Simultaneous Diagonalization of Commuting Operators

Can two different diagonalizable operators be diagonalized by the same basis? This question sounds like a minor technicality. But it is the hinge on which entire theories turn. Examples include the simultaneous measurement of observables in quantum mechanics. They also include the classification of finite abelian groups acting on vector spaces. If two operators commute, they never obstruct each other. If they do not commute, one operator's eigenvectors typically get scrambled by the other. The precise dividing line between these two behaviors is exactly what this article makes rigorous. The big picture is this: diagonalizability of a single operator says a vector space decomposes into eigenlines. On these eigenlines, that operator acts by scalars. Simultaneous diagonalizability asks for a single basis of eigenlines. This basis must work for an entire family of operators at once. Each vector in the basis is an eigenvector for every operator in the family. Different operators may assign different eigenvalues to that same vector. This is a strictly stronger request than each operator individually being diagonalizable. The theorem of this article identifies commutativity as precisely the extra ingredient needed to guarantee it. This matters because commuting families of operators arise constantly. Examples include powers and polynomials of a single diagonalizable operator. They also include commuting families of symmetries in physics. Another example is commuting families of self-adjoint operators encoding compatible observables. Commuting matrices arising from representations of abelian groups are a further case. Whenever such a family shows up, simultaneous diagonalization becomes useful. It lets us replace an entangled system of linear maps with a basis of coordinates. In this basis, every operator in the family is simply a diagonal scaling. This is the single most useful normal form in linear algebra. This article gives the full statement and proof of the Simultaneous Diagonalization Theorem. It covers two commuting diagonalizable operators, and its extension to finite families. It builds the intuition through eigenspace decompositions and invariant subspaces. It exhibits two fully worked numerical examples. It also shows exactly why commutativity cannot be dropped.

Key takeaways
  • A diagonalizable operator gives an eigenbasis; simultaneous diagonalization asks for one basis that is an eigenbasis for every operator in a family at once.
  • The theorem: finitely many commuting diagonalizable operators on a finite-dimensional space are simultaneously diagonalizable, and conversely.
  • The proof rests on one key fact: if SS and TT commute, every eigenspace of SS is TT-invariant — Invariant Subspaces are the real engine here.
  • Restricting TT to each eigenspace of SS and diagonalizing there produces the common eigenbasis, by induction on the number of operators.
  • Without commutativity the result is false in general: a rotation and a reflection can each be diagonalizable while sharing no common eigenbasis.
  • Applications range from quantum observables to decomposing representations of abelian groups to simplifying systems of linear ODEs.
  • Key idea: commuting operators cannot mix each other's eigenspaces, so diagonalizing one operator "respects" the eigenspace structure of the other.

Where this lives today

In quantum mechanics, commuting Hermitian operators represent compatible observables, and the Simultaneous Diagonalization Theorem (in its self-adjoint/unitary form) is the precise mathematical statement that compatible observables have a common basis of simultaneous eigenstates — the basis in which both quantities have sharply defined values at once. In representation theory, a finite abelian group acting on a finite-dimensional complex vector space gives a family of pairwise-commuting diagonalizable operators (roots of unity as eigenvalues, by finite order), and simultaneous diagonalization decomposes the representation into one-dimensional irreducible pieces indexed by characters of the group. In numerical linear algebra and statistics, commuting families such as a covariance matrix and a mean-shift or symmetry operator can sometimes be jointly diagonalized to simplify a data analysis pipeline, and joint diagonalization algorithms (approximate simultaneous diagonalization) underpin blind source separation techniques in signal processing. In the theory of linear ordinary differential equation systems, if the coefficient matrices of a family of commuting linear systems are simultaneously diagonalizable, the coupled system decouples into independent scalar exponential equations in the common eigenbasis, drastically simplifying both analysis and numerical solution.

Ask the tutor: How does this theorem change if we only assume the operators are triangularizable rather than diagonalizable?

DefinitionSimultaneous Diagonalizability

Let VV be a finite-dimensional vector space over a field F\mathbb{F}, and let F={T1,T2,…,Tk}\mathcal{F} = \{T_1, T_2, \dots, T_k\} be a finite collection of linear operators on VV. The family F\mathcal{F} is simultaneously diagonalizable if there exists a basis β=(v1,…,vn)\beta = (v_1, \dots, v_n) of VV such that every vector vi∈βv_i \in \beta is a simultaneous eigenvector for all operators in F\mathcal{F}: that is, for each i∈{1,…,n}i \in \{1,\dots,n\} and each j∈{1,…,k}j \in \{1,\dots,k\} there is a scalar λij∈F\lambda_{ij} \in \mathbb{F} with

Tjvi=λijvi.T_j v_i = \lambda_{ij} v_i.

