Diagonalizability and Its Criteria

Suppose you are given a linear operator on a finite-dimensional vector space. You are asked to compute its hundredth power. Written in an arbitrary basis, the matrix is an opaque grid of numbers. Hundred-fold matrix multiplication is then a nightmare of arithmetic. But suppose you could find a basis in which the operator acts simply. In that basis, it just scales each basis vector. Then the hundredth power would just scale each vector by the hundredth power of its scalar. This is the promise of diagonalization. It replaces an incomprehensible matrix with a list of numbers on a diagonal. The cost is viewing the space through a cleverer lens. Diagonalizability is the question of when such a lens exists. Not every operator admits one. The failure is not a technicality but a genuine structural obstruction. This is already visible in 2×22\times 2 examples. Understanding exactly when a basis of eigenvectors exists is essential. So is having several equivalent ways to check for it. This is one of the load-bearing results of linear algebra. It underlies the spectral theorem and the theory of Markov chains. It also underlies the classification of quadratic forms and differential equation solving. Finally, it underlies the entire apparatus of eigenvalue-based numerical methods. The big picture is this: an operator is diagonalizable exactly when its space decomposes as a Direct Sum of Eigenspaces. This single geometric fact has several algebraic faces. These include a dimension count comparing Geometric vs Algebraic Multiplicity. They also include a polynomial condition and a basis condition. Part of the power of the theory is that these faces are logically equivalent. Whichever is easiest to check in a given problem may be used. This article gives the precise definition of diagonalizability. It proves the central equivalence theorem linking eigenbases, eigenspace decomposition, and multiplicity equality. It also works through asymmetric examples where multiplicities must actually be computed, rather than read off. It isolates, via a counterexample, exactly why algebraic multiplicity strictly exceeding geometric multiplicity is fatal. Finally, it points forward to the polynomial refinement of this theory. This includes the Minimal Polynomial criterion and simultaneous diagonalization of commuting operators.

Key takeaways
  • A linear operator is diagonalizable if some basis consists entirely of its eigenvectors.
  • Equivalently, the space is the direct sum of the eigenspaces of the operator.
  • Equivalently, every eigenvalue's geometric multiplicity equals its algebraic multiplicity, and the characteristic polynomial splits (factors into linear factors) over the field.
  • Distinct eigenvalues always give linearly independent eigenvectors — this is what makes the eigenspace sum direct.
  • A defective eigenvalue (geometric multiplicity below algebraic multiplicity) is the only way diagonalizability can fail once the characteristic polynomial splits.
  • Key idea: diagonalizability is precisely the statement that eigenvectors alone suffice to fill up the whole space.

Where this lives today

Diagonalization is the computational engine behind solving linear systems of ordinary differential equations: writing x˙=Ax\dot{x} = Ax with AA diagonalizable as A=QDQ−1A = QDQ^{-1} reduces the system to nn decoupled scalar equations y˙i=λiyi\dot y_i = \lambda_i y_i via y=Q−1xy = Q^{-1}x, each solved by yi(t)=yi(0)eλity_i(t) = y_i(0)e^{\lambda_i t}. In statistics and data science, Principal Component Analysis diagonalizes a covariance matrix (always possible by the spectral theorem, a special case of this theory for symmetric matrices) to find directions of maximal variance. In graph theory and network science, the diagonalizability of the adjacency or Laplacian matrix underlies spectral clustering and PageRank-style algorithms, where eigenvector bases reveal community structure. In quantum mechanics, diagonalizing a Hamiltonian operator gives the stationary energy states of a system directly as its eigenbasis. In numerical linear algebra, diagonalizability (or its failure) governs the stability and convergence behavior of iterative eigenvalue algorithms such as the QR algorithm and power iteration, and Markov chain long-run behavior is analyzed by diagonalizing the transition matrix to compute high matrix powers efficiently. The theory foreshadows two deeper refinements: Simultaneous Diagonalization, which asks when a whole commuting family of operators can share one eigenbasis, and the Minimal Polynomial criterion, which recasts diagonalizability as the minimal polynomial having no repeated roots — a test that avoids computing individual eigenspace dimensions altogether.

Ask the tutor: Why does having a repeated eigenvalue not automatically break diagonalizability, and what exactly must be checked instead?

DefinitionDiagonalizable operator

Let VV be a finite-dimensional vector space over a field F\mathbb{F}, with dim⁡V=n≥1\dim V = n \geq 1, and let T:V→VT : V \to V be a linear operator. Say TT is diagonalizable if there exists a basis β=(v1,…,vn)\beta = (v_1, \dots, v_n) of VV such that each viv_i is an eigenvector of TT; that is, for each ii there is a scalar λi∈F\lambda_i \in \mathbb{F} with

Tvi=λivi.T v_i = \lambda_i v_i.

