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The first index of the printed Levi-Civita symbol is fixed at $1$. Thus the matrix in the first orthonormal basis isThe specified matrix in the other basis is $I$, with determinant one. If this array were a Cartesian second-rank tensor, an orthogonal basis change would give $A^{\prime }=RA{R}^T$, andThe determinants disagree, proving the array is not a tensor. The fixed index matters: the TeX aid drops it, but the original PDF clearly prints ${ϵ}_{1ij}$.

Use the Cartesian second-rank tensor transformation convention $T^{\prime }=RT{R}^T$, where $R$ is the orthogonal matrix converting components to the new orthonormal basis. The coordinate half-turns areFor a diagonal rotation $R=\mathrm{diag}(s_1,s_2,s_3)$, invariance gives $t_{ij}=s_i{s}_j{t}_{ij}$, with no summation in this equation. For each $i\ne j$, one of the listed half-turns makes $s_i{s}_j=-1$, so $t_{ij}=-t_{ij}=0$. ThusThis is coordinate half-turn invariance of a second-rank tensor. It does not force equal diagonal entries; invariance under all rotations would be the stronger isotropic second-rank tensor condition.

The continuous harmonic function has $m\le \varphi (\mathbf{x})\le M$ on the compact sphere. Integrate these inequalities over its area $4\pi r^2$ and use the mean value property for harmonic functions:The centre value is an average with positive weights, so it lies between the smallest and largest boundary-sphere values.

Put $\mathbf{x}=r\mathbf{\omega }$, with $|\mathbf{\omega }|=1$. The Jacobian determinant in spherical coordinates gives $dS=r^{2}d\omega =r^2\sin \theta d\theta d\phi$, soDifferentiation under this fixed-domain integral yieldsThe last step is the divergence theorem and the Laplace equation. Hence $f$ is constant on radii for which the closed ball lies inside $V$. Continuity at the origin gives ${\lim }_{r\to 0}f(r)=\varphi (0)$, proving the mean value property for harmonic functions:

For the usual Dirichlet problem on a bounded regular volume, suppose ${\varphi }_1,{\varphi }_2$ both satisfy the stated data. Their difference $u={\varphi }_1-{\varphi }_2$ is a harmonic function, with ${\nabla }^{2}u=0$ in $V$ and $u=0$ on $\partial V$. In Green's first identity, choose both functions equal to $u$:The nonnegative continuous integrand forces $\nabla u=0$. Thus $u$ is constant on each connected component, and its zero boundary values force each constant to vanish. This proves uniqueness for the Dirichlet problem.

If the boundary value is a constant $c$, the constant function $c$ is itself a harmonic function with those data. Uniqueness therefore yieldsBoundedness, or an appropriate condition at infinity replacing it, is essential. On the upper half-space both $u=0$ and $u=z$ are harmonic functions with zero data on its boundary plane. They show why unrestricted uniqueness must not be asserted for an unbounded volume.

Apply the divergence theorem to $\mathbf{F}=\psi \nabla \varphi$, for functions smooth enough on the closure of the bounded piecewise smooth volume. The product rule givesThus Green's first theorem, also called Green's first identity, isThe normal is outward. This identifies the boundary term and its sign directly from the divergence theorem.

Suppose $\mathbf{v}=\alpha {\mathbf{e}}_1+\beta {\mathbf{e}}_2$. Applying $M-{\lambda }_{2}I$ would give ${\mathbf{e}}_2=\alpha ({\lambda }_1-{\lambda }_2){\mathbf{e}}_1$, contradicting linear independence of the two eigenvectors. Therefore the generalized eigenvector $\mathbf{v}$ lies outside their span and $({\mathbf{e}}_1,{\mathbf{e}}_2,\mathbf{v})$ is a basis. Write $\mathbf{x}=a{\mathbf{e}}_1+b{\mathbf{e}}_2+d\mathbf{v}$. Since $M\mathbf{v}={\lambda }_2\mathbf{v}+{\mathbf{e}}_2$, the linear system of ordinary differential equations givesSolving these equations gives $a=A_1{e}^{{\lambda }_{1}t}$, $d=A_3{e}^{{\lambda }_{2}t}$ and $b=(A_2+A_{3}t)e^{{\lambda }_{2}t}$. ThusThe length-two Jordan chain solution for the Jordan chain $({\mathbf{e}}_2,\mathbf{v})$ produces the $t{e}^{{\lambda }_{2}t}$ mode. The three arbitrary constants span all initial data, so this is the full general solution.

