Triangularization of Operators

Every matrix over the complex numbers can be simplified. But not every matrix can be diagonalized. A shear-like operator such as one sending e1↦0e_1 \mapsto 0 and e2↦e1e_2 \mapsto e_1 has no chance of being diagonal in any basis. It is not even invertible in the right way. It is nilpotent, and nilpotent operators other than 00 are never diagonalizable. So the natural question is: what is the best possible simplification available to every operator on a complex vector space? The answer is triangularization. This means finding a basis in which the matrix has zeros everywhere below the diagonal. This question matters because triangular matrices retain almost all the computational virtues of diagonal matrices. This holds even though they are less symmetric than diagonal ones. Their determinant and trace are visible by inspection. Their eigenvalues are exactly the diagonal entries. Polynomial expressions in them are easy to control. Triangularization is the structural fact that makes possible the later, sharper theorems of the subject. It is the raw material from which the Cayley-Hamilton Theorem and the theory of Generalized Eigenvectors are built. The big picture is that triangularization is a statement about chains of invariant subspaces, called flags. A single Invariant Subspace tells you an operator respects some subspace. A full flag — a nested sequence of invariant subspaces of every dimension from 00 up to nn — tells you the operator can be built up one dimension at a time. The existence of such a flag is guaranteed for every operator over C\mathbb{C} by the Existence of Eigenvalues over C. This is exactly equivalent to the existence of an upper-triangular matrix representation. This article gives the precise definition of triangularizability. It proves the fundamental existence theorem by induction on dimension. It works two concrete asymmetric examples by hand. It shows exactly why the complex numbers (or more precisely, splitting of the characteristic polynomial) cannot be dispensed with.

Key takeaways
  • An operator is triangularizable if some basis represents it by an upper-triangular matrix.
  • Equivalently, there is a flag {0}=V0⊂V1⊂⋯⊂Vn=V\{0\} = V_0 \subset V_1 \subset \cdots \subset V_n = V of invariant subspaces with dim⁡Vk=k\dim V_k = k.
  • Over C\mathbb{C} (or any algebraically closed field), every operator on a finite-dimensional space is triangularizable.
  • The diagonal entries of an upper-triangular representation are always exactly the eigenvalues of the operator, with multiplicity.
  • Triangularizability is strictly weaker than diagonalizability — nilpotent shifts are triangularizable but not diagonalizable.
  • Key idea: the existence of one eigenvector plus induction on the quotient space builds a full flag, hence an upper-triangular matrix.

Where this lives today

Triangularization is the load-bearing beam beneath much of matrix analysis. In numerical linear algebra, the Schur decomposition — writing A=QUQ−1A = Q U Q^{-1} with QQ unitary and UU upper triangular — is the numerically stable computational cousin of this theorem, avoiding the instability of computing eigenvector bases directly; it underlies most industrial-strength eigenvalue algorithms (the QR algorithm converges toward exactly such a triangular form). In control theory and systems engineering, upper-triangular state-space realizations reveal cascade structure: later states feed forward from earlier ones, exactly as the flag interpretation predicts. In algebra, triangularization is the first step toward the Cayley-Hamilton Theorem, whose cleanest proof triangularizes an operator and then evaluates its own characteristic polynomial diagonal-block by diagonal-block. It is also the scaffolding for the theory of Generalized Eigenvectors and Jordan form, since a triangular matrix is only one refinement away from block-diagonal Jordan structure. In quantum mechanics and representation theory, simultaneous triangularization of families of commuting operators (Lie's theorem for solvable Lie algebras) generalizes this single-operator result to whole algebras of operators acting compatibly on the same flag.

:::quiz[If [T]β[T]_\beta is upper triangular with diagonal entries 5,5,−25, 5, -2, what is det⁡(T)\det(T)?]

  • 88
  • −50-50 ✓
  • 5050
  • −2-2

:::quiz[A basis β=(v1,…,vn)\beta = (v_1,\dots,v_n) makes [T]β[T]_\beta upper triangular if and only if which condition holds for every jj?]

