Eigenvalues, Eigenvectors, and the Characteristic Equation

Hand a linear operator $A$ a generic vector and it does two things at once — turns the vector and rescales it — so that input and output point in different directions and the geometry is hard to read. A privileged few directions escape the turning: along an eigenvector $v$ the operator only stretches, $Av=\lambda v$ , and the whole complicated action collapses to multiplication by a single number $\lambda$ . Those directions are the intrinsic grain of the transformation, the axes along which it acts most simply, and they are visible no matter which basis an accident of coordinates happens to impose. Finding them looks at first like an impossible search over infinitely many candidate vectors, and the characteristic equation is the trick that makes it finite. Rewrite $Av=\lambda v$ as $(A-\lambda I)v=0$ : a nonzero solution exists exactly when $A-\lambda I$ crushes some direction to zero, i.e. is singular, i.e. has vanishing determinant. So the eigenvalues are the roots of one scalar polynomial, $\det(A-\lambda I)=0$ , and the fundamental theorem of algebra guarantees an $n\times n$ matrix has $n$ of them over $\mathbb{C}$ , counted with multiplicity. The search over vectors has become a search over polynomial roots — geometry converted into algebra. Each root then carries a second question: how many independent directions does it actually supply? The gap between what the polynomial promises (algebraic multiplicity) and what the nullspace delivers (geometric multiplicity) decides whether $A$ can be diagonalized or is stuck as a defective matrix needing the Jordan form. Because the entire construction uses only $A-\lambda I$ and a determinant — both immune to change of basis — the spectrum is a genuine property of the operator, which is why eigenanalysis is the backbone of quantum mechanics, normal modes, stability theory, and the principal axes of every physical tensor.

Key takeaways

An eigenvector of $A$ is a nonzero direction the operator only rescales: $Av=\lambda v$ , with the scale factor $\lambda$ its eigenvalue.
Eigenvalues are the roots of the characteristic polynomial $p(\lambda)=\det(A-\lambda I)$ , so an $n\times n$ matrix has exactly $n$ of them over $\mathbb{C}$ , counted with algebraic multiplicity.
For each eigenvalue the eigenvectors fill the eigenspace $E_\lambda=\ker(A-\lambda I)$ , whose dimension (the geometric multiplicity) never exceeds the algebraic multiplicity, $1\le g_\lambda\le m_\lambda$ .
Two invariants come for free: $\operatorname{tr}A=\sum_k\lambda_k$ and $\det A=\prod_k\lambda_k$ , so you can sanity-check eigenvalues without ever finding an eigenvector.
Eigenvalues are basis-independent — similar matrices $B=S^{-1}AS$ share the characteristic polynomial, hence the same spectrum, trace, and determinant.
When the eigenvectors form a basis the operator diagonalizes; when the geometric multiplicity falls short it is defective and needs the Jordan canonical form.
Key idea: solving $Av=\lambda v$ splits into two stages — find the eigenvalues as roots of $\det(A-\lambda I)=0$ , then for each root solve the singular system $(A-\lambda I)v=0$ for its eigenvectors. The first stage is a polynomial problem; the second is a nullspace problem.

The core insight

Solving $Av=\lambda v$ splits into two stages: find the eigenvalues as the roots of $\det(A-\lambda I)=0$ , then for each root solve the singular linear system $(A-\lambda I)v=0$ for its eigenvectors. The first stage is a polynomial problem; the second is a nullspace problem.

Where this lives today

The characteristic equation is beautiful but the roots of a large determinant are catastrophic to compute directly, so almost every modern eigenvalue is found another way entirely. The workhorse is the QR algorithm of Francis and Kublanovskaya (1961) — repeatedly factoring a matrix and reversing the factors until it drifts to triangular form with the eigenvalues on the diagonal — and it is fast and stable enough that it hides inside `numpy.linalg.eig`, MATLAB, and every scientific stack, ranked among the top ten algorithms of the twentieth century. On top of it sits an enormous applied edifice: quantum chemistry diagonalizes molecular Hamiltonians to predict spectra; structural and aerospace engineers run modal analysis to find the resonant frequencies of bridges, wings, and buildings and keep them clear of driving loads; control theory reads the eigenvalues of a linearized system to certify stability. Google's original PageRank was nothing but the dominant eigenvector of a web-link matrix, principal component analysis diagonalizes a data covariance matrix to find the directions of greatest variance, and spectral clustering and graph neural networks lean on the eigenvectors of a graph Laplacian. Dynamic mode decomposition extracts eigenvalues from streaming fluid and video data to expose coherent structures. In every case the same two-stage logic runs underneath — spectrum first, then eigenvectors — even when no determinant is ever expanded.

