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Choose the following consistent sign convention. The Givens rotation $J(i,j,\theta )$ equals the identity matrix except for the block on coordinates $i,j$:Thus it changes only the $i$th and $j$th coordinates of a vector. Write $c=\cos \theta$, $s=\sin \theta$. Its two-coordinate block satisfiesAll remaining coordinates are unchanged, so $J^{T}J=I_n$. Hence every such Givens rotation is an orthogonal matrix. With this convention the two-coordinate rotation matrix rotates through angle $-\theta$.

Apply the product from right to left, choosing every Givens rotation from the current first column. First $J(1,2)$ replaces $(a_1,a_2)$ by $(\sqrt{a_1^2+a_2^2},0)$. Next $J(1,3)$ combines that retained entry with $a_3$ and sets the third entry to zero. Inductively, after eliminating coordinates $2,\dots ,k$, the first coordinate isand coordinates $2,\dots ,k$ are zero. Later Givens rotations involve only coordinate one and a new coordinate, so those zeros remain zero. Thereforewhere $e_1$ is the first standard basis vector. The zero vector is included by choosing the identity matrix for every zero pair. The nonnegative sign follows from the convention fixed in part (c)(ii). This is the first-column stage of QR decomposition by Givens rotations.

Take the identity matrixBoth $B_1$ and $B_2$ have matrix rank one, andThey are distinct and both attain the minimum spectral norm error from the preceding part. Hence a best low-rank approximation of rank one need not be unique. More generally, every orthogonal projection $u{u}^T$ with $\Vert u{\Vert }_2=1$ gives another such minimizer.

The requested base point is the identity matrix $I$, not the zero matrix. Since $I$ commutes with every $H$,The matrix power series justifies the quadratic remainder near zero. ConsequentlyAt zero the derivative would instead be $H$; replacing the printed base point would lose the scalar factor.

Use dimensionless physical helicity $h=\mathbf{S}\cdot \hat{\mathbf{p}}$, with spin eigenvalues $\pm 1/2$ in natural units. In the Dirac representation,The entries $1$ in the supplied transformation denote two-by-two identities. Since its matrix acts only on the upper/lower spinor blocks, it commutes with the repeated Pauli block. Therefore $u^{-1}=u$ andThe eigenvalues remain the physical spin projection along the direction of motion, $+1/2$ or $-1/2$. A convention omitting the factor one half instead labels them $\pm 1$; this changes no measurement probabilities. A change of gamma-matrix representation cannot change the spin observable.

Since $A^2+I=0$, its minimal polynomial $m_A$ divides $X^2+1$. The polynomial $X^2+1$ is an irreducible polynomial over the rational numbers: a reducible quadratic would have a rational root, whereas $r^2+1>0$ for every $r\in \mathbb{Q}$.

Because $n>0$, the constant polynomial one evaluates to the nonzero identity matrix $I$, so $m_A\ne 1$. Both $m_A$ and $X^2+1$ are monic polynomials; hence the only remaining divisor is

Write the first strain invariant in components of the deformation gradient:Here lower-case and upper-case indices refer to the current configuration and reference configuration, respectively. Differentiating gives $\partial I_1/\partial F_{iI}=2F_{iI}$. The chain rule therefore givesSubstitution into the given hyperelastic material law, with its indicated matrix transpose, yieldsThe left Cauchy-Green deformation tensor is $\mathbf{B}=\mathbf{F}{\mathbf{F}}^T$. The scalar $p$ is the Lagrange multiplier enforcing incompressibility, rather than a function fixed by the strain energy density alone. The identity matrix $1$ makes this an isotropic constraint reaction in hyperelastic stress.

