟩$. Its two Gram operators have the same nonzero spectrum: one is $\overline{\rho }$, and the other, $G$, has diagonal $w$. Relative entropy against $\mathrm{diag}w$ gives $0\le D(G\Vert \mathrm{diag}w)=H(w)-S(G)$. Since $H(w)=H(p)+{\sum }_i{p}_{i}S({\rho }_i)$ and $S(G)=S(\overline{\rho })$, the bound follows. Zero-weight columns are omitted.

Finally the support of $\rho$ is contained in $\mathrm{supp}{\rho }_1\otimes \mathrm{supp}{\rho }_2$. Indeed, a vector in a marginal kernel gives a sum of nonnegative product-vector expectations equal to zero; positivity forces the corresponding product vectors into the kernel of $\rho$. The operator identity $\log ({\rho }_1\otimes {\rho }_2)=\log {\rho }_1\otimes I+I\otimes \log {\rho }_2$ on this support and the partial-trace identities giveThis proves Subadditivity of Von Neumann entropy, with equality exactly for a product state. Purify ${\rho }_{12}$ by adjoining a third system, using its spectral decomposition. Equality of pure-state marginal entropies gives $S({\rho }_{13})=S({\rho }_2)$ and $S({\rho }_3)=S({\rho }_{12})$. Subadditivity on systems $1,3$ then gives $S({\rho }_2)\le S({\rho }_1)+S({\rho }_{12})$; swapping $1,2$ gives the Araki–Lieb inequality $\boxed{|S({\rho }_1)-S({\rho }_2)|\le S({\rho }_{12})}$. These finite-dimensional proofs establish the usual basic entropy bounds, rather than attributing all mixed-state behavior to pure-state Schmidt decompositions.

Use logarithms to base two, so all entropies are in bits. For a discrete probability distribution $p$ and a density operator $\rho$ with eigenvalues ${\lambda }_j$, the Shannon entropy and Von Neumann entropy are respectivelyUse $0\log 0=0$. The second definition is basis-independent by the spectral theorem; for a density operator diagonal with probabilities $p_x$, it agrees with the first.

The classical relative entropy, or Kullback-Leibler divergence, of $p$ from $q$ isIt is $+\infty$ if some $p_x>0$ has $q_x=0$. The quantum relative entropy iswhen the support of $\rho$ is contained in that of $\sigma$, and is $+\infty$ otherwise. Matrix logarithms are taken on the corresponding positive supports. In a common eigenbasis this reduces to classical relative entropy; both divergences are nonnegative and are generally asymmetric.

To prove Fano's inequality, let the finite alphabet size be $m\ge 2$ and let $E$ indicate an incorrect guess. Since $E$ is determined by $(X,Y)$, the chain rule for conditional entropy first givesExpanding in the opposite order givesConditioning reduces entropy, so $H(E\mid Y)\le H(E)=h_2(p_e)$, the binary entropy function. When $E=0$, knowledge of $Y$ determines $X=f(Y)$ and the conditional entropy is zero. When $E=1$, $X$ can lie in at most the $m-1$ alphabet values other than $f(Y)$. Entropy on $m-1$ values is at most ${\log }_2(m-1)$: its deficit from that number is the nonnegative relative entropy from the uniform distribution. Averaging the two cases therefore givesCombining the two expansions provesThis proof of Fano's inequality via an error indicator works for any alphabet-valued guessing rule, not just an optimal one. If $m=1$, there is no guessing uncertainty and the claim is simply $H(X\mid Y)=0$; that case is treated separately rather than writing $\log 0$.

For a classical error-correcting code of length $n$, alphabet size $q$, minimum Hamming distance $d$ and $M$ codewords, the Singleton bound isFor a linear $[n,k,d{]}_q$ code this is $n-k\ge d-1$. Deleting any $d-1$ coordinates is injective on the codewords, since two words agreeing on the remaining coordinates would differ in fewer than $d$ positions; this also explains the classical bound.

For an n-qubit quantum code carrying $k>0$ logical qubits and having quantum code distance $d$, the quantum Singleton bound isMore generally replace $k$ by ${\log }_{2}K$ for code dimension $K>1$. The proof applies to degenerate codes too.

Let $r=d-1$. Purify the maximally mixed encoded input by a reference $R$ of dimension $K$:Distance $d$ makes erasure of any subset of at most $r$ qubits correctable. The allowed quantum erasure correction condition says that a correctable erased set $A$ is decoupled from $R$:This is the matrix-element correction condition in another form: all operators on $A$ have constant diagonal matrix elements and zero off-diagonal matrix elements in the logical basis, so tracing the other physical qubits in $|\Omega ⟩⟨\Omega |$ yields the displayed product.