  • TvjTv_j is a scalar multiple of vjv_j alone
  • Tvj∈span⁡(vj,vj+1,…,vn)Tv_j \in \operatorname{span}(v_j, v_{j+1}, \dots, v_n)
  • Tvj∈span⁡(v1,…,vj)Tv_j \in \operatorname{span}(v_1, \dots, v_j) ✓
  • TvjTv_j has all coordinates equal in the basis β\beta

Ask the tutor: How does refining the invariant flag further, using generalized eigenvectors, upgrade a triangular matrix into full Jordan canonical form?

DefinitionUpper-triangular matrix and triangularizable operator

Let VV be a finite-dimensional vector space over a field F\mathbb{F} with dim⁡V=n≥1\dim V = n \ge 1, and let T:V→VT : V \to V be a linear operator.

Upper-triangular matrix. An n×nn \times n matrix A=(aij)A = (a_{ij}) is upper triangular if aij=0a_{ij} = 0 whenever i>ji > j; that is, every entry strictly below the main diagonal is 00.

Matrix of TT with respect to a basis. Given an ordered basis β=(v1,…,vn)\beta = (v_1, \dots, v_n) of VV, the matrix of TT with respect to β\beta, written [T]β[T]_\beta, is the matrix whose jj-th column records the coordinates of TvjTv_j in the basis β\beta:

Tvj=∑i=1naijvi.Tv_j = \sum_{i=1}^n a_{ij} v_i .

Triangularizable. TT is triangularizable if there exists an ordered basis β\beta of VV such that [T]β[T]_\beta is upper triangular. Equivalently — as will be proved — TT is triangularizable if and only if there is a chain of subspaces

{0}=V0⊊V1⊊⋯⊊Vn=V,dim⁡Vk=k,\{0\} = V_0 \subsetneq V_1 \subsetneq \cdots \subsetneq V_n = V, \qquad \dim V_k = k,

such that T(Vk)⊆VkT(V_k) \subseteq V_k for every k=0,1,…,nk = 0, 1, \dots, n. Such a chain is called a flag for TT, and it is invariant if every term VkV_k is a TT-Invariant Subspace.

Intuition

Think of an upper-triangular matrix as encoding a construction process rather than a static array of numbers. The equation Tvj=∑i≤jaijviTv_j = \sum_{i \le j} a_{ij} v_i says that TT sends vjv_j into the span of v1,…,vjv_1, \dots, v_j — never further out. So if you build VV one basis vector at a time, TT never needs information about vectors "later" in the list to describe where the "earlier" ones go. This is exactly the statement that the subspaces Vk=span⁡(v1,…,vk)V_k = \operatorname{span}(v_1, \dots, v_k) are nested and invariant: TT maps each stage of the construction into itself. Diagonalization asks for something stronger: that TT acts on each vjv_j by pure scaling, with no contribution from the other basis vectors at all. Triangularization only asks that TT not reach forward; it is allowed to reach backward, mixing vjv_j's image with all of v1,…,vj−1v_1, \dots, v_{j-1}. This laxity is precisely what makes triangularization achievable so generally, while diagonalization is not. The picture below shows the "staircase" support pattern of an upper-triangular matrix, and the flag of invariant subspaces it corresponds to.

Figure
An upper-triangular matrix and its associated flag of invariant subspaces
Matrix [T]_β (n = 4) a₁₁ a₁₂ a₁₃ a₁₄ 0 a₂₂ a₂₃ a₂₄ 0 0 a₃₃ a₃₄ 0 0 0 a₄₄ zeros below diagonal ⇒ Tv₁ ∈ span(v₁) Tv₂ ∈ span(v₁,v₂) Tv₃ ∈ span(v₁,v₂,v₃) Tv₄ ∈ span(v₁,v₂,v₃,v₄) Flag: {0} ⊂ V₁ ⊂ V₂ ⊂ V₃ ⊂ V₄ = V, dim Vₖ = k, T(Vₖ) ⊆ Vₖ

Second figure: eigenvalues visible on the diagonal

Once a basis triangularizes TT, the eigenvalues of TT can be read directly off the diagonal, because det⁡(T−λI)\det(T - \lambda I) factors as a product of the diagonal entries minus λ\lambda.