Equivalently, writing [Tj]β[T_j]_\beta for the matrix of TjT_j with respect to β\beta, the family F\mathcal{F} is simultaneously diagonalizable exactly when there is a single invertible matrix PP such that P−1AjPP^{-1} A_j P is diagonal for every matrix AjA_j representing TjT_j in some fixed original basis, and j=1,…,kj = 1, \dots, k.

Intuition

Diagonalizing one operator TT means finding an eigenbasis: a basis along which TT acts by independent scalar stretches, one per basis vector. If a second operator SS is also diagonalizable, it has its own eigenbasis, which in general has nothing to do with TT's eigenbasis — a generic pair of diagonalizable matrices does not share eigenvectors at all. The special situation where SS and TT do share a full eigenbasis is extremely rigid, and the theorem below identifies the algebraic condition that forces it: ST=TSST = TS. The reason commutativity is the right condition can be seen structurally. If SS and TT commute and vv is an eigenvector of SS with eigenvalue λ\lambda, then

S(Tv)=T(Sv)=T(λv)=λ(Tv),S(Tv) = T(Sv) = T(\lambda v) = \lambda (Tv),

so TvTv is again an eigenvector of SS with the same eigenvalue λ\lambda (or is the zero vector). In other words, TT maps the λ\lambda-eigenspace of SS into itself: each eigenspace of SS is an invariant subspace for TT. This is the crucial link to Invariant Subspaces. Since TT never sends a λ\lambda-eigenvector of SS outside the λ\lambda-eigenspace, we may study TT eigenspace-by-eigenspace, diagonalizing the restriction of TT to each piece separately, and reassembling a single basis for all of VV at the end. Picture VV sliced into SS-eigenspaces, like horizontal strata. Commutativity guarantees TT never punches through a stratum boundary — it acts within each stratum. Diagonalizing TT stratum by stratum, and using that each stratum is itself diagonalizable for TT (a hypothesis we must assume), yields a basis that diagonalizes both operators simultaneously.

Figure
SS-eigenspaces as TT-invariant strata: TT acts within each stratum, never crossing
V = eigenspace of S for λ₁ ⊕ eigenspace of S for λ₂ ⊕ eigenspace of S for λ₃ E(S, λ₁) — dimension 2 — T-invariant E(S, λ₂) — dimension 1 — T-invariant E(S, λ₃) — dimension 2 — T-invariant T restricted to each stratum can be diagonalized there, giving a shared eigenbasis for S and T on all of V.

Central results

TheoremSimultaneous Diagonalization Theorem, two operators

Let VV be a finite-dimensional vector space over a field F\mathbb{F}, and let S,T:V→VS, T : V \to V be linear operators. Then SS and TT are simultaneously diagonalizable if and only if: 1. SS is diagonalizable, 2. TT is diagonalizable, and 3. ST=TSST = TS.

Proof

We prove both directions. The technique is a structural argument using invariant subspaces and induction on dimension; no circularity arises because we only use the single-operator diagonalizability criterion (a basis of eigenvectors exists iff the eigenspaces span VV) from Diagonalizability and Its Criteria, never the theorem being proved.

(⇒\Rightarrow) Simultaneous diagonalizability implies (1), (2), (3). Suppose β=(v1,…,vn)\beta = (v_1, \dots, v_n) is a basis of VV with Svi=λiviSv_i = \lambda_i v_i and Tvi=μiviT v_i = \mu_i v_i for scalars λi,μi∈F\lambda_i, \mu_i \in \mathbb{F}. Then β\beta is an eigenbasis for SS, so SS is diagonalizable, and likewise β\beta is an eigenbasis for TT, so TT is diagonalizable. For commutativity, it suffices to check ST=TSST = TS on the basis β\beta, since two linear maps agreeing on a basis are equal. For each ii,

STvi=S(μivi)=μiSvi=μiλivi,TSvi=T(λivi)=λiTvi=λiμivi.ST v_i = S(\mu_i v_i) = \mu_i S v_i = \mu_i \lambda_i v_i, \qquad TS v_i = T(\lambda_i v_i) = \lambda_i T v_i = \lambda_i \mu_i v_i.

Since F\mathbb{F} is commutative as a field, μiλi=λiμi\mu_i \lambda_i = \lambda_i \mu_i, so STvi=TSviST v_i = TS v_i for every ii, and hence ST=TSST = TS on all of VV.

