Eigenvalues and Eigenvectors

An eigenvector of a linear operator is a nonzero vector whose one-dimensional span is preserved by that operator. The corresponding eigenvalue is the scalar by which the vector is multiplied. Eigenspaces collect all vectors with the same eigenvalue and expose invariant directions or subspaces of the map.

Eigenvalues are properties of the operator, not of one matrix representation. Coordinates and eigenvector components change with basis; the eigenvalues and eigenspace dimensions do not.

Why Eigenproblems Matter

Eigenproblems identify modes on which a linear transformation acts without mixing directions. They appear whenever one asks for:

• stationary modes of a Hamiltonian or differential equation;
• principal axes or normal modes;
• invariant subspaces of a symmetry;
• decay, growth, or oscillation rates in linear dynamics;
• a basis that simplifies powers, exponentials, or functions of an operator;
• possible sharp values of a quantum observable.

The final item requires additional physical structure. An arbitrary linear operator can have eigenvalues without representing an observable.

Definition

Let $A:V\to V$ be a linear operator on a vector space over a field $\mathbb F$. A scalar $\lambda\in\mathbb F$ is an eigenvalue when there exists a nonzero vector $v\in V$ such that

$$Av=\lambda v.$$

The vector $v$ is an eigenvector belonging to $\lambda$. Equivalently,

$$(A-\lambda I)v=0.$$

Thus $\lambda$ is an eigenvalue exactly when $A-\lambda I$ has a nontrivial kernel.

The zero vector is excluded because

$$A0=\lambda0$$

holds for every scalar $\lambda$ and therefore distinguishes nothing. By contrast, zero is a valid eigenvalue when a nonzero vector lies in $\ker A$.

Eigenspaces

For a fixed eigenvalue $\lambda$, the eigenspace is

$$\mathcal E_\lambda = \ker(A-\lambda I).$$

It is a vector subspace containing the zero vector and all eigenvectors with eigenvalue $\lambda$. If $u,v\in\mathcal E_\lambda$, then

$$A(au+bv) = \lambda(au+bv).$$

Every nonzero linear combination within one eigenspace remains an eigenvector with the same eigenvalue. A sum of eigenvectors from distinct eigenspaces is generally not an eigenvector.

Eigenvectors are directions, not normalized objects. If $v$ is an eigenvector, then every $cv$ with $c\ne0$ is another eigenvector for the same eigenvalue. Normalization becomes convenient after an inner product is supplied, but it is not part of the eigenvalue definition.

Characteristic Polynomial

In a $d$-dimensional space, choose a basis and represent $A$ by a $d\times d$ matrix. A nonzero solution of

$$(A-\lambda I)v=0$$

exists exactly when the matrix is singular. Therefore the eigenvalues are the roots of the characteristic polynomial

$$p_A(\lambda) = \det(\lambda I-A).$$

This is a monic polynomial of degree $d$. Over $\mathbb C$, it factors as

$$p_A(\lambda) = \prod_{j=1}^{r} (\lambda-\lambda_j)^{m_j},$$

where $\lambda_1,\ldots,\lambda_r$ are the distinct eigenvalues and $m_j$ are their algebraic multiplicities. Counting multiplicity,

$$\sum_{j=1}^{r}m_j=d.$$

For small symbolic matrices, the determinant equation is useful. For numerical matrices, explicitly forming a characteristic polynomial is usually a poor algorithm because its coefficients can be sensitive and root finding can magnify error. Stable eigensolvers work with the matrix more directly.

Basis Invariance

If a change of basis replaces $A$ by the similar matrix

$$A'=S^{-1}AS,$$

then

$$\begin{aligned} p_{A'}(\lambda) &= \det(\lambda I-S^{-1}AS) \\ &= \det\left( S^{-1}(\lambda I-A)S \right) \\ &= p_A(\lambda). \end{aligned}$$

Similar matrices therefore have the same eigenvalues with the same algebraic multiplicities. Their eigenvector coordinate columns are related by $v'=S^{-1}v$.

