The Polar Decomposition

Every complex number zz can be written as z=eiθrz = e^{i\theta} r. Here r=∣z∣≥0r = |z| \ge 0 and eiθe^{i\theta} lies on the unit circle. This factorization splits zz into a "size" and a "direction." It is one of the most useful facts in elementary complex analysis. A natural question drives this article. Can an arbitrary linear operator on an inner product space be split, in the same spirit? The split would be into a "rotation-like" piece and a "size" piece. The answer is yes. The resulting theorem — the polar decomposition — is one of the most structurally satisfying results in operator theory on finite-dimensional inner product spaces. It says that any operator TT can be written as T=ST∗TT = S\sqrt{T^*T}. Here T∗T\sqrt{T^*T} is the unique positive square root of the positive operator T∗TT^*T. And SS is an isometry on the range of T∗T\sqrt{T^*T}. This can be extended to a full isometry when TT is invertible. More generally, it extends to an isometry of the whole space when dimensions match. The operator T∗T\sqrt{T^*T} plays the role of ∣z∣|z|. It measures how TT stretches vectors. The operator SS plays the role of eiθe^{i\theta}. It captures the purely rotational, length-preserving part of the action. This theorem sits at the confluence of two major ideas already established. These are positive operators and their square roots, and the singular value decomposition. Indeed, the polar decomposition is really the SVD written in operator form. This is rather than in terms of orthonormal bases and singular values. The two results are logically equivalent. Each can be derived from the other. Understanding the polar decomposition sharpens one's intuition for what an arbitrary linear map "does" geometrically. It always factors as a positive stretch followed by an isometric repositioning. This article gives the precise definition of an isometry and of T∗T\sqrt{T^*T}. It states and proves the polar decomposition theorem in full. This includes the uniqueness of the positive factor. It also includes the characterization of when the isometric factor is unique. The article works through two asymmetric numerical examples. It shows exactly which hypothesis is essential. It does this by exhibiting what breaks without it.

Key takeaways
  • The polar decomposition writes any operator TT on a finite-dimensional inner product space as T=ST∗TT = S\sqrt{T^*T}.
  • T∗T\sqrt{T^*T} is the unique positive square root of the positive operator T∗TT^*T; it is uniquely determined by TT.
  • SS is an isometry from range⁡T∗T\operatorname{range}\sqrt{T^*T} onto range⁡T\operatorname{range} T, extended (non-uniquely in general) to the whole space.
  • SS is unique on the whole space exactly when TT is invertible (equivalently, T∗T\sqrt{T^*T} is invertible).
  • The theorem is the coordinate-free operator form of the Singular Value Decomposition; each implies the other.
  • Key idea: every linear operator factors as an isometry composed with a positive "stretching" operator, exactly as z=eiθ∣z∣z = e^{i\theta}|z| for complex numbers.

Where this lives today

The polar decomposition is far more than a theoretical curiosity; it is a working computational tool.

Numerical linear algebra. The polar decomposition of a matrix is computed in practice via the Singular Value Decomposition (as shown in the Corollary above) or via efficient iterative algorithms (Newton's method applied to matrix square roots), and is the standard way to find the "closest" isometry (in operator norm or Frobenius norm) to a given matrix — a fact used in orthogonal Procrustes problems in statistics and shape analysis.

Continuum mechanics and elasticity. In continuum mechanics, the deformation gradient tensor FF describing how a material deforms is factored via polar decomposition as F=RUF = R U, where U=FTFU = \sqrt{F^{\mathsf T}F} is the (symmetric, positive-definite) "stretch tensor" describing local stretching/compression, and RR is a rotation describing local rigid rotation. This is the direct physical incarnation of the mathematical theorem: separating "how much a material element stretches" from "how it rotates."

Computer graphics and robotics. Extracting a "clean" rotation matrix from a noisy or skewed linear transformation (e.g. after repeated floating-point matrix multiplications accumulate numerical drift away from orthogonality) is done via polar decomposition: the isometric factor SS is exactly the nearest rotation matrix to the given (possibly non-orthogonal) matrix.

Quantum mechanics and operator algebras. As foreshadowed by von Neumann's original motivation, the polar decomposition of possibly unbounded operators is fundamental to the mathematical formulation of quantum mechanics, where it underlies the theory of self-adjoint extensions and the structure of von Neumann algebras.

Statistics. In multivariate statistics, the polar decomposition of a data matrix is closely related to Procrustes analysis, canonical correlation analysis, and the computation of matrix square roots used in whitening transformations.

Ask the tutor: How does the polar decomposition change if we replace T∗TT^*T with TT∗TT^*, and when do the two versions coincide?

DefinitionIsometry and the Polar Decomposition

Let VV be a finite-dimensional inner product space over F\mathbb{F} (F=R\mathbb{F} = \mathbb{R} or C\mathbb{C}), and let S∈L(V)S \in \mathcal{L}(V) be a linear operator.

Isometry. SS is called an isometry if

∥Sv∥=∥v∥for all v∈V.\|Sv\| = \|v\| \quad \text{for all } v \in V.

Equivalently (as shown below), SS is an isometry if and only if S∗S=IS^*S = I.

Polar decomposition. Let T∈L(V)T \in \mathcal{L}(V). A polar decomposition of TT is a factorization

T=ST∗T,T = S\sqrt{T^*T},

where T∗T\sqrt{T^*T} denotes the unique positive square root of the positive operator T∗TT^*T (guaranteed to exist by Positive Operators and Square Roots), and S∈L(V)S \in \mathcal{L}(V) is an isometry.

