Independence of Eigenvectors for Distinct Eigenvalues

Suppose a linear operator stretches one vector by a factor of 2. Suppose it stretches a completely different vector by a factor of 5. Must these two vectors point in genuinely different directions? Specifically, must they be linearly independent? This question sits at the heart of diagonalization. To build a basis of eigenvectors, we need many eigenvectors. These eigenvectors must not collapse onto each other's span. The remarkable fact is this: distinctness of eigenvalues alone forces this independence. This holds with no assumption about the dimension of the space. It also holds regardless of the field, or the diagonalizability of the operator. This result matters because it is the load-bearing lemma behind almost everything that follows in the theory of eigenvalues. It explains why an operator with nn distinct eigenvalues on an nn-dimensional space is automatically diagonalizable. It explains why eigenspaces for different eigenvalues only intersect at the zero vector. It also explains why the sum of eigenspaces is always a direct sum. Without this fact, the entire machinery of building eigenbases would have no foundation. The big picture is that eigenvalues partition the behavior of an operator into independent "modes." This theorem is the precise statement that these modes cannot secretly overlap. It converts an algebraic hypothesis into a geometric conclusion. The hypothesis is eigenvalues being numerically distinct. The conclusion is vectors spanning independent directions. This transfer from algebra to geometry is a recurring theme in linear algebra. Here, it appears in its cleanest form. This article gives the precise definitions. It offers a complete induction proof, due essentially to the classical treatment found in every rigorous linear algebra course. It includes a worked geometric picture. It also provides two fully computed examples over different fields and dimensions. Finally, it gives a counterexample showing exactly where the argument fails if eigenvalues coincide. It closes with the immediate consequences: eigenspaces for distinct eigenvalues intersect trivially, and their sum is direct. This sets up the Diagonalizability Criterion.

Key takeaways
  • If v1,…,vkv_1,\dots,v_k are eigenvectors of an operator TT for pairwise distinct eigenvalues λ1,…,λk\lambda_1,\dots,\lambda_k, then v1,…,vkv_1,\dots,v_k are linearly independent.
  • The proof is by strong induction on kk, using the operator TT itself to "kill" one eigenvalue at a time.
  • No hypothesis on dimension, field, or diagonalizability is needed — this holds for any vector space over any field.
  • Distinctness of eigenvalues is essential: two eigenvectors for the same eigenvalue can easily be dependent (e.g. equal up to scalar).
  • This lemma immediately implies eigenspaces for distinct eigenvalues intersect only at 00, and their sum is a direct sum.
  • It is the engine behind the theorem that nn distinct eigenvalues in dimension nn force diagonalizability.
  • Key idea: distinct eigenvalues cannot conspire to produce a nontrivial linear dependence, because applying TT to a dependence relation and subtracting a scaled copy strictly reduces the number of terms involved.

Where this lives today

The independence of eigenvectors for distinct eigenvalues is the load-bearing lemma for diagonalization: an n×nn \times n matrix with nn distinct eigenvalues has nn linearly independent eigenvectors, hence is diagonalizable, feeding directly into the Diagonalizability Criterion. It also underlies the fact that the Sum of Eigenspaces is Direct, a key structural fact used when eigenvalues repeat and eigenspaces must be assembled carefully. In differential equations, solving linear systems x˙=Ax\dot{x} = Ax decomposes the solution into modes eλitvie^{\lambda_i t} v_i; the independence theorem guarantees these modes do not secretly overlap, so the general solution is a genuine linear combination of independent exponential behaviors. In quantum mechanics, observables are represented by self-adjoint operators, and distinct eigenvalues (distinct measurement outcomes) correspond to orthogonal — in particular independent — eigenstates, a physical manifestation of this algebraic fact (strengthened to orthogonality via the Spectral Theorem in the self-adjoint case). In Google's PageRank and other Markov-chain analyses, distinct eigenvalues of the transition matrix guarantee a full set of independent eigenvectors used to analyze long-run behavior. In numerical linear algebra, algorithms such as the power method and QR algorithm rely on distinctness (or well-separation) of eigenvalues to guarantee convergence to independent eigendirections, and ill-conditioning of eigenvector computations is often traced to eigenvalues that are close together, nearly violating the hypothesis of this theorem.

Ask the tutor: How does this independence lemma get strengthened to orthogonality when the operator is symmetric or self-adjoint?