The physical picture

Think of $A$ as a deformation of the plane. A generic arrow is picked up, rotated, and stretched, so it lands somewhere off its own line — its direction is not preserved. But scattered through the plane are special lines that the deformation leaves alone as directions: an arrow lying along such a line is only lengthened or flipped, never turned. Those lines are the eigen-directions, and the factor by which the arrow along one of them grows is the eigenvalue $\lambda$ . The first figure shows this for the symmetric map $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ : the diagonal direction $(1,1)$ is stretched threefold, the anti-diagonal $(1,-1)$ is left unchanged, and a horizontal arrow gets visibly twisted.

Figure

A symmetric map A = ((2,1),(1,2)) stretches the plane along its eigen-directions: the eigenvector v₁ along (1,1) is scaled by λ₁ = 3 and v₂ along (1,−1) by λ₂ = 1, while a generic horizontal arrow is rotated off its own line — the unit circle is carried to an ellipse with semi-axes 3 and 1

The characteristic equation is the piece of bookkeeping that finds these lines without trial and error. Instead of testing candidate vectors, ask for which numbers $\lambda$ the shifted operator $A-\lambda I$ manages to send some nonzero vector to zero — because that is precisely when $\lambda$ is an eigenvalue. An operator collapses a direction exactly when its determinant vanishes, so the whole problem condenses into a single polynomial equation $\det(A-\lambda I)=0$ . The second figure draws that polynomial for the same matrix: a parabola whose two zero-crossings are the eigenvalues, and whose coefficients already carry the trace and determinant.

Figure

The characteristic polynomial p(λ) = det(A − λI) = λ² − 4λ + 3 crosses zero exactly at the eigenvalues λ = 1 and λ = 3; their sum 4 is the trace of A and their product 3 is its determinant, so one scalar polynomial encodes the whole spectrum

Eigenvalues also carry the dynamics. If $A$ generates the evolution of a system, the magnitude and phase of each $\lambda$ decide the fate of the corresponding mode — growth when $\operatorname{Re}\lambda>0$ (or $|\lambda|>1$ for iterated maps), decay when it is negative, and pure oscillation when $\lambda$ is imaginary. The eigenvectors are the natural coordinates in which a tangled, coupled system falls apart into independent one-dimensional motions, each evolving by its own $\lambda$ . That decoupling — turning a matrix problem into $n$ scalar problems — is the reason eigenanalysis appears everywhere from normal modes to quantum stationary states.

The core equations

The definition is the whole subject in one line: an eigenvector is a nonzero direction the operator merely rescales. Everything downstream is machinery for finding the pairs $(\lambda,v)$ that satisfy it.

Av=\lambda v,\qquad v\neq 0 \qquad (1)

where: $A$ — a linear operator, or its $n\times n$ matrix in some basis; $v$ — the eigenvector, a nonzero element of the vector space; $\lambda$ — the eigenvalue, a scalar in $\mathbb{C}$ (or $\mathbb{R}$ ). Reading: applying $A$ to $v$ has the same effect as multiplying $v$ by the number $\lambda$ — the operator acts, on this one direction, like a scalar. Collecting both terms on one side turns the definition into a homogeneous linear system. This is the pivotal rewrite, because a homogeneous system has a nonzero solution only under a sharp algebraic condition.

(A-\lambda I)v=0 \qquad (2)

where: $I$ — the identity operator; $A-\lambda I$ — the operator $A$ shifted by $\lambda$ along the diagonal. Reading: $v$ is a nonzero vector in the kernel of $A-\lambda I$ , so this operator must fail to be invertible — it must be singular. A square operator is singular exactly when its determinant vanishes, and that condition, written out, is a polynomial in $\lambda$ of degree $n$ . Its roots are the eigenvalues; this is the equation that converts an infinite search into a finite one.

p(\lambda)=\det(A-\lambda I)=0 \qquad (3)

where: $p(\lambda)$ — the characteristic polynomial, of degree $n$ with leading term $(-\lambda)^n$ . Reading: the eigenvalues of $A$ are precisely the roots of $p$ ; over $\mathbb{C}$ there are exactly $n$ of them, counted with algebraic multiplicity, by the fundamental theorem of algebra. Two coefficients of that polynomial are worth reading off directly, because they give the simplest checks on any eigenvalue computation. The sum of the roots is the trace and their product is the determinant.