Intuition

Think of T∗TT^*T as measuring how TT distorts lengths and inner products: for any vv,

⟨T∗Tv,v⟩=⟨Tv,Tv⟩=∥Tv∥2≥0,\langle T^*Tv, v\rangle = \langle Tv, Tv \rangle = \|Tv\|^2 \ge 0,

so T∗TT^*T is positive, and it has a unique positive square root T∗T\sqrt{T^*T}, itself a positive (hence self-adjoint) operator. The key computation is that T∗T\sqrt{T^*T} and TT send vectors to targets of the same length:

∥T∗T v∥2=⟨T∗T v,T∗T v⟩=⟨T∗Tv,v⟩=∥Tv∥2.\|\sqrt{T^*T}\,v\|^2 = \langle \sqrt{T^*T}\,v, \sqrt{T^*T}\,v\rangle = \langle T^*Tv, v \rangle = \|Tv\|^2.

This is the heart of the matter: the map T∗T v↦Tv\sqrt{T^*T}\,v \mapsto Tv is length-preserving on the range of T∗T\sqrt{T^*T}, so it extends to a genuine isometry SS, and T=ST∗TT = S\sqrt{T^*T}. Picture T∗T\sqrt{T^*T} as a "positive stretch" along an orthonormal eigenbasis: it scales the jj-th eigenvector by its (nonnegative) eigenvalue sjs_j, which will turn out to be exactly the jj-th singular value of TT. Then SS takes each stretched vector sjejs_j e_j and rigidly relocates it to TejTe_j without further distortion — a rotation/reflection-type repositioning, exactly as multiplying by eiθe^{i\theta} rotates a positive real number r=∣z∣r=|z| to the point zz in the complex plane.

Figure
The complex-number analogy: z=eiθrz = e^{i\theta}r versus T=ST∗TT = S\sqrt{T^*T}
r = |z| z = e^{iθ} r θ z = e^{iθ} · r √(T*T) v Tv = S(√(T*T) v) T = S · √(T*T) same length, repositioned by isometry S

Central theorem

TheoremPolar Decomposition

Let VV be a finite-dimensional inner product space over F\mathbb{F}, and let T∈L(V)T \in \mathcal{L}(V). Then: 1. Existence. There exists an isometry S∈L(V)S \in \mathcal{L}(V) such that

T=ST∗T.T = S\sqrt{T^*T}.

2. Uniqueness of the positive factor. The operator T∗T\sqrt{T^*T} appearing above is the unique positive operator P∈L(V)P \in \mathcal{L}(V) such that T=SPT = SP for some isometry SS. That is, PP must equal T∗T\sqrt{T^*T}. 3. Uniqueness of SS. SS can always be chosen so that

S(range⁡T∗T)=range⁡T,S\big(\operatorname{range}\sqrt{T^*T}\big) = \operatorname{range} T,

and this restriction of SS to range⁡T∗T\operatorname{range}\sqrt{T^*T} is uniquely determined by TT. The full operator SS on all of VV is unique if and only if TT is invertible, equivalently if and only if T∗T\sqrt{T^*T} is invertible.

LemmaBasic facts about isometries

Let VV be a finite-dimensional inner product space and S∈L(V)S \in \mathcal{L}(V). The following are equivalent: (a) SS is an isometry, i.e. ∥Sv∥=∥v∥\|Sv\| = \|v\| for all v∈Vv \in V. (b) ⟨Sv,Sw⟩=⟨v,w⟩\langle Sv, Sw \rangle = \langle v, w \rangle for all v,w∈Vv, w \in V. (c) S∗S=IS^*S = I. (d) SS maps every orthonormal basis of VV to an orthonormal list in VV.

Proof

(a) ⇒\Rightarrow (b). This is the polarization identity argument. Over C\mathbb{C}, for all v,w∈Vv,w \in V,

⟨v,w⟩=14∑k=03ik∥v+ikw∥2,\langle v, w\rangle = \frac{1}{4}\sum_{k=0}^{3} i^k \|v + i^k w\|^2,

a standard identity verified by direct expansion of each ∥v+ikw∥2=∥v∥2+∥w∥2+ik⟨v,w⟩+ik‾⟨w,v⟩\|v+i^kw\|^2 = \|v\|^2 + \|w\|^2 + i^k\langle v,w\rangle + \overline{i^k}\langle w,v\rangle summed over k=0,1,2,3k=0,1,2,3 (the ∥v∥2,∥w∥2\|v\|^2,\|w\|^2 terms cancel via ∑ik=0\sum i^k = 0, and one collects the cross terms). Applying this same identity to Sv,SwSv, Sw and using ∥Sv+ikSw∥=∥S(v+ikw)∥=∥v+ikw∥\|Sv+i^kSw\| = \|S(v+i^kw)\| = \|v+i^kw\| (by hypothesis (a), applied to the vector v+ikwv + i^k w) gives ⟨Sv,Sw⟩=⟨v,w⟩\langle Sv,Sw\rangle = \langle v,w\rangle. (Over R\mathbb{R} the analogous identity ⟨v,w⟩=14(∥v+w∥2−∥v−w∥2)\langle v,w\rangle = \frac14(\|v+w\|^2 - \|v-w\|^2) gives the same conclusion.)

(b) ⇒\Rightarrow (c). For all v,wv,w, ⟨S∗Sv,w⟩=⟨Sv,Sw⟩=⟨v,w⟩=⟨Iv,w⟩\langle S^*Sv, w\rangle = \langle Sv, Sw\rangle = \langle v,w\rangle = \langle Iv, w\rangle. Since this holds for all ww and the inner product is nondegenerate, S∗Sv=IvS^*Sv = Iv for all vv, so S∗S=IS^*S = I.

(c) ⇒\Rightarrow (d). Let e1,…,ene_1,\dots,e_n be an orthonormal basis of VV. Then ⟨Sej,Sek⟩=⟨S∗Sej,ek⟩=⟨ej,ek⟩=δjk\langle Se_j, Se_k\rangle = \langle S^*Se_j, e_k\rangle = \langle e_j, e_k\rangle = \delta_{jk}, so Se1,…,SenSe_1,\dots,Se_n is orthonormal.