(⇐\Leftarrow) (1), (2), (3) imply simultaneous diagonalizability. Assume SS, TT are diagonalizable and ST=TSST = TS. Let λ1,…,λm\lambda_1, \dots, \lambda_m be the distinct eigenvalues of SS, and let

Ei:={v∈V:Sv=λiv}E_i := \{ v \in V : Sv = \lambda_i v \}

be the corresponding eigenspaces, i=1,…,mi = 1, \dots, m. Since SS is diagonalizable, the single-operator diagonalizability criterion gives

V=E1⊕E2⊕⋯⊕Em.(∗)V = E_1 \oplus E_2 \oplus \cdots \oplus E_m. \tag{$\ast$}

Step 1: each EiE_i is TT-invariant. Let v∈Eiv \in E_i, so Sv=λivSv = \lambda_i v. Using ST=TSST = TS,

S(Tv)=T(Sv)=T(λiv)=λi(Tv).S(Tv) = T(Sv) = T(\lambda_i v) = \lambda_i (Tv).

Hence Tv∈EiTv \in E_i as well. This shows T(Ei)⊆EiT(E_i) \subseteq E_i, i.e. EiE_i is an invariant subspace for TT, exactly as anticipated in the Intuition section and in Invariant Subspaces.

Step 2: T∣EiT|_{E_i}, the restriction of TT to EiE_i, is diagonalizable, for each ii. This is the one place genuine work is needed, because diagonalizability of TT on VV does not automatically hand us diagonalizability of an arbitrary restriction — but the eigenspace decomposition (∗)(\ast) is exactly the structure that makes it work here. Since TT is diagonalizable on VV, there is a basis of VV consisting of TT-eigenvectors; equivalently, letting μ1,…,μp\mu_1, \dots, \mu_p denote the distinct eigenvalues of TT and Fj={v∈V:Tv=μjv}F_j = \{v \in V : Tv = \mu_j v\} the corresponding eigenspaces,

V=F1⊕F2⊕⋯⊕Fp.V = F_1 \oplus F_2 \oplus \cdots \oplus F_p.

We claim

Ei=(Ei∩F1)⊕(Ei∩F2)⊕⋯⊕(Ei∩Fp).(∗∗)E_i = (E_i \cap F_1) \oplus (E_i \cap F_2) \oplus \cdots \oplus (E_i \cap F_p). \tag{$\ast\ast$}

Granting (∗∗)(\ast\ast) for a moment: each Ei∩FjE_i \cap F_j consists of vectors that are TT-eigenvectors with eigenvalue μj\mu_j (or is {0}\{0\}), so (∗∗)(\ast\ast) exhibits EiE_i as a direct sum of TT-eigenspaces intersected with EiE_i; picking a basis of each nonzero summand Ei∩FjE_i \cap F_j and concatenating gives a basis of EiE_i consisting entirely of TT-eigenvectors. By the diagonalizability criterion applied to T∣EiT|_{E_i} (using invariance from Step 1 to make sense of this restriction as an operator on EiE_i), this shows T∣EiT|_{E_i} is diagonalizable. Proof of (∗∗)(\ast\ast). Let v∈Eiv \in E_i. Since V=F1⊕⋯⊕FpV = F_1 \oplus \cdots \oplus F_p, write uniquely

v=w1+w2+⋯+wp,wj∈Fj.v = w_1 + w_2 + \cdots + w_p, \qquad w_j \in F_j.

We must show each wjw_j actually lies in EiE_i, i.e. Swj=λiwjS w_j = \lambda_i w_j. Apply SS to both sides:

Sv=Sw1+Sw2+⋯+Swp.Sv = S w_1 + S w_2 + \cdots + S w_p.

By Step 1 applied with the roles of S,TS,T noted symmetrically — since ST=TSST=TS also gives that each FjF_j is SS-invariant by the identical computation T(Sw)=S(Tw)=S(μjw)=μj(Sw)T(Sw) = S(Tw) = S(\mu_j w) = \mu_j(Sw) — we get Swj∈FjSw_j \in F_j for each jj. But Sv=λiv=λiw1+⋯+λiwpSv = \lambda_i v = \lambda_i w_1 + \cdots + \lambda_i w_p, and λiwj∈Fj\lambda_i w_j \in F_j too (as FjF_j is a subspace). By uniqueness of the decomposition of SvSv as a sum of elements of F1,…,FpF_1, \dots, F_p (direct sum), we may equate the FjF_j-components:

Swj=λiwjfor every j.S w_j = \lambda_i w_j \quad \text{for every } j.

