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Incoming links: Matrix inverse
To invert the matrix, row-reduce the augmented matrix . Subtracting row one from rows two and three gives left rows and , so the pivots are nonzero and back-substitution is possible. The resulting matrix inverse isMultiplying by verifies the identity matrix. This inverse is not ; indeedThus is not orthogonal.
If every eigenvalue is nonzero, define , where . Using the mutually orthogonal projection identity ,The same calculation gives . Thus the matrix inverse is
Since , the matrix inverse is . The orthogonal matrix condition also gives , so . Thus is skew-symmetric, forcing and . Its columns must be unit vectors, giving . Therefore the two possibilities areBoth satisfy the original conditions by direct multiplication. They are the two orthogonal skew-symmetric matrices in two dimensions, representing opposite quarter-turn rotations.
For the choice of in part (b)(i), the two-by-two matrix inverse formula givesMultiplication gives , providing a direct check.
Subtracting the second row from the third gives a third row , so expansion along it shows . Thus the matrix inverse exists and every right-hand side has a unique solution.
In Cartesian coordinates a rank- tensor is an array whose components transform under a change of orthonormal axes by one rotation matrix factor per index:An isotropic tensor has unchanged components under every proper rotation. The Levi-Civita symbol obeysfor , proving rotation invariance of the Levi-Civita symbol. Under an orthogonal transformation of determinant its sign reverses; including such transformations requires the pseudotensor distinction. Here isotropy is with respect to proper rotations.
Contract the determinant identity with . Since , one obtainsEvery index on the right is contracted, so the tensor component transformation law leaves no free rotation factor: the determinant is a scalar. Equivalently, gives .
For the matrix inverse, defineThis is the cofactor matrix. To verify the inverse formula rather than just recognize it, contract with and use the determinant identity again:Here the contraction of two Levi-Civita symbols gives . Therefore and, when ,These are the Levi-Civita formulas for a three-dimensional determinant and inverse.
Now differentiate the cubic determinant formula with respect to the individual entry . The three differentiated factors give equal contributions: exchanging two paired indices changes the sign of both epsilon symbols, leaving their product unchanged. ThusSubstituting the inverse expression proves the entrywise derivative of the determinant:The cofactor version remains valid for singular matrices; only the inverse version requires invertibility.
For the time-dependent Jacobian matrix, the stated smooth expansion impliesApplying the chain rule and the entrywise determinant derivative gives Jacobi's formula near :At zero, , so . A Taylor expansion therefore yieldsThe Jacobian determinant measures local volume change, so this establishes volume preservation to first order. An arbitrary family specified only through its first-order expansion need not preserve volume exactly at higher orders.
Using part (i) and in the Cayley transform with argument givesThe factors commute because and are polynomials in the same matrix, and an invertible commuting matrix also commutes with the other factor's matrix inverse. Both inverses exist by part (ii). Thus is unitary.
Taking the conjugate transpose of and reverses the product order and givesThus is the two-sided matrix inverse of , proving
A finite-dimensional representation of a group on a vector space is a group homomorphism . Thus each is invertible andHere the group identity and the identity linear map are different objects, represented by corresponding symbols. Applying the multiplication property to givesBy the defining property of a matrix inverse,If the definition only states the multiplication law together with invertibility, already forces after multiplication by its inverse.
Since , the QR factorization givesFirst compute by a matrix-vector product, costing scalar multiplications and additions for dense . Then solve the upper triangular matrix system by back substitution:Every denominator is nonzero because is invertible. The back substitution costs multiplications and subtractions together, plus divisions, for arithmetic operations. Including , the total is arithmetic operations under this scalar counting convention, hence , with denoting Big O notation. The already supplied QR factorization is reused; no matrix inverse needs to be computed.
Since every , the diagonal matrix is invertible. SetThe division is by the diagonal entry of row . Thus is an upper triangular matrix with diagonal entries one and .
To use symmetry without assuming the conclusion, multiply by the matrix inverse of and its transpose:The matrix is symmetric because is a symmetric matrix. It is also an upper triangular matrix, since every factor on the right is an upper triangular matrix. A symmetric upper triangular matrix is diagonal. Its diagonal is exactly , because and have diagonal entries one. Hence , givingThis is the LDL decomposition.
Take the positive square roots of the diagonal entries:The matrix is a lower triangular matrix with positive diagonal, andThis is a Cholesky decomposition. For a nonzero vector , the vector is nonzero because is invertible. ThereforeHere is the Euclidean norm. Thus is a positive-definite matrix.
For a real square matrix, a QR factorization writes with an orthogonal matrix, , and an upper triangular matrix. If is invertible, so is , andCompute and use back substitution; no matrix inverse needs to be formed.
Apply the Gram-Schmidt process to the columns . The first column has Euclidean norm two, giving and . For the second column,For the third column,This residual already has Euclidean norm one. Set and . ThusThe columns of are orthonormal, and matrix multiplication verifies .