(d) ⇒\Rightarrow (a). Let e1,…,ene_1,\dots,e_n be an orthonormal basis with Se1,…,SenSe_1,\dots,Se_n orthonormal. Write v=∑jcjejv = \sum_j c_j e_j; then by orthonormality (Parseval), ∥v∥2=∑j∣cj∣2\|v\|^2 = \sum_j |c_j|^2 and Sv=∑jcjSejSv = \sum_j c_j Se_j gives ∥Sv∥2=∑j∣cj∣2\|Sv\|^2 = \sum_j |c_j|^2 since SejSe_j are also orthonormal. Hence ∥Sv∥=∥v∥\|Sv\| = \|v\|. ■\blacksquare

∎
Proof

Proof of the Polar Decomposition Theorem. Step 1: Setup. Since T∗T∈L(V)T^*T \in \mathcal{L}(V) satisfies ⟨T∗Tv,v⟩=∥Tv∥2≥0\langle T^*Tv,v\rangle = \|Tv\|^2 \ge 0 for all vv, and (T∗T)∗=T∗T∗∗=T∗T(T^*T)^* = T^*T^{**} = T^*T, the operator T∗TT^*T is positive (self-adjoint with nonnegative eigenvalues). By Positive Operators and Square Roots, T∗TT^*T has a unique positive square root, which we denote T∗T\sqrt{T^*T}; write P:=T∗TP := \sqrt{T^*T} for brevity. Since PP is positive it is self-adjoint, and P2=T∗TP^2 = T^*T. Step 2: The key length identity. For every v∈Vv \in V,

∥Pv∥2=⟨Pv,Pv⟩=⟨P∗Pv,v⟩=⟨P2v,v⟩=⟨T∗Tv,v⟩=⟨Tv,Tv⟩=∥Tv∥2.\|Pv\|^2 = \langle Pv, Pv\rangle = \langle P^*Pv, v\rangle = \langle P^2 v, v\rangle = \langle T^*Tv, v\rangle = \langle Tv, Tv\rangle = \|Tv\|^2.

Here P∗=PP^* = P was used since PP is self-adjoint, and P2=T∗TP^2 = T^*T from Step 1. Hence

∥Pv∥=∥Tv∥for all v∈V.(∗)\|Pv\| = \|Tv\| \quad \text{for all } v \in V. \tag{$\ast$}

Step 3: Constructing an isometry on the range of PP. Define a map S0:range⁡P→VS_0 : \operatorname{range} P \to V by

S0(Pv):=Tv,v∈V.S_0(Pv) := Tv, \qquad v \in V.

We must check S0S_0 is well defined: if Pv1=Pv2Pv_1 = Pv_2 for v1,v2∈Vv_1, v_2 \in V, then P(v1−v2)=0P(v_1 - v_2) = 0, so by (∗)(\ast), ∥T(v1−v2)∥=∥P(v1−v2)∥=0\|T(v_1-v_2)\| = \|P(v_1-v_2)\| = 0, forcing Tv1=Tv2Tv_1 = Tv_2. Thus S0S_0 is a well-defined function on range⁡P\operatorname{range} P, and it is linear because TT and PP are linear (linearity of S0S_0 follows directly from the defining formula applied to linear combinations). By construction, S0S_0 satisfies

∥S0(Pv)∥=∥Tv∥=∥Pv∥for all v,\|S_0(Pv)\| = \|Tv\| = \|Pv\| \quad \text{for all } v,

using (∗)(\ast) again; so S0S_0 preserves the norm of every vector in range⁡P\operatorname{range} P. Also, by definition, S0∘P=TS_0 \circ P = T, i.e. S0(Pv)=TvS_0(Pv) = Tv for all v∈Vv \in V. Step 4: Extending S0S_0 to an isometry SS on all of VV. We have range⁡P⊆V\operatorname{range} P \subseteq V, a subspace. Let (range⁡P)⊥(\operatorname{range}P)^\perp be its orthogonal complement, so V=range⁡P⊕(range⁡P)⊥V = \operatorname{range}P \oplus (\operatorname{range}P)^\perp. Since S0S_0 is a linear isometry from range⁡P\operatorname{range}P into VV, it maps range⁡P\operatorname{range}P isometrically onto its image S0(range⁡P)=range⁡TS_0(\operatorname{range}P) = \operatorname{range}T (equality holds because S0(Pv)=TvS_0(Pv)=Tv ranges exactly over range⁡T\operatorname{range}T as vv ranges over VV), and in particular S0(range⁡P)S_0(\operatorname{range}P) is a subspace of VV of the same dimension as range⁡P\operatorname{range}P (an isometry is injective, since ∥S0u∥=0⇒∥u∥=0⇒u=0\|S_0 u\| = 0 \Rightarrow \|u\|=0 \Rightarrow u = 0 by (∗)(\ast)-type reasoning restricted to range⁡P\operatorname{range} P, hence injective linear maps preserve dimension of image). Consequently (range⁡T)⊥(\operatorname{range}T)^\perp has dimension dim⁡V−dim⁡(range⁡P)\dim V - \dim(\operatorname{range}P), which equals dim⁡(range⁡P)⊥\dim (\operatorname{range}P)^\perp. Choose orthonormal bases f1,…,fkf_1,\dots,f_k of (range⁡P)⊥(\operatorname{range}P)^\perp and g1,…,gkg_1,\dots,g_k of (range⁡T)⊥(\operatorname{range}T)^\perp (same kk by the dimension count just made), and define S:V→VS : V \to V by

