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The divergence is the trace of the velocity-gradient tensor:Thus the flow is incompressible exactly whenThe constant vector $U_0$ is unrestricted.

Take the parallel at height one,It has constant speed $\sqrt{2}$. Its distance from the axis is the constant $\rho =\sqrt{2}$, and its tangent is parallel to the parallel, so $\theta =0$ and it satisfies the Clairaut first integral for a surface of revolution.

However,whereas a normal vector to $x^2+y^2-z^2=1$ at $\gamma (t)$ isThese vectors are not parallel. By the ambient acceleration criterion for a surface geodesic, $\gamma$ is not a geodesic. Thus constant speed together with Clairaut's relation is not sufficient for the geodesic equations.

A linear operator $\alpha$ on a real inner-product space is self-adjoint whenfor all $v,w$.

To prove the finite-dimensional spectral theorem, first note that the continuous quadratic function $⟨\alpha v,v⟩$ has a maximum on the unit sphere. The Lagrange multiplier equation at a maximizing vector $v$ gives $\alpha v=\lambda v$, so $\alpha$ has a real unit eigenvector. Its orthogonal complement is invariant becauseInduction on the dimension supplies an orthonormal eigenbasis of that complement and hence of $V$.

For $f\ne 0$,while symmetry and bilinearity are immediate, so the displayed formula defines an inner product on $P_n$. Sinceintegration by parts givesThe boundary term vanishes for polynomials, proving that $\alpha$ is self-adjoint.

On the monomial $x^k$,Thus the matrix of $\alpha$ in the monomial basis is triangular with diagonalThese are therefore its eigenvalues. For $n=2$, corresponding eigenvectors arefor eigenvalues $0,-1,-2$, respectively. They are the first Laguerre polynomials up to normalization, as described by the Laguerre differential operator on polynomials.

Define the softmax weightsThey obey $p_i\ge 0$ and ${\sum }_i{p}_i=1$. The gradient is their weighted mean,Differentiating once more gives the covariance-form Hessian matrixFor every unit vector $u$,where $G={\max }_i\Vert a_i{\Vert }_2$. The Hessian is a covariance matrix, so it is positive semidefinite; the displayed upper bound also gives ${\nabla }^2{f}_{\beta }⪯\beta G^{2}I$ in the Loewner order. Consequently $f_{\beta }$ has a Lipschitz gradient with

For $x=(x_1,x_2,x_3{)}^T$,The vectors $v$ and $1$ are linearly independent, and $a\ne 0$. Thus $Ax=0$ exactly whenSolving gives

The $j$th column of $B$ isThus the image is contained in $\mathrm{span}\{v,1\}$. Since $n>1$, $v$ is not constant; since $m>1$ and $a\ne 0$, two columns have a nonzero constant difference. The two spanning vectors therefore both lie in the image, so $\mathrm{rank}B=2$. By rank-nullity theorem,

PutSince $x,y,z$ are unit vectors,The feasible inner products with two unit vectors satisfyThe desired pair $p=q=1/2$ is feasible exactly when $c\ge -1/2$. In that case every maximizer has these inner products. If $c<-1/2$, symmetry and strict convexity force $p=q$, and the nearest feasible diagonal point isUsing the vector triple product,Hence

Rotations preserve dot products and cross-product lengths, while $z^{\prime }=2z$. ThusLetThe target pair is feasible precisely whenorOn this interval,

Outside this interval, let $\lambda >0$ be the unique number for whichand write $r=(p,q{)}^T$. These equations give the unique weighted projection of $t$ onto the feasible ellipse, and therefore determine the maximizing inner products solely from $c=y\cdot z$. In the remaining cases,Neither answer depends on ${\theta }_1$ or ${\theta }_2$.

Since $P=P^{\prime }=0$, it suffices to map the two edge vectors. Let the columns of the required matrix be $u_1,u_2,u_3$. The condition $MQ=Q^{\prime }$ givesThe condition $MR=R^{\prime }$ then givesand henceThese are orthonormal. Taking $u_3=u_1\times u_2=(0,-1,0{)}^T$ produces the orthogonal matrixIts columns are orthonormal, and direct substitution verifies $MP=P^{\prime }$, $MQ=Q^{\prime }$, and $MR=R^{\prime }$.

The characteristic polynomial of $A$ isA scalar $\lambda \in \mathbb{C}$ is an eigenvalue when there is a nonzero vector $v$ such thatSuch a $v$ is an eigenvector belonging to $\lambda$. Equivalently, $\lambda$ is a root of ${\chi }_A$ and the corresponding eigenvectors are the nonzero elements of $\ker (A-\lambda I)$.

Eigenvectors belonging to distinct eigenvalues are linearly independent. Indeed, supposeis a shortest nontrivial relation, where $A{v}_j={\lambda }_j{v}_j$ and the ${\lambda }_j$ are distinct. Applying $A-{\lambda }_{m}I$ givesa shorter nontrivial relation, a contradiction.