The three linearly independent eigenvectors still form a basis, so the same coefficient calculation applies to this diagonalizable matrix matrix. The general solution isA repeated eigenvalue alone does not create a polynomial factor in time; that factor is needed only when a nontrivial Jordan block prevents a full eigenvector basis.

The independence of eigenvectors for distinct eigenvalues ensures that such eigenvectors form a linearly independent set. For completeness, apply ${\prod }_{j\ne k}(M-{\lambda }_{j}I)$ to a proposed relation ${\sum }_i{c}_i{\mathbf{e}}_i=0$. It leaves $c_k{\prod }_{j\ne k}({\lambda }_k-{\lambda }_j){\mathbf{e}}_k=0$, so every $c_k=0$. Thus the three eigenvectors form a basis over the relevant field and $\mathbf{x}={\sum }_i{a}_i(t){\mathbf{e}}_i$. Substitution into the linear system of ordinary differential equations gives ${\sum }_i({\dot{a}}_i-{\lambda }_i{a}_i){\mathbf{e}}_i=0$; linear independence implies ${\dot{a}}_i={\lambda }_i{a}_i$. HenceA real matrix may have a complex conjugate pair of eigenvalues. In that case the calculation is over $\mathbb{C}$ and conjugate coefficients yield real solutions; equivalently use real and imaginary parts of the complex modes. No unstated assumption that all three eigenvalues are real is needed.

Condition on the fair coin, taken independent of $U$. For the squared uniform distribution, the cumulative distribution function of $U^2$ is $\sqrt{x}$ on $0<x<1$, giving probability density function $1/(2\sqrt{x})$. Reflection about $1/2$ gives the probability density function of $1-U^2$ as $1/(2\sqrt{1-x})$. Averaging the two conditional probability density functions produces this mixture model:There are no atoms at the endpoints; each integrable singularity contributes half the total mass. Equivalently the cumulative distribution function is $[\sqrt{x}+1-\sqrt{1-x}]/2$ on the interval.

For $x\ge 0$, the monotonicity of the logarithm and the uniform distribution giveFor $x<0$ the cumulative distribution function is zero. Differentiation gives the probability density function $f(x)=\lambda e^{-\lambda x}$ for $x>0$, and zero elsewhere. The distribution is $\boxed{\mathrm{Exp}(\lambda )}$, with rate $\lambda$. This is inverse transform sampling of an exponential distribution.

For a unitary matrix, $A^{†}=A^{-1}$, so both defining products for a normal matrix are $I$. The matrix is invertible, hence no eigenvalue is zero. If $Ax=\lambda x$, then $A^{-1}x={\lambda }^{-1}x$, while part (a)(iii) gives $A^{†}x=\overline{\lambda }x$. Therefore $\overline{\lambda }={\lambda }^{-1}$, andThus all eigenvalues have unit modulus.

For a skew-Hermitian matrix, $A^{†}=-A$, so both $A^{†}A$ and $A{A}^{†}$ equal $-A^2$. It is therefore a normal matrix. On an eigenvector with eigenvalue $\lambda$, part (a)(iii) givesThus $\overline{\lambda }=-\lambda$, and every eigenvalue lies in $i\mathbb{R}$, including the possible eigenvalue zero.

For a Hermitian matrix, $A^{†}=A$, so $A^{†}A=A^2=A{A}^{†}$ and $A$ is a normal matrix. If $Ax=\lambda x$ with $x\ne 0$, part (a)(iii) gives $A^{†}x=\overline{\lambda }x$. Since also $A^{†}x=Ax=\lambda x$, we obtain $\lambda =\overline{\lambda }$. Thus every eigenvalue is real.

Direct multiplication givesTake $D=B$ and $C=A$ in the preceding result: $B$ has the three distinct eigenvalues $2,4,6$ and the matrices commute. The common eigenbasis from part (i) givesIndeed $AS=S\mathrm{diag}(1,2,2)$ and $BS=S\mathrm{diag}(2,4,6)$, and $\det S=2\ne 0$. These equalities directly verify the simultaneous diagonalization.