S(u+w):=S0(u)+∑i=1k⟨w,fi⟩gi,u∈range⁡P, w∈(range⁡P)⊥,S(u + w) := S_0(u) + \sum_{i=1}^k \langle w, f_i\rangle g_i, \qquad u \in \operatorname{range}P,\ w \in (\operatorname{range}P)^\perp,

using the direct sum decomposition of every vector. This SS is linear (it is built from linear pieces on a direct sum decomposition) and, by the Pythagorean-type orthogonality of the two summands (image of S0S_0 lies in range⁡T\operatorname{range}T; the second piece lies in (range⁡T)⊥(\operatorname{range}T)^\perp, which is orthogonal to range⁡T\operatorname{range}T), satisfies

∥S(u+w)∥2=∥S0(u)∥2+∥∑i⟨w,fi⟩gi∥2=∥u∥2+∥w∥2=∥u+w∥2,\|S(u+w)\|^2 = \|S_0(u)\|^2 + \Big\|\sum_i \langle w,f_i\rangle g_i\Big\|^2 = \|u\|^2 + \|w\|^2 = \|u+w\|^2,

where ∥∑i⟨w,fi⟩gi∥2=∑i∣⟨w,fi⟩∣2=∥w∥2\|\sum_i \langle w,f_i\rangle g_i\|^2 = \sum_i |\langle w,f_i\rangle|^2 = \|w\|^2 by Parseval's identity applied to the orthonormal basis f1,…,fkf_1,\dots,f_k of (range⁡P)⊥(\operatorname{range}P)^\perp. So SS is an isometry on VV, and SS restricted to range⁡P\operatorname{range}P equals S0S_0, so SP=S0P=TSP = S_0 P = T. This proves existence: T=SP=ST∗TT = SP = S\sqrt{T^*T}. Step 5: Uniqueness of the positive factor PP. Suppose T=S′P′T = S'P' for some isometry S′S' and positive operator P′P'. Then

T∗T=(S′P′)∗(S′P′)=P′∗S′∗S′P′=P′IP′=P′2,T^*T = (S'P')^*(S'P') = P'^*S'^*S'P' = P' I P' = P'^2,

using P′∗=P′P'^* = P' (self-adjointness of the positive operator P′P') and S′∗S′=IS'^*S' = I (Lemma, (a)⇔\Leftrightarrow(c), since S′S' is an isometry). So P′P' is a positive operator with P′2=T∗TP'^2 = T^*T; by the uniqueness of positive square roots (established in Positive Operators and Square Roots), P′=T∗T=PP' = \sqrt{T^*T} = P. This proves the positive factor is forced to be PP. Step 6: Uniqueness of SS on range⁡P\operatorname{range} P, and full uniqueness iff TT invertible. From Step 5, any valid factorization has P′=PP' = P, so T=S′PT = S'P for the isometry S′S'. For any vv, S′(Pv)=TvS'(Pv) = Tv, which is exactly the defining formula for S0S_0; so S′S' agrees with S0S_0 on range⁡P\operatorname{range}P, i.e. the restriction of SS to range⁡P\operatorname{range}P is uniquely determined by TT (it must equal S0S_0). This proves the uniqueness claim on range⁡P\operatorname{range}P. If TT is invertible: since ∥Pv∥=∥Tv∥\|Pv\| = \|Tv\| for all vv by (∗)(\ast), and TT invertible means Tv=0⇒v=0Tv=0 \Rightarrow v=0, we get Pv=0⇒∥Tv∥=0⇒v=0Pv = 0 \Rightarrow \|Tv\|=0 \Rightarrow v = 0, so PP is injective, hence (as VV is finite-dimensional) bijective, i.e. range⁡P=V\operatorname{range}P = V. Then the isometry SS is entirely determined by its values on range⁡P=V\operatorname{range}P = V, i.e. S=S0S = S_0 is unique on all of VV. Conversely, if TT is not invertible, then (again by (∗)(\ast)) PP is not invertible either, since Pv=0  ⟺  Tv=0Pv=0 \iff Tv = 0 pointwise identifies the kernels of PP and TT as equal. So range⁡P≠V\operatorname{range}P \ne V, meaning (range⁡P)⊥≠{0}(\operatorname{range}P)^\perp \ne \{0\}. In the construction of Step 4, one is free to choose any isometry from (range⁡P)⊥(\operatorname{range}P)^\perp onto (range⁡T)⊥(\operatorname{range}T)^\perp (there is more than one such isometry whenever (range⁡P)⊥(\operatorname{range}P)^\perp has dimension ≥1\ge 1, since one may compose any fixed choice with any orthogonal/unitary operator on (range⁡P)⊥(\operatorname{range}P)^\perp, of which there is more than one map when the dimension is positive — e.g. multiplication by −1-1 on a one-dimensional space, or any rotation in a higher-dimensional real space). Each such choice yields a different full operator SS on VV, all of which satisfy T=SPT = S P. Hence SS is not unique on all of VV when TT is not invertible. ■\blacksquare

∎
CorollaryRelation to the SVD

Let T∈L(V)T \in \mathcal{L}(V) have singular values s1≥⋯≥sn≥0s_1 \ge \cdots \ge s_n \ge 0 (with n=dim⁡Vn = \dim V), so that by the Singular Value Decomposition there exist orthonormal bases e1,…,ene_1,\dots,e_n and f1,…,fnf_1,\dots,f_n of VV with

Tv=∑j=1nsj⟨v,ej⟩fj.Tv = \sum_{j=1}^n s_j \langle v, e_j\rangle f_j.