The characteristic polynomial also connects eigenvalues to familiar invariants. Over $\mathbb C$,

$$\operatorname{tr}A = \sum_{j=1}^{d}\lambda_j, \qquad \det A = \prod_{j=1}^{d}\lambda_j,$$

with eigenvalues repeated according to algebraic multiplicity.

Algebraic and Geometric Multiplicity

For an eigenvalue $\lambda$:

• its algebraic multiplicity $m_\lambda$ is its multiplicity as a root of $p_A$;
• its geometric multiplicity is

$$g_\lambda = \dim\ker(A-\lambda I).$$

They satisfy

$$1\leq g_\lambda\leq m_\lambda.$$

An eigenvalue with $m_\lambda>1$ is a repeated root. In quantum mechanics, “degenerate eigenvalue” usually means that its eigenspace has dimension $g_\lambda>1$. For Hermitian matrices, algebraic and geometric multiplicities agree, so this distinction causes no conflict. For a general matrix, a repeated eigenvalue can have too few eigenvectors.

The total degeneracy relevant to a sharp quantum outcome is the dimension of its eigenspace. A basis inside that eigenspace is not unique; the subspace is the invariant object.

Distinct Eigenvalues Give Independent Eigenvectors

Eigenvectors belonging to distinct eigenvalues are linearly independent. A short induction proof shows why.

Suppose $v_1,\ldots,v_n$ have distinct eigenvalues $\lambda_1,\ldots,\lambda_n$ and

$$\sum_{j=1}^{n}c_jv_j=0.$$

Apply $A-\lambda_n I$:

$$\sum_{j=1}^{n-1} c_j(\lambda_j-\lambda_n)v_j =0.$$

By the induction hypothesis, $v_1,\ldots,v_{n-1}$ are independent. Because $\lambda_j-\lambda_n\ne0$, one obtains $c_1=\cdots=c_{n-1}=0$, and the original relation then gives $c_n=0$.

This result guarantees enough eigenvectors when a $d\times d$ matrix has $d$ distinct eigenvalues. Repeated eigenvalues require inspection of their eigenspaces.

Relation to Diagonalization

An operator is diagonalizable when its eigenspaces together span the whole space:

$$V = \bigoplus_{\lambda} \mathcal E_\lambda.$$

Equivalently,

$$\sum_{\lambda}g_\lambda=d.$$

A characteristic polynomial that splits over $\mathbb C$ guarantees eigenvalues, but not enough independent eigenvectors. A defective matrix has at least one $g_\lambda<m_\lambda$ and cannot be diagonalized.

The complete criterion, Jordan-block counterexamples, and change-of-basis construction belong to Diagonalization. Normal operators form the especially important class whose eigenvectors can be chosen orthonormal.

Worked Example: A Coupled Two-Level Matrix

Consider

$$A = \begin{pmatrix} 2&1\\ 1&2 \end{pmatrix}.$$

The characteristic polynomial is

$$\begin{aligned} p_A(\lambda) &= \det \begin{pmatrix} \lambda-2&-1\\ -1&\lambda-2 \end{pmatrix} \\ &= (\lambda-2)^2-1 \\ &= (\lambda-1)(\lambda-3). \end{aligned}$$

For $\lambda=3$,

$$(A-3I)v=0$$

gives $v_1=v_2$. For $\lambda=1$, it gives $v_1=-v_2$. Normalized eigenvectors are therefore

$$v_+ = \frac{1}{\sqrt2} \begin{pmatrix} 1\\ 1 \end{pmatrix}, \qquad v_- = \frac{1}{\sqrt2} \begin{pmatrix} 1\\ -1 \end{pmatrix},$$

with

$$Av_+=3v_+, \qquad Av_-=v_-.$$

The matrix is Hermitian, so the eigenvalues are real and the two eigenspaces are orthogonal. Those conclusions are not consequences of the determinant calculation alone; they follow from Hermiticity.

