0.7 Eigenvalues and eigenvectors

Given an operator A, an eigenvector is a non-zero vector |v\rangle such that A|v\rangle = \lambda|v\rangle for some \lambda\in\mathbb{C} (which is called the corresponding eigenvalue). We call the pair (\lambda,|v\rangle) an eigenpair, and we call the set of eigenvalues the spectrum of A, denoted by \sigma(A). It is a surprising (but incredibly useful) fact that every operator has at least one eigenpair.15 Geometrically, an eigenvector of an operator A is a vector upon which A simply acts by “stretching”. Note that eigenvectors can be scaled by an arbitrary non-zero length: if A|v\rangle=\lambda|v\rangle then \begin{aligned} A(\mu|v\rangle) &= \mu(A|v\rangle) \\&= \mu\lambda|v\rangle \\&= \lambda(\mu|v\rangle) \end{aligned} for any \mu\neq0. Because of this, we usually assume all eigenvectors to be of length 1.

Rewriting the defining property of an eigenpair (\lambda,|v\rangle), we see that (A-\lambda\mathbf{1})|v\rangle = 0 which tells us that the operator A-\lambda\mathbf{1} has a non-zero kernel, and is thus non-invertible. This gives a useful characterisation of the spectrum in terms of a determinant: \sigma(A) = \{\lambda\in\mathbb{C} \mid \det(A-\lambda\mathbf{1})=0\}.

The spectrum \sigma(A) allows us to recover information about the operator A. For example, the trace of A is equal to the sum of all its eigenvalues, and the determinant of A is equal to the product of all its eigenvalues. We can show this easily for normal operators using the fact16 that their eigenvectors are orthogonal: they satisfy \langle v|w\rangle=0 for v\neq w. Because eigenvectors can always be scaled, this means that we can assume the eigenvectors of a normal operator to be orthonormal. If we write the eigenpairs as (\lambda_i,|v_i\rangle), we can define U = \sum_i |i\rangle\langle v_i| for an orthonormal basis \{|1\rangle,\ldots,|n\rangle\}, which is an orthogonal matrix (since the eigenvectors are also assumed to be orthonormal). Then we see that \begin{aligned} UAU^\dagger &= \sum_{i,j} |i\rangle\langle v_i| A |v_j\rangle\langle j| \\&= \sum_{i,j} |i\rangle\langle v_i| \lambda_j|v_j\rangle\langle j| \\&= \sum_{i,j} \lambda_j|i\rangle(\langle v_i|v_j\rangle)\langle j| \\&= \sum_i \lambda_i |i\rangle\langle i| \end{aligned} (where we again use this hypothesis that the eigenvectors of A are all orthonormal) which is the diagonal matrix D consisting of the eigenvalues \lambda_i of A along its diagonal. Then, since UAU^\dagger=D, we can equally write A=U^\dagger DU, which gives us A = \sum_i \lambda_i |v_i\rangle\langle v_i|. We call each \lambda_i|v_i\rangle\langle v_i| the eigenspace projector, since a projector is defined to be any operator P that satisfies P=P^\dagger and P^2=P. Note that projectors can only have eigenvalues equal to 0 or 1, since if |v\rangle is an eigenvector of P then, using the fact that P^2=P, \begin{aligned} 0 &= (P^2-P)|v\rangle \\&= (\lambda^2-\lambda)|v\rangle \\\implies \lambda(\lambda-1)=0 \end{aligned} and so \lambda must be equal to either 0 or 1.

Finally we can now return to the relationship between eigenvalues and the trace and determinant, using this fact that any normal operator A gives a unitary operator U such that A=U^\dagger DU for the diagonal matrix D of eigenvalues of A. Indeed, \begin{aligned} \operatorname{tr}(A) &= \operatorname{tr}(U^\dagger DU) \\&= \operatorname{tr}(DUU^\dagger) \\&= \operatorname{tr}(D) \\&= \sum_i \lambda_i \end{aligned} which proves that the trace is equal to the sum of the eigenvalues, and \begin{aligned} \det(A) &= \det(U^\dagger DU) \\&= \det(U^\dagger)\det(D)\det(U) \\&= \det(D)\det(U^\dagger)\det(U) \\&= \det(D)\det(U^\dagger U) \\&= \det(D) \\&= \prod_i \lambda_i \end{aligned} which proves that the determinant is equal to the product of the eigenvalues.

The eigenspace projectors give us the spectral decomposition of A, which is where we write A = \sum_i \lambda_i|v_i\rangle\langle v_i|.

The spectral decomposition of a normal operator gives an effective way of calculating the action of a function on a matrix. If f\colon\mathbb{C}\to\mathbb{C} is a function, then we can define f(A) = \sum_i f(\lambda_i)|v_i\rangle\langle v_i|. For example, if f(x)=x^2, then, by this definition, f(A)=\sum_i \lambda_i^2|v_i\rangle\langle v_i|. But this is consistent with the definition of A^2 that you expect: \begin{aligned} A^2 &= \left(\sum_i \lambda_i|v_i\rangle\langle v_i|\right)\left(\sum_i \lambda_i|v_i\rangle\langle v_i|\right) \\&= \sum_{i,j} \lambda_i\lambda_j |v_i\rangle\langle v_i||v_j\rangle\langle v_j| \\&= \sum_i \lambda_i^2 |v_i\rangle\langle v_i| \end{aligned} using the fact that the eigenvectors |v_i\rangle are orthonormal and that projectors P=|v_i\rangle\langle v_i| satisfy P^2=P.

  1. You can prove this for an (n\times n) matrix A by considering the set \{|v\rangle,A|v\rangle,A^2|v\rangle,\ldots,A^n|v\rangle\} of vectors in \mathbb{C}^n. Since this has n+1 elements, it must be linearly dependent, and so (after some lengthy algebra) we can construct an eigenpair.↩︎

  2. Exercise. Prove this! Hint: start by showing that, if (\lambda,|v\rangle) is an eigenpair of A, then (\lambda^\star,|v\rangle) is an eigenpair of A^\dagger.↩︎