Part 2 · Theory of deformation

2.4 Polar decomposition

Source: Carboni’s notes for lesson 3, Teoria della deformazione 2, pp. 7–10, and his correction on Classroom (5 Oct 2026)

Objectives

Theorem. Every admissible $F$ is a pure stretch followed by a rotation, $F = RU$ , or the same rotation followed by another stretch, $F = VR$ .
Picture. Sphere → ellipsoid along $m_{k}$ → turned into place; or sphere → turned → ellipsoid along $\overset{˘}{m}_{k}$ .
Theorem. $C = U^{2}$ : the eigenvalues of $U$ are the principal stretches, and $V$ has the same ones on the rotated directions.
Recipe. Compute $U$ , $R$ and $V$ from $F$ .

Key observation 2.2.10 said that $C$ sees every length and angle but not a rigid rotation. The polar decomposition makes that split explicit: it factors $F$ into a part that deforms and a part that only turns.

The theorem

A matrix $R$ is orthogonal if $R^{T} R = I$ , that is $R^{T} = R^{- 1}$ : it keeps lengths and angles. With $det R = 1$ it is a rotation (with $- 1$ , a reflection).

Theorem 2.4.1Cauchy’s polar decompositionnotes L3, p. 7

A real non-singular $F$ can be written as $F = R U = V R$ with $R$ orthogonal and $U$ , $V$ symmetric and positive definite.

$U$ is the right stretch tensor (tensore destro), $V$ the left stretch tensor (tensore sinistro).
Since $det F > 0$ , $R$ can be chosen with $det R = 1$ : a rotation.

The notes also recall $R^{T} = R^{- 1} = \frac{1}{d e t R} R^{(a)}$ , with $R^{(a)}$ the adjugate: the inverse of an orthogonal matrix is its transpose.

Key observation 2.4.2stretch, then rotatenotes L3, p. 7

Locally every deformation is a pure stretch followed by a rigid rotation, or the same rotation followed by a different pure stretch.

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Figure 2.4.1 The polar decomposition in two steps. On the left, drag

ν_{1}

and

ν_{2}

to set

F

. On the right, run the step slider (or play): along

RU

the circle is stretched along the principal directions

m_{k}

and then turned; along

VR

it is turned first and then stretched along

\overset{˘}{m}_{k}

. Both end on the dotted ellipse of

F

Where the theorem comes frombeyond the lectures

Build $U$ first: $C = F^{T} F$ is symmetric positive definite, with eigenpairs $(λ_{k}, m_{k})$ , $λ_{k} > 0$ . Its square root $U = \sum_{k} λ_{k} m_{k} m_{k}^{T}$ is symmetric positive definite with $U^{2} = C$ . Then set $R = F U^{- 1}$ and check it is orthogonal: $R^{T} R = U^{- T} F^{T} F U^{- 1} = U^{- 1} U^{2} U^{- 1} = I .$ $det R = det F / det U > 0$ , so it is a rotation. Finally $V = RU R^{T}$ gives $VR = RU R^{T} R = RU = F$ .

C and U

Proposition 2.4.3C = U²notes L3, pp. 7–8

$C = U^{2} .$

Proof

Substitute $F = RU$ , using $U^{T} = U$ and $R^{T} R = I$ : $C = F^{T} F = (RU)^{T} RU = U^{T} R^{T} R U = U^{T} I U = UU = U^{2} .$

The rotation drops out: $C$ only sees the stretch part.

Corollary 2.4.4eigenvalues of Unotes L3, p. 8

$C$ and $U$ have the same eigenvectors, and the eigenvalues of $C$ are the squares of those of $U$ : $U m_{k} = ν_{k} m_{k}, σ_{k}^{2} = λ_{k}, σ_{k} = ν_{k} = λ_{k} .$ The eigenvalues of $U$ are the principal stretches; their squares are the eigenvalues of $C$ .

Proof

Let $U m_{k} = σ_{k} m_{k}$ . Apply $U$ again: $C m_{k} = UU m_{k} = U σ_{k} m_{k} = σ_{k} U m_{k} = σ_{k}^{2} m_{k} .$ So $m_{k}$ is an eigenvector of $C$ with eigenvalue $λ_{k} = σ_{k}^{2}$ , which by Theorem 2.3.2 is $ν_{k}^{2}$ .

