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Suppose an invertible real matrix $C$ intertwines symmetric orthogonal tile-reflection matrices: $C{M}_s=N_{s}C$. Then $C^{T}C$ commutes with every $M_s$, so its positive inverse square root does too. The polar decomposition of an invertible real matrix therefore gives an orthogonal matrix $O=C(C^{T}C{)}^{-1/2}$ satisfying $O{M}_s=N_{s}O$. On a disjoint union of congruent tiles, $O$ preserves both the $L^2$ inner product and Dirichlet energy. The intertwining identities preserve the face trace conditions defining the assembled Sobolev space, so $O$ gives a unitary equivalence of the assembled Laplacians.

For a real matrix $A$ with positive determinant, the polar decomposition of an invertible real matrix gives unique factorizationswhere $U$ and $V$ are the right stretch tensor and left stretch tensor, respectively. Positivity here means positive definiteness, not merely nonnegative entries.

To prove existence, $C=A^{T}A$ is symmetric and $v^{T}Cv=|Av{|}^2>0$ for every nonzero $v$, because $A$ is invertible. The spectral theorem for real symmetric matrices supplies an orthogonal matrix $Q$ and positive eigenvalues $c_j$ such that $C=Q\mathrm{diag}(c_1,c_2,c_3)Q^T$. Define $U$ using the principal square root of a positive semidefinite matrix construction:Then $R^{T}R=U^{-1}{A}^{T}A{U}^{-1}=I$, so $R$ is orthogonal. Also $\det R=\det A/\det U>0$; an orthogonal matrix has determinant plus or minus one, hence $\det R=1$. Set $V=RU{R}^T$. It is symmetric positive definite, $V^2=A{A}^T$, and $VR=RU=A$.

For uniqueness, any second factorization $A=R_1{U}_1$ with the same properties gives $U_1^2=A^{T}A=C$. Since $U_1$ commutes with its square, it preserves every eigenspace of $C$. On a $c$-eigenspace its own positive eigenvalues must square to $c$, so it equals $\sqrt{c}$ times the identity there. Thus $U_1=U$, then $R_1=A{U}^{-1}=R$, and finally $V$ is fixed too. This proves both uniqueness and the polar decomposition in continuum mechanics, with the same proper rotation in its left and right forms.

For the polar decomposition of an invertible real matrix, start with $C=F^{T}F$. For nonzero $y$, $y^{T}Cy=|Fy{|}^2>0$, since $\det F>0$ makes $F$ invertible. The spectral theorem for real symmetric matrices gives an orthonormal basis $n_a$ with $C{n}_a={\lambda }_a^2{n}_a$ and ${\lambda }_a>0$. DefineThen $U$ is a positive-definite symmetric matrix, $U^2=C$, andbecause $\det U=\sqrt{\det (F^{T}F)}=\det F$. Thus $R$ is an orthogonal matrix with positive determinant. Set $V=RU{R}^T$; it too is a positive-definite symmetric matrix, and $VR=RU=F$. Moreover $V^2=F{F}^T$. This provesThe positive square root is unique: any positive symmetric square root of $C$ commutes with $C$, preserves its eigenspaces and has the positive scalar square root on each of them. Hence $U$, then $R$ and $V$, are unique. These are the right stretch tensor, rotation and left stretch tensor of the polar decomposition in continuum mechanics.

A standard spectral strain measure isTake $f\in C^1(0,\infty )$, $f(1)=0$ and $f^{\prime }(1)=1$, with $f$ strictly increasing; the customary stronger condition $f^{\prime }(\lambda )>0$ gives a regular one-to-one parameterization of all positive principal stretches. The zero condition makes a rigid rotation strain free, monotonicity distinguishes extension from contraction and prevents nonunit stretch from being strain free, and the derivative normalization gives the ordinary infinitesimal strain. Indeed, for $F=I+H$,Under a superposed spatial rotation $F\mapsto QF$, $F^{T}F$ and hence $E^f$ remain unchanged, giving material frame indifference. Choosing $f(\lambda )=\lambda -1$ satisfies all these conditions and gives the Biot strain tensor