Figure
Reading eigenvalues off the diagonal of a triangular matrix
[T]_β − λI upper triangular ⇒ det = ∏ (aₖₖ − λ) 3−λ 5 −2 0 1−λ 7 0 0 4−λ det = (3−λ)(1−λ)(4−λ) eigenvalues: λ = 3, 1, 4 Off-diagonal entries never affect this determinant.
TheoremTriangularization Theorem over an algebraically closed field

Let F\mathbb{F} be an algebraically closed field (for instance F=C\mathbb{F} = \mathbb{C}), let VV be a vector space over F\mathbb{F} with 1≤dim⁡V=n<∞1 \le \dim V = n < \infty, and let T:V→VT : V \to V be a linear operator. Then: 1. There exists a flag {0}=V0⊊V1⊊⋯⊊Vn=V\{0\} = V_0 \subsetneq V_1 \subsetneq \cdots \subsetneq V_n = V of TT-invariant subspaces with dim⁡Vk=k\dim V_k = k for each kk. 2. Consequently, there exists an ordered basis β\beta of VV such that [T]β[T]_\beta is upper triangular. 3. Conversely, if some ordered basis β\beta of VV (over any field) makes [T]β[T]_\beta upper triangular with diagonal entries a11,…,anna_{11}, \dots, a_{nn}, then the characteristic polynomial of TT factors as

det⁡(T−λI)=∏k=1n(akk−λ),\det(T - \lambda I) = \prod_{k=1}^n (a_{kk} - \lambda),

so the multiset {a11,…,ann}\{a_{11}, \dots, a_{nn}\} is exactly the multiset of eigenvalues of TT counted with algebraic multiplicity.

Proof

Proof of (1) and (2), by strong induction on n=dim⁡Vn = \dim V. Technique: induction on dimension, using the existence of an eigenvector to descend to a quotient space of dimension n−1n-1. Base case n=1n = 1. Any operator on a 11-dimensional space VV is multiplication by a scalar λ\lambda (fix any nonzero v1∈Vv_1 \in V; then Tv1=λv1Tv_1 = \lambda v_1 for some scalar λ\lambda, since Tv1∈V=span⁡(v1)Tv_1 \in V = \operatorname{span}(v_1)). Set V0={0}V_0 = \{0\}, V1=VV_1 = V. Both are trivially TT-invariant, dim⁡V0=0\dim V_0 = 0, dim⁡V1=1\dim V_1 = 1, and this is the required flag. The single basis vector β=(v1)\beta = (v_1) gives [T]β=(λ)[T]_\beta = (\lambda), a 1×11\times 1 upper-triangular matrix. Inductive step. Let n≥2n \ge 2 and assume the result (both parts (1) and (2)) holds for every operator on every vector space of dimension strictly less than nn, over the same algebraically closed field F\mathbb{F}. Since F\mathbb{F} is algebraically closed and dim⁡V=n≥1\dim V = n \ge 1, the Existence of Eigenvalues over C (which holds verbatim over any algebraically closed field, by the same argument using that the characteristic polynomial, or the polynomial det⁡(T−λI)\det(T-\lambda I), has a root) guarantees that TT has at least one eigenvalue λ1∈F\lambda_1 \in \mathbb{F}, with a corresponding eigenvector v1≠0v_1 \neq 0. Set V1=span⁡(v1)V_1 = \operatorname{span}(v_1), a 11-dimensional subspace. It is TT-invariant: for any scalar cc, T(cv1)=cλ1v1∈V1T(cv_1) = c\lambda_1 v_1 \in V_1. Form the quotient space W=V/V1W = V / V_1, which has dimension n−1n - 1 (by the Rank-Nullity Theorem applied to the quotient map, or directly since dim⁡V=dim⁡V1+dim⁡(V/V1)\dim V = \dim V_1 + \dim(V/V_1)). Define the induced operator

T‾:W→W,T‾(v+V1)=Tv+V1.\overline{T} : W \to W, \qquad \overline{T}(v + V_1) = Tv + V_1.

