# Eigenvalues of tensor product of matrices : question about general properties

I would like to refresh my mind on eigenvalues/eigenvector of tensor product of operator (for Quantum Mechanics purposes).

Consider two hermitic matrices $$A$$ and $$B$$ living in Hilbert spaces of finite dimensions $$N_A$$ and $$N_B$$ respectively.

I would like to know the most general properties about the eigenvalues of the operator $$U=A \otimes B$$.

Let's assume I know the spectrum of $$A$$ and $$B$$:

$$A |a^i\rangle = a |a^i\rangle$$ $$B |b^j\rangle = b |b^j\rangle$$

Where the indices $$i$$ and $$j$$ are here to take in account possible degeneracy in the eigenvalues.

Proposition 1

A possible eigenbasis of $$U$$ is the set of all $$|a^i\rangle \otimes |b^j\rangle$$ with eigenvalues $$ab$$.

Proof:

$$|a^i\rangle \otimes |b^j\rangle$$ is an eigenvector of $$U$$ with eigenvalue $$ab$$.

And as I have $$N_A$$ orthogonal vectors $$|a^i\rangle$$ and $$N_B$$ orthogonal vectors $$|b^j\rangle$$, I have found $$N_A*N_B$$ orthogonal vectors that are eigenstates of $$U$$.

However, there might have quantum state of $$U$$ that are not of the form of a tensor product.

Proposition 2

$$U=A \otimes B$$ admits one degenerated eigenvalue if and only if there exist eigenvectors of $$U$$ that are not a tensor product.

Proof:

Necessary condition:

As $$\{ |a^i\rangle \otimes |b^j\rangle \}$$ are basis of eigenstate of $$U$$, and there is a degenerated eigenvalue of $$U$$, then there exist $$(a,b) \neq (a',b')$$ such that $$|a\rangle \otimes |b\rangle$$ and $$|a'\rangle \otimes |b'\rangle$$ are eigenstates of $$U$$ with the same eigenvalue. Thus $$|a\rangle \otimes |b\rangle + |a'\rangle \otimes |b'\rangle$$ also. Then I found an eigenstate of $$U$$ that is not a product state.

Sufficient condition:

If there is an eigenvector of $$U$$ that is not a tensor product, then it must be a linear combination of different $$|a^i\rangle \otimes |b^j \rangle$$ as they diagonalise $$U$$. And if a linear combination of eigenvector is an eigenvector, then the two initial eigenvector must have the same eigenvalue.

Then, $$U$$ has degenerated eigenvalues.

Do you agree ?

Especially with my second proposition (I am quite confident about the first one).

I use my notations.

i) Let $$A\in M_n,B\in M_m$$ be diagonalizable matrices over $$\mathbb{C}$$. Let $$(u_i)_{i\leq n},(v_j)_{j\leq m})$$ be bases of eigenvectors associated to the eigenvalues $$spectrum(A)=(\lambda_i),spectrum(B)=(\mu_j)$$. Then $$(u_i\otimes v_j)_{i,j}$$ is a basis of eigenvector of $$A\otimes B$$ because $$(A\otimes B)(u_i\otimes v_j)=A(u_i)\otimes B(v_j)=\lambda_i\mu_j u_i\otimes v_j$$.

ii) Assume that $$\tau=\lambda_i\mu_j=\lambda_k\mu_l$$ where $$(i,j)\not= (k,l)$$.

Then $$(A\otimes B)(u_i\otimes v_j+u_k\otimes v_l)=\tau(u_i\otimes v_j+u_k\otimes v_l)$$. Assume that $$u_i\otimes v_j+u_k\otimes v_l$$ is a tensor product $$a\otimes b$$; then, $$ab^T$$, the associated $$n\times m$$ matrix has rank $$1$$ and is the sum of two matrices of rank $$1$$: $$u_i{v_j}^T+u_k{v_l}^T$$ (one has the same result for the transposes). This can only happen if the images are the same, that is $$span(u_i)=span(u_k)$$ or, in the same way, if $$span(v_j)=span(v_l)$$; in other words, $$u_i\otimes v_j+u_k\otimes v_l$$ is a tensor product iff $$i=k,j\not=l$$ or $$j=l,i\not= k$$ iff $$\lambda_i$$ is a multiple eigenvalue of $$A$$ or $$\mu_j$$ is a multiple eigenvalue of $$B$$.

iii) Let $$z$$ be an eigenvector of $$A\otimes B$$ that is not a tensor product. Then $$z=\sum_{(i,j)\in K}z_{i,j}u_i\otimes v_j$$ where $$z_{i,j}\not= 0$$. There is $$\tau$$ s.t.

$$(A\otimes B)(z)=\sum_{i,j} z_{i,j}\lambda_i\mu_j u_i\otimes v_j=\sum_{i,j}\tau z_{i,j}u_i\otimes v_j$$, that implies

for every $$(i,j)\in K$$, $$\tau=\lambda_i\mu_j$$, and $$A\otimes B$$ has a multiple eigenvalue.

$$\textbf{Conclusion.}$$. Finally, your equivalence is valid when $$A,B$$ have simple eigenvalues.

For example, for $$A=diag(1,2),B=I_2$$, then all the eigenvectors of $$A\otimes B$$ are tensor products.

• Oh ok I think I see my mistake for the necessary condition given your proof. If $(a,b) \neq (a',b')$ my vector $|a\rangle | b \rangle$ and $|a'\rangle | b' \rangle$ will indeed be different, but their sum might still be a tensor product (for example if $a=a'$). Do you agree it is the exact place where my proof failed for the necessary condition ? Sep 6 '19 at 19:45