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The linear map $\pi (A)=(A+A^T)/2$ is the projection onto the space $\mathcal{S}$ of symmetric matrices, with kernel $\mathcal{A}$. Define $F=\pi \circ L$ on $V$. Since $DL$ is invertible everywhere on $V$ and $\pi$ is onto, $DF$ is onto everywhere. By part (ii),The regular level set theorem makes this a submanifold of codimension $\dim \mathcal{S}=n(n+1)/2$, and henceThis neighborhood is $V$ around $I$; the source's final occurrence of $U$ in this step is a neighborhood-label slip. For any $Q\in O(n)$, left multiplication by $Q$ is an invertible linear map of the ambient matrix space and carries $V\cap O(n)$ onto $QV\cap O(n)$. These translated charts establish the submanifold property at every group element.

The following R code implements the given recursion for the Cholesky decomposition. The input is assumed to be a real symmetric matrix that is a positive-definite matrix.

mychol <- function(A) {
  n <- nrow(A)
  if (n == 1L) return(matrix(sqrt(A[1L, 1L]), 1L, 1L))
  j <- 2L:n
  L <- matrix(0, n, n)
  L[1L, 1L] <- sqrt(A[1L, 1L])
  b <- A[j, 1L, drop = FALSE] / L[1L, 1L]
  L[j, 1L] <- b
  S <- A[j, j, drop = FALSE] - tcrossprod(b)
  L[j, j] <- mychol(S)
  L
}

Here tcrossprod(b) is the outer product $b{b}^T$. The drop = FALSE arguments preserve matrix dimensions, including the final one-by-one subproblem, and the base case also returns a matrix.

To verify correctness, partition $A=\left(\begin{array}{cc}a& r^T\\ r& B\end{array}\right)$. The code sets $\ell =\sqrt{a}$, $b=r/\ell$ and recursively factors $S=B-b{b}^T$. This Schur complement is a positive-definite matrix: for $z\ne 0$,If $C{C}^T=S$ is the recursive answer, then the returned lower triangular matrix satisfiesThe one-dimensional identity starts this mathematical induction proof. Each recursive call reduces the matrix size by one and therefore terminates.

Each state has exactly $k(n-k)$ neighbours: choose one occupied coordinate and one unoccupied coordinate to swap. The normalized proposal distribution is thereforeThe swap relation is symmetric and the number of neighbours is constant, so $q(v\mid u)=q(u\mid v)$. The discrete uniform distribution used as the prior distribution is also constant on $S_k$, with mass ${\left(\genfrac{}{}{0ex}{}{n}{k}\right)}^{-1}$. Both factors cancel in the Metropolis–Hastings acceptance probability.

Let $Q={\Sigma }^{-1}$ and $d=v-u$. The displayed multivariate normal distribution presupposes that the covariance matrix $\Sigma$ is a positive-definite matrix, so $Q$ is a symmetric matrix. Expanding the difference of quadratic forms givesConsequentlyThe truncation at zero is essential: a log acceptance ratio can be positive, but the logarithm of an acceptance probability cannot.

For the particular proposed swap, let $a=i_1$ be the coordinate changing from zero to one and $b=i_2$ the coordinate changing from one to zero. Then $d=e_a-e_b$, and the untruncated scalar simplifies further toOnly two rows and four entries of $Q$ are needed for this evaluation. This is Metropolis sampling on fixed-cardinality subsets.

For $R=|x|>0$, $DR=\theta$ and $D\theta =(I-\theta \otimes \theta )/R$. Differentiating the spherically symmetric deformation givesFor an orientation-preserving spherically symmetric deformation with positive radius, $r>0$ and $r^{\prime }>0$. The radial direction and its perpendicular plane are eigenspaces of this positive symmetric matrix, soAt a positive diagonal matrix $D$, every diagonal rotation with two minus signs commutes with $D$. The supplied equivariance relation therefore forces $\hat{T}(D)$ to be a diagonal matrix: each off-diagonal entry changes sign under one such rotation while $D$ stays fixed. Varying a positive diagonal entry shows that the diagonal entries of $\hat{T}(D)$ are ${\Phi }_{,j}$. Permutation symmetry of $\Phi$ implies ${\Phi }_{,2}={\Phi }_{,3}$ when ${\lambda }_2={\lambda }_3$.

Choose a rotation whose first column is $\theta$. Applying equivariance to $F=QD{Q}^T$ now giveswith the derivatives evaluated at $(r^{\prime },r/R,r/R)$. Thus the stress has one radial and one repeated tangential principal value, independently of the choice of tangent basis.