\operatorname{tr}A=\sum_{k}\lambda_k,\qquad \det A=\prod_{k}\lambda_k \qquad (4)

where: $\lambda_k$ — the eigenvalues counted with algebraic multiplicity; $\operatorname{tr}A=\sum_i A_{ii}$ — the sum of the diagonal entries. Reading: these follow from Vieta's formulas applied to $p(\lambda)$ , so you can confirm candidate eigenvalues arithmetically without ever solving for an eigenvector. For each eigenvalue the eigenvectors do not stand alone — together with the zero vector they fill a whole subspace, and its dimension records how many independent directions the eigenvalue supplies. That dimension is bounded by the eigenvalue's multiplicity as a root.

E_\lambda=\ker(A-\lambda I),\qquad 1\le g_\lambda=\dim E_\lambda\le m_\lambda \qquad (5)

where: $E_\lambda$ — the eigenspace of $\lambda$ ; $g_\lambda$ — its geometric multiplicity (independent eigenvectors); $m_\lambda$ — its algebraic multiplicity (order of the root in $p$ ). Reading: equality $g_\lambda=m_\lambda$ for every eigenvalue means $A$ has a full eigenbasis and diagonalizes; any strict deficit $g_\lambda<m_\lambda$ marks $A$ as defective. Finally, the construction never referred to a choice of basis, and the last identity makes that explicit: a similarity transformation leaves the characteristic polynomial — hence the entire spectrum — untouched.

\det(S^{-1}AS-\lambda I)=\det(A-\lambda I) \qquad (6)

where: $S$ — any invertible change-of-basis matrix; $S^{-1}AS$ — the matrix of the same operator in the new basis. Reading: because $\lambda$ , trace, and determinant survive similarity, the eigenvalues are properties of the operator itself, not of the coordinates used to write it down.

Derivation AFrom the eigenvalue equation to the characteristic polynomial

The goal is to prove that the vector problem $Av=\lambda v$ is exactly equivalent to a scalar polynomial equation, and to pin down that polynomial's degree and root count. This is the single most important passage in the subject: it explains why an operator has finitely many eigenvalues, why there are exactly $n$ of them over the complex numbers, and why every eigenvalue calculation begins by expanding a determinant. We start from the demand that some nonzero vector be merely scaled, and follow the logic until it lands on $\det(A-\lambda I)=0$ .

Step 1. Write the defining relation $Av=\lambda v$ and insist $v\neq 0$ , then move everything to one side using $\lambda v=\lambda I v$ to get $(A-\lambda I)v=0$ . Why: the zero vector satisfies $Av=\lambda v$ for every $\lambda$ , so it must be excluded or the notion of eigenvalue would be vacuous; folding $\lambda$ into the operator turns a scaling statement into a kernel statement.

Step 2. Recognize $(A-\lambda I)v=0$ as a homogeneous linear system, which always has the trivial solution $v=0$ and admits a nonzero solution only when $A-\lambda I$ has a nontrivial kernel. Why: by rank–nullity, a nonzero kernel means the operator is not injective, hence not invertible — the eigenvalue condition is exactly the failure of $A-\lambda I$ to be invertible.

Step 3. Convert "not invertible" into a scalar test: a square matrix is singular if and only if its determinant is zero, giving $p(\lambda)=\det(A-\lambda I)=0$ . Why: the determinant is the bridge that turns a question about vectors and kernels into a single equation in the unknown $\lambda$ — this is where geometry becomes algebra.

Step 4. Expand the determinant by cofactors and count degrees: each of the $n$ factors $(A_{ii}-\lambda)$ on the diagonal contributes one power of $\lambda$ , so $p$ is a polynomial of degree exactly $n$ with leading term $(-\lambda)^n$ . Why: the degree fixes the maximum possible number of distinct eigenvalues and guarantees the leading coefficient $(-1)^n$ , a useful check on any hand computation.