Equivalently, TT is diagonalizable if the matrix [T]β[T]_\beta of TT with respect to some basis β\beta of VV is a diagonal matrix. For a square matrix A∈Fn×nA \in \mathbb{F}^{n\times n}, say AA is diagonalizable if the operator T=LAT = L_A (left multiplication by AA) on Fn\mathbb{F}^n is diagonalizable; equivalently, AA is diagonalizable if there is an invertible matrix Q∈Fn×nQ \in \mathbb{F}^{n\times n} and a diagonal matrix DD with A=QDQ−1A = QDQ^{-1}.

Intuition

Picture a linear operator as a machine that takes vectors in and produces vectors out. In an arbitrary basis, this machine looks like it shuffles and blends coordinates in an inscrutable way. An eigenvector is a direction the machine treats specially: it doesn't rotate or mix that direction with any other, it merely rescales it by the eigenvalue. If you are lucky enough to find nn linearly independent such special directions in an nn-dimensional space, you can use them as your coordinate axes. In these axes the machine's action becomes transparent — each axis is stretched or shrunk independently, with no mixing at all. That is exactly what a diagonal matrix records. The obstruction to this dream is when eigenvectors are simply too scarce. An eigenvalue λ\lambda might have a large algebraic multiplicity (it appears as a repeated root of the characteristic polynomial) while its eigenspace — the actual geometric collection of vectors it scales — is thin, of smaller dimension. When this happens the machine has a "shear" component in that generalized direction that no single scaling factor can capture; no basis of eigenvectors can exist, because there simply are not enough eigenvectors to go around.

Figure
Direct sum of eigenspaces filling the space, versus a defective eigenvalue leaving a gap
Case A: diagonalizable, dim V = 4 E(λ₁), dim = 2 E(λ₂), dim = 1 E(λ₃), dim = 1 V = E(λ₁) ⊕ E(λ₂) ⊕ E(λ₃), 2+1+1 = 4 = dim V Case B: defective, dim V = 4 E(λ₁), geo mult = 1 gap alg mult(λ₁) = 3 > geo mult(λ₁) = 1: no eigenbasis Sum of eigenspaces has dim < 4
TheoremDiagonalizability Criterion

Let VV be a finite-dimensional vector space over a field F\mathbb{F} with dim⁡V=n\dim V = n, and let T:V→VT: V \to V be linear. Let λ1,…,λk\lambda_1, \dots, \lambda_k be the distinct eigenvalues of TT lying in F\mathbb{F}, with eigenspaces E(λi)=ker⁡(T−λiI)E(\lambda_i) = \ker(T - \lambda_i I). Let mim_i denote the algebraic multiplicity of λi\lambda_i (its multiplicity as a root of the characteristic polynomial χT\chi_T) and di=dim⁡E(λi)d_i = \dim E(\lambda_i) its geometric multiplicity. The following are equivalent: 1. TT is diagonalizable. 2. V=E(λ1)⊕E(λ2)⊕⋯⊕E(λk)V = E(\lambda_1) \oplus E(\lambda_2) \oplus \cdots \oplus E(\lambda_k). 3. ∑i=1kdi=n\displaystyle \sum_{i=1}^{k} d_i = n. 4. χT\chi_T splits into linear factors over F\mathbb{F}, i.e. χT(x)=±∏i=1k(x−λi)mi\chi_T(x) = \pm\prod_{i=1}^k (x-\lambda_i)^{m_i} with ∑mi=n\sum m_i = n, and di=mid_i = m_i for every ii.

LemmaIndependence of eigenvectors from distinct eigenvalues

Let T:V→VT: V \to V be linear and let λ1,…,λk\lambda_1, \dots, \lambda_k be distinct eigenvalues of TT with corresponding eigenvectors v1,…,vkv_1, \dots, v_k (so vi≠0v_i \ne 0 and Tvi=λiviTv_i = \lambda_i v_i). Then v1,…,vkv_1, \dots, v_k are linearly independent. Consequently, the sum E(λ1)+⋯+E(λk)E(\lambda_1) + \cdots + E(\lambda_k) is a direct sum.