This is well defined: if v+V1=v′+V1v + V_1 = v' + V_1, then v−v′∈V1v - v' \in V_1, so T(v−v′)∈V1T(v - v') \in V_1 (by invariance of V1V_1), hence Tv+V1=Tv′+V1Tv + V_1 = Tv' + V_1. Linearity of T‾\overline{T} follows immediately from linearity of TT and linearity of the quotient map π:V→W\pi: V \to W, π(v)=v+V1\pi(v) = v + V_1. By the inductive hypothesis applied to T‾\overline{T} on the (n−1)(n-1)-dimensional space WW, there is a flag of T‾\overline{T}-invariant subspaces

{0}=W0⊊W1⊊⋯⊊Wn−1=W,dim⁡Wk=k,\{0\} = W_0 \subsetneq W_1 \subsetneq \cdots \subsetneq W_{n-1} = W, \qquad \dim W_k = k,

and an ordered basis (vˉ2,…,vˉn)(\bar v_2, \dots, \bar v_n) of WW with T‾(vˉj)∈span⁡(vˉ2,…,vˉj)\overline{T}(\bar v_j) \in \operatorname{span}(\bar v_2, \dots, \bar v_j) for each jj, giving an upper-triangular matrix for T‾\overline T. Now pull this structure back to VV along the quotient map π\pi. For each k=0,…,n−1k = 0, \dots, n-1, define

Vk+1:=π−1(Wk)={v∈V:π(v)∈Wk}.V_{k+1} := \pi^{-1}(W_k) = \{ v \in V : \pi(v) \in W_k \}.

Since π\pi is linear and surjective and WkW_k is a subspace, Vk+1V_{k+1} is a subspace of VV containing V1=π−1({0})V_1 = \pi^{-1}(\{0\}), and by the correspondence between subspaces of V/V1V/V_1 and subspaces of VV containing V1V_1,

dim⁡Vk+1=dim⁡Wk+dim⁡V1=k+1.\dim V_{k+1} = \dim W_k + \dim V_1 = k + 1.

This gives V1=π−1(W0)⊆V2=π−1(W1)⊆⋯⊆Vn=π−1(Wn−1)=VV_1 = \pi^{-1}(W_0) \subseteq V_2 = \pi^{-1}(W_1) \subseteq \cdots \subseteq V_n = \pi^{-1}(W_{n-1}) = V, with strict inclusions because the dimensions strictly increase, and dim⁡Vk=k\dim V_{k} = k for k=1,…,nk = 1, \dots, n. Together with V0={0}V_0 = \{0\} this is exactly a chain of the required dimensions. Each Vk+1V_{k+1} is TT-invariant. Let v∈Vk+1v \in V_{k+1}, so π(v)∈Wk\pi(v) \in W_k. Since WkW_k is T‾\overline{T}-invariant, T‾(π(v))∈Wk\overline{T}(\pi(v)) \in W_k. But T‾(π(v))=π(Tv)\overline{T}(\pi(v)) = \pi(Tv) by definition of the induced operator. Hence π(Tv)∈Wk\pi(Tv) \in W_k, which means precisely Tv∈π−1(Wk)=Vk+1Tv \in \pi^{-1}(W_k) = V_{k+1}. This proves T(Vk+1)⊆Vk+1T(V_{k+1}) \subseteq V_{k+1}, establishing (1).

Basis construction proving (2). Choose vectors v2,…,vn∈Vv_2, \dots, v_n \in V so that π(vj)=vˉj\pi(v_j) = \bar v_j for each j=2,…,nj = 2, \dots, n (possible since π\pi is surjective). We claim β=(v1,v2,…,vn)\beta = (v_1, v_2, \dots, v_n) is a basis of VV making [T]β[T]_\beta upper triangular. Linear independence and spanning. The images π(v2),…,π(vn)\pi(v_2), \dots, \pi(v_n) form a basis of W=V/V1W = V/V_1. A standard fact about quotients (if v1v_1 spans a complement-defining line and (vˉ2,…,vˉn)(\bar v_2,\dots,\bar v_n) is a basis of V/V1V/V_1 then (v1,v2,…,vn)(v_1, v_2, \dots, v_n) is a basis of VV) gives that β\beta is a basis: indeed, if ∑j=1ncjvj=0\sum_{j=1}^n c_j v_j = 0, apply π\pi to get ∑j=2ncjvˉj=0\sum_{j=2}^n c_j \bar v_j = 0 in WW (since π(v1)=0\pi(v_1)=0), forcing c2=⋯=cn=0c_2 = \cdots = c_n = 0 by independence of the vˉj\bar v_j; then c1v1=0c_1 v_1 = 0 forces c1=0c_1 = 0 since v1≠0v_1 \ne 0. This shows nn vectors are linearly independent in an nn-dimensional space, hence β\beta is a basis. Triangularity. We have Tv1=λ1v1∈span⁡(v1)Tv_1 = \lambda_1 v_1 \in \operatorname{span}(v_1). For j≥2j \ge 2: since T‾(vˉj)∈span⁡(vˉ2,…,vˉj)\overline{T}(\bar v_j) \in \operatorname{span}(\bar v_2, \dots, \bar v_j), write

