Existence of Eigenvalues over C

Does every linear operator secretly contain a hidden direction it merely stretches or shrinks? Consider rotating the plane $\mathbb{R}^2$ by $90°$ . No nonzero real vector is sent to a multiple of itself. So as a real operator it has no eigenvalues at all. Yet the moment we reinterpret the same matrix as acting on $\mathbb{C}^2$ , eigenvalues appear — namely $\pm i$ . This article proves that this is not a coincidence but a theorem: every linear operator on a nonzero finite-dimensional complex vector space has at least one eigenvalue. This fact is one of the true load-bearing pillars of linear algebra. It is what makes complex vector spaces so much more hospitable to structure theory than real ones. It guarantees that operators on $\mathbb{C}^n$ always have an invariant line to start peeling off. This is exactly the seed from which triangularization, Jordan form, and the spectral theorem all grow. Nothing analogous is true over $\mathbb{R}$ . The entire subsequent theory of canonical forms for complex operators traces back to this single existence statement. The proof is a genuinely beautiful piece of mathematics. It takes a purely algebraic object — a linear operator on a finite-dimensional space. It produces an eigenvalue using the analytic/algebraic fact that every nonconstant polynomial over $\mathbb{C}$ has a root (the Fundamental Theorem of Algebra). The bridge between these two worlds is built from a simple but clever dimension-counting argument. This argument is applied to the powers of a vector under the operator. This article states the theorem with full hypotheses. It proves the theorem completely from the Fundamental Theorem of Algebra and basic facts about Polynomials applied to operators. It illustrates the phenomenon with two worked examples. It shows exactly which hypothesis (algebraic closedness of the scalar field) cannot be dropped.

Key takeaways

Every operator $T$ on a nonzero finite-dimensional complex vector space $V$ has an eigenvalue.
The proof feeds the vectors $v, Tv, T^2v, \dots, T^n v$ (where $n = \dim V$ ) into a linear dependence argument.
Linear dependence of these $n+1$ vectors produces a nonzero polynomial that "kills" $v$ under $T$ .
Factoring that polynomial over $\mathbb{C}$ (Fundamental Theorem of Algebra) forces $T - \lambda I$ to be non-injective for some $\lambda$ .
This fails completely over $\mathbb{R}$ : rotation by $90°$ on $\mathbb{R}^2$ has no eigenvalue.
Key idea: algebraic closedness of $\mathbb{C}$ turns "some polynomial in $T$ kills a vector" into "some $\lambda I - T$ kills a vector," i.e., an eigenvalue.

Where this lives today

This existence theorem is the entry point to the entire theory of canonical forms for complex operators. It guarantees the invariant line needed to begin an inductive proof of Triangularization of Operators (Schur's theorem): find one eigenvector, restrict $T$ to a complementary quotient space, and repeat by induction on dimension — a strategy that would collapse instantly if eigenvalues could fail to exist. It also underlies why the The Characteristic Polynomial of a complex matrix always has a root, tying the operator-theoretic existence proof given here to the determinant-based approach. Beyond pure linear algebra, this fact is the reason quantum mechanics is formulated with complex Hilbert spaces: physical observables (Hermitian operators) always have real eigenvalues precisely because they act on complex spaces where eigenvalues are guaranteed to exist and are additionally constrained to be real by self-adjointness. In numerical linear algebra and engineering (control theory, vibration analysis, PageRank-style Markov chain analysis), the guaranteed existence of complex eigenvalues underlies the entire apparatus of eigenvalue decomposition algorithms, since real matrices are routinely analyzed by extending scalars to $\mathbb{C}$ to guarantee a full spectrum. In algebraic geometry and representation theory, this same closure phenomenon (every operator has an eigenvector over an algebraically closed field) is a baseline tool for decomposing group representations and studying module structure over polynomial rings.

Ask the tutor: How does this existence-of-eigenvalues argument extend into the inductive proof that every complex operator has an upper-triangular matrix with respect to some basis?

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Intuition

Think of an operator $T$ on a finite-dimensional space $V$ as a dynamical rule: start with a vector $v$ , and repeatedly apply $T$ to generate a sequence $v, Tv, T^2v, T^3v, \dots$ . If $\dim V = n$ , then any $n+1$ vectors in $V$ must be linearly dependent — there simply is not enough "room" for them to be independent. So the vectors $v, Tv, \dots, T^nv$ (that's $n+1$ vectors) must satisfy some nontrivial linear relation. A linear relation among these vectors is exactly a statement that some nonzero polynomial $p$ (of degree at most $n$ ) satisfies $p(T)v = 0$ . This is purely algebraic and uses nothing about $\mathbb{C}$ — it works over any field. The magic happens next: because $\mathbb{C}$ is algebraically closed, $p$ factors completely into linear factors,

p(z) = c(z - \lambda_1)(z - \lambda_2)\cdots(z - \lambda_m), \qquad c \neq 0.