Step 5. Apply the fundamental theorem of algebra to factor $p$ completely over $\mathbb{C}$ , $p(\lambda)=(-1)^n\prod_{k=1}^{n}(\lambda-\lambda_k)$ , so the eigenvalues are the roots $\lambda_k$ with their algebraic multiplicities. Why: this is why every $n\times n$ complex matrix has exactly $n$ eigenvalues counted with multiplicity — the count is a theorem about polynomials, imported wholesale into linear algebra. *As a concrete check: for $\begin{pmatrix}2&1\\1&2\end{pmatrix}$ , cofactor expansion gives $p(\lambda)=\lambda^2-4\lambda+3=(\lambda-1)(\lambda-3)$ , confirming $\lambda=1,3$ with leading coefficient $(-1)^2=1$ .*

Go deeper · Derivation B · 1 minTrace, determinant, and the invariance of the spectrum

Derivation BTrace, determinant, and the invariance of the spectrum

The aim here is to extract the two cheapest and most-used facts about eigenvalues — that they sum to the trace and multiply to the determinant — and then to show that neither the eigenvalues nor these invariants depend on the basis in which $A$ is written. Together these results give a physicist a running arithmetic check on any spectrum and the licence to speak of "the eigenvalues of an operator" without mentioning coordinates at all. We read the invariants off the factored polynomial, then use determinant multiplicativity to kill the basis dependence.

Step 1. Factor the characteristic polynomial as $p(\lambda)=\prod_k(\lambda_k-\lambda)$ and evaluate at $\lambda=0$ : $p(0)=\det(A)=\prod_k\lambda_k$ . Why: setting $\lambda=0$ strips the shift and leaves $\det A$ on one side and the product of roots on the other, proving the determinant is the product of the eigenvalues.

Step 2. Compare the coefficient of $\lambda^{n-1}$ on both sides of the same factorization; on the determinant side it comes only from the product of diagonal terms, giving $\sum_k\lambda_k=\operatorname{tr}A$ . Why: this is Vieta's relation for the second coefficient, and it means the trace — trivially read off the diagonal — already fixes the sum of a spectrum you have not yet computed.

Step 3. For a change of basis $B=S^{-1}AS$ , use $S^{-1}IS=I$ to write $B-\lambda I=S^{-1}(A-\lambda I)S$ . Why: the shift by $\lambda I$ passes cleanly through the similarity because the identity is fixed by conjugation, so the shifted operator transforms exactly like $A$ itself.

Step 4. Take determinants and apply multiplicativity, $\det(S^{-1}(A-\lambda I)S)=\det(S^{-1})\det(A-\lambda I)\det(S)=\det(A-\lambda I)$ , since $\det(S^{-1})\det(S)=1$ . Why: the characteristic polynomial is therefore identical for $A$ and $B$ , so all similar matrices share eigenvalues, trace, and determinant — these are genuine invariants of the operator, not artefacts of coordinates. *As a concrete check: for $\begin{pmatrix}2&1\\1&2\end{pmatrix}$ , $\lambda_1+\lambda_2=1+3=4=\operatorname{tr}A$ and $\lambda_1\lambda_2=1\cdot3=3=\det A$ , confirming both invariants, and the similarity invariance follows from determinant multiplicativity as shown above.*

Go deeper · Derivation C · 1 minEigenvectors of distinct eigenvalues are independent

Derivation CEigenvectors of distinct eigenvalues are independent

The purpose of this argument is to establish the cornerstone of diagonalizability: eigenvectors belonging to different eigenvalues can never be linearly dependent, so an operator with $n$ distinct eigenvalues automatically owns a full eigenbasis. This is the theorem that lets you stop worrying about defectiveness the moment you see $n$ distinct roots, and it is the engine behind the spectral theorem for operators with simple spectra. The proof is a clean induction that plays the operator against the coefficients of a supposed dependence relation until every coefficient is forced to vanish.

Step 1. Take eigenvectors $v_1,\dots,v_r$ with pairwise distinct eigenvalues $\lambda_1,\dots,\lambda_r$ , and assume as induction hypothesis that any fewer than $r$ such eigenvectors are independent; the base case $r=1$ holds because $v_1\neq 0$ . Why: induction on the number of vectors is natural here, since eliminating one eigenvector at a time reduces the problem to a smaller instance of itself.

Step 2. Suppose a dependence $\sum_{k=1}^{r}c_k v_k=0$ and apply $A$ to it, using $Av_k=\lambda_k v_k$ , to obtain $\sum_{k=1}^{r}c_k\lambda_k v_k=0$ . Why: applying the operator weights each term by its own eigenvalue, producing a second relation among the same vectors that differs from the first — the leverage needed to cancel a term.