Proof

Technique: strong induction on the number of vectors, using the operator to kill terms. We prove the eigenvector statement by induction on kk. Base case k=1k=1: a single nonzero vector v1v_1 is linearly independent by definition (a set {v1}\{v_1\} is dependent only if v1=0v_1 = 0 or some nontrivial combination vanishes, impossible for a lone nonzero vector). Inductive step: suppose the claim holds for k−1k-1 distinct eigenvalues, and suppose

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

for scalars ci∈Fc_i \in \mathbb{F}. Apply TT to both sides:

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

Multiply (∗)(*) by λk\lambda_k and subtract from (∗∗)(**):

c1(λ1−λk)v1+c2(λ2−λk)v2+⋯+ck−1(λk−1−λk)vk−1=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} = 0.

This is a linear relation among v1,…,vk−1v_1, \dots, v_{k-1}, eigenvectors for the k−1k-1 distinct eigenvalues λ1,…,λk−1\lambda_1, \dots, \lambda_{k-1}. By the inductive hypothesis these are linearly independent, so every coefficient vanishes:

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

Since the λi\lambda_i are distinct, λi−λk≠0\lambda_i - \lambda_k \ne 0 for i<ki < k, forcing ci=0c_i = 0 for i=1,…,k−1i=1,\dots,k-1. Substituting back into (∗)(*) gives ckvk=0c_k v_k = 0, and since vk≠0v_k \ne 0, ck=0c_k = 0 as well. Hence all coefficients vanish and v1,…,vkv_1,\dots,v_k are linearly independent, completing the induction. For the "consequently" clause: recall that a sum of subspaces W1+⋯+WkW_1 + \cdots + W_k is direct iff the only way to write 0=w1+⋯+wk0 = w_1 + \cdots + w_k with wi∈Wiw_i \in W_i is wi=0w_i = 0 for all ii. If wi∈E(λi)w_i \in E(\lambda_i) are not all zero in such a relation, discard the zero ones; the surviving wiw_i are eigenvectors for distinct eigenvalues satisfying a nontrivial dependence, contradicting the independence just proved. Hence the sum is direct. ■\blacksquare

∎
Proof

Proof of the Diagonalizability Criterion. Technique: cyclic chain of implications (1) ⇒ (2) ⇒ (3) ⇒ (1), then (3) ⇔ (4).

(1) ⇒\Rightarrow (2). Suppose TT is diagonalizable, so VV has a basis β=(v1,…,vn)\beta = (v_1,\dots,v_n) of eigenvectors. Group the basis vectors by eigenvalue: for each distinct eigenvalue λi\lambda_i (i=1,…,ki=1,\dots,k) appearing among the Tvj=λjvjTv_j = \lambda_j v_j, let Si⊆βS_i \subseteq \beta be the subset of basis vectors with eigenvalue λi\lambda_i. Since β\beta spans VV, every v∈Vv \in V is a linear combination of vectors from S1∪⋯∪SkS_1 \cup \cdots \cup S_k, hence v=w1+⋯+wkv = w_1 + \cdots + w_k with wi∈span(Si)⊆E(λi)w_i \in \mathrm{span}(S_i) \subseteq E(\lambda_i). So V=E(λ1)+⋯+E(λk)V = E(\lambda_1) + \cdots + E(\lambda_k). By the Lemma: Independence of eigenvectors from distinct eigenvalues just proved, this sum is direct (any nontrivial relation among nonzero elements of distinct eigenspaces would contradict independence of eigenvectors for distinct eigenvalues). Hence V=E(λ1)⊕⋯⊕E(λk)V = E(\lambda_1)\oplus\cdots\oplus E(\lambda_k).

(2) ⇒\Rightarrow (3). If VV is the direct sum of the E(λi)E(\lambda_i), then by the standard dimension formula for a direct sum of finitely many subspaces (dimension of a direct sum is the sum of dimensions — proved by concatenating bases of the summands and checking the concatenation is a basis of the sum, which uses precisely the "directness" to get linear independence and spans from the spanning property of each piece),

n=dim⁡V=∑i=1kdim⁡E(λi)=∑i=1kdi.n = \dim V = \sum_{i=1}^k \dim E(\lambda_i) = \sum_{i=1}^k d_i.