The radial unit vector in spherical coordinates is ${\mathbf{e}}_R=\mathbf{X}/R$, so $\mathbf{X}\otimes \mathbf{X}/R^2={\mathbf{e}}_R\otimes {\mathbf{e}}_R$. The identity matrix is the sum of the three basis outer products. Substitution in the given deformation gradient givesThe reference and current radial directions coincide for this spherically symmetric deformation. The positive principal stretches are therefore ${\lambda }_r=f^{\prime }(R)$ and ${\lambda }_{\theta }={\lambda }_{\varphi }=f(R)/R$. Incompressibility imposesIntegrating and using $f(A)=a$ givesIn particular $b^3-a^3=B^3-A^3$, expressing conservation of shell volume. Writing $\lambda =r/R$ also gives ${\lambda }_r={\lambda }^{-2}$ and ${\lambda }_{\theta }={\lambda }_{\varphi }=\lambda$.

Restore the particle label suppressed in the question, writing the static family as ${\mathbf{x}}_0(\mathbf{X};\tau )$. At each fixed $\tau$, its Cauchy stress tensor satisfiesThis dynamic continuation of a static incompressible elastic deformation sets $\tau =t$ to form a motion. At every time its deformation gradient is the same as that of the corresponding static member. For an elastic material the elastic part of the Cauchy stress tensor depends on that deformation gradient, while incompressibility leaves an arbitrary scalar constraint reaction in hyperelastic stress. Therefore writeHere $1$ is the identity matrix. Subtracting static balance of linear momentum from dynamic balance of linear momentum givesSince the homogeneous incompressible material has constant mass density, define $\xi =(p-p_0)/\rho$. NecessarilyConversely, suppose this acceleration potential exists. Choose $p=p_0+\rho \xi$. ThenThe added isotropic stress is symmetric, so angular momentum conservation remains satisfied. This proves both directions for the field equations and the elastic constitutive equation. The surface tractions for the motion are those of the adjusted stress; identical prescribed static tractions would impose an additional boundary restriction on $\xi$.

Use the characteristic polynomial convention $p(\lambda )=\det (\lambda I-P)$, with $I$ the identity matrix. Put $u=\lambda -1/2$. ThenThe required characteristic polynomial equation for this matrix is thereforeUsing $\det (P-\lambda I)$ instead negates the cubic polynomial and leaves the eigenvalues unchanged.

A length-zero graph path starts and ends at the same vertex. Therefore $P_0$ is the Boolean identity matrix. A graph path of length at most one is either such a stationary graph path or one original directed edge:The logical disjunction is entrywise. Diagonal entries of $E$ may be false if there are no self-loops, but every diagonal entry of $P_1$ is true because the zero-length graph path is allowed.

A graph path of length at most $k+1$ either already has length at most $k$, or consists of a prefix of length at most $k$ followed by one edge. ThereforeThe forward implication follows by removing the final edge of a nonempty graph path. Conversely, concatenating the indicated prefix and edge gives a walk of length at most $k+1$, and removing repetitions if necessary gives a graph path no longer than that walk.

Writing $\odot$ for Boolean matrix multiplication, the formula is $P_{k+1}=P_k\vee (P_k\odot E)$. Because $P_1$ includes the identity matrix, an equivalent formula isCompute each new matrix from an unchanged copy of the preceding matrix, so the subscript continues to specify exactly the length bound.

Split a graph path of length at most $2k$ into two consecutive parts of length at most $k$, allowing a zero-length part. Conversely, concatenating two such parts gives a walk of length at most $2k$ and hence the same reachability. ThereforeStart at $P_1=I\vee E$, then square to obtain $P_2,P_4,P_8,\dots$. Stop at equality of successive matrices, or after the length bound reaches $n-1$. The invariant is now $P=P_{2^r}$ after $r$ squarings. Equality implies closure under composition, and the included identity matrix ensures reflexivity.

For the same elementary dense Boolean matrix multiplication and equality stopping test, the cost isThe additive constant covers the final stability check and the cases $m=0,1$. Thus, with $m\ge 2$, the number of cubic passes falls from $\Theta (m)$ to $\Theta (\log m)$. In terms of $n$ alone the worst-case bound improves from $\Theta (n^4)$ to $\Theta (n^3\log n)$ for these implementations. A directed chain also requires logarithmically many squarings. This comparison uses ordinary triple-loop multiplication, rather than a faster matrix multiplication method.