For this homogeneous motion, suppose the matrix $\mathbf{A}=\ddot{\mathbf{F}}{\mathbf{F}}^{-1}$ is symmetric. Put $\mathbf{y}=\mathbf{x}-\mathbf{c}(t)$. The acceleration potential of a homogeneous motion iswhere $g(t)$ is arbitrary. When taking the spatial gradient, $t$, $\mathbf{c}$, $\ddot{\mathbf{c}}$, and $\mathbf{A}$ are fixed. Symmetry givesThis verifies the potential directly and also proves sufficiency of the symmetric matrix condition within the static-family setting. Its dimension is ${\mathsf{L}}^2{\mathsf{T}}^{-2}$, so the pressure correction $-\rho \xi$ has the dimension of stress. Combining with part (b)(ii), the corresponding dynamic Cauchy stress tensor iswhere $\tilde{C}(t)=C(t)-\rho g(t)$. Substituting its divergence into the Cauchy momentum equation reproduces $\rho [\mathbf{Ay}+\ddot{\mathbf{c}}]$, as required.

For each pair of adjacent spins, define the transfer matrix for a classical spin chain by $T_{s{s}^{\prime }}=e^{-\beta E_0(s,s^{\prime })}$. Multiplying its entries gives the Boltzmann factor of a whole configuration. The periodic boundary condition identifies the last index with the first, soThe assumed interchange symmetry of $E_0$ makes $T$ a real symmetric matrix, hence it has an orthonormal eigenbasis. At finite positive temperature, every matrix entry is strictly positive, so the Perron–Frobenius theorem supplies a unique positive largest eigenvalue ${\lambda }_{+}$, larger in modulus than every other one. Therefore $L^{-1}\log Z_L\to \log {\lambda }_{+}$.

Local observables become matrix insertions. For example, with $S_{s{s}^{\prime }}=s{\delta }_{s{s}^{\prime }}$,Thus partition functions and correlation functions reduce to a fixed-dimensional eigenvalue problem, rather than a sum over exponentially many microstates.

Put $\beta =(k_{B}T{)}^{-1}$ and label the two spin values in the order $+1,-1$. Assign each bond its Boltzmann weight,The symmetry of the bond energy makes $V$ a real symmetric matrix. For a fixed configuration the product of successive matrix entries is its Boltzmann weight. Summing over all spins and identifying the last index with the first therefore giveswhereThis replaces a sum over $2^L$ configurations by the powers of a two-by-two matrix. For finite bond energies at $T>0$, every matrix entry is positive, so the Perron–Frobenius theorem gives ${\lambda }_{+}>0$ and $|{\lambda }_{-}|<{\lambda }_{+}$. Hence the Helmholtz free energy per spin tends to $-k_{B}T\log {\lambda }_{+}$. A diagonal local observable $O$ can be inserted at one site: $⟨O_1⟩=\mathrm{tr}(O{V}^L)/Z_L$. Its limiting average is the expectation in the normalized eigenvector of ${\lambda }_{+}$.

Let $\beta =1/(k_{B}T)$ and use the two spin values $+1,-1$ as row and column labels. Define the spin-chain transfer matrixIt is a real symmetric matrix because $E_0(s,t)=E_0(t,s)$; every entry is strictly positive for finite bond energies at $T>0$. The Boltzmann factor of a spin configuration factorizes into the product $W_{{\sigma }_1{\sigma }_2}\cdots W_{{\sigma }_L{\sigma }_1}$. Summing the intermediate spin labels is exactly matrix multiplication, and the final closure imposes periodic boundary conditions. ThusWriting $W=\left(\begin{array}{cc}a& b\\ b& d\end{array}\right)$, its eigenvalues areA positive-entry matrix need not have two positive eigenvalues; ${\lambda }_{-}$ can be negative. The Perron–Frobenius theorem ensures ${\lambda }_{+}>|{\lambda }_{-}|$ here. Therefore the Helmholtz free energy is $F_L=-k_{B}T\ln Z_L$, and its value per spin in the thermodynamic limit is $-k_{B}T\ln {\lambda }_{+}$.

