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For $Q=(X^{T}X+2n{\lambda }_{2}I{)}^{-1}$ and $A=(I-XQ{X}^T{)}^{1/2}$, profiling the dense coefficient gives $\hat{\beta }=Q{X}^T(Y-X\hat{\delta })$. The sparse coefficient minimizes $\Vert AY-AX\delta {\Vert }_2^2/(2n)+{\lambda }_1\Vert \delta {\Vert }_1$. The transformation uses the principal square root of a positive semidefinite matrix and remains valid for rank-deficient designs.

Let $B=A^{\mathsf{T}}A$. Since $\det A>0$, $B$ is a positive-definite matrix. The spectral theorem gives a unique principal square root of a positive semidefinite matrixThen $R^{\mathsf{T}}R=U^{-1}B{U}^{-1}=I$ and $\det R=\det A/\det U=1$, so $R$ is a rotation matrix. Define $V=RU{R}^{\mathsf{T}}$. It is a symmetric matrix, is positive definite, and satisfies $V^2=A{A}^{\mathsf{T}}$. Consequently the polar decomposition in continuum mechanics isHere $U$ and $V$ are the right stretch tensor and left stretch tensor.

For the Cauchy stress tensor, use both material isotropy and material frame indifference. Their respective transformation laws areThe first law and $A=VR$ give $\sigma (A)=\sigma (V)$. Combining the two laws gives $\sigma (QV{Q}^{\mathsf{T}})=Q\sigma (V)Q^{\mathsf{T}}$. In a principal stretch basis, $V=\mathrm{diag}({\lambda }_1,{\lambda }_2,{\lambda }_3)$. Each half-turn about a coordinate axis leaves $V$ unchanged; its conjugation changes the signs of the corresponding off-diagonal stress tensor entries. Those entries must therefore vanish. If two principal stretches coincide, rotations within their eigenspace additionally force the stress tensor to be scalar on that eigenspace. Thus the Cauchy stress tensor and left stretch tensor have common principal axes, including at repeated principal stretches. This proves coaxiality of isotropic elastic stress.

The precise symmetry conclusion is equivariance of isotropic principal stresses. Every permutation of the principal stretches can be realized by a signed permutation rotation matrix, sowhen the indexing is permuted consistently. Equivalently, permuting the arguments permutes the labelled principal stresses in the same way. In particular, ${\sigma }_1$ is symmetric in ${\lambda }_2,{\lambda }_3$, and analogous statements hold for ${\sigma }_2,{\sigma }_3$.

For distinct principal stretches, interpolate the three values ${\sigma }_i$ at the three points ${\lambda }_i$:Uniqueness of this interpolation implies that the scalar coefficients ${\beta }_j$ are symmetric functions of the unordered principal stretches; they may be expressed through the elementary symmetric invariants. Smooth isotropic constitutive laws have the corresponding invariant representation across repeated stretches as well. The wording about symmetric functions must be understood in this collective sense: each labelled principal stress need not be symmetric in all three arguments. For example, the objective isotropic law $\sigma =V^2$ has ${\sigma }_i={\lambda }_i^2$, which disproves that stronger interpretation.

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.

A Hermitian matrix satisfies $A=A^{†}$, where $†$ means conjugate transpose. If $Av=\lambda v$ with $v\ne 0$, then $v^{†}Av=\lambda \Vert v{\Vert }^2$ is real, since it equals its complex conjugate $v^{†}{A}^{†}v$. Thus every eigenvalue is real. To prove the finite-dimensional spectral theorem, choose a unit eigenvector $v$; one exists because the complex characteristic polynomial has a root. Its orthogonal complement is invariant, since for $w\perp v$,The restriction to that complement is again Hermitian. Induction on dimension supplies an orthonormal basis of eigenvectors, so $A=U\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)U^{†}$ for a unitary matrix $U$.

In this basis, $x^{†}Ax={\sum }_i{\lambda }_i|(U^{†}x{)}_i{|}^2$. Therefore $\boxed{A>0⟺{\lambda }_i>0\text{ for every }i}$. In that case the square root of a matrixis Hermitian, positive definite and satisfies $C^2=A$. For uniqueness, any positive-definite Hermitian $D$ satisfying $D^2=A$ commutes with $A$ and hence preserves each eigenspace of $A$. Its restriction to the $\lambda$-eigenspace is Hermitian, and its positive eigenvalues all solve ${\mu }^2=\lambda$. Thus that restriction is $\sqrt{\lambda }I$, forcing $D=C$. This is the principal square root of a positive semidefinite matrix, here strictly positive definite.