Substituting $T$ for $z$ , the operator $p(T)$ factors as a product of the commuting operators $(T - \lambda_1 I), \dots, (T - \lambda_m I)$ . Since this product kills the nonzero vector $v$ , at least one of the factors must fail to be injective (if they were all injective, their composition would be injective too, and could not send a nonzero vector to $0$ ). That failing factor $T - \lambda_j I$ has nontrivial kernel — which is precisely an eigenvector for eigenvalue $\lambda_j$ . The picture below shows the chain of vectors collapsing into a dependency, and the resulting polynomial factoring down to reveal a single eigenvalue.

Figure

From a dependent chain of vectors to a factored polynomial revealing an eigenvalue

The central theorem

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v \in V

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Worked examples

Example 1: A $3\times 3$ non-diagonal, asymmetric matrix. Let $V = \mathbb{C}^3$ and let $T$ be given by the matrix

A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \end{pmatrix}.

This is upper triangular, so we already suspect its eigenvalues are its diagonal entries $2, 2, 3$ — but let's verify the existence proof's mechanism rather than shortcut to that fact, using a vector for which the dependency isn't obvious in advance. Take $v = e_1 = (1,0,0)$ . Compute the chain:

v = (1,0,0), \qquad Tv = A e_1 = (2, 0, 0), \qquad T^2 v = A(2,0,0) = (4,0,0).

Already $Tv = 2v$ , so $v$ itself is an eigenvector for $\lambda = 2$ ! The dependency shows up already at length $2$ (well before the guaranteed bound of $n+1 = 4$ vectors): $Tv - 2v = 0$ , giving the annihilating polynomial $p(z) = z - 2$ , already linear, requiring no further factoring. To see a case where more work is genuinely needed, take instead $v = e_3 = (0,0,1)$ :

v = (0,0,1), \quad Tv = (0,1,3), \quad T^2 v = A(0,1,3) = (1, 5, 9).

These three vectors ( $n = 3$ , so we're allowed up to $n+1=4$ ) are linearly independent — one can check the matrix with rows $v, Tv, T^2v$ ,

\begin{pmatrix} 0&0&1\\0&1&3\\1&5&9\end{pmatrix},

has determinant $-1 \neq 0$ (expanding along the first column: $-1 \cdot \det\begin{pmatrix}0&1\\1&3\end{pmatrix} = -1(0\cdot 3 - 1 \cdot 1) = 1$ ; sign from cofactor position gives $-1$ times that, i.e. $-1$ ), so we must go one step further to $T^3 v$ :

T^3 v = A(1,5,9) = (2\cdot1 + 1\cdot5,\ 2\cdot5+1\cdot9,\ 3\cdot9) = (7, 19, 27).

Now $v, Tv, T^2v, T^3v$ is a list of $4$ vectors in $\mathbb{C}^3$ , forced to be dependent. Solving $a_0 v + a_1 Tv + a_2 T^2 v + a_3 T^3 v = 0$ coordinate-by-coordinate:

\begin{aligned} \text{(coord 1)}&: a_2 + 7a_3 = 0\\ \text{(coord 2)}&: a_1 + 5a_2 + 19a_3 = 0\\ \text{(coord 3)}&: a_0 + 3a_1 + 9a_2 + 27a_3 = 0 \end{aligned}

Set $a_3 = 1$ (we know a nontrivial solution exists; normalize the free parameter). Then $a_2 = -7$ . Then $a_1 = -5(-7) - 19 = 35 - 19 = 16$ . Then $a_0 = -3(16) - 9(-7) - 27 = -48 + 63 - 27 = -12$ . So $p(z) = -12 + 16z - 7z^2 + z^3$ annihilates $v$ (one checks $p(T)v=0$ using the coordinate relations just derived). Since $A$ is upper triangular with diagonal $2,2,3$ , its characteristic polynomial is $(z-2)^2(z-3) = z^3 - 7z^2 + 16z - 12$ , matching $p(z)$ exactly — confirming $p$ factors as

p(z) = (z-2)(z-2)(z-3),

so $p(T) = (T-2I)^2(T-3I)$ , and since $p(T)v = 0$ with $v \neq 0$ , at least one factor is non-injective. Indeed $\ker(T - 2I) = \operatorname{span}\{(1,0,0)\}$ and $\ker(T-3I) = \operatorname{span}\{(1,2,1)\}$ (solving $(A-3I)w=0$ : $-w_1+w_2=0,\,-w_2+w_3=0$ gives $w=(1,1,1)$ up to scale — recompute: $-w_1 + w_2 = 0 \Rightarrow w_2 = w_1$ ; $-w_2+w_3=0 \Rightarrow w_3=w_2$ ; so $w=(1,1,1)$ ), both nonzero, exhibiting eigenvalues $2$ and $3$ concretely.