DefinitionEigenvectors, eigenvalues, and linear independence

Let VV be a vector space over a field F\mathbb{F}, and let T:V→VT : V \to V be a linear operator. A scalar λ∈F\lambda \in \mathbb{F} is an eigenvalue of TT if there exists a nonzero vector v∈Vv \in V such that

T(v)=λv.T(v) = \lambda v.

Any such nonzero vv is called an eigenvector of TT corresponding to λ\lambda. A finite list of vectors v1,…,vk∈Vv_1, \dots, v_k \in V is linearly independent if the only scalars c1,…,ck∈Fc_1,\dots,c_k \in \mathbb{F} satisfying

c1v1+c2v2+⋯+ckvk=0c_1 v_1 + c_2 v_2 + \cdots + c_k v_k = 0

are c1=c2=⋯=ck=0c_1 = c_2 = \cdots = c_k = 0. Scalars λ1,…,λk∈F\lambda_1, \dots, \lambda_k \in \mathbb{F} are pairwise distinct if λi≠λj\lambda_i \neq \lambda_j whenever i≠ji \neq j.

Intuition

Picture eigenvectors as arrows pointing along special directions that an operator merely rescales, without rotating. If two eigenvectors correspond to different scaling factors, intuition suggests they cannot point along a common line or plane spanned by the others — because if they did, applying TT would have to rescale that shared direction consistently, but the two different eigenvalues demand two different rescalings simultaneously. That tension is impossible unless the "overlap" vector is zero. The proof formalizes this tension using a subtraction trick: if a dependence relation among eigenvectors exists, apply TT to it (multiplying each term by its own eigenvalue), then subtract a suitably scaled copy of the original relation. This cancels one term completely, producing a shorter dependence relation among fewer eigenvectors — with eigenvalues still distinct. Iterating this collapses the problem down to the base case of a single nonzero eigenvector, which is trivially independent (as a list of one nonzero vector always is). The whole argument is a controlled shrinking process, the linear-algebra analogue of an infinite descent.

Figure
Two eigenvectors with distinct eigenvalues cannot be parallel or share a smaller dependency
v₁ (λ₁=2) v₂ (λ₂=5) If v₁, v₂ were dependent, T would force λ₁=λ₂ T(v) = λv stretches each eigenvector along its own line
TheoremEigenvectors for distinct eigenvalues are linearly independent

Let VV be a vector space over a field F\mathbb{F} and let T:V→VT : V \to V be a linear operator. Let λ1,λ2,…,λk∈F\lambda_1, \lambda_2, \dots, \lambda_k \in \mathbb{F} be pairwise distinct eigenvalues of TT, and let v1,v2,…,vk∈Vv_1, v_2, \dots, v_k \in V be corresponding eigenvectors, so

T(vi)=λivi,vi≠0,i=1,…,k.T(v_i) = \lambda_i v_i, \qquad v_i \neq 0, \qquad i = 1,\dots,k.

Then v1,v2,…,vkv_1, v_2, \dots, v_k are linearly independent.

Proof

We use strong induction on kk, the number of eigenvectors.

Base case (k=1k = 1). A list consisting of a single vector v1v_1 is linearly independent if and only if v1≠0v_1 \neq 0. By hypothesis v1v_1 is an eigenvector, hence nonzero by Eigenvalues and Eigenvectors. So the base case holds.

Inductive step. Fix k≥2k \geq 2 and assume the statement holds for every list of fewer than kk eigenvectors with pairwise distinct eigenvalues (this is the strong induction hypothesis, though here we will only need the case of exactly k−1k-1 vectors). We must show that v1,…,vkv_1, \dots, v_k, with pairwise distinct eigenvalues λ1,…,λk\lambda_1,\dots,\lambda_k, are linearly independent. Suppose, for contradiction, that they are linearly dependent. Then there exist scalars c1,…,ck∈Fc_1, \dots, c_k \in \mathbb{F}, not all zero, such that

c1v1+c2v2+⋯+ckvk=0.(1)c_1 v_1 + c_2 v_2 + \cdots + c_k v_k = 0. \tag{1}

Step 1: Apply TT to both sides. Since TT is linear and T(vi)=λiviT(v_i) = \lambda_i v_i,

T(c1v1+⋯+ckvk)=c1T(v1)+⋯+ckT(vk)=c1λ1v1+c2λ2v2+⋯+ckλkvk.T(c_1v_1 + \cdots + c_kv_k) = c_1 T(v_1) + \cdots + c_k T(v_k) = c_1\lambda_1 v_1 + c_2 \lambda_2 v_2 + \cdots + c_k \lambda_k v_k.