T‾(vˉj)=∑i=2jcivˉi\overline{T}(\bar v_j) = \sum_{i=2}^j c_i \bar v_i

for scalars ci∈Fc_i \in \mathbb{F}. Then π(Tvj)=π(∑i=2jcivi)\pi(Tv_j) = \pi\left(\sum_{i=2}^j c_i v_i\right), so Tvj−∑i=2jcivi∈ker⁡π=V1=span⁡(v1)Tv_j - \sum_{i=2}^j c_i v_i \in \ker \pi = V_1 = \operatorname{span}(v_1). Hence there is a scalar c1c_1 with

Tvj=c1v1+∑i=2jcivi∈span⁡(v1,…,vj).Tv_j = c_1 v_1 + \sum_{i=2}^j c_i v_i \in \operatorname{span}(v_1, \dots, v_j).

This holds for every j=1,…,nj = 1, \dots, n, which is exactly the condition for [T]β[T]_\beta to be upper triangular. This proves (2), completing the induction.

Proof of (3). Suppose [T]β=A=(aij)[T]_\beta = A = (a_{ij}) is upper triangular with respect to some basis β\beta, over any field F\mathbb{F} (this direction needs no algebraic closure). Then A−λIA - \lambda I is also upper triangular, with diagonal entries akk−λa_{kk} - \lambda. The determinant of an upper-triangular matrix equals the product of its diagonal entries: this is proved by cofactor expansion along the first column, since A−λIA - \lambda I has only one possibly-nonzero entry in the first column (the (1,1)(1,1) entry a11−λa_{11}-\lambda), reducing det⁡(A−λI)\det(A-\lambda I) to (a11−λ)(a_{11}-\lambda) times the determinant of the (n−1)×(n−1)(n-1)\times(n-1) upper-triangular submatrix obtained by deleting the first row and column; induction on nn then gives

det⁡(A−λI)=∏k=1n(akk−λ).\det(A - \lambda I) = \prod_{k=1}^n (a_{kk} - \lambda).

Since the characteristic polynomial of TT is det⁡(T−λI)=det⁡(A−λI)\det(T-\lambda I) = \det(A - \lambda I) (a basis-independent quantity), it factors completely with roots exactly a11,…,anna_{11}, \dots, a_{nn}, each appearing with the multiplicity it has as a factor. This is precisely the statement that the diagonal entries are the eigenvalues of TT with algebraic multiplicity, proving (3). ■\blacksquare

∎

Worked examples

Example 1 — a genuinely non-diagonalizable 3×33\times 3 operator, triangularized by hand. Let T:C3→C3T : \mathbb{C}^3 \to \mathbb{C}^3 be given in the standard basis by

A=(210021003).A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \end{pmatrix}.

This matrix is already upper triangular, so by part (3) of the theorem its eigenvalues (with multiplicity) are 2,2,32, 2, 3, read straight off the diagonal — no computation of the characteristic polynomial is needed, though we can check: det⁡(A−λI)=(2−λ)2(3−λ)\det(A - \lambda I) = (2-\lambda)^2(3-\lambda), confirming eigenvalues λ=2\lambda = 2 (multiplicity 22) and λ=3\lambda = 3 (multiplicity 11). We verify TT is not diagonalizable, to show that triangularizability is doing real work here. For λ=2\lambda = 2: solve (A−2I)x=0(A - 2I)x = 0,

A−2I=(010001001).A - 2I = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix}.