Worked Example: Degeneracy

Let

$$B = \begin{pmatrix} 2&0&0\\ 0&2&0\\ 0&0&5 \end{pmatrix}.$$

Its characteristic polynomial is

$$p_B(\lambda) = (\lambda-2)^2(\lambda-5).$$

The eigenspaces are

$$\mathcal E_2 = \operatorname{span} \left\lbrace \begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\1\\0\end{pmatrix} \right\rbrace$$

and

$$\mathcal E_5 = \operatorname{span} \left\lbrace \begin{pmatrix}0\\0\\1\end{pmatrix} \right\rbrace.$$

The eigenvalue $2$ has algebraic and geometric multiplicity two. Every nonzero vector in the plane $\mathcal E_2$ is an eigenvector with eigenvalue $2$. Rotating the basis inside that plane changes the eigenvectors used in a coordinate description but does not change the eigenspace.

Dependence on the Scalar Field

Whether eigenvalues exist can depend on the field. The real rotation matrix

$$R = \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix}$$

has characteristic polynomial

$$p_R(\lambda)=\lambda^2+1.$$

It has no real eigenvalues. On the complexified space, it has eigenvalues $+i$ and $-i$, with eigenvectors proportional to

$$\begin{pmatrix}1\\-i\end{pmatrix}, \qquad \begin{pmatrix}1\\i\end{pmatrix},$$

respectively. Quantum state spaces are complex, so finite-dimensional characteristic polynomials always split into linear factors, although a matrix may still be defective.

Hermitian, Normal, and General Matrices

The eigenvalue equation alone does not guarantee real eigenvalues, orthogonality, or a complete eigenbasis.

Operator class	Eigenvalue facts
General complex matrix	eigenvalues may be complex; eigenvectors may be incomplete or nonorthogonal
Diagonalizable matrix	has a basis of eigenvectors, not necessarily orthogonal
Normal matrix	has an orthonormal eigenbasis
Hermitian matrix	has an orthonormal eigenbasis and real eigenvalues
Unitary matrix	has an orthonormal eigenbasis and eigenvalues of unit modulus

Hermitian Operators and Normal Operators own the proofs and spectral consequences for those classes.

For a non-Hermitian matrix, one may need both right eigenvectors,

$$Av_j=\lambda_jv_j,$$

and left eigenvectors, represented by functionals satisfying

$$w_j^\dagger A = \lambda_jw_j^\dagger.$$

They need not be related by ordinary conjugate transpose, and their sensitivity can be much worse than in the Hermitian case. Do not silently import orthogonality or projector formulas from Hermitian spectral theory.

Quantum Interpretation

In the finite-dimensional projective measurement model, a Hermitian observable $A$ has real eigenvalues that label possible sharp outcomes. A nondegenerate eigenvector determines a one-dimensional outcome ray; a degenerate eigenvalue corresponds to its entire eigenspace and orthogonal projector.

This interpretation has conditions:

• the operator must represent the observable under the quantum postulates;
• eigenvalues carry the physical units of that operator;
• states must be normalized before probabilities are assigned;
• probabilities come from spectral projectors and the Born rule;
• an arbitrary operator’s eigenvalues are not automatically measurable quantities.

The physical treatment is Eigenvalues and Eigenstates. This page supplies the finite-dimensional linear algebra that treatment uses.

Differential and Infinite-Dimensional Eigenproblems

For differential operators, the equation

$$A\psi=\lambda\psi$$

still defines an eigenproblem, but admissible functions, boundary conditions, and the operator domain are part of the problem. A formal solution of the differential equation need not be an allowed eigenvector.

Infinite-dimensional operators can also have continuous spectrum, where a spectral value has no normalizable eigenvector. Characteristic determinants are not the general tool. See Eigenvalue Problems for boundary-value formulations and Discrete and Continuous Spectra for the physical distinction.

Numerical Eigenpairs

For a computed pair $(\widetilde\lambda,\widetilde v)$, check the residual

$$r = A\widetilde v -\widetilde\lambda\widetilde v.$$

A small relative residual confirms that the pair nearly satisfies the matrix equation. It does not by itself guarantee that an eigenvector is accurately determined when eigenvalues are clustered or the matrix is highly nonnormal.

Within a degenerate eigenspace, individual numerical eigenvectors are not unique: a solver may return any orthonormal basis of that subspace. Compare the projector or invariant subspace rather than matching vectors component by component. Numerical algorithms, conditioning, residual scaling, and truncation checks belong to Matrix Diagonalization.

