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Proceed by induction on $\dim V$. The zero-dimensional case has the empty orthonormal basis. In positive dimension, the fundamental theorem of algebra gives a root of the characteristic polynomial, hence a unit eigenvector $e$ of $\alpha$, say with eigenvalue $\lambda$. Part (ii) makes it an eigenvector of ${\alpha }^{\ast }$ as well. Thus the line spanned by $e$ and its orthogonal complement are invariant under both operators by the introductory adjoint argument.

On $\{e{\}}^{\perp }$, the restriction of ${\alpha }^{\ast }$ is the adjoint of the restriction of $\alpha$, and the restrictions still commute. The restricted linear operator is therefore normal. By induction it has an orthonormal basis of eigenvectors. Adjoining $e$ proves an orthonormal eigenbasis exists for $V$, establishing the finite-dimensional spectral theorem by construction.

For the final deduction, use this orthonormal basis $e_j$ and write $\alpha e_j={\lambda }_j{e}_j$. DefineThe real nonnegative diagonal entries make $\beta$ a Hermitian operator, and the modulus-one diagonal entries make $\gamma$ a unitary operator. Both are diagonal in the same orthonormal basis, so they commute and $\beta \gamma e_j={\lambda }_j{e}_j$ even when ${\lambda }_j=0$. This proves the required factorization without assuming invertibility.

Conversely, if $\alpha =\beta \gamma$ with ${\beta }^{\ast }=\beta$, ${\gamma }^{\ast }\gamma =\gamma {\gamma }^{\ast }=I$, and $\beta \gamma =\gamma \beta$, then ${\alpha }^{\ast }={\gamma }^{\ast }\beta$ andThus $\alpha$ is normal. Normality is equivalent to a commuting Hermitian–unitary factorization, with the Hermitian factor even selectable nonnegative. This is the polar decomposition with a unitary extension on the kernel.

For a hyperelastic material $T_R(F)=DW(F)$. Material frame indifference of this first Piola-Kirchhoff law meansThe spatial force components rotate under a superposed rigid rotation, whereas the reference normal does not.

If $W(QF)=W(F)$, differentiate with respect to $F$ in direction $H$ to obtain $DW(QF):(QH)=DW(F):H$, hence the displayed stress transformation. Conversely, assume that stress transformation. For fixed $Q$, the derivative of $W(QF)-W(F)$ with respect to $F$ vanishes. The positive-determinant matrix domain is connected, as follows from polar decomposition and the connectedness of $SO(3)$ and of the positive-definite cone. Thus $W(QF)-W(F)=c(Q)$, independent of $F$. Iterating givesThe assumed lower bound on $W$ forces $c(Q)=0$: either nonzero sign would tend to minus infinity for one direction of integer $n$. This proves the required equivalence and explains the use of the lower-bound hypothesis.

In the frame-indifferent setting, isotropy means invariance under reference rotations as well: $W(FQ)=W(F)$ for $Q\in SO(3)$. The distinction matters: left rotations encode changes of observer, and right rotations encode material symmetry.

An orientation-preserving singular value decomposition is $F=R\mathrm{diag}(v_1,v_2,v_3)S$, with $R,S\in SO(3)$ and $v_i>0$. Material frame indifference and isotropy remove $R,S$, leaving a function $\Phi$ of the stretches. Permuting the stretches using signed permutation rotations does not change the value, so $\Phi$ is symmetric. Conversely a symmetric stretch function is unchanged by both left and right rotations, in particular is isotropic.

The eigenvalues of $B=F{F}^T$ are $v_i^2$. Its three principal invariants areThey determine the unordered positive eigenvalues as the roots of ${\lambda }^3-I{\lambda }^2+II\lambda -III=0$. Therefore a symmetric stretch function defines a well-defined $h(I,II,III)$ on the admissible invariant triples. Conversely a function of these invariants is a symmetric stretch function, proving the equivalence of all three descriptions.

The printed domain of $\Phi$ must be $(0,\infty {)}^3$, since it takes three arguments. Also isotropy alone as right invariance would not imply the spectral representation without material frame indifference: $\mathrm{tr}(AF{F}^T)$ for a fixed nonscalar symmetric positive-definite $A$ is right invariant but not generally objective. The equivalence here uses the frame-indifferent constitutive setting.

For the specific energy, Newton identities for the three eigenvalues gives ${\sum }_i{v}_i^6=I^3-3III+3III$, while $v_1{v}_2{v}_3=J=\sqrt{III}$. ThusTo differentiate it, useThese follow from $I=|F{|}^2$, the displayed trace formula for $II$, and $III=J^2$. The three partial derivatives of $h$ are $h_I=\alpha +3\beta I^2-3\beta II$, $h_{II}=-3\beta I$ and $h_{III}=3\beta -\gamma II{I}^{-3/2}/2$. The chain rule therefore givesFinally $-6\beta II=-3\beta (\mathrm{tr}B{)}^2+3\beta \mathrm{tr}(B^2)$, and $BF=F{F}^{T}F$. This is precisely the requested expression, with the coefficient of $F$ separated from the $BF$ and cofactor terms. The positive sign of the $6\beta J$ term comes from the $3\beta III$ contribution in the energy.