An elementary matrix acts on rows by multiplication on the left. Apply the desired elementary row operation to the identity matrix to construct it. In the order requested, the three elementary matrices and their inverse matrices areThe inverse operations halve row 1, add twice row 1 to row 3, and interchange rows 2 and 3 again, respectively.

Define the characteristic polynomial using $p(\lambda )=\det (\lambda I-P)$, where $I$ is the identity matrix. Expanding the determinant along the first row givesThus the characteristic equation isThe convention $\det (P-\lambda I)$ changes the sign of this cubic polynomial but gives the same roots.

Suppose an invertible matrix has two LU factorizations, $A=L_1{U}_1=L_2{U}_2$, with $L_1,L_2$ unit lower triangular matrices and $U_1,U_2$ upper triangular matrices. All four factors are invertible matrices, since their square products are invertible. Multiplication by the appropriate inverse matrices givesBy the supplied closure properties, the left side is a unit lower triangular matrix and the right side is an upper triangular matrix. A matrix of both types is diagonal; its diagonal entries are all one. Thus both sides are the identity matrix, whenceQualification of the printed wording: this establishes uniqueness if the specified LU factorization exists. Invertibility alone does not guarantee an unpivoted LU factorization. For example $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ is invertible, but in a product $\left(\begin{array}{cc}1& 0\\ \ell & 1\end{array}\right)\left(\begin{array}{cc}{u}_{11}& u_{12}\\ 0& u_{22}\end{array}\right)$, the zero top-left entry forces $u_{11}=0$, making the bottom-left entry $\ell u_{11}=0$ as well, contrary to the required value one. A permutation matrix is needed in general, as in PLU factorization.

Let $e_i$ be standard coordinate columns. The required elementary matrix isIt is the identity matrix with the extra entry $\alpha$ in position $(j,i)$. Left matrix multiplication implements the row operation: all other rows are unchanged. Since $j>i$, $E$ is a lower triangular matrix with every diagonal entry one. Part (a) therefore gives $\det (E)=1$ andHence adding a multiple of one row to another leaves the determinant unchanged, even if $A$ is singular.

Write the determinant in its Leibniz formula for determinants, where $S_n$ is the set of permutations of $\{1,\dots ,n\}$:The permutation sign is justified by the given row-interchange property: a permutation matrix is obtained from the identity matrix by row interchanges, and every interchange changes its determinant by a factor $-1$. Reversing a sequence of interchanges constructs the inverse permutation with the same number of interchanges. Consequently $\mathrm{sgn}({\sigma }^{-1})=\mathrm{sgn}(\sigma )$. The Leibniz formula for determinants can itself be obtained by expanding the rows multilinearly in the standard basis: repeated selected coordinate rows contribute zero, and the remaining terms are precisely these signed permutation matrices.

Since $(A^T{)}_{ij}=a_{ji}$, the same formula givesReindex the product with $j=\sigma (i)$ and reindex the sum with $\tau ={\sigma }^{-1}$. Both reindexings are bijections, soThus $\det (A)=\det (A^T)$ for every real square matrix, including singular matrices. This proof does not assume the analogous column-operation rules in order to establish the transpose identity.

Undo each elementary row operation: divide row 3 by 3, add 3 times row 2 to row 4, and repeat the same swap. HenceMultiplying each inverse by its original matrix gives the identity matrix.

Let the original rows be $r_1,\dots ,r_5$. Begin with rows $r_2$, $r_3-r_2$ and $r_4$:Subtract the third from the second, then subtract that new second row from the first. The resulting rows areThey are linearly independent because their first three coordinates form the identity matrix. All original rows lie in their span: $r_1=2r_2+r_4$ and $r_5=r_2-(r_3-r_2)+r_4$. Thus they form a basis of the row space, and $\mathrm{rank}A=3$. Equivalently, these are the nonzero rows of the reduced row echelon form, with two further zero rows.