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At the angles in part (f),Let the two diagonal matrices be $D=\mathrm{diag}(d_1,d_2,d_3)$ and $D^{\prime }=\mathrm{diag}(e_1,e_2,e_3)$. Multiplication givesThe vector condition is exactly $c{e}_2{d}_3=b$. For $c\ne 0$, the general form is thereforeEvery pair $p,q$ is attainable, for example with $D=I$ and $D^{\prime }=\mathrm{diag}(-p,b/c,-q)$. The transformation need not be orthogonal or invertible; the question asks only for diagonal rescalings with this one vector image. If $c=0$ and $b\ne 0$, no linear map can satisfy the condition. If $c=b=0$, the $(2,3)$ entry is arbitrary as well.

A square matrix is a diagonalizable matrix over the chosen field when there are an invertible matrix $P$ and a diagonal matrix $D$ such that $A=PD{P}^{-1}$. If $v_j$ is column $j$ of $P$ and $D_{jj}={\lambda }_j$, the identity $AP=PD$ gives $A{v}_j={\lambda }_j{v}_j$. Invertibility makes these $n$ columns linearly independent eigenvectors.

Conversely, put $n$ linearly independent eigenvectors in the columns of $P$ and their eigenvalues in $D$. Then $P$ is invertible and $AP=PD$, givingRepeated eigenvalues are allowed; the requirement is a complete eigenbasis, not distinctness of the eigenvalues.

Since every $u_{jj}>0$, the diagonal matrix $D$ is invertible. SetThe division is by the diagonal entry of row $i$. Thus $\hat{U}$ is an upper triangular matrix with diagonal entries one and $A=LD\hat{U}$.

To use symmetry without assuming the conclusion, multiply by the matrix inverse of $L$ and its transpose:The matrix $M$ is symmetric because $A$ 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 $D$, because $\hat{U}$ and $L^{-T}$ have diagonal entries one. Hence $M=D$, givingThis is the LDL decomposition.

Take the positive square roots of the diagonal entries:The matrix $\hat{L}$ is a lower triangular matrix with positive diagonal, andThis is a Cholesky decomposition. For a nonzero vector $x$, the vector ${\hat{L}}^{T}x$ is nonzero because ${\hat{L}}^T$ is invertible. ThereforeHere $\Vert \cdot {\Vert }_2$ is the Euclidean norm. Thus $A$ is a positive-definite matrix.

A real singular value decomposition has the form $A=U\Sigma V^T$, where $U\in {\mathbb{R}}^{m\times m}$ and $V\in {\mathbb{R}}^{n\times n}$ are orthogonal matrices, and $\Sigma$ is a rectangular diagonal matrix with entries ${\sigma }_1\ge \cdots \ge {\sigma }_{\min (m,n)}\ge 0$. These entries are the singular values. Their squares are the eigenvalues of $A^{T}A$, omitting the additional zero eigenvalues when $n>m$. The positive singular values number exactly the matrix rank.

Let $u_j$ and $v_j$ be the columns of the orthogonal matrices $U$ and $V$ in the singular value decomposition. Multiplying by the rectangular diagonal matrix retains its $r$ nonzero diagonal entries:The families $(u_j)$ and $(v_j)$ are orthonormal, hence in particular orthogonal. Each summand is a rank-one outer product, and the remaining terms vanish because their singular values are zero.

Write $A=D+L+R$, where $D$ is the diagonal matrix of diagonal entries of $A$, and $L,R$ are its strictly lower- and upper-triangular parts. Since $A$ is symmetric, $R=L^T$. Subtract the two factors in the SOR iteration:Therefore, for $\omega \ne 0$,Positive definiteness makes each $D_{ii}>0$, so the triangular matrix $B(\omega )$ is an invertible matrix for every $\omega \ne 0$. The original formula gives $C(0)=I$; the expression for $B(\omega )$ itself is undefined there. The affine term ensuring that the desired solution is a fixed point is $c=B(\omega {)}^{-1}b$.

Put $B=A_1^T{A}_1$. It is a symmetric matrix and a positive semidefinite matrix, while $A_2^T{A}_2=I_n-B$ is also a positive semidefinite matrix. By the spectral theorem, choose an orthogonal matrix $V=[v_1,\dots ,v_n]$ such thatThe bounds follow directly from $z^{T}Bz=\Vert A_{1}z{\Vert }_2^2\in [0,\Vert z{\Vert }_2^2]$. Define the diagonal matricesThe columns of $A_{1}V$ are mutually orthogonal, since $(A_1{v}_i{)}^T(A_1{v}_j)={\lambda }_j{\delta }_{ij}$. For each ${\lambda }_j>0$, set $u_{1j}=A_1{v}_j/\sqrt{{\lambda }_j}$. These columns are orthonormal. At a zero ${\lambda }_j$, one has $A_1{v}_j=0$; choose $u_{1j}$ by completing the already prescribed columns to an orthonormal set of $n$ columns in ${\mathbb{R}}^k$. Such a completion exists because $k\ge n$. This gives $A_{1}V=U_1{\Sigma }_1$, including the zero columns.