Since the left side of (1) is the zero vector and T(0)=0T(0) = 0, we obtain

c1λ1v1+c2λ2v2+⋯+ckλkvk=0.(2)c_1 \lambda_1 v_1 + c_2 \lambda_2 v_2 + \cdots + c_k \lambda_k v_k = 0. \tag{2}

Step 2: Eliminate the kk-th term by subtraction. Multiply equation (1) by the scalar λk\lambda_k:

c1λkv1+c2λkv2+⋯+ckλkvk=0.(3)c_1 \lambda_k v_1 + c_2 \lambda_k v_2 + \cdots + c_k \lambda_k v_k = 0. \tag{3}

Subtract (3) from (2):

c1(λ1−λk)v1+c2(λ2−λk)v2+⋯+ck−1(λk−1−λk)vk−1+ck(λk−λk)vk=0.c_1(\lambda_1 - \lambda_k) v_1 + c_2(\lambda_2 - \lambda_k)v_2 + \cdots + c_{k-1}(\lambda_{k-1} - \lambda_k) v_{k-1} + c_k(\lambda_k - \lambda_k)v_k = 0.

The last term vanishes since λk−λk=0\lambda_k - \lambda_k = 0, leaving

c1(λ1−λk)v1+c2(λ2−λk)v2+⋯+ck−1(λk−1−λk)vk−1=0.(4)c_1(\lambda_1 - \lambda_k) v_1 + c_2(\lambda_2 - \lambda_k) v_2 + \cdots + c_{k-1}(\lambda_{k-1} - \lambda_k) v_{k-1} = 0. \tag{4}

This is a linear dependence relation among only the first k−1k-1 eigenvectors v1,…,vk−1v_1,\dots,v_{k-1}, whose eigenvalues λ1,…,λk−1\lambda_1,\dots,\lambda_{k-1} are pairwise distinct (they are a sub-list of the originally distinct λ1,…,λk\lambda_1,\dots,\lambda_k).

Step 3: Apply the induction hypothesis. By the strong induction hypothesis applied to the list of k−1k-1 eigenvectors v1,…,vk−1v_1,\dots,v_{k-1} (with pairwise distinct eigenvalues λ1,…,λk−1\lambda_1,\dots,\lambda_{k-1}), these vectors are linearly independent. Since equation (4) is a linear dependence relation among them, every coefficient must vanish:

ci(λi−λk)=0for i=1,…,k−1.c_i(\lambda_i - \lambda_k) = 0 \qquad \text{for } i = 1, \dots, k-1.

Because λ1,…,λk\lambda_1,\dots,\lambda_k are pairwise distinct, we have λi≠λk\lambda_i \neq \lambda_k for each i=1,…,k−1i = 1,\dots,k-1, so λi−λk≠0\lambda_i - \lambda_k \neq 0 in the field F\mathbb{F}. Since F\mathbb{F} is a field (in particular has no zero divisors), ci(λi−λk)=0c_i(\lambda_i-\lambda_k)=0 forces

ci=0for i=1,…,k−1.c_i = 0 \qquad \text{for } i = 1, \dots, k-1.

Step 4: Conclude ck=0c_k = 0 as well. Substituting c1=⋯=ck−1=0c_1 = \cdots = c_{k-1} = 0 back into the original relation (1) gives

ckvk=0.c_k v_k = 0.