Thus wj∈Eiw_j \in E_i for every jj with wj≠0w_j \neq 0, so v=∑jwjv = \sum_j w_j is a sum of vectors, each lying in Ei∩FjE_i \cap F_j. This proves Ei⊆∑j(Ei∩Fj)E_i \subseteq \sum_j (E_i \cap F_j); the reverse inclusion ∑j(Ei∩Fj)⊆Ei\sum_j(E_i \cap F_j) \subseteq E_i is immediate since Ei∩Fj⊆EiE_i \cap F_j \subseteq E_i. The sum ∑j(Ei∩Fj)\sum_j (E_i \cap F_j) is direct because it is a subsum of the direct sum F1⊕⋯⊕FpF_1 \oplus \cdots \oplus F_p (any dependency among vectors from distinct Ei∩Fj⊆FjE_i \cap F_j \subseteq F_j would already be a dependency among vectors from distinct FjF_j, forcing all terms to vanish). This proves (∗∗)(\ast\ast).

Step 3: assembling the common eigenbasis. For each i=1,…,mi = 1, \dots, m, Step 2 gives a basis γi\gamma_i of EiE_i consisting of TT-eigenvectors; since every vector of EiE_i is by definition an SS-eigenvector with eigenvalue λi\lambda_i, every vector of γi\gamma_i is a simultaneous eigenvector of SS (eigenvalue λi\lambda_i) and TT (some eigenvalue depending on the vector). Let

β:=γ1∪γ2∪⋯∪γm.\beta := \gamma_1 \cup \gamma_2 \cup \cdots \cup \gamma_m.

Because V=E1⊕⋯⊕EmV = E_1 \oplus \cdots \oplus E_m by (∗)(\ast), the concatenation of bases of the summands is a basis of VV (a standard fact about direct sum decompositions: a sequence formed by concatenating bases of the summands of a direct sum is linearly independent, because a dependency relation splits by the summand-components exactly as in the argument for (∗∗)(\ast\ast), and it spans since every vector of VV is a sum of vectors from the EiE_i). Hence β\beta is a basis of VV, and every vector of β\beta is a simultaneous eigenvector of SS and TT. This exhibits S,TS, T as simultaneously diagonalizable, completing the proof. ■\blacksquare

∎
CorollaryFinite families

Let VV be finite-dimensional over F\mathbb{F} and let T1,…,TkT_1, \dots, T_k be linear operators on VV, each diagonalizable, with TiTj=TjTiT_i T_j = T_j T_i for all i,ji,j. Then T1,…,TkT_1, \dots, T_k are simultaneously diagonalizable.

Proof

By induction on kk. The case k=1k = 1 is the hypothesis that T1T_1 is diagonalizable. The case k=2k=2 is the Simultaneous Diagonalization Theorem above. For the inductive step, suppose the result holds for any k−1k-1 pairwise-commuting diagonalizable operators, and let T1,…,TkT_1, \dots, T_k be pairwise-commuting and diagonalizable. By the inductive hypothesis applied to T1,…,Tk−1T_1, \dots, T_{k-1}, there is a common eigenbasis β\beta for these k−1k-1 operators; equivalently, VV decomposes as a direct sum of simultaneous joint eigenspaces

V=⨁λ⃗Eλ⃗,Eλ⃗={v∈V:Tiv=λiv, i=1,…,k−1},V = \bigoplus_{\vec\lambda} E_{\vec\lambda}, \qquad E_{\vec\lambda} = \{v \in V : T_i v = \lambda_i v, \ i = 1,\dots,k-1\},

where λ⃗=(λ1,…,λk−1)\vec\lambda = (\lambda_1,\dots,\lambda_{k-1}) ranges over joint eigenvalue tuples that occur. Exactly as in Step 1 of the proof above, commutativity of TkT_k with each TiT_i (i<ki < k) shows each joint eigenspace Eλ⃗E_{\vec\lambda} is TkT_k-invariant: if v∈Eλ⃗v \in E_{\vec\lambda} then for each i<ki<k,

Ti(Tkv)=Tk(Tiv)=Tk(λiv)=λi(Tkv),T_i(T_k v) = T_k(T_i v) = T_k(\lambda_i v) = \lambda_i (T_k v),

so Tkv∈Eλ⃗T_k v \in E_{\vec\lambda}. The same argument as Step 2 (applied with TkT_k playing the role of TT, and the ambient decomposition of VV into TkT_k-eigenspaces F1,…,FpF_1,\dots,F_p coming from diagonalizability of TkT_k) shows Tk∣Eλ⃗T_k|_{E_{\vec\lambda}} is diagonalizable on each joint eigenspace Eλ⃗E_{\vec\lambda}. Choosing a TkT_k-eigenbasis of each Eλ⃗E_{\vec\lambda} and concatenating over all λ⃗\vec\lambda, exactly as in Step 3, produces a single basis of VV in which every vector is a simultaneous eigenvector of T1,…,TkT_1,\dots,T_k. This completes the induction. ■\blacksquare

∎

Worked examples

Example 1: two genuinely different, asymmetric 3×33\times 3 commuting matrices. Let

A=(200050005),B=(−10003404−3).A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix}, \qquad B = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 3 & 4 \\ 0 & 4 & -3 \end{pmatrix}.