Likewise, the Gram matrix of $A_{2}V$ is $\mathrm{diag}(1-{\lambda }_j)$. Normalize its nonzero columns and complete at its zero columns in ${\mathbb{R}}^{m-k}$, using $m-k\ge n$. The resulting $U_2$ has orthonormal columns and $A_{2}V=U_2{\Sigma }_2$. Combining the identities givesThis is a form of the cosine-sine decomposition. Completing the zero columns is essential: neither block was assumed to have full column rank.

Choose the nonnegative square roots of the eigenvalues of $A^{T}A$ for the diagonal entries of $\Sigma$, in nonincreasing order. Its assumed matrix inverse exists, so these entries are positive. The eigendecomposition givesHence $U$ has orthonormal columns. AlsoTogether with the orthogonal matrix $V$ and the nonnegative diagonal matrix $\Sigma$, these are precisely the properties of a thin singular value decomposition.

Take a singular value decomposition $F=Q_{1}D{Q}_2^T$, with $D=\mathrm{diag}({\lambda }_1,{\lambda }_2,{\lambda }_3)$ positive and $Q_1,Q_2\in SO(3)$. Such a choice is possible because $\det F>0$. Left material frame indifference removes $Q_1$, and right material isotropy removes $Q_2$, giving $W(F)=W(D)$.

Moreover $W(D)$ is unchanged when the positive diagonal entries are permuted. Every permutation can be implemented by a signed permutation matrix of determinant one; the signs disappear in conjugating a diagonal matrix. Thus the energy depends only on the unordered principal stretches.

The three invariants determine precisely this unordered triple: the roots of $t^3-I_B{t}^2+I{I}_{B}t-II{I}_B$ are ${\lambda }_1^2,{\lambda }_2^2,{\lambda }_3^2$, and their positive square roots give the stretches. Consequently, on the set of admissible invariant triples, define $h(I_B,I{I}_B,II{I}_B)$ to be $W(D)$. The preceding argument proves that this is well defined and thatNot every positive triple is realizable by three positive roots. Since the requested statement only asserts existence of a function on $(0,\infty {)}^3$, extend $h$ arbitrarily on the nonrealizable triples; no choice there affects the formula for a deformation in continuum mechanics. Smooth invariant representations, where needed for differentiating the constitutive equation, are understood locally on the realizable set.

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.

Use $-A=(-I_n)A$. The diagonal matrix $-I_n$ has $n$ diagonal entries equal to $-1$, so the product and triangular matrix formulas from part (a) giveEquivalently, multiplying each of the $n$ rows by $-1$ supplies one factor $-1$ per row.

The diagonal matrix entry is $D_{ii}={\sum }_{k=1}^N{B}_{ik}{C}_{ki}$. Only $k=i$ can contribute, so $D_{ii}=B_{ii}{C}_{ii}$. This also shows that the product is a diagonal matrix, with its diagonal obtained by entrywise multiplication of the two diagonals.

By matrix multiplication, $D_{ij}={\sum }_{k=1}^N{B}_{ik}{C}_{kj}$. A potentially nonzero term requires $k=i$ because $B$ is a diagonal matrix, and $k=j$ because $C$ is a diagonal matrix. For $i\ne j$ these requirements cannot both hold. Thus every summand is zero and $D_{ij}=0$ whenever $i\ne j$.

The diagonalization gives $A=SD{S}^{-1}$ and hence $A^{n-1}=S{D}^{n-1}{S}^{-1}$: the intervening factors $S^{-1}S$ cancel. The diagonal matrix power has entries ${\alpha }^{n-1},{\beta }^{n-1}$. Since $1-\beta =\alpha$ and $\alpha -1=-\beta$,Reading the second component gives the Binet formulaThe derivation includes $n=1$, and the expression gives $U_1=U_2=1$, consistent with the specified indexing.

With the two diagonal matrices from the preceding part, write $D_A=PA{P}^{-1}$ and $D_B=PB{P}^{-1}$. Their diagonal products agree entry by entry, so $D_A{D}_B=D_B{D}_A$. ThereforeHence simultaneous diagonalization implies that $A,B$ are commuting operators.