Row reducing: the second and third rows are identical, giving x2=0,x3=0x_2 = 0, x_3 = 0, with x1x_1 free. So the eigenspace for λ=2\lambda = 2 is span⁡(e1)\operatorname{span}(e_1), only 11-dimensional — but the algebraic multiplicity is 22. Since geometric multiplicity (11) is less than algebraic multiplicity (22), TT is not diagonalizable. Yet TT is triangularizable (trivially, since AA is already triangular), consistent with the theorem: triangularizability is guaranteed over C\mathbb{C} regardless of diagonalizability. The flag of invariant subspaces guaranteed by the theorem is visible directly from the columns:

V1=span⁡(e1),V2=span⁡(e1,e2),V3=C3,V_1 = \operatorname{span}(e_1), \qquad V_2 = \operatorname{span}(e_1, e_2), \qquad V_3 = \mathbb{C}^3,

and one checks T(e1)=2e1∈V1T(e_1) = 2e_1 \in V_1, T(e2)=e1+2e2∈V2T(e_2) = e_1 + 2e_2 \in V_2, T(e3)=e2+3e3∈V3T(e_3) = e_2 + 3e_3 \in V_3 — matching the columns of AA exactly.

Example 2 — triangularizing an operator not initially given in triangular form. Let T:C2→C2T : \mathbb{C}^2 \to \mathbb{C}^2 have standard matrix

A=(0−144).A = \begin{pmatrix} 0 & -1 \\ 4 & 4 \end{pmatrix}.

This is not upper triangular. We find a basis that makes it so, following the constructive steps of the proof. Step 1: find an eigenvalue. The characteristic polynomial is

det⁡(A−λI)=det⁡(−λ−144−λ)=−λ(4−λ)−(−1)(4)=λ2−4λ+4=(λ−2)2.\det(A - \lambda I) = \det\begin{pmatrix} -\lambda & -1 \\ 4 & 4-\lambda \end{pmatrix} = -\lambda(4-\lambda) - (-1)(4) = \lambda^2 - 4\lambda + 4 = (\lambda - 2)^2.

So λ1=2\lambda_1 = 2 is the unique eigenvalue, with algebraic multiplicity 22. Step 2: find an eigenvector for λ1=2\lambda_1 = 2. Solve (A−2I)x=0(A - 2I)x = 0:

A−2I=(−2−142).A - 2I = \begin{pmatrix} -2 & -1 \\ 4 & 2 \end{pmatrix}.

Row 2 is −2-2 times row 1, so the system reduces to −2x1−x2=0-2x_1 - x_2 = 0, i.e. x2=−2x1x_2 = -2x_1. Take v1=(1,−2)v_1 = (1, -2). Set V1=span⁡(v1)V_1 = \operatorname{span}(v_1). Step 3: extend to a basis. Choose any vector v2v_2 completing v1v_1 to a basis of C2\mathbb{C}^2; take v2=(0,1)v_2 = (0,1), since v1,v2v_1, v_2 are clearly independent (their matrix (10−21)\begin{pmatrix}1&0\\-2&1\end{pmatrix} has determinant 1≠01 \ne 0). Step 4: compute [T]β[T]_\beta for β=(v1,v2)\beta = (v_1, v_2). We have

Tv1=Av1=(0−144)(1−2)=(2−4)=2(1−2)=2v1,Tv_1 = Av_1 = \begin{pmatrix} 0 & -1 \\ 4 & 4 \end{pmatrix}\begin{pmatrix} 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \end{pmatrix} = 2\begin{pmatrix}1\\-2\end{pmatrix} = 2v_1,

confirming v1v_1 is an eigenvector with eigenvalue 22, as it must be.

Tv2=Av2=(0−144)(01)=(−14).Tv_2 = Av_2 = \begin{pmatrix} 0 & -1 \\ 4 & 4 \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}.

Express this in the basis β\beta: solve c1v1+c2v2=(−1,4)c_1 v_1 + c_2 v_2 = (-1, 4), i.e. c1(1,−2)+c2(0,1)=(−1,4)c_1(1,-2) + c_2(0,1) = (-1,4). From the first coordinate, c1=−1c_1 = -1. From the second, −2c1+c2=4⇒2+c2=4⇒c2=2-2c_1 + c_2 = 4 \Rightarrow 2 + c_2 = 4 \Rightarrow c_2 = 2. So

Tv2=−v1+2v2.Tv_2 = -v_1 + 2v_2.