For an open chain there are $L-1$ bonds, with both endpoint Ising spins independently summed. Retain the same spin-chain transfer matrix, but replace the trace by endpoint vectors. With $1=(1,1{)}^{\mathsf{T}}$,This follows directly by expanding the product of $L-1$ Boltzmann factors. If ${\mathbf{v}}_{\pm }$ are orthonormal eigenvectors of the symmetric matrix $W$, the equivalent spectral decomposition isFixed endpoint Ising spins replace the corresponding $1$ by a basis vector; extra endpoint energies replace its entries by the corresponding Boltzmann factors. The leading bulk Helmholtz free energy per spin agrees with the periodic result, because the boundary changes only subextensive terms. In particular, $L=1$ gives $Z_1^{\mathrm{open}}=1^{\mathsf{T}}1=2$, as required for one free Ising spin.

True. Transposing the inverse identities shows $(X^{-1}{)}^T=(X^T{)}^{-1}$. If $X^T=X$, this becomes $(X^{-1}{)}^T=X^{-1}$, exactly the symmetric matrix condition. The reasoning uses the matrix inverse and transpose identities, without assuming that different matrix factors commute.

They are similar matrices, so their eigenvalues, with algebraic multiplicities, are identical. Explicitly,Moreover their eigenvectors correspond by $Av=\lambda v⟺(D^{-1}AD)(D^{-1}v)=\lambda D^{-1}v$. Since the transformed matrix is a real symmetric matrix, the spectral theorem for real symmetric matrices also shows that $A$ has real eigenvalues and is a diagonalizable matrix.

For the analogous positive off-diagonal entries in size $n$, takeThe same entrywise computation gives $(D^{-1}AD{)}_{ii}=a_i$ and $(D^{-1}AD{)}_{i,i+1}=(D^{-1}AD{)}_{i+1,i}=\sqrt{b_i{c}_i}$, with every other entry zero. Thus the result is a symmetric matrix with a tridiagonal pattern. Positivity of the diagonal entries $a_i$ is not needed for this positive diagonal symmetrization of a tridiagonal matrix; positivity of $b_i,c_i$ ensures the displayed positive, invertible $D$. In particular, the assertion is not an arbitrary-real-coefficient symmetrization claim when the off-diagonal signs differ.

For $D=\mathrm{diag}(d_1,d_2,d_3)$, the upper entry of $D^{-1}AD$ in positions $i,i+1$ is $b_i{d}_{i+1}/d_i$, and the corresponding lower entry is $c_i{d}_i/d_{i+1}$. They are equal when $(d_{i+1}/d_i{)}^2=c_i/b_i$. All coefficients used here are positive, so chooseDirect matrix multiplication gives the symmetric matrixIts pattern remains tridiagonal. This is positive diagonal symmetrization of a tridiagonal matrix.

Split $B$ into its symmetric part and skew-symmetric part:Taking the transpose gives $S^{\top }=S$ and $K^{\top }=-K$. Thus $S$ is a symmetric matrix and $K$ a skew-symmetric matrix. These formulas also show uniqueness, since a real matrix that is both symmetric and skew-symmetric is zero.

Put $h=\Delta x$ and suppose $L=(n+1)h$, so there are $n$ interior grid points $x_j=jh$, $j=1,\dots ,n$. Write $u_j(t)\simeq u(x_j,t)$. The fixed ends impose homogeneous Dirichlet boundary conditions,For a sufficiently smooth displacement field, the two Taylor series at fixed time givewhere all displayed derivatives on the right are evaluated at $(x_j,t)$. Adding cancels the odd terms, leaving the second-order central differenceSubstitute this finite difference into the wave equation and retain continuous time. The resulting method of lines gives coupled second-order ordinary differential equations,In particular, the first and last equations are ${\ddot{u}}_1=(c^2/h^2)(-2u_1+u_2)$ and ${\ddot{u}}_n=(c^2/h^2)(u_{n-1}-2u_n)$ for $n\ge 2$. With the interior displacement vector only,Equivalently, $A_{jj}=-2c^2/h^2$, $A_{j,j+1}=A_{j+1,j}=c^2/h^2$, and all other entries vanish. This entrywise definition also covers small $n$, including $A=(-2c^2/h^2)$ when $n=1$. The matrix is a symmetric matrix and a tridiagonal matrix. The spatial approximation has error $O(h^2)$ in Big O notation; time has not yet been discretized.

For each row, a Gershgorin disc has centre $a_{jj}=-2$ and radius $r_j={\sum }_{k\ne j}|a_{jk}|$. Here $r_1=r_3=1$ and $r_2=2$, so the Gershgorin circle theorem givesSince $A$ is a real symmetric matrix, its eigenvalues are real. Intersecting the union of these Gershgorin discs with the real axis yieldsThis bound permits zero but does not assert a zero eigenvalue. The fixed-end Dirichlet boundary conditions exclude it: for any nonzero real vector $v=(v_1,v_2,v_3{)}^T$, extend by $v_0=v_4=0$ and calculateIndeed, equality would force every adjacent value to agree with the zero endpoint. Thus $-A$ is a symmetric positive-definite matrix, $A$ is a negative-definite matrix, and the eigenvalues satisfy $-4\le \lambda <0$. The same argument works for any number of interior grid points, with prefactor $c^2/h^2>0$ in the sum. The fixed string has no zero-frequency rigid translation.