For $n=2$, takeThey are commuting operators because every matrix commutes with the identity matrix. However $A$ has only the eigenvalue zero, and $\ker A=\mathrm{span}\{(1,0{)}^T\}$ has dimension one. It has no eigenbasis, so it cannot appear in any simultaneous diagonalization. Equivalently, a diagonal matrix similar to this nonzero nilpotent matrix would have to be zero, an impossibility. For any prescribed $n\ge 2$, use $A=E_{12}$ and $B=I_n$; its zero eigenspace has dimension $n-1$. For $n=1$ no such example exists.

The all-ones matrix is diagonalizable for every $n\ge 1$. The vector $u$ together with the $n-1$ vectors from part (ii) is an eigenbasis. Independence follows because a multiple of $u$ has coordinate sum $n$ times that multiple, whereas every other basis vector has coordinate sum zero. In this basis the diagonal matrix is $\mathrm{diag}(n,0,\dots ,0)$. At $n=1$ the one-vector basis suffices. Alternatively, the spectral theorem for real symmetric matrices applies. If dimension zero is allowed by the convention for $\mathbb{N}$, the empty $0\times 0$ matrix has no eigenvalues, an empty basis and a vacuous diagonalization.

Write $B=QD{Q}^{-1}$ with $D$ a diagonal matrix. Then $0=B^k=Q{D}^k{Q}^{-1}$, so each diagonal entry ${\lambda }_i$ satisfies ${\lambda }_i^k=0$ and therefore ${\lambda }_i=0$. Hence $D=0$ and $B=0$. Conversely the zero matrix is diagonalizable and nilpotent. This proves the existing result that a diagonalizable nilpotent matrix is zero; nilpotence alone does not imply a zero matrix, because nonzero nilpotent Jordan blocks exist.

Reference wave phases label complex amplitudes without changing measured powers. If $\mathbf{b}=S\mathbf{a}$, choose diagonal matrices $D_{\mathrm{in}},D_{\mathrm{out}}$ whose entries have modulus one, and set ${\mathbf{a}}^{\prime }=D_{\mathrm{in}}\mathbf{a}$, ${\mathbf{b}}^{\prime }=D_{\mathrm{out}}\mathbf{b}$. Then $S^{\prime }=D_{\mathrm{out}}S{D}_{\mathrm{in}}^{-1}$ and each component has unchanged squared modulus. For a beam splitter scattering matrix, transforming $S$ without transforming input complex amplitudes changes the stated physical input.

We fix the convention $PA=LU$: the permutation matrix $P$ acts on the rows of $A$, and the same row ordering must be applied to $b$ before the Triangular solves for LU decomposition. Equivalently, $A=P^{T}LU$, since $P^{-1}=P^T$. LAPACK documents its factors as $A=PLU$ instead, so check the defining equation rather than just the letter $P$.^[ref] Partial pivoting selects the largest absolute candidate in the current column, among rows $k,\dots ,n$, before forming elimination multipliers.^[ref]

A zero first pivot element does not imply a singular matrix. For example,Unpivoted Gaussian elimination would divide $2$ by zero. Indeed, no $A_0=LU$ with a unit lower triangular matrix $L$ exists: $u_{11}=0$ would force $(LU{)}_{21}=l_{21}{u}_{11}=0$, contradicting $(A_0{)}_{21}=2$. One row interchange resolves the obstruction. For $b_0=(1,5{)}^T$, use $P{b}_0=(5,1{)}^T$; back substitution then gives $x=(1,1{)}^T$. Using the original $b_0$ with these factors would solve a different linear system.

Partial pivoting bounds each elimination multiplier by one in absolute value in exact arithmetic. This helps control roundoff error, but intermediate entries can still grow; it is not an unconditional guarantee of numerical stability for every matrix.^[ref] It also does not remove the original problem's condition number. For the diagonal matrix $D=\mathrm{diag}(1,\epsilon )$ with $0<\epsilon <1$, the condition number in the Euclidean norm is ${\kappa }_2(D)=1/\epsilon$. Row permutations preserve this condition number, because they preserve Euclidean norms. Even an exact solve changes from $(1,1{)}^T$ to $(1,1+\eta /\epsilon {)}^T$ when the right-hand side changes from $(1,\epsilon {)}^T$ to $(1,\epsilon +\eta {)}^T$. This sensitivity is a property of the linear system, not a zero-pivot element failure.^[ref]