Since vkv_k is an eigenvector, vk≠0v_k \neq 0. In a vector space, ckvk=0c_k v_k = 0 with vk≠0v_k \neq 0 forces ck=0c_k = 0 (a standard consequence of the vector space axioms: if ck≠0c_k \neq 0 then vk=ck−1(ckvk)=ck−1⋅0=0v_k = c_k^{-1}(c_k v_k) = c_k^{-1} \cdot 0 = 0, a contradiction). Hence all coefficients c1,…,ckc_1, \dots, c_k are zero, contradicting our assumption that not all of them were zero. This contradiction shows that the dependence relation (1) with not-all-zero coefficients cannot exist, so v1,…,vkv_1, \dots, v_k are linearly independent. By strong induction, the theorem holds for every k≥1k \geq 1. ■\blacksquare

∎
CorollaryEigenspaces for distinct eigenvalues intersect trivially

Let T:V→VT : V \to V be linear, and let λ≠μ\lambda \neq \mu be two distinct eigenvalues of TT, with eigenspaces

Eλ={v∈V:T(v)=λv},Eμ={v∈V:T(v)=μv}.E_\lambda = \{v \in V : T(v) = \lambda v\}, \qquad E_\mu = \{v \in V : T(v) = \mu v\}.

Then Eλ∩Eμ={0}E_\lambda \cap E_\mu = \{0\}.

Proof

Suppose v∈Eλ∩Eμv \in E_\lambda \cap E_\mu with v≠0v \neq 0. Then vv is an eigenvector for λ\lambda and simultaneously an eigenvector for μ\mu, with λ≠μ\lambda \neq \mu. Applying the Independence of Eigenvectors for Distinct Eigenvalues theorem with k=2k=2 to the eigenvectors v1=vv_1 = v (for λ1=λ\lambda_1 = \lambda) and v2=vv_2 = v (for λ2=μ\lambda_2 = \mu) would force v1,v2v_1, v_2 to be linearly independent; but a list containing the same nonzero vector twice is always linearly dependent (take c1=1,c2=−1c_1 = 1, c_2 = -1), a contradiction. Hence no nonzero vv lies in both eigenspaces, so Eλ∩Eμ={0}E_\lambda \cap E_\mu = \{0\}. ■\blacksquare

∎

Worked examples

Example 1 (asymmetric real 3×33\times 3 matrix, three distinct eigenvalues). Let T:R3→R3T : \mathbb{R}^3 \to \mathbb{R}^3 be given by the matrix

A=(410030−212).A = \begin{pmatrix} 4 & 1 & 0 \\ 0 & 3 & 0 \\ -2 & 1 & 2 \end{pmatrix}.

This matrix is lower-triangular in a permuted sense but let's just compute directly. Its characteristic polynomial is

det⁡(A−xI)=det⁡(4−x1003−x0−212−x).\det(A - xI) = \det\begin{pmatrix} 4-x & 1 & 0 \\ 0 & 3-x & 0 \\ -2 & 1 & 2-x \end{pmatrix}.

Expanding along the third column (only the bottom-right entry 2−x2-x is nonzero there):

det⁡(A−xI)=(2−x)det⁡(4−x103−x)=(2−x)(4−x)(3−x).\det(A-xI) = (2-x)\det\begin{pmatrix}4-x & 1 \\ 0 & 3-x\end{pmatrix} = (2-x)(4-x)(3-x).

So the eigenvalues are λ1=2\lambda_1 = 2, λ2=4\lambda_2 = 4, λ3=3\lambda_3 = 3 — pairwise distinct. Eigenvector for λ1=2\lambda_1 = 2: solve (A−2I)v=0(A - 2I)v = 0:

A−2I=(210010−210).A - 2I = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 1 & 0 \\ -2 & 1 & 0 \end{pmatrix}.

Row 2 gives y=0y = 0. Row 1 gives 2x+y=0⇒x=02x + y = 0 \Rightarrow x = 0. Row 3 gives −2x+y=0-2x+y=0, consistent. Then zz is free. So

v1=(0,0,1).v_1 = (0,0,1).

Eigenvector for λ2=4\lambda_2 = 4: solve (A−4I)v=0(A-4I)v=0:

A−4I=(0100−10−21−2).A - 4I = \begin{pmatrix} 0 & 1 & 0 \\ 0 & -1 & 0 \\ -2 & 1 & -2 \end{pmatrix}.

Row 1 gives y=0y = 0. Row 3: −2x+y−2z=0⇒−2x−2z=0⇒x=−z-2x + y - 2z = 0 \Rightarrow -2x - 2z = 0 \Rightarrow x = -z. Let z=1z=1, so x=−1x = -1:

v2=(−1,0,1).v_2 = (-1, 0, 1).

Eigenvector for λ3=3\lambda_3 = 3: solve (A−3I)v=0(A-3I)v=0:

A−3I=(110000−21−1).A - 3I = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 0 & 0 \\ -2 & 1 & -1 \end{pmatrix}.