Step 3. Subtract $\lambda_r$ times the original relation from this one, so the $k=r$ term cancels and $\sum_{k=1}^{r-1}c_k(\lambda_k-\lambda_r)v_k=0$ . Why: eliminating $v_r$ drops the count to $r-1$ eigenvectors, exactly the situation the induction hypothesis covers.

Step 4. By the hypothesis those $r-1$ eigenvectors are independent, so $c_k(\lambda_k-\lambda_r)=0$ for $k<r$ ; since the eigenvalues are distinct, $\lambda_k-\lambda_r\neq 0$ , forcing $c_k=0$ . The original relation then reduces to $c_r v_r=0$ , so $c_r=0$ as well. Why: distinctness is precisely what makes each factor $\lambda_k-\lambda_r$ nonzero, and without it the argument — and diagonalizability — can fail. *The induction above implies that $n$ distinct eigenvalues guarantee $n$ independent eigenvectors, hence diagonalizability.*

Limiting cases

The structure of the spectrum mirrors the structure of the operator, and the special classes below are the ones physics leans on constantly:

Case	Result	Meaning
Hermitian / symmetric	$A=A^\dagger$	All eigenvalues real; eigenvectors of distinct eigenvalues orthogonal; always diagonalizable by a unitary. This is the spectral theorem, and it is why observables have real spectra.
Unitary / orthogonal	$A^\dagger A=I$	Eigenvalues lie on the unit circle, $\|\lambda\|=1$ ; eigenvectors orthogonal. Norms are preserved, so these operators are rotations of the state space.
Distinct eigenvalues	$\lambda_i\neq\lambda_j$	$n$ distinct roots force $n$ independent eigenvectors, hence diagonalizability — no need to check multiplicities.
Defective	$g_\lambda<m_\lambda$	Too few eigenvectors; no eigenbasis exists and the Jordan canonical form is required to describe the operator.
Triangular	$A_{ij}=0,\ i>j$	Eigenvalues are just the diagonal entries, since $\det(A-\lambda I)=\prod_i(A_{ii}-\lambda)$ — read them straight off.

Common trapAlgebraic multiplicity is not geometric multiplicity

The most common eigenvalue error is assuming a repeated eigenvalue automatically supplies as many independent eigenvectors as its multiplicity — that every matrix diagonalizes. It does not. A root can appear twice in the characteristic polynomial yet leave the nullspace of $A-\lambda I$ only one-dimensional. The shear $\begin{pmatrix}2&1\\0&2\end{pmatrix}$ is the canonical trap: $p(\lambda)=(\lambda-2)^2$ gives $\lambda=2$ with algebraic multiplicity $2$ , but $A-2I=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ has rank $1$ , so its kernel is only spanned by $(1,0)^{\mathsf T}$ — a single eigenvector. The matrix is defective, cannot be diagonalized, and needs a Jordan block. Always confirm $g_\lambda=m_\lambda$ before assuming an eigenbasis exists; the safe shortcut is that $n$ distinct eigenvalues guarantee diagonalizability, while repeated ones demand a genuine nullspace check.

Worked examples

1 · A symmetric matrix and its invariant checks (confidence builder)

Find the eigenvalues and eigenvectors of $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ and confirm them against the trace and determinant.

Step 1. Form the characteristic polynomial. $\det(A-\lambda I)=\det\begin{pmatrix}2-\lambda&1\\1&2-\lambda\end{pmatrix}=(2-\lambda)^2-1$ . Expanding, $p(\lambda)=\lambda^2-4\lambda+3=(\lambda-1)(\lambda-3)$ , so the eigenvalues are $\boldsymbol{\lambda_1=1}$ and $\boldsymbol{\lambda_2=3}$ — two distinct real roots, as the spectral theorem promises for a symmetric matrix.

Step 2. Solve $(A-3I)v=0$ . Here $A-3I=\begin{pmatrix}-1&1\\1&-1\end{pmatrix}$ , whose rows both say $-v_1+v_2=0$ , so $v_1=v_2$ and $v_2\propto(1,1)^{\mathsf T}$ . Normalizing, $\boldsymbol{v_2=\tfrac{1}{\sqrt2}(1,1)^{\mathsf T}}$ — the direction stretched by $3$ in the first figure.