Step 1: check they commute. Since A=diag(2,5,5)A = \mathrm{diag}(2,5,5) acts as scalar 22 on the first coordinate and scalar 55 on the last two, and BB is block-diagonal with blocks (−1)(-1) and (344−3)\begin{pmatrix}3&4\\4&-3\end{pmatrix} matching that same block structure, we compute directly:

AB=(−20001520020−15),BA=(−20001520020−15).AB = \begin{pmatrix} -2 & 0 & 0 \\ 0 & 15 & 20 \\ 0 & 20 & -15 \end{pmatrix}, \qquad BA = \begin{pmatrix} -2 & 0 & 0 \\ 0 & 15 & 20 \\ 0 & 20 & -15 \end{pmatrix}.

So AB=BAAB = BA. Step 2: diagonalize AA. AA is already diagonal with eigenvalues 22 (multiplicity 1, eigenspace E2=span(e1)E_2 = \mathrm{span}(e_1)) and 55 (multiplicity 2, eigenspace E5=span(e2,e3)E_5 = \mathrm{span}(e_2, e_3)). Step 3: restrict BB to each eigenspace of AA and diagonalize there. On E2=span(e1)E_2 = \mathrm{span}(e_1): Be1=−e1B e_1 = -e_1, so e1e_1 is already a BB-eigenvector with eigenvalue −1-1. No work needed — this is forced since E2E_2 is one-dimensional. On E5=span(e2,e3)E_5 = \mathrm{span}(e_2, e_3): the restriction of BB is the 2×22\times 2 block

B′=(344−3).B' = \begin{pmatrix} 3 & 4 \\ 4 & -3 \end{pmatrix}.

Find its eigenvalues: characteristic polynomial

det⁡(B′−tI)=(3−t)(−3−t)−16=t2−9−16=t2−25,\det(B' - tI) = (3-t)(-3-t) - 16 = t^2 - 9 - 16 = t^2 - 25,

so eigenvalues t=5t = 5 and t=−5t = -5. For t=5t=5: solve (B′−5I)w=0(B' - 5I)w = 0, i.e. (−244−8)w=0\begin{pmatrix}-2&4\\4&-8\end{pmatrix} w = 0, giving w=(2,1)w = (2,1) up to scale, i.e. in VV-coordinates the vector 2e2+e32e_2 + e_3. For t=−5t=-5: solve (B′+5I)w=0(B'+5I)w=0, i.e. (8442)w=0\begin{pmatrix}8&4\\4&2\end{pmatrix} w = 0, giving w=(1,−2)w=(1,-2) up to scale, i.e. e2−2e3e_2 - 2e_3. Step 4: assemble. The common eigenbasis is

v1=e1,v2=2e2+e3,v3=e2−2e3,v_1 = e_1, \qquad v_2 = 2e_2 + e_3, \qquad v_3 = e_2 - 2e_3,

with

Av1=2v1, Av2=5v2, Av3=5v3,Bv1=−v1, Bv2=5v2, Bv3=−5v3.A v_1 = 2 v_1,\ A v_2 = 5 v_2,\ A v_3 = 5 v_3, \qquad B v_1 = -v_1,\ B v_2 = 5 v_2,\ B v_3 = -5 v_3.

Let's verify Bv2Bv_2 directly: B(2e2+e3)=2(3e2+4e3)+(4e2−3e3)=(6+4)e2+(8−3)e3=10e2+5e3=5(2e2+e3)=5v2B(2e_2+e_3) = 2(3e_2+4e_3) + (4e_2 - 3e_3) = (6+4)e_2 + (8-3)e_3 = 10e_2 + 5e_3 = 5(2e_2+e_3) = 5v_2. ✓. And Bv3=B(e2−2e3)=(3e2+4e3)−2(4e2−3e3)=(3−8)e2+(4+6)e3=−5e2+10e3=−5(e2−2e3)=−5v3Bv_3 = B(e_2 - 2e_3) = (3e_2+4e_3) - 2(4e_2-3e_3) = (3-8)e_2 + (4+6)e_3 = -5e_2+10e_3 = -5(e_2-2e_3) = -5v_3. ✓. Setting P=(10002101−2)P = \begin{pmatrix} 1&0&0\\0&2&1\\0&1&-2\end{pmatrix}, we get P−1AP=diag(2,5,5)P^{-1}AP = \mathrm{diag}(2,5,5) and P−1BP=diag(−1,5,−5)P^{-1}BP = \mathrm{diag}(-1,5,-5) simultaneously, exactly as the theorem promises — note the eigenvalues of AA repeat while the eigenvalues of BB that come from the same AA-eigenspace split apart, illustrating why we needed Step 2's finer analysis on the repeated-eigenvalue block.