Using precisely this QR factorization, transform the right-hand side:Back substitution in givesTherefore
Choose a shift which is not an eigenvalue. Inverse iteration applies power iteration to , implemented by a linear system solve rather than by forming an explicit matrix inverse:A single LU decomposition of can be reused for all the solves at this fixed shift. An eigenpair transforms intoThus an eigenvalue uniquely closest to becomes the dominant eigenvalue of . For a diagonalizable matrix, the argument in the preceding subpart gives convergence of diagonalizable fixed-shift inverse iteration, provided the starting coefficient in the target eigenspace is nonzero, with power iteration direction error , whereFor any prescribed simple eigenvalue, taking a sufficiently close but unequal shift makes the numerator small while the denominator stays close to its positive separation from the remaining eigenvalues. This makes small and gives very fast convergence per iteration. The Rayleigh quotient of estimates directly. If estimates the associated eigenvalue of , the corresponding estimate is .
The selection principle also applies when the other eigenvalues have nontrivial blocks in the Jordan normal form: their contributions acquire powers of , multiplying strictly smaller geometric factors. A simple uniquely closest eigenvalue still attracts the direction when the Riesz projection of onto its generalized eigenspace is nonzero. For nonreal target eigenvalues, allow complex shifts and arithmetic; a real shift cannot distinguish a conjugate pair of equal distances. Setting makes the solve singular, and an exactly zero starting target component prevents convergence to that target in exact arithmetic. When is real, account for the sign behavior of fixed-shift inverse iteration.
Past exam of the numerical linear algebra course of the University of Oxford 2011 2 a Solution 2026-10-31
The spectral theorem for real symmetric matrices gives with an orthogonal matrix . Positive definiteness makes every eigenvalue positive. Define the matrix square rootIt is symmetric, its square is , and for any nonzero vector ,Thus it is also positive-definite, and its matrix inverse is .
Past exam of the numerical linear algebra course of the University of Oxford 2025 1 b i Solution 2026-10-25
Choose the nonnegative square roots of the eigenvalues of for the diagonal entries of , in nonincreasing order. Its assumed matrix inverse exists, so these entries are positive. The eigendecomposition givesHence has orthonormal columns. AlsoTogether with the orthogonal matrix and the nonnegative diagonal matrix , these are precisely the properties of a thin singular value decomposition.
Past exam of the rings and modules course of the University of Oxford 2023 3 a iv Solution 2026-10-25
In the basis , the linear map given by sends to the second basis vector and sends to . Equivalently,Multiplying on the left by the matrix inverse givesThis change of basis produces the companion matrix found in part (a)(ii).
The Cholesky decomposition gives . If , reversal of factors in a matrix inverse givesHence the required R implementation is
invsigma <- function(sigma) {
L <- mychol(sigma)
B <- ltinv(L)
crossprod(B)
}Choose distinct coordinates uniformly for the starting ones; each element of then has starting probability . At each iteration choose one current one and one current zero uniformly and swap them. This implements the normalized proposal distribution from (c)(i), remains in , and needs just one precomputed matrix inverse.
mymcmc <- function(x, k, sigma) {
n <- length(x)
stopifnot(length(k) == 1L, is.finite(k), k == floor(k),
k > 0L, k < n, is.matrix(sigma),
all(dim(sigma) == c(n, n)))
sigma_inv <- invsigma(sigma)
u <- integer(n)
u[sample.int(n, k, replace = FALSE)] <- 1L
draws <- matrix(0, 100L, n)
for (iteration in seq_len(100L)) {
ones <- which(u == 1L)
zeros <- which(u == 0L)
i <- ones[sample.int(length(ones), 1L)]
j <- zeros[sample.int(length(zeros), 1L)]
v <- u
v[i] <- 0L
v[j] <- 1L
alpha <- mylogapp(x, u, v, sigma_inv)
if (runif(1L) < alpha) u <- v
draws[iteration, ] <- u
}
colMeans(draws)
}Sampling positions with
sample.int and then indexing into ones or zeros also works when or . In R, a call such as sample(ones, 1) would incorrectly sample from 1:ones when ones is a single numeric value greater than one.The code records the state after each of exactly transitions, including repetitions after rejection. Its answer is the vector Monte Carlo estimatorThis estimates the posterior mean coordinate by coordinate; because each coordinate is binary, its entries estimate the posterior probabilities of inclusion. Every stored state has coordinate sum , so the estimated means lie in and sum to .
For neighbouring states, the symmetry of the proposal and the Metropolis–Hastings acceptance probability give detailed balance:Thus the posterior distribution is a stationary distribution. Any two states in can be connected by moving the ones that occupy the wrong positions, and the multivariate normal distribution gives strictly positive target masses. The resulting Markov chain is therefore an irreducible Markov chain, and its long-run state averages estimate the posterior mean.
The requested iterations are a finite approximation, without a convergence guarantee or discarded burn-in. A randomly chosen starting subset has the prior distribution, rather than generally the posterior distribution. No aperiodicity assumption is needed for the finite-state Markov chain ergodic theorem for averages; indeed a uniform two-state target with gives a period-two alternating chain. This does not invalidate either the code or the state-average estimator.
Transpose both inverse identities, using reversal of multiplication under matrix transposition:Thus , by uniqueness of the matrix inverse.
True. Transposing the inverse identities shows . If , this becomes , exactly the symmetric matrix condition. The reasoning uses the matrix inverse and transpose identities, without assuming that different matrix factors commute.