Step 3. Solve $(A-I)v=0$ . Now $A-I=\begin{pmatrix}1&1\\1&1\end{pmatrix}$ gives $v_1+v_2=0$ , so $v_1\propto(1,-1)^{\mathsf T}$ and $\boldsymbol{v_1=\tfrac{1}{\sqrt2}(1,-1)^{\mathsf T}}$ . Note $v_1\cdot v_2=0$ : the eigenvectors are orthogonal, exactly as a symmetric operator requires.

Step 4. Check with the invariants. The sum $\lambda_1+\lambda_2=1+3=4$ matches $\operatorname{tr}A=2+2=4$ , and the product $\lambda_1\lambda_2=1\cdot3=3$ matches $\det A=2\cdot2-1\cdot1=3$ . Both invariants confirm the spectrum $\{1,3\}$ with orthogonal eigenvectors along $(1,-1)$ and $(1,1)$ .

2 · Normal modes of two coupled masses (fully worked)

Two equal masses $m$ sit on a line, joined to each other and to two fixed walls by three identical springs of stiffness $k$ . Find the normal-mode frequencies and mode shapes.

Step 1. Write the equations of motion. With displacements $x_1,x_2$ , the central spring couples the masses: $m\ddot x_1=-kx_1+k(x_2-x_1)=-2kx_1+kx_2$ and $m\ddot x_2=k x_1-2k x_2$ . In matrix form $m\ddot{\mathbf x}=-K\mathbf x$ with the stiffness matrix $K=\begin{pmatrix}2k&-k\\-k&2k\end{pmatrix}$ .

Step 2. Seek oscillating solutions $\mathbf x=\mathbf v\,e^{i\omega t}$ . Substituting gives $-m\omega^2\mathbf v=-K\mathbf v$ , i.e. the eigenvalue problem $K\mathbf v=m\omega^2\mathbf v$ — the squared frequencies $\omega^2$ are the eigenvalues of $K/m$ , and the eigenvectors are the mode shapes.

Step 3. Diagonalize $K$ . Its characteristic polynomial is $(2k-\Lambda)^2-k^2=0$ , so $2k-\Lambda=\pm k$ and the eigenvalues are $\Lambda=k$ and $\Lambda=3k$ . Hence $\boldsymbol{\omega_1^2=k/m}$ and $\boldsymbol{\omega_2^2=3k/m}$ .

Step 4. Read the mode shapes from the nullspaces. For $\Lambda=k$ , $(K-kI)\mathbf v=\begin{pmatrix}k&-k\\-k&k\end{pmatrix}\mathbf v=0$ gives $\mathbf v\propto(1,1)^{\mathsf T}$ : the masses move in phase, the central spring never stretches, so the frequency is the lowest. For $\Lambda=3k$ , $\mathbf v\propto(1,-1)^{\mathsf T}$ : the masses move out of phase, compressing the central spring hardest, so the frequency is higher.

Step 5. Put in numbers. Taking $k=1\ \mathrm{N/m}$ and $m=1\ \mathrm{kg}$ , the two normal-mode frequencies are $\omega_1=1\ \mathrm{rad/s}$ and $\omega_2=\sqrt3\approx\mathbf{1.732}\ \mathrm{rad/s}$ . The symmetric mode $(1,1)$ oscillates at $\sqrt{k/m}$ and the antisymmetric mode $(1,-1)$ at $\sqrt{3k/m}$ — the eigenvectors decouple two coupled masses into two independent oscillators.

3 · A real matrix with a complex spectrum

Find the eigenvalues of the rotation $A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ , which turns every real vector by $90^\circ$ .

Step 1. The characteristic polynomial is $\det\begin{pmatrix}-\lambda&-1\\1&-\lambda\end{pmatrix}=\lambda^2+1=0$ , so $\boldsymbol{\lambda=\pm i}$ — a real matrix with no real eigenvalues, because no real direction survives a quarter-turn unrotated.

Step 2. The eigenvectors live in $\mathbb{C}^2$ : $(A-iI)v=0$ gives $v\propto(1,i)^{\mathsf T}$ , and $\lambda=-i$ gives $(1,-i)^{\mathsf T}$ . Check the invariants: $\lambda_1+\lambda_2=i+(-i)=0=\operatorname{tr}A$ and $\lambda_1\lambda_2=(i)(-i)=1=\det A$ . The purely imaginary eigenvalues $\pm i$ encode pure oscillation — the hallmark of a rotation — and are visible only over $\mathbb{C}$ .