Hence

[T]β=(2−102),[T]_\beta = \begin{pmatrix} 2 & -1 \\ 0 & 2 \end{pmatrix},

which is upper triangular, with diagonal entries 2,22, 2 matching the double eigenvalue found above. This operator is a scalar multiple of the identity plus a nonzero nilpotent piece (its Jordan-block shape foreshadows the theory of Generalized Eigenvectors) and is not diagonalizable — the eigenspace for λ=2\lambda=2 here is only 11-dimensional, since (A−2I)(A-2I) has rank 11 — yet it is triangularizable, exactly as the theorem promises.

CounterexampleWhy algebraic closure (splitting) is needed

Drop the hypothesis that F\mathbb{F} is algebraically closed, and work instead over R\mathbb{R}. Let V=R2V = \mathbb{R}^2 and let TT be the rotation by 90°90°:

A=(0−110).A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}.

The characteristic polynomial is det⁡(A−λI)=λ2+1\det(A - \lambda I) = \lambda^2 + 1, which has no real roots. So TT has no eigenvalue in R\mathbb{R}, hence no eigenvector in R2\mathbb{R}^2. Suppose, for contradiction, that some real basis β=(v1,v2)\beta = (v_1, v_2) made [T]β[T]_\beta upper triangular, say [T]β=(ab0d)[T]_\beta = \begin{pmatrix} a & b \\ 0 & d \end{pmatrix}. Then Tv1=av1Tv_1 = a v_1, forcing v1v_1 to be a real eigenvector of TT with real eigenvalue aa — contradicting the fact just established that TT has no real eigenvalue. So TT is not triangularizable over R\mathbb{R}, even though dim⁡V=2\dim V = 2 is finite and TT is a perfectly good linear operator. The proof of the Triangularization Theorem breaks down at the very first inductive step: the appeal to the Existence of Eigenvalues over C (or, more precisely, existence of a root of the characteristic polynomial in F\mathbb{F}) simply fails when F=R\mathbb{F} = \mathbb{R}, because x2+1x^2+1 does not split over R\mathbb{R}. This shows the hypothesis "algebraically closed field" (or "characteristic polynomial splits into linear factors over F\mathbb{F}") is not a technical convenience — it is load-bearing. Over C\mathbb{C}, the very same TT, viewed as an operator on C2\mathbb{C}^2, does have eigenvalues ±i\pm i and is triangularizable (in fact diagonalizable) there.

Scope, edge cases & common traps

SituationWhat happens
F=C\mathbb{F} = \mathbb{C}, any TT, finite dimensionAlways triangularizable (theorem applies unconditionally)
F=R\mathbb{F} = \mathbb{R}, characteristic polynomial splits over R\mathbb{R}Triangularizable (the proof only needs the field to contain all the roots, not full algebraic closure)
F=R\mathbb{F} = \mathbb{R}, characteristic polynomial does not splitMay fail to be triangularizable — see the rotation counterexample
TT diagonalizableAutomatically triangularizable, with off-diagonal entries all zero
TT triangularizableNot necessarily diagonalizable (Examples 1 and 2 above)
Infinite-dimensional VVThe inductive proof does not apply; the very notion of a finite basis/flag needs reformulation
Non-unique triangularizing basisThe theorem asserts existence, not uniqueness — many bases can triangularize the same TT

The mathematician behind it

Origins · 1 minCamille Jordan and how this was discovered
CJ
Camille Jordan
1838–1922 · 19th-century France, structural theory of linear operators

Jordan's foundational 1870 treatise Traité des substitutions et des équations algébriques organized the emerging theory of what we now call linear operators, developing canonical forms for matrices well before the modern axiomatic language of vector spaces existed. His name attaches to the finest possible triangularization — the Jordan Canonical Form — in which the flag of invariant subspaces is refined all the way down to blocks recording exactly how far an operator is from being diagonalizable. Triangularization is the coarser, more robust ancestor of that refinement: it needs only the existence of eigenvalues, while the full Jordan form needs the theory of generalized eigenvectors built on top of it.

