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Here $\exp (I)=eI$. Applying part (a) at that matrix gives $D{s}_{eI}(K)=e(K^t+K)$, while part (b) gives $D{\exp }_I(H)=eH$. ThereforeBoth the outer evaluation point and the inner derivative contribute a factor of $e$.

The total derivative is the unique linear map $D{f}_a:{\mathbb{R}}^m\to {\mathbb{R}}^n$ satisfyingThis describes the best first-order linear approximation without choosing coordinates. The chain rule isThe evaluation point $f(a)$ in the outer derivative matters, particularly when composing a quadratic matrix map with the exponential.

Condition on the state after $m$ steps and use the Markov property and time-homogeneous Markov property:This is the Chapman-Kolmogorov equation, or $P^{m+n}=P^m{P}^n$. A stationary distribution is a probability row vector with nonnegative entries summing to one and $\boxed{\pi P=\pi }$. Inductively $\pi P^n=\pi$, so a chain started with that distribution retains it at every time.

For the given matrix, its first stationary equation is ${\pi }_1={\pi }_1/2+({\pi }_2+{\pi }_3)/3$. Normalization gives ${\pi }_1=2/5$. The third equation gives ${\pi }_3=2{\pi }_2/3+{\pi }_3/3$, hence ${\pi }_3={\pi }_2$. ThereforeDirect multiplication verifies the second equation as well. The finite chain is irreducible, so this stationary probability distribution is unique.

The first-column largest pivot element is four, so eliminate with multipliers $1/2,0,1/4$ below it. The augmented matrix becomesPartial pivoting in the second column swaps rows two and three because two is largest in modulus. The next elimination multipliers are $1/4$ and $1/2$. This givesThe third pivot element is already largest and the final multiplier below it is zero. Back substitution gives $z=1$, $y=3$, $x=-6$, $w=-1/2$. ThereforeThe row swap must also reorder the previously stored first-column multipliers in $L$. The factors areMultiplying these factors verifies $PA=LU$. This is the required elimination with partial pivoting, not an alternative solution method.

The $j$th column of the matrix of $Df(a)$ in the standard bases is $Df(a)e_j$. Its $i$th component is $\partial f_i/\partial x_j(a)$, found by taking increments $t{e}_j$. Thus the matrix of the total derivative is the Jacobian matrix, with entry $(i,j)$ equal to $\partial f_i/\partial x_j(a)$.

For a complex matrix $C=(c_{ij})$, Gershgorin's theorem puts every eigenvalue in the union of the row discs $D_i=\{z:|z-c_{ii}|\le R_i\}$, where $R_i={\sum }_{j\ne i}|c_{ij}|$. To prove it, choose an eigenvector $v$ for an eigenvalue $\lambda$ and an index $i$ maximizing $|v_i|>0$. Its $i$th equation givesDivision proves $\lambda \in D_i$.

The usual counting refinement says that a connected component made of $k$ discs contains exactly $k$ eigenvalues, with algebraic multiplicity. For completeness, deform the matrix from its diagonal part to the original by multiplying all off-diagonal entries by $t\in [0,1]$. Its discs have radii $t{R}_i$ and stay inside the original discs. Continuous variation of the roots of the characteristic polynomial keeps the number inside any contour separating one original component from the others constant. At $t=0$, the diagonal entries give exactly $k$ roots there, proving the refinement.

For the application put $s=-\mu >0$. The matrix isIts three row discs have positive real centres and leftmost real coordinatesAll are strictly positive, so zero lies in none of the discs. It cannot be an eigenvalue. Thus $A-\mu B$ is nonsingular for every real $\mu <0$.

At the first step, the chain enters state two with probability $p$ and state four with probability $1-p$. The classes $\{2,3\}$ and $\{4,5\}$ are closed and disjoint, so only the first branch can reach state three. On $\{2,3\}$ the transition matrix is exactly the two-state matrix from part (b). Thus after the remaining $n-1$ steps, the probability of the second state in that class isFor $n=1$ it is zero, as the original matrix requires; this statement covers the zero-base exponent case without ambiguity.

The inverse function theorem states that a $C^1$ map on an open subset of ${\mathbb{R}}^2$ has a $C^1$ local inverse near a point where its derivative is invertible; the inverse derivative at the image point is the inverse matrix of the original derivative.

Put $s=1+x+y$. Quotient differentiation givesThe entries are smooth and the determinant is nonzero everywhere on the stated domain, so the theorem supplies the required local inverse at every point. The point $(1/3,1/3)$ is the image of $(1,1)$, whereConsequentlyDirectly solving the two rational equations also gives $f^{-1}(u,v)=(u/(1-u-v),v/(1-u-v))$ on $u+v\ne 1$, consistent with the derivative calculation.