Example 2: three commuting operators via polynomials of one operator, on R2\mathbb{R}^2. Let TT be the operator on R2\mathbb{R}^2 given by

T=(4−211).T = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix}.

Check diagonalizability: characteristic polynomial det⁡(T−tI)=(4−t)(1−t)−(−2)(1)=t2−5t+4+2=t2−5t+6=(t−2)(t−3)\det(T - tI) = (4-t)(1-t) - (-2)(1) = t^2 -5t+4+2 = t^2-5t+6=(t-2)(t-3), so eigenvalues 2,32,3, distinct, hence TT is diagonalizable (two distinct eigenvalues in dimension 2 automatically give an eigenbasis). For t=2t=2: (T−2I)v=0⇒(2−21−1)v=0⇒v=(1,1)(T-2I)v=0 \Rightarrow \begin{pmatrix}2&-2\\1&-1\end{pmatrix}v=0 \Rightarrow v = (1,1). For t=3t=3: (T−3I)v=0⇒(1−21−2)v=0⇒v=(2,1)(T-3I)v=0 \Rightarrow \begin{pmatrix}1&-2\\1&-2\end{pmatrix}v=0 \Rightarrow v=(2,1). Now consider the family F={T, T2, 3I−T}\mathcal{F} = \{T,\ T^2,\ 3I - T\}. Every polynomial in TT commutes with TT and with every other polynomial in TT, because scalar multiples and powers of a fixed matrix always commute with each other: T⋅T2=T2⋅T=T3T \cdot T^2 = T^2 \cdot T = T^3, and T(3I−T)=3T−T2=(3I−T)TT(3I-T) = 3T - T^2 = (3I-T)T. Each operator in F\mathcal{F} is diagonalizable: TT as shown; T2T^2 has eigenvalues 22=4,32=92^2=4, 3^2=9 on the same eigenvectors (since T2v=T(Tv)=T(λv)=λ2vT^2 v = T(Tv) = T(\lambda v) = \lambda^2 v whenever Tv=λvTv=\lambda v), hence is diagonalizable with the same eigenbasis; similarly 3I−T3I-T has eigenvalues 3−2=13-2=1 and 3−3=03-3=0 on the same eigenvectors. By the Corollary, all three operators are simultaneously diagonalizable, and indeed we see explicitly that the single eigenbasis ((1,1), (2,1))\big((1,1),\,(2,1)\big) works for all three at once:

T(11)=2(11),T2(11)=4(11),(3I−T)(11)=1(11),T\begin{pmatrix}1\\1\end{pmatrix}=2\begin{pmatrix}1\\1\end{pmatrix},\quad T^2\begin{pmatrix}1\\1\end{pmatrix}=4\begin{pmatrix}1\\1\end{pmatrix},\quad (3I-T)\begin{pmatrix}1\\1\end{pmatrix}=1\begin{pmatrix}1\\1\end{pmatrix},
T(21)=3(21),T2(21)=9(21),(3I−T)(21)=0(21).T\begin{pmatrix}2\\1\end{pmatrix}=3\begin{pmatrix}2\\1\end{pmatrix},\quad T^2\begin{pmatrix}2\\1\end{pmatrix}=9\begin{pmatrix}2\\1\end{pmatrix},\quad (3I-T)\begin{pmatrix}2\\1\end{pmatrix}=0\begin{pmatrix}2\\1\end{pmatrix}.

This example also shows the theorem's converse direction is automatic whenever operators are all polynomials in a common diagonalizable TT: commutativity here is not a coincidence but a structural certainty.

CounterexampleDropping commutativity

The hypothesis ST=TSST=TS cannot be removed: two individually diagonalizable operators that fail to commute need not share any eigenbasis. Take V=R2V = \mathbb{R}^2 and let

S=(110−1),T=(0110).S = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}, \qquad T = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.