Then in the polar decomposition T=ST∗TT = S\sqrt{T^*T}, the operator T∗T\sqrt{T^*T} acts by

T∗T v=∑j=1nsj⟨v,ej⟩ej,\sqrt{T^*T}\, v = \sum_{j=1}^n s_j \langle v, e_j\rangle e_j,

i.e. it has eigenvalues exactly the singular values sjs_j with eigenvectors eje_j; and one may take SS to be the isometry sending ej↦fje_j \mapsto f_j for each jj.

Proof

By definition of the SVD, T∗Tv=∑jsj2⟨v,ej⟩ejT^*Tv = \sum_j s_j^2 \langle v,e_j\rangle e_j for all vv (this is the standard computation T∗fi=sieiT^*f_i = s_i e_i, giving T∗T=∑jsj2⟨⋅,ej⟩ejT^*T = \sum_j s_j^2 \langle \cdot, e_j\rangle e_j; it is part of the content of the SVD construction). This operator is positive with eigenvalues sj2≥0s_j^2 \ge 0 on the orthonormal eigenbasis eje_j. The operator QQ defined by Qv=∑jsj⟨v,ej⟩ejQv = \sum_j s_j\langle v,e_j\rangle e_j is manifestly positive (self-adjoint, since it is diagonal in an orthonormal basis with real nonnegative diagonal entries, and ⟨Qv,v⟩=∑jsj∣⟨v,ej⟩∣2≥0\langle Qv,v\rangle = \sum_j s_j|\langle v,e_j\rangle|^2 \ge 0) and satisfies Q2=T∗TQ^2 = T^*T by direct computation using orthonormality of eje_j. By the uniqueness of positive square roots, Q=T∗TQ = \sqrt{T^*T}. Now define SS by Sej=fjSe_j = f_j for each jj; since both {ej}\{e_j\} and {fj}\{f_j\} are orthonormal bases, SS is an isometry by (d)⇒\Rightarrow(a) of the Lemma. Then

S(T∗T v)=S(∑jsj⟨v,ej⟩ej)=∑jsj⟨v,ej⟩fj=Tv,S(\sqrt{T^*T}\,v) = S\Big(\sum_j s_j\langle v,e_j\rangle e_j\Big) = \sum_j s_j \langle v,e_j\rangle f_j = Tv,

so ST∗T=TS\sqrt{T^*T} = T, matching the theorem. ■\blacksquare

∎

Worked examples

Example 1: A non-invertible, non-symmetric 2×22\times 2 real matrix. Let V=R2V = \mathbb{R}^2 with the standard inner product, and let TT have matrix

T=(3040)T = \begin{pmatrix} 3 & 0 \\ 4 & 0 \end{pmatrix}

with respect to the standard basis. Note TT is deliberately asymmetric and singular (det⁡T=0\det T = 0). Step 1: compute T∗TT^*T. Since TT is real, T∗=TTT^* = T^{\mathsf T}:

TT=(3400),TTT=(3400)(3040)=(9+16000)=(25000).T^{\mathsf T} = \begin{pmatrix} 3 & 4 \\ 0 & 0\end{pmatrix}, \qquad T^{\mathsf T}T = \begin{pmatrix} 3 & 4 \\ 0 & 0\end{pmatrix}\begin{pmatrix} 3 & 0 \\ 4 & 0 \end{pmatrix} = \begin{pmatrix} 9+16 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 25 & 0 \\ 0 & 0\end{pmatrix}.

Step 2: find T∗T\sqrt{T^*T}. Since T∗TT^*T is already diagonal with nonnegative entries 25,025, 0 on the standard orthonormal basis, its unique positive square root is obtained by taking nonnegative square roots of the eigenvalues in the same eigenbasis:

P:=T∗T=(5000).P := \sqrt{T^*T} = \begin{pmatrix} 5 & 0 \\ 0 & 0 \end{pmatrix}.

(Check: P2=(25000)=T∗TP^2 = \begin{pmatrix}25&0\\0&0\end{pmatrix} = T^*T, and PP is symmetric with eigenvalues 5,0≥05,0\ge 0, so PP is positive. Uniqueness guarantees this is the square root.) Step 3: find an isometry SS with T=SPT = SP. We have range⁡P=span⁡{(1,0)}\operatorname{range}P = \operatorname{span}\{(1,0)\} (the image of e1=(1,0)e_1=(1,0) under PP is (5,0)(5,0); the image of e2e_2 is 00). Define S0S_0 on range⁡P\operatorname{range}P by S0(Pe1)=Te1S_0(Pe_1) = Te_1, i.e. S0(5,0)=(3,4)S_0(5,0) = (3,4), so S0(1,0)=(3/5,4/5)S_0(1,0) = (3/5,4/5). This is indeed a unit vector, consistent with S0S_0 being an isometry there (∥(3/5,4/5)∥=1=∥(1,0)∥\|(3/5,4/5)\|=1=\|(1,0)\|). Now (range⁡P)⊥=span⁡{(0,1)}(\operatorname{range}P)^\perp = \operatorname{span}\{(0,1)\} and range⁡T=span⁡{(3,4)}\operatorname{range}T = \operatorname{span}\{(3,4)\} so (range⁡T)⊥=span⁡{(−4,5)}(\operatorname{range}T)^\perp = \operatorname{span}\{(-4,5)\} (normalized: (−4/5,3/5)(-4/5,3/5), since (3,4)⋅(−4,3)=0(3,4)\cdot(-4,3)=0). Choose the isometry sending (0,1)↦(−4/5,3/5)(0,1) \mapsto (-4/5,3/5). So

S=(3/5−4/54/53/5),S = \begin{pmatrix} 3/5 & -4/5 \\ 4/5 & 3/5\end{pmatrix},

which is a rotation matrix (check: columns are orthonormal, det⁡S=9/25+16/25=1\det S = 9/25+16/25=1). Verification:

SP=(3/5−4/54/53/5)(5000)=(3040)=T.✓SP = \begin{pmatrix} 3/5 & -4/5 \\ 4/5 & 3/5\end{pmatrix}\begin{pmatrix}5&0\\0&0\end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 4 & 0\end{pmatrix} = T. \checkmark

As guaranteed by the theorem, since TT is not invertible, there was a genuine choice in the second column of SS (any unit vector orthogonal to (3/5,4/5)(3/5,4/5) would have worked, e.g. (4/5,−3/5)(4/5,-3/5) instead, giving a reflection matrix instead of a rotation) — confirming non-uniqueness of SS when TT is singular.