The meaning of work-conjugate stress and strain here is that a symmetric material tensor $T^f$ satisfiesfor every admissible rate, with both powers measured per reference volume. To keep the nominal-stress index convention explicit, write $M_{iI}=P_{Ii}$, so $M=P^T$ and the power is $M:\dot{F}$. Differentiating $F=RU$ gives $\dot{F}=R(\Omega U+\dot{U})$, where $\Omega =R^T\dot{R}$ is skew. Conservation of angular momentum gives symmetry of $M{F}^T$, or equivalently of $R^{T}MU$. ThereforeThe first term vanishes because a symmetric tensor has zero contraction with a skew tensor. Since ${\dot{E}}^{(1)}=\dot{U}$, the symmetric Biot stress isIf the nominal stress is instead stored spatial-index first as $M$, the same answer is $\mathrm{sym}(R^{T}M)$. The symmetrization is essential: $PR$ need not itself be symmetric.

The space of real $2\times 2$ matrices has a determinant form of signature $(2,2)$: writeThe action $(A,B):X\mapsto AX{B}^{-1}$ of $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ preserves this determinant. Each factor is connected by polar decomposition of an invertible real matrix: its orthogonal factor lies in connected $SO(2)$ and its positive factor is an exponential of a symmetric traceless matrix. Thus the image is contained in $S{O}_0(2,2)$.

If the action is trivial, $X=I$ gives $A=B$, and fixing every $X$ makes $A$ a scalar matrix. Determinant one gives $A=B=\pm I$. The kernel is therefore the diagonal ${\mathbb{Z}}_2$. Its discreteness makes the differential injective; domain and target both have dimension six. An open subgroup of connected $S{O}_0(2,2)$ is the whole group, soAgain the kernel is diagonal, and the target is the identity component.

The two unheaded requests are addressed here; the individual group-cover calculations are in the scoped parts below. A quaternion $q=a+bi+cj+dk$ has norm $|q{|}^2=a^2+b^2+c^2+d^2$. Thus the unit quaternions form exactly the sphere $S^3\subset {\mathbb{R}}^4$, with quaternion multiplication giving its smooth group structure. The explicit identification with the special unitary group $SU(2)$ isThe matrices representing $i,j,k$ satisfy the quaternion multiplication relations. The displayed matrix obeys $U^{†}U=|q{|}^{2}I$ and $\det U=|q{|}^2$; conversely every $SU(2)$ matrix has that form with $|q|=1$. Hence $S^3$ is the manifold of unit quaternions, and its group is $SU(2)$.

For the exterior-square double cover of SO(3,3), fix a basis $e_1,\dots ,e_4$ and volume element $v=e_1\wedge e_2\wedge e_3\wedge e_4$. Define the symmetric bilinear form $Q$ on the six-dimensional exterior square byIt is symmetric because two-forms commute under the wedge product. PutThen $Q(a_i,a_j)=Q(b_i,b_j)=0$ and $Q(a_i,b_j)={\delta }_{ij}$. The Gram matrix is $\left(\begin{array}{cc}0& I_3\\ I_3& 0\end{array}\right)$: the vectors $(a_i+b_i)/\sqrt{2}$ have norm $+1$, while $(a_i-b_i)/\sqrt{2}$ have norm $-1$. Therefore the quadratic form $\omega \mapsto \omega \wedge \omega$ has signature $(3,3)$.

For $A\in SL(4,\mathbb{R})$,Thus $\rho (A)={\Lambda }^{2}A$ preserves $Q$. The special linear group is connected: polar decomposition of an invertible real matrix writes $A=OP$ with $O\in SO(4)$ and $P$ positive definite of determinant one, and both factors can be joined to the identity. The image consequently lies in $S{O}_0(3,3)$.

If ${\Lambda }^{2}A=I$, then $Au\wedge Aw=u\wedge w$ for every pair, so $A$ preserves every two-plane. A line is an intersection of two such planes, hence every line is preserved. A linear map preserving every line is scalar: apply it to basis vectors and then to their pairwise sums. Thus $A=\lambda I$, and ${\Lambda }^{2}A=I$ forces ${\lambda }^2=1$. The kernel is exactly $\{\pm I\}$. It is discrete, so the derivative of $\rho$ is injective. Both Lie algebras have dimension $15$, making the image open; an open subgroup of a connected group is the whole group. HenceThe identity-component restriction is essential. The same principle of an explicit form-preserving representation, kernel calculation and dimension argument supplies the five remaining covers.

A real Lie group is a group $G$ equipped with a finite-dimensional real smooth manifold structure, conventionally Hausdorff and second countable, for which multiplication $G\times G\to G$ and inversion $G\to G$ are smooth.