A QR decomposition is a factorization $A=QR$ with $Q^{T}Q=I$ and $R$ upper triangular. An orthogonal transformation preserves the Euclidean norm becauseFor the basic QR algorithm, start with $A_0=A$ and repeatedly formIn a convergent diagonal form, the diagonal entries give the eigenvalues; practical symmetric-matrix implementations use shifts and deflation to obtain and isolate such convergence. The basic unshifted iteration alone need not converge diagonally for every symmetric input.

At every step, however,Thus each iterate is orthogonally similar to its predecessor. Similar matrices have the same characteristic polynomial and eigenvalues, including multiplicities, so induction proves that every iterate has exactly the original eigenvalues. Orthogonal similarity also preserves symmetry.

The elementary elimination step replaces row $i$ by row $i$ minus a multiple $m_{ij}$ of an earlier pivot row $j$. With coordinate vectors $e_i$, its premultiplication matrix isand its inverse is $I+m_{ij}{e}_i{e}_j^T$, since $(e_i{e}_j^T{)}^2=0$. Choose $m_{ij}$ as the current entry below the pivot divided by the current pivot. Required pivots are assumed nonzero, consistently with elimination without row interchanges.

For dimension 4, first eliminate positions $(2,1),(3,1),(4,1)$, then $(3,2),(4,2)$, then $(4,3)$. ThusThe rightmost operation acts first, and every multiplier is computed from the matrix at its own stage. For general $n$, the numbers of eliminations in successive columns are $n-1,n-2,\dots ,1$, totaling $n(n-1)/2$; a zero multiplier simply gives an identity operation.

The product $E$ is unit lower triangular and $U=EA$ is upper triangular. Its inverse $L=E^{-1}$ is unit lower triangular, soThe stored multiplier in position $(i,j)$ is the corresponding subdiagonal entry of $L$, as follows by reversing the pivot-row subtractions column by column.

If $A$ is upper Hessenberg, only the first subdiagonal can be nonzero below the diagonal. At stage $j$ only position $(j+1,j)$ needs elimination. Subtracting row $j$ cannot create lower entries further down because its entries before its pivot are already zero. Hence only $n-1$ operations are needed. The inverse factors occur in increasing column order, and their cross products vanish, givingwhich is lower bidiagonal; $U$ is upper triangular, possibly dense above its diagonal.

If the Hessenberg matrix is also symmetric, it is tridiagonal. At each elimination step the pivot row has no entries past its immediate superdiagonal, so the updated matrix retains that upper bandwidth. Thus $L$ is lower bidiagonal and $U$ is upper bidiagonal, with at most two nonzero entries in each row, as required.

For small $\Vert \Delta \Vert$, the matrix $A+\Delta =A(I+A^{-1}\Delta )$ remains invertible and its inverse is locally bounded, for example by the convergent Neumann series when $\Vert A^{-1}\Delta \Vert <1$. Applying the resolvent identity twice givesThe last term is $O(\Vert \Delta {\Vert }^2)$ by that local bound. Hence the inverse map is differentiable, withThis verifies the differentiability and its remainder, rather than only formally differentiating the identity $A{A}^{-1}=I$.

The Fréchet derivative at $a$ is the linear map $D{f}_a:{\mathbb{R}}^n\to {\mathbb{R}}^k$ satisfyingIt is unique. When both derivatives exist, the chain rule is $D(g\circ f{)}_a=D{g}_{f(a)}\circ D{f}_a$.

For the square map,The remainder has norm at most $\Vert \Delta {\Vert }^2$ in a submultiplicative matrix norm, hence is $o(\Vert \Delta \Vert )$. ThereforeMatrix multiplication is not commutative, so the two terms cannot generally be combined as $2A\Delta$.

A real square matrix is orthogonal if $Q^{T}Q=I$. Then $\Vert Qx{\Vert }^2=x^T{Q}^{T}Qx=x^{T}x=\Vert x{\Vert }^2$, proving length preservation.

Define $J(i,j,\theta )$ to be identity outside rows and columns $i,j$, with blockThe block's transpose times itself is identity because ${\cos }^2\theta +{\sin }^2\theta =1$ and the off-diagonal terms cancel. Thus $J$ is orthogonal, with determinant one.

For $r=\sqrt{x_i^2+x_j^2}>0$, choose $\cos \theta =x_i/r$, $\sin \theta =x_j/r$. Then $y_i=r$, $y_j=0$ and every other coordinate is unchanged. If $r=0$, choose $\theta =0$. This fixes the useful nonnegative pivot convention.