Row 1: x+y=0⇒y=−xx + y = 0 \Rightarrow y = -x. Row 3: −2x+y−z=0⇒−2x−x−z=0⇒z=−3x-2x + y - z = 0 \Rightarrow -2x - x - z = 0 \Rightarrow z = -3x. Let x=1x = 1:

v3=(1,−1,−3).v_3 = (1, -1, -3).

Now check independence directly (as the theorem guarantees, but let's verify by hand for concreteness). Form the matrix with these as columns:

M=(0−1100−111−3).M = \begin{pmatrix} 0 & -1 & 1 \\ 0 & 0 & -1 \\ 1 & 1 & -3 \end{pmatrix}.

Compute det⁡M\det M by expanding along the first column: only the entry in row 3 is nonzero there, contributing 1⋅det⁡(−110−1)1 \cdot \det\begin{pmatrix}-1 & 1 \\ 0 & -1\end{pmatrix} with sign (−1)3+1=+1(-1)^{3+1}=+1:

det⁡M=1⋅((−1)(−1)−(1)(0))=1⋅1=1≠0.\det M = 1 \cdot \big((-1)(-1) - (1)(0)\big) = 1 \cdot 1 = 1 \neq 0.

Since det⁡M≠0\det M \neq 0, the columns v1,v2,v3v_1, v_2, v_3 are linearly independent — exactly as the theorem predicted from the three distinct eigenvalues 2,3,42, 3, 4, with no computation of the determinant strictly necessary to know this in advance.

Example 2 (complex 2×22\times 2 matrix, distinct complex eigenvalues, illustrating the theorem over C\mathbb{C}). Let T:C2→C2T : \mathbb{C}^2 \to \mathbb{C}^2 be given by

B=(1−211).B = \begin{pmatrix} 1 & -2 \\ 1 & 1 \end{pmatrix}.

The characteristic polynomial is

det⁡(B−xI)=(1−x)2+2=x2−2x+3.\det(B - xI) = (1-x)^2 + 2 = x^2 - 2x + 3.

By the quadratic formula, x=2±4−122=1±i2x = \dfrac{2 \pm \sqrt{4 - 12}}{2} = 1 \pm i\sqrt{2}. So the eigenvalues are

λ1=1+i2,λ2=1−i2,\lambda_1 = 1 + i\sqrt{2}, \qquad \lambda_2 = 1 - i\sqrt{2},

which are distinct complex numbers (they differ by 2i2≠02i\sqrt2 \ne 0). Eigenvector for λ1\lambda_1: solve (B−λ1I)v=0(B - \lambda_1 I)v = 0, i.e.

(−i2−21−i2)(xy)=0.\begin{pmatrix} -i\sqrt2 & -2 \\ 1 & -i\sqrt2 \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = 0.

The second row gives x=i2 yx = i\sqrt2\, y. Take y=1y = 1, so

v1=(i2, 1).v_1 = (i\sqrt2,\ 1).

(Check first row: −i2(i2)−2(1)=−i2⋅2−2=2−2=0-i\sqrt2 (i\sqrt2) - 2(1) = -i^2 \cdot 2 - 2 = 2 - 2 = 0. Confirmed.) Eigenvector for λ2\lambda_2: by the conjugate structure (real matrix, conjugate eigenvalue), v2v_2 is the complex conjugate of v1v_1:

v2=(−i2, 1).v_2 = (-i\sqrt2,\ 1).

(Direct check: solve (B−λ2I)v=0(B-\lambda_2 I)v=0 with λ2=1−i2\lambda_2 = 1 - i\sqrt2: second row gives x=−i2 yx = -i\sqrt2\,y, matching.) Now verify independence: suppose c1v1+c2v2=0c_1 v_1 + c_2 v_2 = 0 in C2\mathbb{C}^2:

c1(i2,1)+c2(−i2,1)=(0,0).c_1(i\sqrt2, 1) + c_2(-i\sqrt2, 1) = (0,0).