The material symmetry group consists of invertible admissible reference transformations $H$ for which $W(FH)=W(F)$ for every admissible $F$. Its rotational part lies in $SO(3)$; a determinant-one reference convention restricts the full group to the corresponding special linear group. This right action is distinct from the spatial left action $W(QF)=W(F)$ of material frame indifference.

The proper cube-rotation group has $\boxed{24}$ elements. Choose the image of one outward face normal in six ways, then the image of an adjacent perpendicular normal in four ways; orientation fixes the third direction. No reflections are included. For an isotropic material every proper rotation belongs to the material symmetry group.

Combine material isotropy with the physical frame-indifference condition. The singular value decomposition may be written $F=R\mathrm{diag}(v_1,v_2,v_3)Q$ with $R,Q\in SO(3)$ because $\det F>0$. Left and right rotational invariance remove $R,Q$. Signed permutation rotations permit any permutation of the three diagonal stretches, soRight invariance alone, without the usual material frame indifference assumption, would not suffice to remove spatial orientation.

The strict Baker-Ericksen inequalities say that larger distinct principal stretches carry larger principal Cauchy stresses:A natural state of a hyperelastic material has zero stress. If two stretches differed, their equal zero stresses would contradict the strict inequality. Thus all three equal some $v>0$, and polar decomposition gives $\boxed{F=vQ}$, $Q\in SO(3)$. A weak non-strict inequality alone would not prove this conclusion, so its strict convention is explicit here.

For the radial deformation put $f(R)=r(R)/R$, $R=|\mathbf{x}|$, $\mathbf{\theta }=\mathbf{x}/R$. Since ${\partial }_{j}R={\theta }_j$,This symmetric tensor has radial eigenvalue $r^{\prime }$ and the repeated tangential eigenvalue $r/R$. Both are positive in the open annulus under the stated monotonicity/nonnegativity conditions. Therefore these are its principal stretches. The two cases below establish radial zero-energy rigidity by different arguments.

A homogeneous hyperelastic material has a stored energy per reference volume $W(F)$ depending on the local deformation gradient but not explicitly on the material point. Its first Piola-Kirchhoff response isA superposed rigid rotation changes the observer's description without changing the energy of the deformation. Material frame indifference is thereforeThis left-rotation invariance is distinct from a material isotropy assumption on right rotations. Differentiate it in an arbitrary matrix direction $H$:The Frobenius inner product then gives $Q^T\hat{T}(QF)=\hat{T}(F)$, proving $\hat{T}(QF)=Q\hat{T}(F)$.

For the converse, that stress covariance makes the derivative in $F$ of $W(QF)-W(F)$ zero. The positive-determinant matrices form a connected set, as also follows from polar decomposition, so this difference is a constant $c(Q)$ independent of $F$. Composition gives $c(Q_1{Q}_2)=c(Q_1)+c(Q_2)$. For fixed $F$, continuity of $W$ and compactness of $\mathrm{SO}(3)$ make $c$ bounded. But $c(Q^n)=nc(Q)$ for every positive integer, forcing $c(Q)=0$. Thus stress covariance and frame indifference of the energy are equivalent. Compactness removes the possible additive energy constant in the converse.

A deformation maps a material point in the reference body to its current position, $y:\Omega \to {\mathbb{R}}^3$, with deformation gradient $F=Dy$. A homogeneous deformation has constant gradient and is affine, $y(X)=FX+b$.

If $\det F>0$, the matrix $F^{T}F$ is symmetric positive definite. The matrix-square-root theorem says that any such matrix has a unique symmetric positive definite square root, obtained by taking positive square roots in its orthogonal spectral decomposition. SetThen $R^{T}R=U^{-1}{F}^{T}F{U}^{-1}=I$ and $\det R=\det F/\det U>0$, so $R\in \mathrm{SO}(3)$ and $F=RU$. If also $F=\tilde{R}\tilde{U}$, then $F^{T}F={\tilde{U}}^2$; uniqueness of the positive square root gives $\tilde{U}=U$ and then $\tilde{R}=R$. This proves existence and uniqueness of polar decomposition.

The principal stretches are the positive eigenvalues of $U$, equivalently the singular values of $F$ or square roots of the eigenvalues of $F^{T}F$.

Every bounded operator on a Hilbert space has a polar decomposition $T=U|T|$, where $|T|=(T^{\ast }T{)}^{1/2}$ and $U$ is a partial isometry with $\ker U=\ker T$. On $\mathrm{im}|T|$, it is defined by $U(|T|x)=Tx$.

This is the bounded-operator case of polar decomposition.

Every invertible real matrix has the polar decompositionwhere $P$ is symmetric positive definite and $R$ is orthogonal.

This is the invertible real-matrix case of polar decomposition.