(3) ⇒\Rightarrow (1). Suppose ∑di=n\sum d_i = n. For each ii choose a basis Bi\mathcal{B}_i of E(λi)E(\lambda_i), so ∣Bi∣=di|\mathcal{B}_i| = d_i. Let β=B1∪⋯∪Bk\beta = \mathcal{B}_1 \cup \cdots \cup \mathcal{B}_k; this is a list of ∑di=n\sum d_i = n eigenvectors of TT. We claim β\beta is linearly independent. Suppose a linear combination of vectors of β\beta vanishes; group the terms by which Bi\mathcal{B}_i they come from, obtaining w1+⋯+wk=0w_1 + \cdots + w_k = 0 with wi∈E(λi)w_i \in E(\lambda_i) (a linear combination of vectors from Bi\mathcal{B}_i, hence lying in span(Bi)⊆E(λi)\mathrm{span}(\mathcal B_i) \subseteq E(\lambda_i)). Since eigenspaces for distinct eigenvalues sum directly by the Lemma above (independence of eigenvectors gives directness of any subfamily of eigenspaces exactly as argued in (1)⇒\Rightarrow(2)), each wi=0w_i = 0. But wi=0w_i = 0 is a linear combination of the linearly independent set Bi\mathcal{B}_i equaling zero, forcing all its coefficients to be zero. Hence all coefficients in the original combination are zero, so β\beta is linearly independent. A linearly independent list of n=dim⁡Vn = \dim V vectors in VV is a basis of VV (standard fact: any linearly independent list of size dim⁡V\dim V is automatically spanning, since one can extend a linearly independent set to a basis and a basis of VV has exactly dim⁡V\dim V elements, leaving no room to add more without creating dependence). Thus β\beta is a basis of VV consisting of eigenvectors of TT, so TT is diagonalizable.

(3) ⇔\Leftrightarrow (4). First, recall the general multiplicity inequality (a standard fact assumed as prerequisite, Geometric vs Algebraic Multiplicity): for every eigenvalue λi\lambda_i of TT,

1≤di≤mi.1 \le d_i \le m_i.

Also recall that the characteristic polynomial χT\chi_T has degree exactly nn, and if it does not split over F\mathbb{F} then the sum of algebraic multiplicities of the eigenvalues lying in F\mathbb{F} is strictly less than nn (some irreducible factor of degree ≥2\ge 2 over F\mathbb{F} remains, contributing no eigenvalue in F\mathbb{F}). (⇐\Leftarrow) Suppose χT\chi_T splits and di=mid_i = m_i for all ii. Splitting gives ∑imi=n\sum_i m_i = n (the degree of χT\chi_T, distributed entirely among the linear factors). Combined with di=mid_i = m_i, we get ∑idi=∑imi=n\sum_i d_i = \sum_i m_i = n, which is (3). (⇒\Rightarrow) Suppose ∑idi=n\sum_i d_i = n. Using di≤mid_i \le m_i termwise,

n=∑i=1kdi≤∑i=1kmi≤n,n = \sum_{i=1}^k d_i \le \sum_{i=1}^k m_i \le n,

where the last inequality holds because ∑imi\sum_i m_i is the total multiplicity of all roots of χT\chi_T lying in F\mathbb{F}, which cannot exceed deg⁡χT=n\deg \chi_T = n regardless of splitting. Since the two ends of this chain are both nn, equality holds throughout: ∑mi=n\sum m_i = n (forcing χT\chi_T to split completely, as every root is accounted for among λ1,…,λk\lambda_1,\dots,\lambda_k with total multiplicity equal to the full degree, leaving no leftover non-linear factor) and, since di≤mid_i \le m_i for each ii individually with ∑di=∑mi\sum d_i = \sum m_i, none of these individual inequalities can be strict (a sum of nonnegative "slacks" mi−di≥0m_i - d_i \ge 0 that totals zero forces each slack to be zero). Hence di=mid_i = m_i for every ii, giving (4). This closes the chain (1) ⇒\Rightarrow (2) ⇒\Rightarrow (3) ⇔\Leftrightarrow (4) ⇒\Rightarrow (3) ⇒\Rightarrow (1), so all four statements are equivalent. ■\blacksquare

∎

Worked examples

Example 1 (asymmetric 3×33\times 3 matrix, distinct eigenvalues forced by computation). Let

A=(411242114),A∈R3×3.A = \begin{pmatrix} 4 & 1 & 1 \\ 2 & 4 & 2 \\ 1 & 1 & 4 \end{pmatrix}, \qquad A \in \mathbb{R}^{3\times 3}.

Compute the characteristic polynomial χA(x)=det⁡(xI−A)\chi_A(x) = \det(xI - A):

xI−A=(x−4−1−1−2x−4−2−1−1x−4).xI - A = \begin{pmatrix} x-4 & -1 & -1 \\ -2 & x-4 & -2 \\ -1 & -1 & x-4 \end{pmatrix}.

Expand along the first row:

det⁡=(x−4)[(x−4)2−2]−(−1)[−2(x−4)−2]+(−1)[2+(x−4)].\det = (x-4)\big[(x-4)^2 - 2\big] - (-1)\big[-2(x-4) - 2\big] + (-1)\big[2 + (x-4)\big].