Common Mistakes

• Allowing the zero vector as an eigenvector. It satisfies every formal eigenvalue equation and therefore carries no eigenvalue information.
• Treating zero as an invalid eigenvalue. It is valid when the kernel is nontrivial.
• Finding roots but not eigenspaces. The characteristic polynomial alone does not provide eigenvectors or geometric multiplicities.
• Confusing repeated roots with complete degeneracy data. Distinguish algebraic from geometric multiplicity.
• Assuming every matrix is diagonalizable. Defective matrices lack enough independent eigenvectors.
• Assuming eigenvectors are automatically orthogonal. This is guaranteed for normal operators, not general matrices.
• Treating eigenvector coordinates as unique. Scaling, phase, basis choice, and rotations within a degenerate eigenspace all change them.
• Assigning measurement meaning to every eigenvalue. The observable postulate and spectral projectors supply the physical interpretation.
• Using characteristic polynomials as numerical eigensolvers. Use stable matrix algorithms and residual checks.

Exercises

1. Find normalized eigenvectors as well as eigenvalues of

$$\sigma_x = \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}.$$

Solution

The characteristic equation is

$$\det(\lambda I-\sigma_x) = \lambda^2-1 =0.$$

Thus $\lambda_\pm=\pm1$. For $\lambda=+1$, the equation gives $v_1=v_2$; for $\lambda=-1$, it gives $v_1=-v_2$. Normalized choices are

$$v_+ = \frac{1}{\sqrt2} \begin{pmatrix}1\\1\end{pmatrix}, \qquad v_- = \frac{1}{\sqrt2} \begin{pmatrix}1\\-1\end{pmatrix}.$$

They are orthogonal, as expected because $\sigma_x$ is Hermitian.

2. For

$$A = \begin{pmatrix} 4&1&0\\ 0&4&0\\ 0&0&2 \end{pmatrix},$$

find the algebraic and geometric multiplicities and decide whether $A$ is diagonalizable.

Solution

The matrix is triangular, so

$$p_A(\lambda) = (\lambda-4)^2(\lambda-2).$$

Thus $m_4=2$ and $m_2=1$. For $\lambda=4$,

$$A-4I = \begin{pmatrix} 0&1&0\\ 0&0&0\\ 0&0&-2 \end{pmatrix},$$

so the kernel is spanned by $(1,0,0)^{\mathsf T}$ and $g_4=1$. For $\lambda=2$, the eigenspace is spanned by $(0,0,1)^{\mathsf T}$, so $g_2=1$.

Because

$$g_4+g_2=2<3,$$

the matrix does not have an eigenbasis and is not diagonalizable.

3. Let $v_1$ and $v_2$ be eigenvectors with distinct eigenvalues $\lambda_1\ne\lambda_2$. Prove directly that they are linearly independent.

Solution

Assume

$$c_1v_1+c_2v_2=0.$$

Apply $A-\lambda_2I$:

$$c_1(\lambda_1-\lambda_2)v_1=0.$$

Because $v_1\ne0$ and $\lambda_1-\lambda_2\ne0$, one has $c_1=0$. The original relation then gives $c_2=0$. Hence the vectors are linearly independent.

4. A numerical diagonalization of a Hermitian matrix with a two-fold degenerate eigenvalue returns different normalized eigenvectors on two machines, but both pairs span the same two-dimensional subspace. Is this a disagreement?

Solution

No. An eigenspace of dimension two has infinitely many orthonormal bases. Floating-point details may select different bases within that same invariant subspace. The basis-independent comparison is the orthogonal projector

$$P = \lvert v_1\rangle\langle v_1\rvert +\lvert v_2\rangle\langle v_2\rvert,$$

or an equivalent subspace-distance diagnostic. If the projectors agree within numerical tolerance and the residuals are small, the eigenspace results agree.

References

• S. Axler, Linear Algebra Done Right, 3rd ed., Springer, 2015.
• R. A. Horn and C. R. Johnson, Matrix Analysis, 2nd ed., Cambridge University Press, 2013.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• L. N. Trefethen and D. Bau III, Numerical Linear Algebra, SIAM, 1997.