Physics of the portrait

The eigenvalue problem was born in the sky. Long before matrices existed, astronomers studying how the planets slowly perturb one another's orbits met a determinant equation — the "secular equation" — whose roots governed whether the Solar System was stable, and it was the analysis of that equation that first isolated what we now call eigenvalues.

Origins · 1 minAugustin-Louis Cauchy and how this was discovered

Augustin-Louis Cauchy

1789–1857 · the secular equation and real spectra of symmetric matrices

Augustin-Louis Cauchy, the most prolific analyst of the nineteenth century, gave the eigenvalue problem its first general theorem. The problem itself was older and thoroughly physical: in the eighteenth century Euler had reduced the free rotation of a rigid body to finding the principal axes of its inertia — an eigenvalue problem in disguise — and Lagrange and Laplace, studying the "secular" (slow) perturbations of planetary orbits, had arrived at a determinant equation whose roots decided the long-term stability of the Solar System. That équation séculaire was the characteristic equation before anyone knew to call it one, and its roots could in principle have been complex, which would have signalled runaway instability. In an 1829 memoir Cauchy proved the decisive fact that they cannot be: a real symmetric matrix has only real eigenvalues, and its principal axes are mutually orthogonal. This was, in modern language, the spectral theorem for symmetric operators — the guarantee that the inertia tensor, the stress tensor, and later every quantum observable have real spectra and orthogonal eigenvectors. The vocabulary settled slowly around him. James Joseph Sylvester spoke of "latent roots" in 1883; only in 1904 did David Hilbert, working on integral equations, coin the German Eigenwert — "proper value" — that gives us "eigenvalue" today. Cauchy did not have the word, but he had the theorem that made the word matter, turning a fragile question about planetary orbits into a permanent structural fact about symmetric operators.

History

1743

— d'Alembert reduces a system of coupled masses to equations whose special frequencies are eigenvalues

1751

— Euler finds the principal axes of a rotating rigid body, an eigenvalue problem for the inertia tensor

1829

— Cauchy proves a real symmetric matrix has only real eigenvalues, from the planetary secular equation

1858

— Cayley founds matrix algebra and states the Cayley–Hamilton theorem

1904

— Hilbert coins the term Eigenwert in his theory of integral equations

Going further

Texts. Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 3 (matrix eigenvalue problems) · Stone & Goldbart, Mathematics for Physics, Ch. on linear operators and their spectra · Hassani, Mathematical Physics, chapters on spectral decomposition · Byron & Fuller, Mathematics of Classical and Quantum Physics, Ch. 3–4 · Strang, Linear Algebra and Its Applications, Ch. 5 · Horn & Johnson, Matrix Analysis, Ch. 1. Key numerical anchors for all results: $p(\lambda)=\lambda^2-4\lambda+3$ for $\begin{pmatrix}2&1\\1&2\end{pmatrix}$ gives $\lambda=1,3$ with $\operatorname{tr}=4$ , $\det=3$ and orthogonal eigenvectors $(1,-1),(1,1)$ ; the coupled-oscillator eigenvalues are $\Lambda=k,3k$ giving $\omega=\sqrt{k/m},\sqrt{3k/m}$ (with $k=m=1$ , $\omega=1,\sqrt3\approx1.732$ ); the rotation $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ has $\lambda=\pm i$ with $\operatorname{tr}=0$ , $\det=1$ ; the defective shear $\begin{pmatrix}2&1\\0&2\end{pmatrix}$ has $m=2$ , $g=1$ ; and similarity invariance follows from determinant multiplicativity.

Prerequisites. Linear Operators and Matrix Representations · Determinants and Their Properties · Rank, Nullity, and the Fundamental Theorem of Linear Maps · Linear Independence, Basis, and Dimension. Leads to. Diagonalization and the Spectral Decomposition · Jordan Canonical Form and Defective Matrices · Adjoint and Hermitian Operators · Change of Basis and Similarity Transformations.

Ask the tutor: Ask me to walk through the characteristic-equation computation for any matrix you have in mind, or to show why a Hermitian operator always has real eigenvalues and orthogonal eigenvectors.

Why must an eigenvector be nonzero?

Because the zero vector satisfies $A\mathbf 0=\lambda\mathbf 0$ for every scalar $\lambda$ whatsoever. If $\mathbf 0$ counted as an eigenvector, every number would qualify as an eigenvalue and the concept would carry no information. The whole point of an eigenvalue is that it labels a genuine direction the operator preserves, so we demand $v\neq 0$ ; the eigenvalues are then exactly the scalars $\lambda$ for which $\ker(A-\lambda I)$ contains something besides zero. The eigenspace $E_\lambda$ does include $\mathbf 0$ — it must, to be a subspace — but $\mathbf 0$ is never called an eigenvector.