Compute each bracket. First, (x−4)2−2=x2−8x+14(x-4)^2 - 2 = x^2 - 8x + 14, so the first term is (x−4)(x2−8x+14)=x3−12x2+46x−56(x-4)(x^2-8x+14) = x^3 - 12x^2 + 46x - 56 (multiplying out: (x−4)(x2−8x+14)=x3−8x2+14x−4x2+32x−56=x3−12x2+46x−56(x-4)(x^2 - 8x+14) = x^3 - 8x^2+14x -4x^2+32x-56 = x^3 -12x^2+46x-56). Second term: −(−1)[−2(x−4)−2]=1⋅(−2x+8−2)=−2x+6-(-1)\big[-2(x-4)-2\big] = 1\cdot(-2x+8-2) = -2x+6. Third term: (−1)[2+x−4]=(−1)(x−2)=−x+2(-1)[2 + x - 4] = (-1)(x-2) = -x+2. Sum:

χA(x)=(x3−12x2+46x−56)+(−2x+6)+(−x+2)=x3−12x2+43x−48.\chi_A(x) = (x^3 - 12x^2+46x-56) + (-2x+6) + (-x+2) = x^3 - 12x^2 + 43x - 48.

Test x=3x=3: 27−108+129−48=027 - 108 + 129 - 48 = 0. So x=3x=3 is a root. Divide:

x3−12x2+43x−48=(x−3)(x2−9x+16).x^3 - 12x^2+43x-48 = (x-3)(x^2 - 9x + 16).

The quadratic x2−9x+16x^2-9x+16 has discriminant 81−64=1781 - 64 = 17, giving roots x=9±172x = \frac{9 \pm \sqrt{17}}{2}, both real and irrational, and distinct from 33 and from each other since 1717 is not a perfect square. So AA has three distinct real eigenvalues: λ1=3\lambda_1 = 3, λ2=9+172\lambda_2 = \frac{9+\sqrt{17}}{2}, λ3=9−172\lambda_3 = \frac{9-\sqrt{17}}{2}, each with algebraic multiplicity 11. By the multiplicity inequality, geometric multiplicity is always at least 11 and at most algebraic multiplicity, so each di=1=mid_i = 1 = m_i. Condition (4) of the theorem holds (χA\chi_A splits over R\mathbb{R} and every di=mid_i = m_i), so AA is diagonalizable over R\mathbb{R}, with

A=QDQ−1,D=diag(3,9+172,9−172),A = QDQ^{-1}, \qquad D = \mathrm{diag}\left(3, \tfrac{9+\sqrt{17}}{2}, \tfrac{9-\sqrt{17}}{2}\right),

for some invertible QQ whose columns are eigenvectors of AA for these three eigenvalues respectively. Note we did not need to compute a single eigenvector to conclude diagonalizability — distinctness of roots was enough, illustrating the corollary that any operator with nn distinct eigenvalues in F\mathbb{F} on an nn-dimensional space is automatically diagonalizable.

Example 2 (repeated eigenvalue where geometric multiplicity must be checked directly). Let

B=(5−3−13−1−16−60)∈R3×3.B = \begin{pmatrix} 5 & -3 & -1 \\ 3 & -1 & -1 \\ 6 & -6 & 0 \end{pmatrix} \in \mathbb{R}^{3\times 3}.

Compute χB(x)=det⁡(xI−B)\chi_B(x) = \det(xI - B):

xI−B=(x−531−3x+11−66x).xI - B = \begin{pmatrix} x-5 & 3 & 1 \\ -3 & x+1 & 1 \\ -6 & 6 & x \end{pmatrix}.

Expand along the first row:

det⁡=(x−5)[(x+1)x−6]−3[(−3)x−(−6)]+1[(−18)−(−6)(x+1)].\det = (x-5)\big[(x+1)x - 6\big] - 3\big[(-3)x - (-6)\big] + 1\big[(-18) - (-6)(x+1)\big].

First bracket: (x+1)x−6=x2+x−6(x+1)x - 6 = x^2+x-6, so term one is (x−5)(x2+x−6)=x3+x2−6x−5x2−5x+30=x3−4x2−11x+30(x-5)(x^2+x-6) = x^3 + x^2 - 6x -5x^2 -5x+30 = x^3 -4x^2 -11x+30. Second term: −3[−3x+6]=9x−18-3[-3x+6] = 9x - 18. Third term: 1[−18+6x+6]=6x−121[-18 +6x+6] = 6x - 12. Sum:

χB(x)=(x3−4x2−11x+30)+(9x−18)+(6x−12)=x3−4x2+4x+0=x3−4x2+4x.\chi_B(x) = (x^3-4x^2-11x+30) + (9x-18) + (6x-12) = x^3 - 4x^2 + 4x + 0 = x^3 - 4x^2 + 4x.