Example 2: A real rotation, reinterpreted over $\mathbb{C}$ . Let $T$ be the operator on $\mathbb{C}^2$ given by the rotation matrix

A = \begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix},

i.e. rotation by $90°$ . Over $\mathbb{R}$ this operator sends every nonzero real vector to a vector perpendicular to it, so no real eigenvector exists. We verify the theorem's mechanism produces a genuine complex eigenvalue. Take $v = e_1 = (1,0) \in \mathbb{C}^2$ , $n=2$ . Compute

Tv = (0,1), \qquad T^2 v = A(0,1) = (-1,0) = -v.

So already at length $3 \le n+1 = 3$ , the vectors $v, Tv, T^2v$ are dependent: $T^2 v + v = 0$ , giving the annihilating polynomial

p(z) = z^2 + 1,

of degree exactly $n=2$ (the maximum allowed). Factor over $\mathbb{C}$ :

p(z) = (z - i)(z+i).

Then $p(T) = (T - iI)(T+iI)$ and $p(T)v = 0$ . Check directly whether each factor is injective: solve $(T - iI)w = 0$ , i.e. $Aw = iw$ . Writing $w = (w_1, w_2)$ :

-w_2 = i w_1, \qquad w_1 = i w_2.

The first equation gives $w_2 = -iw_1$ ; substituting into $w_1 = i w_2 = i(-iw_1) = w_1$ , an identity, so $w_1$ is free. Taking $w_1 = 1$ , $w_2 = -i$ , we get the eigenvector

w = (1, -i), \qquad Tw = A(1,-i) = (0\cdot1 + (-1)(-i),\ 1\cdot 1 + 0\cdot(-i)) = (i, 1) = i(1,-i) = iw.

Indeed $Tw = iw$ , confirming $\lambda = i$ is an eigenvalue with eigenvector $(1,-i)$ — a genuinely complex phenomenon invisible over $\mathbb{R}$ , exactly as the theorem promises.

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Scope, edge cases & common traps

Situation	Does the theorem apply?	Why
$T$ on nonzero finite-dim $\mathbb{C}^n$	Yes	exactly the hypotheses
$T$ on nonzero finite-dim $\mathbb{R}^n$	Not guaranteed	$\mathbb{R}$ not algebraically closed (see counterexample)
$T$ on the zero space $\{0\}$	Vacuously excluded	no nonzero vector exists to be an eigenvector; theorem explicitly assumes $V \neq \{0\}$
$T$ on infinite-dimensional $\mathbb{C}$ -space	Not covered	the pigeonhole dependence argument needs finite dimension; e.g. the shift/derivative operators may have empty spectrum in some infinite-dim settings
$T$ on finite-dim space over an arbitrary algebraically closed field $\mathbb{F}$	Yes	Corollary — the proof only used algebraic closure and finite dimension
$T$ is the zero operator on $\mathbb{C}^n$ , $n\ge1$	Yes, eigenvalue $0$	every nonzero vector satisfies $Tv = 0 = 0\cdot v$

The mathematician behind it

Origins · 1 minJean-Robert Argand and Carl Friedrich Gauss and how this was discovered

Jean-Robert Argand and Carl Friedrich Gauss

1768–1822 (Argand), 1777–1855 (Gauss) · Fundamental Theorem of Algebra and the geometry of ℂ

While the linear-algebraic packaging of this theorem (as a statement about operators, via the pigeonhole dimension-counting argument) is a modern development — closely associated with the style of proof popularized in 20th-century linear algebra texts — its analytic heart is the Fundamental Theorem of Algebra, whose first widely accepted rigorous proofs are due to Carl Friedrich Gauss, who gave four increasingly rigorous proofs between 1799 and 1849. Jean-Robert Argand, a self-taught Swiss-French mathematician, gave in 1806 an influential geometric proof and, independently of Gauss and Wessel, popularized the geometric representation of complex numbers as points in the plane — the very picture that makes "every operator has an eigenvalue over $\mathbb{C}$ " feel geometrically inevitable rather than a strange coincidence of algebra.