Warningslip in the notes, corrected by Carboni

L3 p. 9 says the eigenvalues of $C$ are the squares of the principal stretches “e coincidono con gli autovalori di $U$ ”, which reads as if $C$ and $U$ had the same eigenvalues. Carboni corrected this on Classroom on 5 Oct 2026: the eigenvalues of $U$ are the principal stretches, and their squares are the eigenvalues of $C$ . The derivation on pp. 7–8 already says so.

In the principal basis both are diagonal: $C = ν_{1}^{2} 00 0 ν_{2}^{2} 0 00 ν_{3}^{2}, U = ν_{1} 00 0 ν_{2} 0 00 ν_{3} .$

What U and R do to a fibre

Split the action of $F$ on a fibre: $\overset{˘}{P} \overset{˘}{Q} = F PQ = R (U PQ) = R PQ^{'}, PQ^{'} = U PQ .$

$PQ^{'}$ is the fibre only stretched: $U$ is the stretch tensor.
$R$ then turns it rigidly into place.

Along a principal direction, $PQ = ∣ PQ ∣ m_{k}$ : $PQ^{'} = U m_{k} ∣ PQ ∣ = ν_{k} ∣ PQ ∣ m_{k} .$ The fibre keeps its direction and its length is multiplied by $ν_{k}$ . A fibre that is not principal is also turned a little by $U$ (beyond the lectures), since $Um$ is parallel to $m$ only for eigenvectors: watch a diagonal of the square in Figure 2.4.1 during the first step.

Warningslip in the notes

L3 p. 7 writes $PQ^{'} = U m ∣ PQ ∣ = ν_{m} m$ . This holds only for a principal direction $m$ (an eigenvector of $U$ ), and the right side has lost the factor $∣ PQ ∣$ .

The principal directions after the deformation

Proposition 2.4.5the left stretch tensornotes L3, p. 9

$\overset{˘}{m}_{k} = R m_{k}, V \overset{˘}{m}_{k} = ν_{k} \overset{˘}{m}_{k} .$ The deformed principal directions are the reference ones turned by $R$ ; $V$ has the same eigenvalues as $U$ , on the turned directions.

Proof

Apply $F = RU$ to a principal direction: $ν_{k} = F m_{k} = R U m_{k} = ν_{k} R m_{k},$ and $ν_{k} = ν_{k} \overset{˘}{m}_{k}$ by Definition 2.2.6, so $\overset{˘}{m}_{k} = R m_{k}$ . Now apply $F = VR$ to the same vector: $ν_{k} = F m_{k} = V R m_{k} = V \overset{˘}{m}_{k},$ and this equals $ν_{k} \overset{˘}{m}_{k}$ . So $(ν_{k}, \overset{˘}{m}_{k})$ is an eigenpair of $V$ .

Proposition 2.4.6spectral formsnotes L3, p. 9

$U = k = 1 \sum 3 ν_{k} m_{k} m_{k}^{T} = ν_{1} m_{1} m_{1}^{T} + ν_{2} m_{2} m_{2}^{T} + ν_{3} m_{3} m_{3}^{T}, V = k = 1 \sum 3 ν_{k} \overset{˘}{m}_{k} \overset{˘}{m}_{k}^{T} .$

Proofbeyond the lectures

$m_{k} m_{k}^{T}$ (a column times a row: a $3 \times 3$ matrix, a dyad) is the projector onto $m_{k}$ : $(m_{k} m_{k}^{T}) a = m_{k} (m_{k} \cdot a)$ . Applied to $m_{j}$ , the sum gives $\sum_{k} ν_{k} m_{k} (m_{k}^{T} m_{j}) = ν_{j} m_{j}$ , because $m_{k}^{T} m_{j}$ is 1 for $k = j$ and 0 otherwise. Both sides act the same way on a basis, so they are the same matrix. The same argument with $\overset{˘}{m}_{k}$ gives $V$ .

Warningslip in the notes

L3 p. 9 writes the terms as $ν_{k} m_{k}^{T} m_{k}$ , each preceded by a stray basis vector. $m_{k}^{T} m_{k} = 1$ is a number; the tensor is the dyad $m_{k} m_{k}^{T}$ , and there is no basis-vector prefix. Same for $V$ .