Can a matrix of real numbers have complex eigenvalues?

Yes, and it happens whenever the operator rotates. The characteristic polynomial of a real matrix has real coefficients, but real polynomials can have complex roots, which then arrive in conjugate pairs. The clean example is the $90^\circ$ rotation $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ , whose eigenvalues are $\pm i$ : no real direction survives a quarter-turn unrotated, so no real eigenvector exists, and the eigenvectors live in $\mathbb{C}^2$ . Physically the imaginary part of an eigenvalue signals oscillation — in a dynamical system $\dot{\mathbf x}=A\mathbf x$ a complex pair $\alpha\pm i\beta$ produces spirals of frequency $\beta$ and growth or decay set by $\alpha$ . Only special classes are guaranteed real spectra: Hermitian and symmetric operators, by the spectral theorem.

What is the difference between algebraic and geometric multiplicity?

Algebraic multiplicity $m_\lambda$ is how many times $\lambda$ appears as a root of the characteristic polynomial — a bookkeeping count from the algebra. Geometric multiplicity $g_\lambda=\dim\ker(A-\lambda I)$ is how many independent eigenvectors the eigenvalue actually delivers — a count from the geometry. They always obey $1\le g_\lambda\le m_\lambda$ , but the inequality can be strict: the shear $\begin{pmatrix}2&1\\0&2\end{pmatrix}$ has $m=2$ yet $g=1$ . When $g_\lambda=m_\lambda$ for every eigenvalue the operator has a full eigenbasis and diagonalizes; when some $g_\lambda<m_\lambda$ it is defective and the missing directions are supplied by generalized eigenvectors in the Jordan canonical form. The distinction is the single most important subtlety in the subject.

How can I check candidate eigenvalues without finding eigenvectors?

Use the two invariants that fall out of the characteristic polynomial for free. The eigenvalues must sum to the trace, $\sum_k\lambda_k=\operatorname{tr}A$ , and multiply to the determinant, $\prod_k\lambda_k=\det A$ . Both are one-line computations from the matrix entries — the trace is just the diagonal sum — so they catch most arithmetic slips instantly. For the symmetric example, $1+3=4=\operatorname{tr}A$ and $1\cdot3=3=\det A$ confirm the spectrum without touching a nullspace. For larger matrices there are further checks: the sum of the $2\times2$ principal minors equals the sum of pairwise products $\sum_{i<j}\lambda_i\lambda_j$ , and so on down the coefficients of $p(\lambda)$ . Trace and determinant are simply the first and last of these.

What exactly makes a matrix diagonalizable?

A matrix is diagonalizable when its eigenvectors span the whole space — equivalently, when the geometric multiplicity equals the algebraic multiplicity for every eigenvalue. Then the eigenvectors form a basis, stacking them into a matrix $S$ gives $S^{-1}AS=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$ , and the operator acts as pure scaling in those coordinates. Two convenient sufficient conditions cover most cases: having $n$ distinct eigenvalues guarantees diagonalizability (distinct eigenvalues force independent eigenvectors), and being Hermitian or symmetric guarantees it even with repeats (the spectral theorem provides an orthonormal eigenbasis). Diagonalization fails only for defective matrices, where a repeated eigenvalue is short of eigenvectors; those are handled by the Jordan form instead.

How does the eigenvalue problem connect to quantum mechanics?

Directly and literally. In quantum mechanics every observable is a Hermitian operator, its possible measured values are exactly its eigenvalues, and the states of definite value are its eigenvectors. The time-independent Schrödinger equation $\hat H\psi=E\psi$ is an eigenvalue problem word for word, with the allowed energies $E$ the eigenvalues of the Hamiltonian and the stationary states $\psi$ its eigenvectors. Cauchy's theorem is what makes this consistent: because $\hat H$ is Hermitian, its eigenvalues are real, so measured energies are real numbers, and eigenstates of different energies are orthogonal, so they represent physically distinguishable states. The same structure — Hermitian operator, real spectrum, orthogonal eigenbasis — underlies angular momentum, spin, and every other observable, which is why linear-algebra eigenanalysis is the mathematical spine of the whole theory.