TT is diagonalizable: its characteristic polynomial is t2−1t^2 - 1, roots ±1\pm 1, distinct, so TT has an eigenbasis, namely (1,1)(1,1) with eigenvalue 11 and (1,−1)(1,-1) with eigenvalue −1-1. SS is diagonalizable: its characteristic polynomial is det⁡(S−tI)=(1−t)(−1−t)−0=t2−1\det(S - tI) = (1-t)(-1-t) - 0 = t^2 - 1, again roots ±1\pm 1, distinct, so SS too has an eigenbasis. For eigenvalue 11: (S−I)v=0⇒(010−2)v=0⇒v=(1,0)(S-I)v=0 \Rightarrow \begin{pmatrix}0&1\\0&-2\end{pmatrix}v = 0 \Rightarrow v=(1,0). For eigenvalue −1-1: (S+I)v=0⇒(2100)v=0⇒v=(1,−2)(S+I)v=0 \Rightarrow \begin{pmatrix}2&1\\0&0\end{pmatrix}v=0 \Rightarrow v=(1,-2). They do not commute: compute

ST=(110−1)(0110)=(11−10),TS=(0110)(110−1)=(0−111).ST = \begin{pmatrix}1&1\\0&-1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix} = \begin{pmatrix}1&1\\-1&0\end{pmatrix}, \qquad TS = \begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1&1\\0&-1\end{pmatrix} = \begin{pmatrix}0&-1\\1&1\end{pmatrix}.

Since ST≠TSST \ne TS (their (1,1)(1,1)-entries already differ, 1≠01 \ne 0), the operators do not commute. No common eigenbasis exists: the eigenvectors of SS are (up to scale) (1,0)(1,0) and (1,−2)(1,-2); the eigenvectors of TT are (up to scale) (1,1)(1,1) and (1,−1)(1,-1). Any common eigenvector would have to be a scalar multiple of both an SS-eigenvector and a TT-eigenvector; but (1,0)(1,0) is not a scalar multiple of (1,1)(1,1) or (1,−1)(1,-1), and (1,−2)(1,-2) is not a scalar multiple of (1,1)(1,1) or (1,−1)(1,-1) either (check: (1,−2)=c(1,1)(1,-2) = c(1,1) forces c=1c=1 and −2=1-2=1, false; (1,−2)=c(1,−1)(1,-2)=c(1,-1) forces c=1c=1 and −2=−1-2=-1, false). So SS and TT share no common eigenvector at all, let alone a full common eigenbasis — the strongest possible failure of the conclusion. This shows commutativity is doing indispensable work: individual diagonalizability of each operator, without more, guarantees nothing about shared structure between them.

Scope, edge cases & common traps

SituationWhat holdsWhat can fail
S,TS,T diagonalizable and commutingSimultaneously diagonalizable (theorem)—
S,TS,T commuting but not both diagonalizableNeed not share full eigenbasis; may still share invariant subspaces or generalized eigenspaces, but simultaneous diagonalization can fail entirelyNilpotent commuting operators (e.g. S=T=(0100)S=T=\begin{pmatrix}0&1\\0&0\end{pmatrix}) have no eigenbasis at all
S,TS,T diagonalizable but not commutingNo guarantee of shared eigenvectorsSee Counterexample above
Infinite family, pairwise commuting, each diagonalizableSimultaneously diagonalizable if VV is finite-dimensional (same proof: only finitely many joint eigenspaces can be nonzero)Care needed in infinite dimensions (spectral theory required)
Repeated eigenvalues within one operatorMust diagonalize the restriction of the other operator on that eigenspace — cannot just "read off" a basisForgetting this step and picking an arbitrary basis of a repeated eigenspace, which generally is not an eigenbasis for the second operator

The mathematician behind it

Origins · 1 minCamille Jordan and how this was discovered
CJ
Camille Jordan
1838–1922 · 19th-century France, foundational linear algebra and group theory

Camille Jordan was a French mathematician whose work on canonical forms of linear operators — culminating in the Jordan Normal Form — sits at the historical root of the entire theory of diagonalization and its refinements. Jordan's investigations into the structure of matrices under similarity transformations, alongside contemporaries developing the theory of commuting linear substitutions, established the framework in which questions like simultaneous diagonalizability could even be precisely posed: what invariants classify an operator up to change of basis, and how do these invariants interact when several operators must be considered together?