Key observation 2.4.7the commutative diagramnotes L3, p. 10

The notes close lesson 3 with a square diagram:

top left, a small sphere at $P$ with the principal directions $m_{1}, m_{2}, m_{3}$ ;
across the top, $U$ : the sphere becomes an ellipsoid with semi-axes $ν_{k} m_{k}$ , still along the original directions;
down the right, $R$ : the ellipsoid is turned, its axes becoming $ν_{k} \overset{˘}{m}_{k}$ ;
the other way round: $R$ down the left (axes $\overset{˘}{m}_{k}$ ), then $V$ along the bottom.

Both paths end at the same ellipsoid: $RU = VR$ . The two buttons of Figure 2.4.1 are the two paths.

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Figure 2.4.2 The same in 3D. The principal stretches and frame set

U

; the last slider is a rotation

R

about

e_{3}

. Turning

R

moves the ellipsoid but leaves

C

, its eigenvalues and the dashed

m_{k}

alone.

Computing the decomposition

Recipe 2.4.8polar decompositionbeyond the lectures

$C = F^{T} F$ , its eigenvalues $λ_{k}$ and orthonormal eigenvectors $m_{k}$ (Recipe 2.3.6).
$U = \sum_{k} λ_{k} m_{k} m_{k}^{T}$ . In the principal basis it is $diag (ν_{1}, ν_{2}, ν_{3})$ .
$R = F U^{- 1}$ , with $U^{- 1} = \sum_{k} \frac{1}{ν _{k}} m_{k} m_{k}^{T}$ . Check $R^{T} R = I$ and $det R = 1$ .
$V = RU R^{T}$ (multiply $RU = VR$ on the right by $R^{T}$ ).

Example 2.4.9a stretch and a quarter turnbeyond the lectures

$F = 030 - 2 00 001$ .

$C = F^{T} F = diag (9, 4, 1)$ , already diagonal: the principal directions are $e_{1}, e_{2}, e_{3}$ , with stretches $3, 2, 1$ .
$U = diag (3, 2, 1)$ .
$R = F U^{- 1} = 010 - 1 00 001$ , a $90°$ rotation about $e_{3}$ , $det R = 1$ .
$V = RU R^{T} = diag (2, 3, 1)$ . Check: $VR = F$ .

In words: stretch by 3 along $e_{1}$ and by 2 along $e_{2}$ , then turn by $90°$ . Or turn first, then stretch by 3 along the turned $e_{1}$ (now $e_{2}$ ) and by 2 along the turned $e_{2}$ (now along $- e_{1}$ ). $V$ has the eigenvalues of $U$ on $\overset{˘}{m}_{k} = R m_{k}$ .

Exercises

Exercise 2.4.1

Find $U$ , $R$ and $V$ for the plane $F = [11 - 1 1]$ (in the $e_{1} e_{2}$ plane). Describe it in words.

Solution

$C = F^{T} F = [2002] = 2 I$ : every direction is principal, with $ν = 2$ .
$U = 2 I$ , $R = F / 2 = [cos 45° sin 45° - sin 45° cos 45°]$ , $V = 2 I$ .
An equal stretch by $2$ in every direction and a $45°$ turn, in either order: with a spherical stretch the order does not matter.

Exercise 2.4.2

Show that if $F$ is symmetric and positive definite, then $R = I$ and $U = V = F$ . What does that say about simple shear, whose $F$ is not symmetric?

Solution

If $F = F^{T}$ is positive definite, $C = F^{2}$ , so its positive square root is $U = F$ and $R = F U^{- 1} = I$ ; then $V = U$ . Simple shear has a non-symmetric $F$ , so $R \neq = I$ : it contains a rotation as well as a stretch, even though it moves no point of the line $X_{2} = 0$ .

Exercise 2.4.3

For $F$ with $C = diag (9, 4, 1)$ and $det F = 6$ , what are the eigenvalues of $U$ , of $V$ and of $C$ ? Which tensor’s eigenvectors are the deformed principal directions?

Solution

$C$ : $9, 4, 1$ . $U$ and $V$ : $3, 2, 1$ (and $3 \cdot 2 \cdot 1 = 6 = det F$ ). The eigenvectors of $U$ (and $C$ ) are the reference principal directions $m_{k}$ ; those of $V$ are the deformed ones, $\overset{˘}{m}_{k} = R m_{k}$ .