Factor: χB(x)=x(x2−4x+4)=x(x−2)2\chi_B(x) = x(x^2 - 4x+4) = x(x-2)^2. So the eigenvalues are λ1=0\lambda_1 = 0 with algebraic multiplicity 11, and λ2=2\lambda_2 = 2 with algebraic multiplicity 22. Since m1=1m_1 = 1, automatically d1=1d_1 = 1. The interesting question is whether d2=2d_2 = 2. Compute E(2)=ker⁡(B−2I)E(2) = \ker(B - 2I):

B−2I=(3−3−13−3−16−6−2).B - 2I = \begin{pmatrix} 3 & -3 & -1 \\ 3 & -3 & -1 \\ 6 & -6 & -2 \end{pmatrix}.

All three rows are scalar multiples of (3,−3,−1)(3,-3,-1) (row 2 equals row 1, row 3 is twice row 1), so rank(B−2I)=1\mathrm{rank}(B-2I) = 1. By rank–nullity,

d2=dim⁡ker⁡(B−2I)=3−1=2.d_2 = \dim \ker(B-2I) = 3 - 1 = 2.

So d2=2=m2d_2 = 2 = m_2. Combined with d1=1=m1d_1 = 1 = m_1 and the fact that χB\chi_B splits over R\mathbb{R}, condition (4) holds, so BB is diagonalizable. Let's exhibit the eigenbasis. Solving (B−2I)v=0(B-2I)v=0: the single independent equation is 3v1−3v2−v3=03v_1 - 3v_2 - v_3 = 0, i.e. v3=3v1−3v2v_3 = 3v_1 - 3v_2. Taking (v1,v2)=(1,0)(v_1,v_2)=(1,0) gives u1=(1,0,3)u_1 = (1,0,3); taking (v1,v2)=(0,1)(v_1,v_2)=(0,1) gives u2=(0,1,−3)u_2 = (0,1,-3). Check u1u_1: Bu1=(5⋅1+0−3,3⋅1+0−3,6⋅1+0+0)=(2,0,6)=2(1,0,3)Bu_1 = (5\cdot1 + 0 -3, 3\cdot1+0-3, 6\cdot1+0+0) = (2,0,6) = 2(1,0,3). ✓. Check u2u_2: Bu2=(0−0−(−1)⋅(−3)…)Bu_2 = (0-0-(-1)\cdot(-3)\ldots); more carefully, Bu2=(5(0)−3(1)−1(−3),3(0)−1(1)−1(−3),6(0)−6(1)+0(−3))=(−3+3,−1+3,−6)=(0,2,−6)=2(0,1,−3)Bu_2 = (5(0)-3(1)-1(-3), 3(0)-1(1)-1(-3), 6(0)-6(1)+0(-3)) = (-3+3, -1+3, -6) = (0,2,-6) = 2(0,1,-3). ✓. For λ1=0\lambda_1=0: solve Bv=0Bv=0. Row reduce BB: rows are (5,−3,−1),(3,−1,−1),(6,−6,0)(5,-3,-1), (3,-1,-1), (6,-6,0). From row 3: 6v1−6v2=0⇒v1=v26v_1 - 6v_2 = 0 \Rightarrow v_1 = v_2. Substitute into row 2: 3v1−v1−v3=0⇒2v1=v33v_1 - v_1 - v_3 = 0 \Rightarrow 2v_1 = v_3. Check row 1: 5v1−3v1−2v1=05v_1 - 3v_1 - 2v_1 = 0 ✓ consistent. So eigenvector u3=(1,1,2)u_3 = (1,1,2). Check: Bu3=(5−3−2,3−1−2,6−6+0)=(0,0,0)Bu_3 = (5-3-2, 3-1-2, 6-6+0) = (0,0,0). ✓. So β=((1,0,3),(0,1,−3),(1,1,2))\beta = \big((1,0,3),(0,1,-3),(1,1,2)\big) is a basis of R3\mathbb{R}^3 of eigenvectors (independence follows since u1,u2u_1,u_2 span E(2)E(2), a 2-dim space, and u3∉E(2)u_3 \notin E(2) as it has a different eigenvalue, so together they are independent by the Lemma), and with Q=(1010113−32)Q = \begin{pmatrix}1&0&1\\0&1&1\\3&-3&2\end{pmatrix},

Q−1BQ=(200020000).Q^{-1}BQ = \begin{pmatrix} 2&0&0\\0&2&0\\0&0&0 \end{pmatrix}.
CounterexampleDropping "geometric multiplicity equals algebraic multiplicity": a defective matrix

The theorem's condition (4) requires χT\chi_T to split and di=mid_i = m_i for every eigenvalue. Drop the equality part and keep only "χT\chi_T splits" — does the operator remain diagonalizable? Consider

N=(2102)∈R2×2.N = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix} \in \mathbb{R}^{2\times 2}.