Example 2: An invertible 2×22\times 2 complex matrix. Let V=C2V = \mathbb{C}^2 with the standard Hermitian inner product, and

T=(20i1).T = \begin{pmatrix} 2 & 0 \\ i & 1 \end{pmatrix}.

This TT is invertible (det⁡T=2≠0\det T = 2 \ne 0), non-normal, and has complex entries — a genuinely asymmetric test case. Step 1: compute T∗TT^*T. Here T∗=T‾T=(2−i01)T^* = \overline{T}^{\mathsf T} = \begin{pmatrix} 2 & -i \\ 0 & 1\end{pmatrix}. Then

T∗T=(2−i01)(20i1)=(4+1−ii1)=(5−ii1).T^*T = \begin{pmatrix} 2 & -i \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ i & 1\end{pmatrix} = \begin{pmatrix} 4 + 1 & -i \\ i & 1 \end{pmatrix} = \begin{pmatrix} 5 & -i \\ i & 1\end{pmatrix}.

(Check Hermitian: the (1,2)(1,2) entry is −i-i, the conjugate of the (2,1)(2,1) entry ii; correct, and diagonal entries 5,15,1 are real.) Step 2: diagonalize T∗TT^*T. Characteristic polynomial:

det⁡(T∗T−λI)=(5−λ)(1−λ)−(−i)(i)=(5−λ)(1−λ)−1.\det(T^*T - \lambda I) = (5-\lambda)(1-\lambda) - (-i)(i) = (5-\lambda)(1-\lambda) - 1.

Expand: (5−λ)(1−λ)=5−5λ−λ+λ2=λ2−6λ+5(5-\lambda)(1-\lambda) = 5 - 5\lambda - \lambda + \lambda^2 = \lambda^2 - 6\lambda + 5. So

det⁡(T∗T−λI)=λ2−6λ+5−1=λ2−6λ+4.\det(T^*T-\lambda I) = \lambda^2 - 6\lambda + 5 - 1 = \lambda^2 - 6\lambda + 4.

Roots: λ=6±36−162=6±202=3±5\lambda = \dfrac{6 \pm \sqrt{36-16}}{2} = \dfrac{6\pm\sqrt{20}}{2} = 3 \pm \sqrt5. Both eigenvalues 3+5≈5.2363+\sqrt5 \approx 5.236 and 3−5≈0.7643-\sqrt5 \approx 0.764 are positive, consistent with T∗TT^*T being positive (indeed positive definite, since TT is invertible). Eigenvectors. For λ1=3+5\lambda_1 = 3+\sqrt5: solve (5−λ1)x−iy=0(5-\lambda_1)x - iy = 0, i.e. x(2−5)=iyx(2-\sqrt5) = iy, so y=−i(2−5)x=i(5−2)xy = -i(2-\sqrt5)x = i(\sqrt5-2)x. Take x=1x=1: eigenvector u1=(1, i(5−2))u_1 = (1,\ i(\sqrt5-2)), with norm2^2 =1+(5−2)2=1+5−45+4=10−45= 1 + (\sqrt5-2)^2 = 1 + 5 - 4\sqrt5+4 = 10-4\sqrt5. For λ2=3−5\lambda_2 = 3-\sqrt5: similarly x(2+5)=iy⇒y=−i(2+5)xx(2+\sqrt5) = iy \Rightarrow y = -i(2+\sqrt5)x, eigenvector u2=(1, −i(2+5))u_2 = (1,\,-i(2+\sqrt5)), norm2=1+(2+5)2=1+4+45+5=10+45^2 = 1+(2+\sqrt5)^2 = 1+4+4\sqrt5+5 = 10+4\sqrt5. (Orthogonality check: ⟨u1,u2⟩=1⋅1‾+[i(5−2)][−i(2+5)]‾=1+i(5−2)⋅i(2+5)=1+i2(5−2)(5+2)=1−(5−4)=0\langle u_1,u_2\rangle = 1\cdot\overline{1} + [i(\sqrt5-2)]\overline{[-i(2+\sqrt5)]} = 1 + i(\sqrt5-2)\cdot i(2+\sqrt5) = 1 + i^2(\sqrt5-2)(\sqrt5+2) = 1 - (5-4) = 0. ✓\checkmark Orthogonal, as required for eigenvectors of a Hermitian operator with distinct eigenvalues.) Step 3: Assemble P=T∗TP = \sqrt{T^*T}. Normalizing, let u^1=u1/∥u1∥\hat u_1 = u_1/\|u_1\|, u^2=u2/∥u2∥\hat u_2 = u_2/\|u_2\|. Then in this orthonormal eigenbasis,

P=λ1 u^1u^1∗+λ2 u^2u^2∗,λ1=3+5≈2.288,  λ2=3−5≈0.874.P = \sqrt{\lambda_1}\,\hat u_1 \hat u_1^* + \sqrt{\lambda_2}\,\hat u_2\hat u_2^*, \qquad \sqrt{\lambda_1} = \sqrt{3+\sqrt5}\approx 2.288,\ \ \sqrt{\lambda_2} = \sqrt{3-\sqrt5}\approx 0.874.