Second coordinate: c1+c2=0⇒c2=−c1c_1 + c_2 = 0 \Rightarrow c_2 = -c_1. First coordinate: i2 c1−i2 c2=0⇒i2(c1−c2)=0⇒c1=c2i\sqrt2\, c_1 - i\sqrt2\, c_2 = 0 \Rightarrow i\sqrt2(c_1 - c_2) = 0 \Rightarrow c_1 = c_2 (since i2≠0i\sqrt2 \neq 0). Combined with c2=−c1c_2 = -c_1, we get c1=−c1⇒2c1=0⇒c1=0c_1 = -c_1 \Rightarrow 2c_1 = 0 \Rightarrow c_1 = 0, and then c2=0c_2 = 0. So v1,v2v_1, v_2 are linearly independent over C\mathbb{C}, confirming the theorem even though the eigenvalues and eigenvectors involve non-real complex numbers — the theorem's proof only used field axioms (existence of (λi−λk)−1(\lambda_i - \lambda_k)^{-1}), which hold equally well in C\mathbb{C}.

CounterexampleDropping distinctness: repeated eigenvalue can force dependence

The hypothesis that λ1,…,λk\lambda_1,\dots,\lambda_k are pairwise distinct is essential; without it the conclusion can fail. Consider T:R2→R2T : \mathbb{R}^2 \to \mathbb{R}^2 given by the identity matrix

I=(1001).I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.

Every nonzero vector in R2\mathbb{R}^2 is an eigenvector with eigenvalue λ=1\lambda = 1. Take

v1=(1,0),v2=(2,0).v_1 = (1, 0), \qquad v_2 = (2, 0).

Both satisfy T(vi)=1⋅viT(v_i) = 1\cdot v_i, so both are eigenvectors for the same eigenvalue λ1=λ2=1\lambda_1 = \lambda_2 = 1 — the eigenvalues are not pairwise distinct. Yet

2v1−v2=(2,0)−(2,0)=(0,0),2v_1 - v_2 = (2,0) - (2,0) = (0,0),

a nontrivial dependence relation (c1=2,c2=−1c_1 = 2, c_2 = -1, not both zero). So v1,v2v_1, v_2 are linearly dependent, even though both are genuine eigenvectors of TT. Tracing through the proof shows exactly where it breaks: in Step 3, we needed λi−λk≠0\lambda_i - \lambda_k \neq 0 to divide it out and conclude ci=0c_i = 0. With λ1=λ2\lambda_1 = \lambda_2, the quantity λ1−λ2=0\lambda_1 - \lambda_2 = 0, so equation (4) becomes the vacuous statement 0=00 = 0, giving no information about c1c_1 at all. The induction collapses and no contradiction is reached — as indeed no contradiction should be reached, since the vectors genuinely can be dependent.

Scope, edge cases & common traps

SituationDoes the theorem apply?Comment
Eigenvalues pairwise distinct, any field, any dimension (finite or infinite)YesProof uses only field axioms and linearity; no dimension bound needed.
Two eigenvectors share the same eigenvalueNo conclusion guaranteedThey may be dependent (e.g. scalar multiples) or independent (e.g. two vectors spanning a 2-dim eigenspace); the theorem is silent here.
Eigenvalues distinct but eigenvectors chosen as 00Vacuous / excludedEigenvectors are by definition nonzero, so this never arises for genuine eigenvectors.
Infinite list of eigenvectors for pairwise distinct eigenvaluesYes, applies to every finite sub-listLinear independence of an infinite list means every finite subset is independent, which follows from the theorem applied to each finite subset.
Field of characteristic ppStill holdsThe proof only needs λi−λk≠0⇒(λi−λk)−1\lambda_i - \lambda_k \neq 0 \Rightarrow (\lambda_i-\lambda_k)^{-1} exists, true in any field regardless of characteristic.

The mathematician behind it

Origins · 1 minAugustin-Louis Cauchy and how this was discovered
AC
Augustin-Louis Cauchy
1789–1857 · Early 19th-century French mathematical analysis and algebra

Cauchy's foundational work on determinants and quadratic forms in the 1820s–1840s laid much of the groundwork for the modern theory of eigenvalues (then called "characteristic roots"), including recognizing that distinct roots of the characteristic equation correspond to independent principal axes in problems of mechanics and the theory of surfaces. His investigations into the diagonalization of quadratic forms — reducing a form to a sum of squares along principal axes — are a direct ancestor of the modern eigenvector independence argument presented here.