History

- 1799 — Gauss's doctoral dissertation gives the first widely circulated (though topologically incomplete) proof of the Fundamental Theorem of Algebra.

- 1806 — Argand publishes a geometric proof of the Fundamental Theorem of Algebra and develops the complex plane picture.

- 1816 — Gauss publishes a second, more rigorous proof of the Fundamental Theorem of Algebra.

- 1849 — Gauss gives a fourth proof, allowing complex coefficients, closing remaining gaps in rigor.

- 20th century — the polynomial/pigeonhole style proof of existence of eigenvalues via annihilating polynomials becomes the standard modern linear-algebra treatment, replacing older determinant-based derivations.

Why does the proof need finite-dimensionality?

Finite-dimensionality is what guarantees that any $n+1$ vectors in $V$ must be linearly dependent — this is the pigeonhole step producing a nonzero annihilating polynomial for a chosen vector $v$ . In infinite dimensions, the vectors $v, Tv, T^2v, \dots$ can all be linearly independent (this happens, for instance, with certain shift operators), so no finite polynomial relation need exist, and the argument simply has no starting point. Infinite-dimensional operator theory has its own separate machinery (spectral theory on Banach or Hilbert spaces) to discuss eigenvalues and more general spectral values, but it is a genuinely different and harder subject.

Is the eigenvalue produced by the theorem unique?

No — the theorem only asserts existence of at least one eigenvalue, not uniqueness. Many operators have several distinct eigenvalues (up to $n = \dim V$ of them, counted without multiplicity), and the specific eigenvalue produced by the proof can depend on which starting vector $v$ was chosen, since different vectors can be annihilated by different polynomials. What is always true is that the set of all eigenvalues (the spectrum) is nonempty and finite, with at most $n$ distinct values.

Does this theorem also work for infinite lists of vectors or general linear maps between different spaces?

It is specifically about a linear operator, meaning a linear map from $V$ to itself — eigenvalues are meaningless for maps between different spaces $T: V \to W$ with $V \ne W$ , since $Tv = \lambda v$ would not even typecheck unless $V=W$ . The theorem also genuinely requires $V$ finite-dimensional; for infinite-dimensional spaces, one instead studies the "spectrum" of an operator using tools from functional analysis, and the spectrum can be empty, all of $\mathbb{C}$ , or something in between depending on the operator.

How does this relate to the characteristic polynomial approach to eigenvalues?

The characteristic polynomial approach defines eigenvalues as roots of $\det(T - \lambda I) = 0$ , then invokes the Fundamental Theorem of Algebra directly on that specific polynomial to get a root. The proof given here is more constructive and conceptual: it builds a possibly different annihilating polynomial directly from the dynamics of $T$ on a single vector, without ever computing a determinant. Both routes ultimately rely on the same algebraic fact (existence of roots over $\mathbb{C}$ ), but the vector-chain proof generalizes more easily to infinite-dimensional contexts involving minimal polynomials and is the standard modern approach in operator-first treatments of linear algebra.

Why does the argument fail if we only assume the annihilating polynomial has degree exactly 1?

It doesn't fail — that is actually the easiest case, immediately giving $Tv = \lambda v$ directly. The subtlety the proof handles is the opposite: the annihilating polynomial found by the pigeonhole argument might have degree as large as $n$ , and a priori factors into as many as $n$ linear pieces. The nontrivial content of the proof is showing that even in that worse case, some single linear factor $T - \lambda_j I$ must still be non-injective, which is where the composition-of-injective-maps argument is essential.

Does every complex matrix actually have a real trace or determinant related to these eigenvalues?

Not necessarily real, but always meaningful: for an $n\times n$ complex matrix, the sum of all eigenvalues (counted with multiplicity) always equals the trace, and the product of all eigenvalues (with multiplicity) always equals the determinant, regardless of whether the eigenvalues themselves are real. This existence theorem is what guarantees there are always exactly $n$ eigenvalues (with multiplicity) to sum and multiply in the first place — a fact made precise later via The Characteristic Polynomial.

What role does commutativity of powers of T play in the proof?

It is used twice: first, to identify the linear combination $a_0 v + a_1 Tv + \cdots$ with $p(T)v$ for a specific polynomial $p$ (routine from the definition of $p(T)$ ), and second, and more importantly, to justify that the factored form $c(T-\lambda_1 I)\cdots(T-\lambda_m I)$ can be composed in any order and still equal $p(T)$ . This relies on the fact that any two polynomials in the same operator $T$ commute with each other, since $T^j T^k = T^{j+k} = T^k T^j$ — a simple but essential structural fact.