How Do Eigenvectors of Tensor Product Operators Work?

For operators acting on separate Hilbert spaces, the tensor product combines their actions. The basic eigenvector rule is simple: tensor an eigenvector of each operator, and multiply the eigenvalues. The main qualification is that these simple tensor vectors do not always make up every eigenvector of the combined operator.

Assume throughout that $A:H\to H$ and $B:K\to K$ are bounded linear operators. Their tensor product $A\otimes B$ acts on the Hilbert-space tensor product $H\otimes K$.

The basic eigenvector rule

Suppose $x\in H$ and $y\in K$ are nonzero vectors satisfying

$$Ax=\lambda x, \qquad By=\mu y.$$

The tensor product operator acts on a simple tensor by applying each factor to its own vector:

$$(A\otimes B)(x\otimes y)=(Ax)\otimes(By).$$

Substituting the eigenvector equations gives

$$(A\otimes B)(x\otimes y) =(\lambda x)\otimes(\mu y) =\lambda\mu(x\otimes y).$$

Thus $x\otimes y$ is an eigenvector of $A\otimes B$ with eigenvalue $\lambda\mu$. The vector $x\otimes y$ is nonzero because both factors are nonzero.

A closely related special case is $A\otimes I_K$. If $Ax=\lambda x$, then $x\otimes y$ is an eigenvector of $A\otimes I_K$ with eigenvalue $\lambda$ for every nonzero $y\in K$. The identity factor does not change the eigenvalue.

Worked example with matrices

Let $H=K=\mathbb{C}^2$ and take

$$A=\begin{pmatrix}2&0\\0&3\end{pmatrix}, \qquad B=\begin{pmatrix}5&0\\0&7\end{pmatrix}.$$

With the standard basis vectors $e_1=(1,0)$ and $e_2=(0,1)$, we have $Ae_1=2e_1$, $Ae_2=3e_2$, $Be_1=5e_1$, and $Be_2=7e_2$. The tensor-product basis vectors therefore satisfy

$$\begin{aligned} (A\otimes B)(e_1\otimes e_1)&=10(e_1\otimes e_1),\\ (A\otimes B)(e_1\otimes e_2)&=14(e_1\otimes e_2),\\ (A\otimes B)(e_2\otimes e_1)&=15(e_2\otimes e_1),\\ (A\otimes B)(e_2\otimes e_2)&=21(e_2\otimes e_2). \end{aligned}$$

In this example, the four tensor products form a basis, so they give all eigenvectors up to linear combinations within each eigenspace. The eigenvalues are $10$, $14$, $15$, and $21$.

Why not every eigenvector is a simple tensor

An arbitrary vector in $H\otimes K$ need not have the form $x\otimes y$. Such vectors are often called entangled (when the Hilbert spaces describe quantum systems). Even when the operators have simple tensor eigenvectors, their linear combinations may be additional eigenvectors.

For example, let $A=B=\operatorname{diag}(1,-1)$ on $\mathbb{C}^2$. Then $A\otimes B$ has eigenvalue $1$ on both $e_1\otimes e_1$ and $e_2\otimes e_2$, since the corresponding products are $1\cdot1$ and $(-1)\cdot(-1)$. Consequently,

$$\frac{e_1\otimes e_1+e_2\otimes e_2}{\sqrt{2}}$$

is also an eigenvector with eigenvalue $1$. It is not a simple tensor: its coefficient matrix has rank $2$, whereas a simple tensor has a rank-$1$ coefficient matrix.

More generally, if $A$ and $B$ are diagonalizable finite-dimensional operators, tensor products of eigenbases form a basis of $H\otimes K$. In that case, the eigenvalues of $A\otimes B$ are the products $\lambda\mu$ of eigenvalues from the two factors. If several pairs give the same product, the eigenspace includes the span of all corresponding basis tensors. Without diagonalizability or additional assumptions, the product-vector rule remains valid, but it is not by itself a complete description of the spectrum or eigenspaces.

Common mistakes

• Adding instead of multiplying eigenvalues: For $A\otimes B$, the eigenvalue on $x\otimes y$ is $\lambda\mu$, not $\lambda+\mu$. A sum appears for a different operator, such as $A\otimes I+I\otimes B$; on $x\otimes y$, that operator has eigenvalue $\lambda+\mu$.
• Treating every vector as a simple tensor: The tensor-product space contains sums of simple tensors, and some eigenvectors cannot be written as one tensor product.
• Assuming the product rule lists everything: It guarantees eigenvectors from pairs of factor eigenvectors. A full characterization of all eigenvectors requires more information about the operators, such as finite-dimensional diagonalizability.
• Ignoring domains for unbounded operators: The formulas above are stated for bounded operators. For unbounded operators, domains and the definition of the tensor-product operator require separate care; not every formal tensor expression is automatically in the operator’s domain.