This is the exact numerical form of T∗T\sqrt{T^*T}; while the explicit 2×22\times2 matrix entries could be written out by expanding the rank-one projections, the essential content — and the reason this example is included — is: Step 4: uniqueness of SS. Since TT is invertible, PP is invertible (both eigenvalues are strictly positive, as verified above), so by the theorem the isometry SS in T=SPT = SP is uniquely determined, given explicitly by S=TP−1S = TP^{-1}. Unlike Example 1, there is no ambiguity: no orthogonal complement of range⁡P\operatorname{range}P exists (it is all of C2\mathbb{C}^2), so there is no freedom left in defining SS. This illustrates precisely the dichotomy proved in Step 6 of the theorem's proof: singular TT (Example 1) ⇒\Rightarrow non-unique SS; invertible TT (Example 2) ⇒\Rightarrow unique SS.

CounterexampleWhy finite-dimensionality (or at least the existence of T∗T\sqrt{T^*T}) is essential

The proof of existence leaned on two finite-dimensional facts: (i) the positive square root T∗T\sqrt{T^*T} exists (from Positive Operators and Square Roots, itself proved using the spectral theorem for self-adjoint operators, which in the form used requires finite dimension), and (ii) the dimension-counting argument in Step 4 that lets us extend S0S_0 to a full isometry, which uses dim⁡(range⁡P)⊥=dim⁡(range⁡T)⊥<∞\dim(\operatorname{range}P)^\perp = \dim(\operatorname{range}T)^\perp < \infty. To see concretely what can go wrong without a well-behaved notion of "positive square root," consider dropping self-adjointness from the hypothesis on the second factor rather than working infinite-dimensionally (a cleaner way to see the necessity of positivity, not just some factor). Take V=R2V = \mathbb{R}^2 and try to factor

T=(1101)T = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}

as T=SQT = SQ where SS is an isometry (equivalently orthogonal, since we are over R\mathbb{R}) and QQ is merely invertible but not required to be positive. This is easy: take S=IS = I and Q=TQ = T itself — after all TT is already invertible. This "decomposition" trivially exists for any invertible TT, no matter how badly behaved, and carries no information: it does not tell us how TT stretches space, because Q=TQ=T could be anything. It also is wildly non-unique: for any invertible QQ we can set S=TQ−1S = TQ^{-1} provided TQ−1TQ^{-1} happens to be an isometry, but if we drop positivity there is no constraint pinning QQ down as we found in Step 5 of the proof (which crucially used P′∗=P′P'^*=P' to conclude P′2=T∗TP'^2 = T^*T forces P′=T∗TP' = \sqrt{T^*T}). Concretely: both

Q1=T∗T=(11/21/25/2)  (the genuine positive root, computed the same way as Example 2’s method)Q_1 = \sqrt{T^*T} = \begin{pmatrix} 1 & 1/2 \\ 1/2 & \sqrt{5}/2 \end{pmatrix} \ \ (\text{the genuine positive root, computed the same way as Example 2's method})

and, say, Q2=(1011)Q_2 = \begin{pmatrix} 1 & 0\\ 1 & 1\end{pmatrix} (not symmetric, hence not positive) can serve as the second factor for some choice of first factor if we only demand the first factor be invertible rather than isometric — e.g. T=(TQ2−1)Q2T = (TQ_2^{-1})Q_2 trivially. This shows that without insisting the "size" factor be positive, the entire content of the theorem — a unique, geometrically meaningful measure of stretching — collapses: any invertible operator admits infinitely many "factorizations" T=ABT = AB with AA invertible and BB arbitrary invertible, none of them canonical. Positivity of T∗T\sqrt{T^*T} (equivalently, self-adjointness plus nonnegative spectrum) is exactly the hypothesis that rules out this degeneracy and pins down a unique factor, as proved rigorously in Step 5.

Scope, edge cases & common traps

SituationWhat happens
TT invertibleP=T∗TP = \sqrt{T^*T} invertible; S=TP−1S = TP^{-1} is the unique isometry with T=SPT=SP
TT singular, T≠0T\ne 0PP singular; SS unique only on range⁡P\operatorname{range}P, free (non-unique) on (range⁡P)⊥(\operatorname{range}P)^\perp
T=0T = 0P=0P = 0; any isometry SS works, since S⋅0=0=TS\cdot 0 = 0 = T for every isometry SS
TT self-adjoint, not positivePolar decomposition still exists; SS need not be II (e.g. TT negative definite gives S=−IS=-I on relevant eigenspaces)
F=R\mathbb{F}=\mathbb{R} vs C\mathbb{C}Theorem holds verbatim in both cases; "isometry" = orthogonal operator over R\mathbb{R}, unitary-type over C\mathbb{C} (though SS need not be globally unitary unless TT invertible and dim⁡=n\dim=n; it is always at least an isometry)
Infinite dimensionsAnalogous polar decomposition exists for bounded operators on Hilbert space, but the proof of existence of T∗T\sqrt{T^*T} and the extension step require operator-theoretic machinery (spectral theorem for bounded self-adjoint operators) beyond this finite-dimensional treatment

The mathematician behind it

Origins · 1 minJohn von Neumann and how this was discovered
JV
John von Neumann
1903–1957 · Mathematical foundations of operator theory and quantum mechanics

John von Neumann developed the rigorous mathematical foundations of operator theory on Hilbert spaces in the late 1920s, motivated in large part by the need for a precise footing for quantum mechanics. The polar decomposition, in its general (including infinite-dimensional, unbounded-operator) form, is a cornerstone of his theory of operator algebras and self-adjoint operators, generalizing the elementary matrix fact to arbitrary bounded — and even unbounded, densely defined — linear operators on Hilbert space.