Its tangent space at the identity $e$ is defined by smooth curves through $e$: two curves represent the same tangent vector when their derivatives in a local chart agree at zero. For $X\in T_{e}G$, left translation $L_g(h)=gh$ determines the left-invariant vector fieldThe commutator of these derivations on smooth functions is another left-invariant vector field, so define the Lie bracketThis construction supplies the Lie algebra structure. In a Matrix Lie group, $X^L(g)=gX$ and differentiating the two fields gives $[X,Y]=XY-YX$.

For the special linear group, differentiate the determinant along a curve $g(t)$ through $I$. Expansion of the determinant, or its differential $D{\det }_A(H)=\det (A)\mathrm{tr}(A^{-1}H)$, givesThe determinant differential is surjective at $I$, so the level set $\det =1$ has tangent space equal to its differential's kernel. Equivalently, every tangent matrix has zero trace, and every zero-trace matrix supplies a curve $e^{tX}$ of determinant $e^{t\mathrm{tr}X}=1$. Hencewith the matrix commutator bracket. The same calculation over $\mathbb{C}$ gives the complex special linear Lie algebra ${\mathfrak{sl}}_n(\mathbb{C})$; as a real group, the complex group has that space viewed as a real Lie algebra.

The matrix exponential is the everywhere-convergent serieswhose inverse matrix is $\exp (-X)$. The matrix logarithm is locally defined near $I$ byThese maps are inverse on suitable neighbourhoods of zero and $I$, giving a logarithmic chart of a matrix Lie group. A logarithm is not a globally single-valued inverse of the exponential.

Every invertible complex matrix nevertheless has at least one matrix logarithm. Put it in Jordan normal form. For a block $J=\lambda I+N$, $\lambda \ne 0$ and $N^s=0$, choose any complex scalar logarithm $\ell$ of $\lambda$ and setThe finite logarithm and exponential identities in the nilpotent variable give $e^L=\lambda (I+N/\lambda )=J$. Combine the blocks and conjugate back. Thus the exponential map is surjective on ${\mathrm{GL}}_n(\mathbb{C})$, by existence of a logarithm for every invertible complex matrix.

A connected counterexample is ${\mathrm{SL}}_2(\mathbb{R})$. It is connected: polar decomposition of an invertible real matrix writes each element as $KP$, with $K\in \mathrm{SO}(2)$ and $P$ positive definite symmetric of determinant one; $\mathrm{SO}(2)$ is connected and $P$ is connected to $I$ through $P^t$.

But $g=\mathrm{diag}(-2,-1/2)$ belongs to this group and is not $e^X$ for any real $X$. Such an $X$ would commute with $e^X=g$. Since $g$ has two distinct real eigenvalues, direct commutation makes $X$ a real diagonal matrix. Its exponential has positive diagonal entries, a contradiction. ThereforeThis is an exponential-surjectivity obstruction from distinct negative eigenvalues; connectedness does not eliminate the obstruction.

By the real spectral theorem, a real symmetric matrix has an orthogonal diagonalizationWriting $y=Q^{T}x$ givesThis is nonnegative for every $x$ exactly when every ${\lambda }_i\ge 0$. Thus the quadratic-form and eigenvalue definitions of a positive semidefinite matrix agree.

When $A$ is positive semidefinite, defineThis matrix is symmetric and positive semidefinite, and its square is $A$, so it is the principal square root of a positive semidefinite matrix.

For nonsingular $M$, the matrix $M^{T}M$ is symmetric andfor every nonzero $x$. Thus $M^{T}M$ is positive definite, andexists and is nonsingular. Let $R=M{P}^{-1}$. Since $P$ is symmetric,Hence $R$ is an orthogonal matrix andthe polar decomposition of an invertible real matrix.

In three dimensions, $P$ stretches or contracts along three mutually perpendicular eigenvector directions by its positive eigenvalues. The orthogonal map $R$ is then applied: it is a rotation when $\det M>0$, and a rotation combined with a reflection when $\det M<0$.

For an orientation-preserving invertible deformation gradient, the polar decomposition of an invertible real matrix separates a rotation $R$ from the right stretch tensor $U$ or the left stretch tensor $V$. The rotation is shared by the two factorizations, with $V=RU{R}^T$.