History
- 1870: Jordan publishes Traité des substitutions et des équations algébriques, developing canonical forms for linear substitutions and groundwork for eigenvalue theory.
- 1880s: Frobenius and others develop the theory of commuting matrices and simultaneous reduction, motivated by problems in the theory of algebraic forms and group representations.
- 1900s: The theory of group representations (Frobenius, Schur) makes systematic use of simultaneous diagonalization for commuting families arising from abelian group actions.
- 1920s–1930s: Quantum mechanics (Heisenberg, Dirac, von Neumann) reinterprets simultaneous diagonalizability as the mathematical statement that commuting observables can be measured simultaneously with definite, compatible values.
- Mid-20th century: The result becomes a standard component of linear algebra curricula, stated cleanly in terms of invariant eigenspaces, as presented in modern textbook treatments.
Does the theorem require the field to be algebraically closed, like C\mathbb{C}?

No. The theorem as stated works over any field F\mathbb{F}, because diagonalizability is already assumed as a hypothesis rather than derived from a characteristic polynomial splitting into linear factors. The proof only uses the eigenspace decomposition guaranteed by diagonalizability, and this decomposition is a hypothesis, not something the proof needs to derive from properties of F\mathbb{F}. Algebraic closedness becomes relevant only if one wants a sufficient condition guaranteeing diagonalizability itself (such as "distinct eigenvalues" or "minimal polynomial splits into distinct linear factors"), which is a separate question from simultaneous diagonalization.

If SS and TT commute but neither is individually diagonalizable, can they still be simultaneously diagonalized?

No — simultaneous diagonalizability requires each individual operator to be diagonalizable, since restricting a common eigenbasis to a single operator already gives that operator an eigenbasis. So diagonalizability of each operator is a necessary condition, not just part of a sufficient package. A concrete example: two equal nonzero nilpotent operators trivially commute with each other, but a nilpotent operator has no eigenbasis (its only eigenvalue is 00, and if it were diagonalizable it would have to be the zero operator, a contradiction unless it already is zero).

Is simultaneous diagonalizability the same as saying the operators can be represented by matrices that commute as matrices?

Not quite — commuting as matrices in some basis is the same as commuting as operators, since commutativity of the abstract composition ST=TSST=TS is basis-independent. But that alone is not equivalent to simultaneous diagonalizability; you also need each operator to be individually diagonalizable. Commuting matrices that are not diagonalizable (such as two equal Jordan blocks with eigenvalue 00) certainly commute, yet cannot be jointly diagonalized because they cannot even be individually diagonalized.

How does this connect to invariant subspaces more generally?

The entire proof hinges on the single fact that eigenspaces of one commuting operator are invariant under the other, a special case of the general theory in Invariant Subspaces. More broadly, when operators commute, all of the natural subspaces attached to one operator (eigenspaces, generalized eigenspaces, kernels of polynomials in that operator) become invariant under the other. This is why commuting families are so much easier to analyze jointly than arbitrary families: commutativity systematically preserves each other's natural subspace structure.

Does the theorem generalize to infinite-dimensional vector spaces or to infinite families of operators?

The finite-dimensional statement generalizes cleanly to any finite commuting family of diagonalizable operators, exactly as the Corollary shows, by finite induction. Infinite families or infinite-dimensional spaces require more care: one typically needs additional structure such as compactness or self-adjointness (as in the spectral theorem for commuting families of self-adjoint operators on a Hilbert space) to make sense of joint diagonalization, and the elementary linear-algebraic argument given here does not directly extend without such analytic hypotheses.

If AA and BB are simultaneously diagonalizable, is A+BA+B automatically diagonalizable too?

Yes, and this is one of the most useful consequences. If P−1AP=D1P^{-1}AP = D_1 and P−1BP=D2P^{-1}BP=D_2 are both diagonal for the same PP, then P−1(A+B)P=D1+D2P^{-1}(A+B)P = D_1+D_2 is diagonal, since the sum of diagonal matrices is diagonal. The same argument shows any polynomial combination p(A,B)p(A,B) built from A,BA,B using addition, scalar multiplication, and multiplication is diagonalized by the same PP, since all these operations commute with conjugation by PP and diagonal matrices are closed under them (as long as multiplication order does not matter, which is automatic here since A,BA,B commute).

Why does the proof need to look at Ei∩FjE_i \cap F_j instead of just EiE_i directly?

Because TT restricted to EiE_i can still have several distinct eigenvalues if dim⁡Ei>1\dim E_i > 1; simply asserting "TT is diagonalizable on VV" does not by itself tell us how the eigenvectors of TT distribute across the different SS-eigenspaces. Intersecting each FjF_j (a TT-eigenspace of VV) with EiE_i isolates exactly the vectors in EiE_i that are also TT-eigenvectors for the specific eigenvalue μj\mu_j, and the direct sum decomposition (∗∗)(\ast\ast) is precisely the statement that these intersections exhaust EiE_i — which is what lets us conclude T∣EiT|_{E_i} is diagonalizable.

Before this