Regard $A$ as a linear map ${\mathbb{R}}^4\to {\mathbb{R}}^4$. By the rank-nullity theorem and matrix rank $3$,All three are vector subspaces of ${\mathbb{R}}^4$, with the row space identified with four-component row vectors in the usual way. The null space lies in the input vector space of concentrations; the column space lies in the output vector space of concentration-change rates. Both have four coordinates, although they play different physical roles.

In this particular matrix, the row sums and column sums both vanish. Put $1=(1,1,1,1{)}^T$. Then $A1=0$ and $1^{T}A=0$, and the two three-dimensional vector subspaces coincide:This equality follows here from both sum identities and the dimensions of a vector space; it does not make $A$ a symmetric matrix and is not true for every square matrix.

The strict positivity of a quadratic form on every nonzero vector. A real symmetric matrix has positive definiteness exactly when its associated quadratic form does, making it a positive-definite matrix. A Hermitian form is positive definite when its value on $(v,v)$ is real and strictly positive for every nonzero vector.

Think of a positive-definite quadratic form as an energy cost for a displacement: every direction costs something, and zero cost identifies the origin. For example, the real symmetric matrixis a positive-definite matrix because the two squares can vanish together only at $(0,0)$. The mixed term couples the coordinates, but cannot overwhelm their positive contributions: $q(1,1)=8$ and $q(1,-1)=4$. The graph is a bowl, and its positive level sets are ellipses with axes along the eigenvectors, rather than along the original coordinates.

The real and complex settings require different symmetry assumptions. For a real symmetric matrix use $q(x)=x^{T}Ax$ with $x\in {\mathbb{R}}^n$; for a complex Hermitian matrix use $q(z)=z^{\ast }Az$ with $z\in {\mathbb{C}}^n$, where $z^{\ast }$ is the conjugate transpose. The latter expression is real, so strict positivity is meaningful. A complex quadratic form such as $z_1^2+z_2^2$ has no such positivity: it vanishes at $(1,i)$ and takes negative real values at $(i,0)$. A Hermitian form uses complex conjugation, which changes this example to $|z_1{|}^2+|z_2{|}^2$.

For a real matrix $M$ without a symmetry hypothesis, its quadratic form depends only on its symmetric part of a matrix:Positive eigenvalues of $M$ alone do not establish positive definiteness of this quadratic form. For example, $M=\left(\begin{array}{cc}1& 4\\ 0& 1\end{array}\right)$ has both eigenvalues equal to $1$, yet $q(1,-1)=1-4+1=-2$. The relevant test is positivity of the symmetric part of a matrix, not the spectrum of an arbitrary matrix.^[ref]

If a real matrix $B$ has a positive diagonal matrix $W$ such that $WB$ is a symmetric matrix, then $H=W^{1/2}B{W}^{-1/2}$ is a symmetric matrix related to $B$ by a similarity transformation: $B^{T}W=WB$ gives $H^T=H$. Thus $B$ is a diagonalizable matrix with real eigenvalues. If ${\mathbf{z}}^{T}WB\mathbf{z}\le 0$ for all real vectors $\mathbf{z}$, its eigenvalues are nonpositive. This converts symmetry in a weighted inner product into Euclidean symmetry.

Let a real tridiagonal matrix have diagonal entries $a_i$, positive upper entries $b_i$, and positive lower entries $c_i$. Set $d_1=1$, $d_{i+1}=d_i\sqrt{c_i/b_i}$ and $D=\mathrm{diag}(d_i)$. Then $D^{-1}AD$ is a symmetric matrix with diagonal $a_i$ and adjacent entries $\sqrt{b_i{c}_i}$. Indeed its two adjacent entries are $b_i{d}_{i+1}/d_i$ and $c_i{d}_i/d_{i+1}$, which are equal by construction. All other zeros are preserved. This matrix similarity implies real eigenvalues and diagonalizability, by the spectral theorem for real symmetric matrices. No sign assumption on $a_i$ is needed. In the weighted formulation of positive diagonal symmetrization of a matrix, the positive diagonal matrix is $W=D^{-2}$.