Its characteristic polynomial is χN(x)=(x−2)2\chi_N(x) = (x-2)^2, which splits over R\mathbb{R} into linear factors — so the "splitting" half of hypothesis (4) holds. The only eigenvalue is λ=2\lambda = 2 with algebraic multiplicity m=2m = 2. Compute the eigenspace: N−2I=(0100)N - 2I = \begin{pmatrix}0&1\\0&0\end{pmatrix}, whose kernel is {(v1,0):v1∈R}\{(v_1,0) : v_1 \in \mathbb{R}\}, one-dimensional. So the geometric multiplicity is d=1<m=2d=1 < m=2. Because d<md < m, condition (4) fails (the equality di=mid_i=m_i is violated), so by the theorem NN is not diagonalizable — and indeed ∑di=1≠2=dim⁡V\sum d_i = 1 \ne 2 = \dim V, violating (3) directly: there is no basis of R2\mathbb{R}^2 consisting of eigenvectors of NN, since every eigenvector of NN is a scalar multiple of (1,0)(1,0), and one cannot extend a single direction to a 2-dimensional basis while staying inside eigenvectors of NN. This shows that splitting of the characteristic polynomial alone is not sufficient; the multiplicity-equality clause is doing indispensable work, and cannot be dropped from hypothesis (4).

Scope, edge cases & common traps

SituationWhat happens
All eigenvalues distinct (nn of them, n=dim⁡Vn=\dim V)Automatically diagonalizable (Example 1); no need to compute eigenspaces
Repeated eigenvalue, but its eigenspace still has full dimensionDiagonalizable (Example 2); must verify via rank of T−λIT-\lambda I
Repeated eigenvalue with a smaller eigenspace (defective)Not diagonalizable (counterexample above)
Characteristic polynomial fails to split over F\mathbb{F} (e.g. real matrix with genuinely complex eigenvalues)Not diagonalizable over F\mathbb{F}; may become diagonalizable after extending scalars to C\mathbb{C}
TT is the identity or a scalar multiple of itDiagonalizable trivially: every vector is already an eigenvector
Nilpotent operator T≠0T \ne 0 with Tk=0T^k = 0Never diagonalizable (its only eigenvalue is 00, but if diagonalizable it would force T=0T=0)

The mathematician behind it

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

Cauchy did not use the modern vocabulary of "diagonalizable operator," but his systematic study of characteristic equations and the algebraic manipulation of what are now called eigenvalues (his term was closer to "racines de l'équation séculaire," roots of the secular equation, from celestial mechanics) laid the groundwork that Jacobi, Cayley, Sylvester, and later Frobenius sharpened into the modern spectral vocabulary. Cauchy's 1829 work on quadratic forms established that a real symmetric matrix's characteristic roots are real, an early ancestor of the diagonalizability theory for self-adjoint operators.

History
- 1829 — Cauchy studies the characteristic equation of a quadratic form and shows its roots are real for symmetric matrices, an early diagonalizability-adjacent result.
- 1858 — Cayley's memoir on matrix theory introduces matrix algebra abstractly, including matrix multiplication and the Cayley–Hamilton theorem's precursor ideas.
- 1878 — Frobenius develops the theory of elementary divisors, giving precise multiplicity-based criteria for similarity to diagonal or near-diagonal (Jordan) form.
- 1892 — Weierstrass and Jordan's earlier (1870) canonical form work becomes standard, clarifying exactly which non-diagonalizable matrices look like via Jordan blocks.
- 1904–1907 — Hilbert's work on integral equations extends eigenvalue/eigenvector language to infinite-dimensional operators, foreshadowing spectral theory beyond finite dimensions.
What is the difference between "diagonalizable" and "has real eigenvalues"?