History
- 1870 — Camille Jordan publishes his treatise organizing canonical forms of linear substitutions, laying groundwork for triangular and block-triangular decompositions.
- 1878 — Georg Frobenius develops further canonical form theory for matrices over general fields, clarifying invariant factors.
- 1898 — Kurt Hensel and others begin systematizing the abstract notion of a field, giving the language in which "algebraically closed" is later phrased precisely.
- 1930s — The Schur decomposition (unitary triangularization) is isolated as the numerically stable analogue of Jordan's triangular forms.
- 1942 — Paul Halmos's Finite-Dimensional Vector Spaces helps fix the modern axiomatic, basis-free treatment of invariant subspaces and triangularization used in this article.
Is every square complex matrix similar to an upper-triangular matrix?

Yes. This is exactly the content of the Triangularization Theorem applied to the operator TT on Cn\mathbb{C}^n represented by that matrix in the standard basis. Since C\mathbb{C} is algebraically closed, the characteristic polynomial of any complex operator splits completely into linear factors, so the inductive construction of an eigenvector, followed by passage to a quotient space, always succeeds. The resulting change-of-basis matrix PP satisfies P−1AP=UP^{-1}AP = U with UU upper triangular. This is a purely existential statement: it guarantees such a PP exists but the proof's inductive method, while explicit in principle, is rarely the most efficient computational route; numerical algorithms use the Schur decomposition instead.

Does triangularizability imply diagonalizability?

No, and this is one of the most important distinctions in the subject. Triangularizability only requires that each TvjTv_j lie in the span of v1,…,vjv_1,\dots,v_j; diagonalizability requires the much stronger condition that TvjTv_j be a scalar multiple of vjv_j alone. Nilpotent operators other than zero are the sharpest illustration: they are always triangularizable (their only eigenvalue is 00), yet never diagonalizable, because a nonzero diagonalizable nilpotent operator would have to be the zero matrix in its eigenbasis, contradicting nonzero-ness. Both worked examples in this article exhibit triangularizable, non-diagonalizable operators explicitly.

Why does the proof need to pass to a quotient space rather than a subspace complement?

Working with a complement WW such that V=V1⊕WV = V_1 \oplus W is tempting, but TT generally does not map WW into itself — only V1V_1 is guaranteed invariant, not any particular complement. The quotient V/V1V/V_1 sidesteps this: the induced map T‾\overline{T} is always well defined on the quotient regardless of which complement one might have chosen, precisely because V1V_1 being invariant is exactly the condition needed for T‾(v+V1)=Tv+V1\overline{T}(v+V_1) = Tv + V_1 to be independent of the choice of representative vv. This makes the induction clean and choice-free.

Is the triangularizing basis unique?

No. The theorem only asserts existence, and in general many different bases triangularize the same operator, often with different orderings or different off-diagonal entries, though the multiset of diagonal entries (the eigenvalues with multiplicity) is an invariant of TT itself, by part (3) of the theorem, and does not depend on which triangularizing basis is chosen. Even the order in which eigenvalues appear along the diagonal can typically be permuted by choosing the eigenvector flag in a different order, subject to compatibility with invariance at each stage.

How does this relate to the Cayley-Hamilton theorem?

One of the cleanest proofs of the Cayley-Hamilton Theorem triangularizes TT first, using exactly this theorem, and then shows that (T−λ1I)(T−λ2I)⋯(T−λnI)(T-\lambda_1 I)(T - \lambda_2 I)\cdots(T-\lambda_n I) annihilates every basis vector one at a time, by induction using the flag structure. Without the guarantee that a triangularizing basis exists, that argument has no foothold. This is a primary reason triangularization is placed before Cayley-Hamilton in the logical development of the subject, rather than after.

What happens for operators over R whose characteristic polynomial does not split?

Such operators are not triangularizable over R\mathbb{R} as shown in the rotation counterexample, but they can always be triangularized after extending scalars to C\mathbb{C} (viewing the same matrix as acting on Cn\mathbb{C}^n). Real linear algebra instead uses real canonical forms with 2×22\times 2 rotation-scaling blocks along the diagonal in place of single complex eigenvalues, reflecting the pairs of complex-conjugate eigenvalues that arise from real matrices with non-real roots of their characteristic polynomial.

Does the theorem require T to be invertible?

No, invertibility plays no role anywhere in the statement or proof. The zero operator, for instance, is (trivially) already upper triangular — indeed diagonal — in every basis, and singular operators are triangularized by the same inductive construction as invertible ones; the only place an eigenvalue is used is to seed the induction, and eigenvalues can certainly be zero.

Before this

Where this leads