First justify the sizes needed for the partition. If $2r>n$, partition all physical qubits into $A,B$, both of size at most $r$. Both sets are correctable. Purity of $RAB$ and the Schmidt decomposition giveAdding forces $k=0$, a contradiction. Thus $2r\le n$, without assuming the bound we aim to prove.

Choose disjoint sets $A,B$ of size $r$, and let $C$ contain the remaining $n-2r$ qubits. The joint state $RABC$ is pure, so decoupling and Subadditivity of Von Neumann entropy giveAdding and cancelling gives $k\le S(C)$. The Von Neumann entropy of $D$-dimensional states is at most ${\log }_{2}D$: for positive eigenvalues $p_i$, concavity of the logarithm gives ${\sum }_i{p}_i{\log }_2(1/p_i)\le {\log }_2{\sum }_i{p}_i/p_i\le {\log }_{2}D$. Here $D=2^{n-2r}$, so $k\le n-2r$, proving the quantum bound. The code is assumed to encode information ($K>1$); a one-dimensional code requires a separate distance convention, since simply re-preparing a known state makes its erasure-correction condition vacuous.

For a density matrix $\rho$ on a finite-dimensional Hilbert space, its Von Neumann entropy in bits iswhere ${\lambda }_j$ are its eigenvalues and $0{\log }_{2}0=0$. It is the Shannon entropy of the spectrum, is zero for a pure state, and is invariant under unitary conjugation.

For a density matrix $\rho$ on a $d$-dimensional Hilbert space, its Von Neumann entropy, in bits, iswhere the ${\lambda }_a$ are its eigenvalues and $0{\log }_{2}0=0$. Each summand is nonnegative, proving $S\ge 0$, with equality exactly when one eigenvalue is one and the rest zero, namely for a pure state.

We first prove the nonnegativity of quantum relative entropy needed below. Diagonalize $\rho$ with eigenvalues ${\lambda }_a$ and eigenvectors $u_a$, and $\sigma$ with eigenvalues ${\mu }_b$ and eigenvectors $v_b$. Put $B_{ab}=|⟨u_a,v_b⟩{|}^2$ and $q_a=⟨u_a|\sigma |u_a⟩={\sum }_b{B}_{ab}{\mu }_b$. Each row of $B$ sums to one. Concavity of the ordinary logarithm gives ${\sum }_b{B}_{ab}{\log }_2{\mu }_b\le {\log }_2{q}_a$. ConsequentlyThe final Gibbs inequality follows directly from ${\sum }_{{\lambda }_a>0}{\lambda }_a\log (q_a/{\lambda }_a)\le \log {\sum }_{{\lambda }_a>0}{q}_a\le 0$. Support failure gives $D=+\infty$; zero eigenvalues can otherwise be handled by restriction or regularization. Equality requires $q_a={\lambda }_a$ and equality in the logarithm inequality on every positive-weight row. Thus $u_a$ is a $\sigma$ eigenvector of eigenvalue ${\lambda }_a$ wherever ${\lambda }_a>0$. The remaining diagonal entries of $\sigma$ vanish and positivity makes the remaining rows and columns zero. Hence equality holds exactly when $\rho =\sigma$.

Taking $\sigma =I/d$ gives $D(\rho \Vert I/d)={\log }_{2}d-S(\rho )$, provingThe lower equality is attained exactly by pure states and the upper exactly by the maximally mixed state $I/d$.

For $\bar{\rho }={\sum }_i{p}_i{\rho }_i$, each positive-weight state has support contained in that of $\bar{\rho }$. Expanding the quantum relative entropy givesThis proves Concavity of Von Neumann entropy, with equality precisely when every positive-weight ${\rho }_i$ is the same state.

For the upper entropy bound for a quantum mixture, diagonalize each component, ${\rho }_i={\sum }_a{\lambda }_{ia}|v_{ia}⟩⟨v_{ia}|$, and set $q_{ia}=p_i{\lambda }_{ia}$. Form the rectangular matrix $B$ whose columns are $\sqrt{q_{ia}}{v}_{ia}$. Then $B{B}^{†}=\bar{\rho }$, while the Gram matrix $\Gamma =B^{†}B$ has diagonal $q_{ia}$. The two matrices have the same nonzero eigenvalues: $B$ and $B^{†}$ map corresponding positive-eigenvalue eigenspaces isomorphically. Thus $S(\Gamma )=S(\bar{\rho })$.