These are independent properties. A matrix can have all real eigenvalues yet still fail to be diagonalizable if some eigenvalue is defective, meaning its geometric multiplicity is strictly less than its algebraic multiplicity — the shear matrix (2102)\begin{pmatrix}2&1\\0&2\end{pmatrix} is the standard example, with the single real eigenvalue 22 but only a one-dimensional eigenspace. Conversely, a matrix can be diagonalizable over the complex numbers without having any real eigenvalues, such as a rotation matrix, which is diagonalizable over C\mathbb{C} but not over R\mathbb{R}. Diagonalizability is about whether eigenvectors — real or complex, matching the field of scalars in use — are numerous enough to form a full basis, which is a strictly stronger requirement than merely having eigenvalues in the field at all.

Is every symmetric real matrix diagonalizable?

Yes, and this is one of the most important special cases of the theory, known as the spectral theorem for symmetric matrices. Every real symmetric matrix has only real eigenvalues, and for every eigenvalue the geometric multiplicity equals the algebraic multiplicity, so condition (4) of the diagonalizability criterion automatically holds. In fact more is true: a real symmetric matrix is diagonalizable by an orthogonal matrix, meaning the eigenbasis can be chosen orthonormal. This stronger orthogonal diagonalizability is proved using the inner product structure and is a genuinely separate (though related) theorem built on top of the general diagonalizability criterion discussed here.

If I only know the characteristic polynomial splits, can I conclude diagonalizability?

No — splitting alone is necessary but not sufficient. The counterexample (2102)\begin{pmatrix}2&1\\0&2\end{pmatrix} has characteristic polynomial (x−2)2(x-2)^2, which splits completely over R\mathbb{R}, yet the matrix is not diagonalizable because the single eigenvalue 22 has geometric multiplicity only 11 while its algebraic multiplicity is 22. You must additionally check, for every repeated eigenvalue, that the eigenspace dimension matches the multiplicity as a root of the characteristic polynomial. Only when both the splitting condition and the multiplicity-equality condition hold simultaneously is diagonalizability guaranteed.

How does diagonalizability relate to the minimal polynomial?

There is a clean equivalent criterion using the Minimal Polynomial: an operator TT on a finite-dimensional space is diagonalizable if and only if its minimal polynomial factors into distinct linear factors, with no repeated roots. This is often faster to check than computing every eigenspace dimension by hand, especially when the minimal polynomial is already known or easy to compute from a recurring pattern in powers of the matrix. The two criteria — eigenspace dimension counting and minimal polynomial factorization — are proved equivalent using the theory of invariant subspaces and are two windows onto the same underlying structural fact.

Can a diagonalizable operator become non-diagonalizable after a change of basis, or vice versa?

No. Diagonalizability is a property of the linear operator itself (or, for matrices, of the similarity class), not of any particular matrix representation. If [T]β[T]_\beta is diagonalizable for one basis β\beta, then for any other basis γ\gamma, the matrix [T]γ[T]_\gamma is similar to [T]β[T]_\beta via the change-of-basis matrix, and similar matrices always share the same eigenvalues, eigenspace dimensions, and characteristic polynomial. So [T]γ[T]_\gamma satisfies exactly the same diagonalizability criteria as [T]β[T]_\beta. Diagonalizability is a similarity invariant, which is precisely why the definition is stated for the operator TT and not tied to any single matrix.

Does diagonalizability depend on the field, e.g. real versus complex?

Yes, critically. A matrix that is not diagonalizable over R\mathbb{R} can become diagonalizable when the same matrix is viewed over C\mathbb{C}, because the characteristic polynomial may fail to split over R\mathbb{R} but always splits over C\mathbb{C} by the Fundamental Theorem of Algebra. A rotation matrix (0−110)\begin{pmatrix}0&-1\\1&0\end{pmatrix} has characteristic polynomial x2+1x^2+1, irreducible over R\mathbb{R}, so it is not diagonalizable over R\mathbb{R}; but over C\mathbb{C} its roots ±i\pm i are distinct, so it becomes diagonalizable there. The multiplicity-equality condition, by contrast, is field-independent once the polynomial does split — it is only the splitting itself that depends on the field.

What is the fastest way to check diagonalizability on an exam without full computation?

First check whether all eigenvalues (roots of the characteristic polynomial) are distinct; if so, stop — the operator is automatically diagonalizable by the corollary of the independence lemma. If some eigenvalue repeats, you must dig further: compute the rank of T−λIT - \lambda I for that repeated eigenvalue and use rank–nullity to get its geometric multiplicity, then compare to the algebraic multiplicity read off from the characteristic polynomial's factorization. If they match for every repeated eigenvalue, and the polynomial splits, the operator is diagonalizable; a single mismatch anywhere is enough to kill it.

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Where this leads