Now let $D=\sqrt{B}$ and $X=C-D$. Expanding without assuming that $C$ and $D$ commute givesThe Hermitian matrix $X$ has $X^2\ge 0$ because $v^{†}{X}^{2}v=\Vert Xv{\Vert }^2$, whereas $A-B>0$. Hence $CX+XC>0$. If $Xv=\mu v$, $v\ne 0$, thenSince $C>0$, every real eigenvalue $\mu$ of $X$ is positive. The positive-eigenvalue criterion now gives the strict finite-dimensional operator monotonicity of the square root:

Use the convention that ridge regression minimizes $\Vert Y-Xb{\Vert }_2^2/(2n)+(\gamma /2)\Vert b{\Vert }_2^2$. Its closed-form ridge regression estimator isThe factors $n$ and $1/2$ here specify the tuning convention. No intercept is needed for the centred responses and predictors.

For the two-component procedure, fix $\delta$ and put $\theta =Y-X\delta$. Differentiating the objective with respect to its dense coefficient $\beta$ givesThus, with $Q=(X^{T}X+2n{\lambda }_2{I}_p{)}^{-1}$,The inverse exists because ${\lambda }_2>0$, even for a rank-deficient design. The sum of the fitted sparse and dense components is the Lava estimator.

Multiplying by $2n$ and completing the square gives the exact profile calculationLet $B=I_n-XQ{X}^T$. A singular value decomposition of $X$ shows that its eigenvalues are $2n{\lambda }_2/(r_i^2+2n{\lambda }_2)$ in nonzero singular directions and one in the orthogonal complement. Hence $B$ is a positive-definite matrix. Take its symmetric principal square root of a positive semidefinite matrix, $A=B^{1/2}$. Profiling out $\beta$ therefore yields the Lasso reduction for the Lava estimator:No restandardization of $AX$ is needed: this is the exact transformed objective with its original penalty.

Write $\tilde{X}=AX$, $h={\hat{\delta }}_{\lambda }-{\delta }^0$, and $\eta =\tilde{X}{\beta }^0+A\epsilon$. Then $AY=\tilde{X}{\delta }^0+\eta$. Comparing the transformed Lasso objective at ${\hat{\delta }}_{\lambda }$ and ${\delta }^0$ gives the Basic inequality for the LassoOn $\Omega$, Holder inequality bounds the score by ${\lambda }_1\Vert h{\Vert }_1$. Using $\Vert h{\Vert }_1\le \Vert {\hat{\delta }}_{\lambda }{\Vert }_1+\Vert {\delta }^0{\Vert }_1$ cancels the fitted penalty and yields the slow-rate prediction bound for the Lasso:Finally, if $\kappa ={\lambda }_{\max }(XQ{X}^T)<1$, then $B⪰(1-\kappa )I_n$, soConsequentlyPositive ${\lambda }_2$ actually guarantees $\kappa <1$ for every finite design; the factor measures how much prediction norm the profiling transformation can remove.

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 $\det M<0$.

Let $A$ and $M$ be real symmetric positive-definite matrices. The principal square root of a positive semidefinite matrix construction gives an invertible matrix for each. SetThen $Ax=b$ is equivalent to $By=c$. This is a matrix congruence, and $B$ is a symmetric positive-definite matrix: $B^T=B$, and for every nonzero vector $v$,since $S^{-1}v\ne 0$. This short proof establishes both requirements for applying the conjugate gradient method to the transformed linear system.

The left transformed matrix has the same eigenvalues becauseIts correct geometry uses the weighted inner product $⟨u,v{⟩}_M=u^{T}Mv$: $⟨u,M^{-1}Av{⟩}_M=u^{T}Av=⟨M^{-1}Au,v{⟩}_M$. Thus it is symmetric in this inner product, even when it is not a symmetric matrix in ordinary coordinates. The spectral condition number governing the transformed symmetric problem is ${\kappa }_2(B)$, not generally ${\kappa }_2(M^{-1}A)$.

The Preconditioned conjugate gradient method realizes this transformation using auxiliary solves $M{z}_k=r_k$, without constructing $S$ or its matrix inverse. With fixed $M$, initialize $r_0=b-A{x}_0$ and $p_0=z_0$; while $r_k\ne 0$, usethen solve $M{z}_{k+1}=r_{k+1}$ and setStop at a zero linear-system residual before forming another ratio. Positive definiteness makes the denominators positive before termination.^[ref]

An invertible matrix alone is insufficient: with $A=I$ and $M=\mathrm{diag}(-1,1)$, the left transformed matrix has eigenvalues $-1,1$, and $M$ has no real symmetric positive-definite matrix square root. Even a valid symmetric positive-definite matrix can worsen conditioning if its scaling is unsuitable, as the next worked example shows.

If $P$ is a real positive semidefinite matrix, then $x^{T}Px=0$ if and only if $Px=0$. Indeed $x^{T}Px=\Vert \sqrt{P}x{\Vert }^2$ by the principal square root of a positive semidefinite matrix, so a zero quadratic form puts $x$ in the kernel of $\sqrt{P}$ and hence of $P$.