Apply the rotations to the current matrix in the displayed order from right to left. First $J(1,2)$ eliminates entry $(2,1)$; then $J(1,3)$ eliminates $(3,1)$ without changing row two. Finally $J(2,3)$ eliminates $(3,2)$ while preserving the zeros in column one. ThusInverting the product gives $A=QR$. Even if $A$ starts lower triangular, the row rotations generally create nonzero entries above the diagonal; there is no further universal simplification. For example $A=\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 1& 1\end{array}\right)$ gives positive-pivotwhose three entries above the diagonal are all nonzero.

If $A$ is orthogonal, $R$ is both upper triangular and orthogonal. Orthogonality of its columns successively forces all off-diagonal entries to vanish and all diagonal entries to have absolute value one. The first two pivots are positive under our convention, and the rotations have determinant one. Therefore the precise result isWithout a specified pivot-sign convention, the invariant statement is a diagonal matrix with entries $\pm 1$ and product $\det A$.

The maximum-likelihood estimator is a maximizerWriting ${\ell }_n=\log L_n$ and the score $S_n={\nabla }_{\theta }{\ell }_n$, the expected Fisher information matrix iswhere the equality uses the usual regularity conditions. For independent identically distributed observations, ${\mathcal{I}}_n=n{\mathcal{I}}_1$.

Under the standard identifiability, interior-parameter, differentiability and nonsingular-information assumptions, the large-sample distribution isor approximately $\hat{\theta }\sim N(\theta ,{\mathcal{I}}_n(\theta {)}^{-1})$. Boundary parameters need not obey this regular approximation.

The inverse derivative formula, evaluated at $G(0,0)=(0,0)$, givesMultiplying the two displayed matrices in either order gives the identity element, verifying the total derivative explicitly.

Let $E(t)$ be the matrix whose columns are the moving orthonormal basis vectors. Then $E{E}^T=I$. Differentiating givesso $W=\dot{E}{E}^T$ is skew-symmetric. Every real skew-symmetric $3\times 3$ matrix has the unique formwhich acts on a vector as $Wv=\mathbf{\omega }\times v$. Since $\dot{E}=WE$, each column satisfiesThe vector $\mathbf{\omega }$ is the instantaneous angular velocity. It may also be written $\mathbf{\omega }=\frac{1}{2}{\sum }_i{\mathbf{e}}_i\times {\dot{\mathbf{e}}}_i$.

A real square matrix $Q$ is orthogonal when $Q^{T}Q=I$, equivalently $Q^{-1}=Q^T$. For $c=\cos \theta$ and $s=\sin \theta$,Thus $M(\theta )$ is orthogonal for every allowed angle, and indeed for every real angle.

Write $J_1,J_2,J_3$ for the actual successive rotations $J(1,2)$, $J(1,3)$, and $J(2,3)$, each selected using the currently transformed matrix. The first eliminates position $(2,1)$; the second eliminates $(3,1)$ while leaving row two unchanged. After these, the first column is $(r_{11},0,0{)}^T$. The final rotation of rows two and three preserves both zeros in that column and eliminates $(3,2)$. HenceA lower-triangular starting matrix does not force any further zeros above the diagonal. For example, the lower-triangular matrix with all entries on and below its diagonal equal to one gives, with the nonnegative pivot convention,which has all upper-triangular entries nonzero.

Inverting the product of rotations yields the QR factorizationThe order reverses when taking the transpose. If $A$ is orthogonal, then $R$ is both orthogonal and upper triangular. Its first column forces $|r_{11}|=1$ and orthogonality with the other columns forces $r_{12}=r_{13}=0$. Applying the same argument to the remaining block makes $R$ diagonal with entries of modulus one.

Our elimination rule makes $r_{11}$ and $r_{22}$ positive when nonzero, so they are both one. Every rotation has determinant one; thus $\det R=\det A$ andWithout a fixed pivot-sign convention one could only specify a diagonal sign matrix. The convention chosen in part (c) determines the precise signs above.

If $F:U\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^m$ is differentiable at $x$, and $G$ is defined on a neighborhood of $F(x)$ and differentiable there, the chain rule statesIn standard coordinates the same statement isThe evaluation point in the first factor is $F(x)$, and the matrix order follows the order in which the linear maps act on increments.

The total derivative of $F$ at $x$ is the linear map $D{F}_x:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe Jacobian matrix is the $m\times n$ matrix $J_F(x)=(\partial F_i/\partial x_j)(x)$. When the total derivative exists, this is its matrix in standard bases: $D{F}_x(h)=J_F(x)h$. Existence of partial derivatives alone need not imply existence of the total derivative.