If there are $L$ columns, let $W=\mathrm{diag}(1,\omega ,\dots ,{\omega }^{L-1})$, with $\omega =e^{2\pi i/L}$. ThenAll the summands have the same Von Neumann entropy. The just-proved concavity therefore implies $S(\bar{\rho })=S(\Gamma )\le S(\mathrm{diag}\Gamma )=H(q)$. Finally,We have proved both requested bounds:For the upper equality, strictness in the concavity argument forces $\Gamma$ to be diagonal. Its entries between two positive-weight component eigenvectors then force those vectors to be orthogonal. Thus upper equality holds exactly when the supports of the positive-weight component states are mutually orthogonal. In that case their classical label is perfectly distinguishable, explaining the extra $H(p)$ bits.

Use base-two logarithms for Von Neumann entropy, so the units are bits, and adopt $0{\log }_{2}0=0$. All systems here may be taken finite-dimensional; the entropy arguments also apply when the displayed Von Neumann entropies are finite. The density operator of the joint pure state is ${\rho }_{AB}=|{\Psi }_{AB}⟩⟨{\Psi }_{AB}|$. It is a rank-one orthogonal projector, whose only nonzero eigenvalue is one. ConsequentlyThe reduced density matrices are ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ and ${\rho }_B={\mathrm{Tr}}_A{\rho }_{AB}$. The quantum conditional entropy is defined by the entropy differenceFor the pure state under consideration this becomes $S(B|A)=-S({\rho }_A)$. To determine when it is negative, write the Schmidt decompositionThe eigenvalues of ${\rho }_A$ are the ${\lambda }_j$, together with possible zeros. Each contribution $-{\lambda }_j{\log }_2{\lambda }_j$ is nonnegative, and their sum vanishes exactly when one eigenvalue equals one. Equivalently the Schmidt rank is one and $|{\Psi }_{AB}⟩$ is a product state. If the pure state is entangled, then $r\ge 2$, every nonzero eigenvalue is strictly below one, and the Von Neumann entropy is strictly positive. Thus the pure-state negative conditional entropy criterion givesThe purity assumption matters: for mixed density operators, negative quantum conditional entropy is not necessary for entanglement.

Let a density operator on a Hilbert space ${\mathcal{H}}_S$ have the spectral decompositionIntroduce an auxiliary Hilbert space ${\mathcal{H}}_R$ with an orthonormal basis containing $r$ vectors $|j{⟩}_R$, and setThis vector has norm one. Its partial trace isThis explicitly constructs a purification of a density operator. The same construction works for an arbitrary trace-class density operator: its nonzero eigenvalues form a finite or countable summable family, and the displayed vector converges in the Hilbert space norm. The entropy inequality below is understood with finite Von Neumann entropies, avoiding undefined differences of infinities.

Apply this construction with $S=AB$ to obtain a joint pure state on $ABC$. The Schmidt decomposition across $AB:C$, $A:BC$ and $B:AC$ respectively givesNow apply Subadditivity of Von Neumann entropy to the reduced density matrices on $BC$ and $AC$:These inequalities say $S(A)-S(B)\le S(C)$ and $S(B)-S(A)\le S(C)$. Replacing $S(C)$ by $S(AB)$ proves the Araki–Lieb inequality:The subadditivity used here concerns the actual reduced density operators, rather than statistical independence. In particular it follows from nonnegativity of quantum relative entropy, sinceand similarly for $AC$. Thus no independence assumption on the purification of a density operator is needed.

Diagonalize the reduced density matrix on Alice's Hilbert space:For every positive eigenvalue, define a vector on Bob's Hilbert space byThe definition of the partial trace givesThus the $|b_j⟩$ are orthonormal. Extend the $|a_j⟩$ to an orthonormal basis of Alice's Hilbert space. If $|a⟩$ lies in the zero-eigenvalue subspace of ${\rho }_A$, thenThere is therefore no component of $|{\Psi }_{AB}⟩$ in those additional basis directions. Expansion in Alice's orthonormal basis yields the Schmidt decompositionIts positive Schmidt coefficients are $s_j=\sqrt{{\lambda }_j}$ and its Schmidt rank is $r$. Taking the two partial traces and using the orthonormality of both families givesHence the two reduced density matrices have exactly the same nonzero eigenvalues, including multiplicities. Their numbers of zero eigenvalues may differ when the local Hilbert spaces have different dimensions. In particular their Von Neumann entropies agree.