History
- 1826: Cauchy studies "characteristic values" of quadratic forms arising from mechanics and elasticity problems.
- 1829: Cauchy proves that a real symmetric quadratic form can be diagonalized via what are now called principal axes, implicitly using the independence of eigenvectors for distinct eigenvalues.
- 1858: Arthur Cayley's memoir on matrix theory formalizes matrices as algebraic objects, setting the stage for the general operator formulation of eigenvalue theory.
- 1904: David Hilbert's work on integral equations extends eigenvalue theory to infinite-dimensional spaces, where the distinct-eigenvalue independence lemma remains essential.
- 1930s: The modern abstract vector-space formulation (Halmos, van der Waerden and others) states and proves the theorem exactly in the field-independent form given here.
Does the theorem require the vector space to be finite-dimensional?

No. The proof only uses linearity of TT, the field axioms (existence of inverses for nonzero scalars), and the vector space axioms (if cv=0cv=0 with v≠0v\neq0 then c=0c=0). None of these require finite dimension. The theorem therefore applies verbatim in infinite-dimensional spaces such as spaces of sequences or functions, where operators like differentiation or shift operators have eigenvectors (e.g. exponential functions eλxe^{\lambda x} for the derivative operator). Any finite collection of eigenvectors with pairwise distinct eigenvalues is linearly independent, regardless of the ambient space's dimension.

Does this theorem tell us anything when two eigenvectors share the same eigenvalue?

No — the theorem's hypothesis explicitly requires the eigenvalues to be pairwise distinct, so it simply does not apply to same-eigenvalue eigenvectors. In that case independence must be checked directly. Vectors sharing an eigenvalue can be dependent (scalar multiples of each other) or independent (if the eigenspace has dimension greater than one and they are chosen to span different directions within it). This is precisely the content explored further when studying the geometric multiplicity of an eigenvalue.

Why does the proof use strong induction rather than ordinary induction?

Strong induction is used because the inductive step invokes the statement for exactly k−1k-1 vectors, which is the immediately preceding case — so in fact ordinary induction (assuming only the (k−1)(k-1)-case) suffices here, and strong induction is simply a safe general framework that includes this as a special case. Some expositions phrase it as ordinary induction on kk for exactly this reason. Either phrasing produces the identical rigorous argument; the important structural feature is that each step reduces cleanly to the immediately smaller case via the elimination trick.

Is this theorem still true if TT is not diagonalizable?

Yes, completely independently of diagonalizability. The theorem concerns any finite list of eigenvectors with distinct eigenvalues, whether or not those eigenvectors happen to span the whole space VV. In fact, this theorem is a key ingredient used to prove the Diagonalizability Criterion (an operator with n=dim⁡Vn = \dim V distinct eigenvalues is diagonalizable), rather than a consequence of diagonalizability. It applies just as well to operators that are far from diagonalizable, such as those with repeated eigenvalues and deficient eigenspaces.

Can this theorem be extended to generalized eigenvectors?

The direct statement as given concerns eigenvectors in the strict sense (Tv=λvT v = \lambda v). A related but distinct fact holds for generalized eigenvectors (associated with Jordan chains): generalized eigenspaces for distinct eigenvalues also intersect trivially and their sum is direct, which is a deeper result proved using the structure of nilpotent operators on generalized eigenspaces, going beyond the elementary argument given here. That extension is typically developed alongside the Sum of Eigenspaces is Direct result and Jordan canonical form theory.

Does the argument depend on the vectors being real or complex?

No. The proof works over any field F\mathbb{F}, since it only uses that F\mathbb{F} has no zero divisors and that every nonzero element has a multiplicative inverse. This is illustrated explicitly in the second worked example, where the eigenvalues and eigenvectors are genuinely complex (non-real), and the identical proof mechanics apply. The theorem equally holds over finite fields such as Fp\mathbb{F}_p used in coding theory and cryptography.

What is the single most common mistake students make when applying this theorem?

The most common mistake is assuming the converse — believing that linearly independent eigenvectors must automatically come with distinct eigenvalues. This is false, as a diagonalizable operator with a repeated eigenvalue of multiplicity m>1m > 1 has an mm-dimensional eigenspace, and any basis of that eigenspace gives mm linearly independent eigenvectors all sharing one eigenvalue. Always check which direction of the implication you actually need before invoking this theorem.

Before this