Von Neumann's broader program treated operators as the natural objects of study for quantum observables, where self-adjoint operators correspond to measurable physical quantities and unitary operators correspond to symmetries and time evolution. The polar decomposition, splitting an arbitrary operator into a "positive-definite content" and a "unitary repositioning," fit precisely into this program: it let von Neumann and his contemporaries analyze arbitrary bounded operators by reducing questions about them to two better-understood classes — positive operators (whose spectral theory was already well developed) and unitary/isometric operators.

Origins · 1 minhow this was discovered
History
- 1929 — von Neumann's foundational papers establish rigorous Hilbert space operator theory, including precursors to the general polar decomposition.
- 1932 — von Neumann's Mathematische Grundlagen der Quantenmechanik systematizes self-adjoint and unitary operators as the language of quantum observables and symmetries.
- 1936 — Murray and von Neumann's work on operator algebras ("rings of operators") further develops polar decomposition techniques for operators in von Neumann algebras.
- 1960s — the finite-dimensional polar decomposition becomes a standard topic in linear algebra textbooks, tied closely to the newly popularized singular value decomposition.
- 2000s–present — the polar decomposition is a standard computational tool in numerical linear algebra, robotics, and computer graphics for extracting a "closest rotation" to a general linear transformation.
Why must T∗T\sqrt{T^*T} specifically appear, rather than some other square root of T∗TT^*T?

Because T∗TT^*T is a positive operator, it has exactly one square root that is itself positive — this is the content of the uniqueness half of the theory of Positive Operators and Square Roots. A self-adjoint operator can have several self-adjoint square roots (for instance, ±\pm the positive root on different eigenspaces), but only one of them has an entirely nonnegative spectrum. The proof of the polar decomposition's uniqueness step shows algebraically that any valid second factor P′P' in T=S′P′T=S'P' must satisfy P′2=T∗TP'^2 = T^*T and be positive, and positivity is exactly what singles out T∗T\sqrt{T^*T} among all self-adjoint square roots of T∗TT^*T.

Is the polar decomposition the same as writing T=UPT = UP versus T=PUT = PU?

These are genuinely different (though closely related) factorizations, sometimes called the left and right polar decompositions. The version proved here, T=ST∗TT = S\sqrt{T^*T}, is the "right" polar decomposition (the positive factor multiplies on the right of SS... actually here T∗T\sqrt{T^*T} is applied first, then SS, so as operators T=S∘T∗TT = S \circ \sqrt{T^*T}). There is an analogous "left" decomposition T=TT∗ S′T = \sqrt{TT^*}\, S' for a possibly different isometry S′S', built by applying the same theorem to T∗T^* and taking adjoints. Both exist simultaneously for any TT, and they generally involve different positive operators, T∗T\sqrt{T^*T} versus TT∗\sqrt{TT^*}, unless TT is normal.

Does the polar decomposition theorem require TT to be invertible?

No. Existence of the polar decomposition T=ST∗TT=S\sqrt{T^*T} holds for every linear operator TT on a finite-dimensional inner product space, invertible or not — this is exactly what Steps 1 through 4 of the proof establish without ever assuming invertibility. What invertibility affects is only the uniqueness of the isometric factor SS: it is uniquely determined on all of VV precisely when TT is invertible, and only uniquely determined on range⁡T∗T\operatorname{range}\sqrt{T^*T} otherwise (Step 6 of the proof). So invertibility sharpens uniqueness, but is never needed for existence.

How does the polar decomposition relate to writing a complex number as reiθre^{i\theta}?

The analogy is precise on one-dimensional spaces: if V=CV=\mathbb{C} (viewed as a one-dimensional complex inner product space) and TT is "multiplication by zz," then T∗T^* is multiplication by zˉ\bar z, so T∗TT^*T is multiplication by ∣z∣2≥0|z|^2 \ge 0, and T∗T\sqrt{T^*T} is multiplication by ∣z∣=r|z| = r. An isometry on C1\mathbb{C}^1 is exactly multiplication by a unit complex number eiθe^{i\theta}. So the polar decomposition T=ST∗TT = S\sqrt{T^*T} becomes literally z=eiθ⋅rz = e^{i\theta} \cdot r, recovering the classical polar form of a complex number as the special one-dimensional case of the general operator theorem.

Why is SS called an isometry rather than "unitary" or "orthogonal"?

In general, the theorem only guarantees SS preserves norms of all vectors in VV — this is what "isometry" (as defined here) means, and by the Lemma this is equivalent to S∗S=IS^*S=I. When TT is invertible, SS additionally satisfies SS∗=ISS^*=I (since then SS is bijective and S∗S=IS^*S=I forces S−1=S∗S^{-1}=S^*, hence SS∗=ISS^*=I too), making SS unitary (over C\mathbb{C}) or orthogonal (over R\mathbb{R}) in the fullest sense. But when TT is singular, the isometry constructed is only guaranteed to satisfy S∗S=IS^*S=I; it is still true that SS is invertible on all of VV by the explicit construction in Step 4 (it maps one orthonormal basis to another), so in finite dimensions "isometry" and "unitary/orthogonal" actually coincide for operators on the same finite-dimensional space — the terminology "isometry" is used mainly for uniformity with contexts (e.g. maps between different spaces) where the distinction matters.

Can the polar decomposition fail to exist in infinite dimensions?

The finite-dimensional proof given here relies on spectral theory results proved for finite-dimensional self-adjoint operators. In infinite-dimensional Hilbert spaces, an analogous polar decomposition does exist for bounded (and even suitably defined unbounded, densely defined) operators, but its proof requires the spectral theorem for possibly unbounded self-adjoint operators, a substantially more delicate piece of functional analysis developed by von Neumann and others. The statement of the theorem generalizes essentially verbatim; the proof technique (dimension counting for the orthogonal complement, existence of eigenbases) must be replaced by measure-theoretic spectral calculus.

Before this