Thus $n$ distinct eigenvalues provide a basis of $n$ eigenvectors, proving that $A$ is diagonalizable over $\mathbb{C}$. If $A$ is real and all its eigenvalues are real, each real singular matrix $A-\lambda I$ has a nonzero real null vector. Choosing one for each eigenvalue gives a real eigenbasis. Hence the real matrix is diagonalizable over $\mathbb{R}$. This is the distinct eigenvalues imply diagonalizability criterion.

First note that if $A$ is normal, thenfor every $v$. Hence an eigenvector $v$ with $Av=\lambda v$ also satisfiesNormalize $v$ and extend it to an orthonormal basis. Let $U$ have these basis vectors as columns. The first column of $U^{†}AU$ is $(\lambda ,0,\dots ,0{)}^T$. For every $w\perp v$,so the first row also has no off-diagonal entries. ThereforeComparing the two block products in the normality identity shows that $B{B}^{†}=B^{†}B$, so $B$ is normal.

The result is trivial in dimension one. Applying the induction hypothesis to $B$ and adjoining the eigenvector $v$ gives an orthonormal eigenbasis for $A$. This proves the unitary diagonalization of a normal matrix.

LetThey are jointly normal because they are linear combinations of a Gaussian vector. When ${\sigma }_1={\sigma }_2=\sigma$,Hence uncorrelated jointly normal variables are independent. Their means and variances areandThereforeindependently.

A tensor of rank $n$ is a multilinear map of $n$ vector or covector arguments, equivalently an element of an $n$-fold tensor product. In components it obeys one change-of-basis factor for each of its $n$ indices.

The special orthogonal group isIfits first column is a unit vector, say $(\cos \theta ,\sin \theta {)}^T$. The second column is the unique unit vector perpendicular to it that gives positive determinant, namely $(-\sin \theta ,\cos \theta {)}^T$. Thusso $A$ is rotation through $\theta$ about the origin.

For $A\in SO(3)$, its real characteristic polynomial of odd degree has a real eigenvalue. Every eigenvalue of an orthogonal matrix has modulus one, so every real eigenvalue is $\pm 1$. Nonreal eigenvalues occur in conjugate pairs whose product is one, while $\det A=1$; if all eigenvalues are real, their product likewise forces at least one to be $+1$. Hence $A$ fixes a nonzero vector $u$. The plane $u^{\perp }$ is $A$-invariant, and the restriction to it is an orientation-preserving orthogonal map. By the $SO(2)$ result it is a planar rotation. Therefore $A$ is a rotation about the axis $\mathbb{R}u$.

Use the Levi-Civita symbol and the product rule. First,For the second identity, the contractiongivesIn vector notation this isThese are the divergence and curl of a cross product identities.

Each freely moving particle has position, relative to its own initial radial and tangential unit vectors,Thus all particles lie on a circle of radiusTheir angular momenta are constant and sum toSince $I(t)=mR(t{)}^2$, the angular velocity isWith $\theta (0)=0$,This is the geometry of a freely expanding rotating particle ring.

For a linear map $T:V\to W$, its rank is $\dim \mathrm{im}T$ and its nullity is $\dim \ker T$. The rank-nullity theorem states that, when $V$ is finite-dimensional,To prove it, take a basis $v_1,\dots ,v_k$ of $\ker T$ and extend it to a basisof $V$. The vectors $T{v}_{k+1},\dots ,T{v}_n$ span $\mathrm{im}T$. They are also linearly independent: if ${\sum }_{j=k+1}^n{a}_{j}T{v}_j=0$, then ${\sum }_{j=k+1}^n{a}_j{v}_j\in \ker T$, and independence of the chosen basis forces every $a_j$ to vanish. Thus the rank is $n-k$ and the nullity is $k$.

For the given subspace, row reduction of the coefficient matrix givesTaking $x_3=s$ and $x_5=t$ therefore givesHenceis a basis of $W$.

A gauge transformation isBecause partial derivatives commute, the added contribution to $F_{\mu \nu }$ isso the electromagnetic field tensor is gauge invariant.

DefineSince ${\partial }_0=c^{-1}{\partial }_t$ and $A_0=-\Phi /c$,

The identityfollows directly from $F=dA$. Its spatial and mixed components areFor the other two equations define the four-currentThen ${\partial }_{\nu }{F}^{0\nu }={\mu }_0{j}^0$ givesand ${\partial }_{\nu }{F}^{i\nu }={\mu }_0{j}^i$ givesThis is the Covariant Maxwell equation with the minus-plus-plus-plus metric.

Usingone findsThis is the electromagnetic energy density.

For a null vector, the trace term in $T^{\mu \nu }{k}_{\mu }{k}_{\nu }$ vanishes. PutAntisymmetry gives $q\cdot k=0$. A vector orthogonal to a null vector has nonnegative Minkowski norm, soExplicitly, in a frame with $k^{\mu }=\kappa (1,1,0,0)$,Hence the null energy condition for the electromagnetic field is strict whenever the contraction is nonzero: