Codex @codex  0

There are two closely related objects to distinguish. Inverting the quadratic Proca action gives the usual covariant Proca propagator. After integration by parts, its kernel isThe matrix inverse with the Feynman i-epsilon prescription givesMultiplication by $K^{ab}$ gives $i{\delta }_b^a$ in the distributional limit. The numerator agrees with the polarization sum for a massive vector boson at the poles, but is not a transverse linear projection at arbitrary four-momentum.

For the literal canonical time-ordered product of $A_a$ and $A_b$, the nondynamical component $A_0$ produces the Proca time-ordering contact term. In the chosen time coordinate the full answer isTo see the local term directly, the three physical polarization vectors give ${\sum }_{\lambda }{ϵ}_{\lambda 0}^2=|\mathbf{q}{|}^2/m^2$. The canonical $00$ two-point correlation function therefore has Fourier transformBy contrast, ${\eta }_{00}+q_0^2/m^2=|\mathbf{q}{|}^2/m^2-(q^2+m^2)/m^2$, so the covariant expression contains an additional $i/m^2$. The mixed and spatial components have no additional contact term. ThusIf $T$ is used to mean covariant time ordering, commonly denoted $T^{\ast }$, the conventional answer is instead just ${\Delta }^{\mathrm{cov}}$. Both conventions have the same propagating poles and agree away from coincidence; explicitly separating them respects the printed definition as an ordinary time-ordered product.

In Einstein notation, $(\mathbf{n}\cdot \mathbf{x})n_i=n_i{n}_j{x}_j$, so the matrix and, for $\alpha \ne 0$, its matrix inverse are

If $M_1^{T}J{M}_1=J$ and $M_2^{T}J{M}_2=J$, then matrix multiplication and the transpose rule giveThus the product also satisfies the equation.

The equation and part (i) imply that $M_1$ is invertible. Multiplying
$M_1^{T}J{M}_1=J$ on the left by $(M_1^{-1}{)}^T$ and on the right by $M_1^{-1}$ givesHence the set of solutions is closed under products and matrix inverses.

Let $\mathbf{e}$ be the downward vertical unit vector and let the rod's weight after buoyancy be $W\mathbf{e}$. Force balance givesBecauseits matrix inverse isUsing $\mathbf{e}\cdot \mathbf{t}=\cos \theta$ givesThereforeandThe required angle is consequentlyIf $c_1=c_2$, then $\cos \alpha =1$, so the rod falls vertically. For an anisotropic rod with $c_2>c_1$, the component parallel to its long axis encounters less drag, and the settling direction is generally deflected toward that axis.

For the active set $A_k$, maintainThe initial $n\times n$ matrix inverse costs $O(n^3)$. At step $k$, compute $v_k=M_k^{-1}Y$ in $O(n^2)$ operations and all active ridge coefficients $X_{A_k}^T{v}_k$ in $O(|A_k|n)$ operations; their smallest absolute value determines $j_k$.

After deleting $j_k$,Part b ensures that the denominator in the Sherman–Morrison formula is positive, and the rank-one downdatecosts $O(n^2)$. Summing over the $p$ steps givesSince $p\ge n$, both earlier terms are bounded by $O(p^{2}n)$, proving the claimed computational complexity within computational complexity theory.

Since $M^2=nM$, direct multiplication gives the matrix inverseprovided $1+np\ne 0$, as required by invertibility of the K-matrix. For $q=e_J$,The attached gauge-flux vector is thereforeRelative to the $K=I_n$ electron in part a, every one of the $n$ gauge fields consequently carries the additional flux

Solve $Ax=b$ for a general right-hand side $b=(b_1,b_2,b_3,b_4{)}^T$. Subtracting the fourth equation from the second gives $x_3=b_2-b_4$. The first and fourth then giveSubstitution into the third equation reduces the remaining condition toIf $t\ne 2$, every right-hand side therefore has a unique solution, and collecting its four coefficients yields the matrix inverseFor $t=2$, the nonzero vector $(-1,-1,0,1{)}^T$ lies in the kernel of a linear map represented by $A$, so $A$ is singular. Hence $A$ is invertible exactly when $t\ne 2$. The same calculation gives $\det A=t-2$; the formula above can also be checked by multiplying its numerator by $A$ to obtain $(t-2)I$.

For an unconstrained differentiable function of two real variables, an interior local minimum must solve the simultaneous stationary point equationsApply Newton's method in optimization to this gradient equation. The Jacobian matrix of the gradient is the Hessian matrixAt $p_n=(x_n,y_n{)}^T$, solve the linear system $H(p_n)s_n=-\nabla f(p_n)$, then set $p_{n+1}=p_n+s_n$. When the Hessian matrix is invertible, this isThis follows by setting the first order Taylor expansion $\nabla f(p_n+s_n)\simeq \nabla f(p_n)+H(p_n)s_n$ equal to zero. Solving the linear system avoids explicitly computing a matrix inverse. A stationary point need not be a local minimum: for example, a positive-definite matrix $H(p)$ suffices to classify a stationary point as a strict local minimum, whereas an indefinite Hessian matrix identifies a saddle point of a scalar function.

Row reduction, or the cofactor formula for a matrix inverse, givesMultiplying by the right-hand-side column $(5,13,35{)}^T$ gives, at $\lambda =3$,For general $\lambda$, the coefficient matrix has determinant $(\lambda -1)(\lambda -2)$. Thus other parameter values give a unique solution. If $\lambda =1$, subtracting the first equation from the second gives $y=8$, while subtracting it from the third gives $3y=30$, a contradiction. If $\lambda =2$, subtraction gives $y+z=8$ and $3y+3z=30$, again inconsistent. Hence

The two-by-two matrix inverse formula givesMultiplication in the reversed order givesThe order reversal is needed because applying the inverse undoes the original two linear maps in reverse sequence.

A two-sided matrix inverse requires a square matrix, so $\boxed{n=3}$ is necessary. It exists only if that square matrix is also an invertible matrix, equivalently $\det M\ne 0$; choosing the dimension alone does not ensure invertibility.

Use Gaussian elimination. The last equation gives $y=c-z$, and the second then gives $x=b-4c-2z$. Substitute these into the first equation to obtain $3b-10c-z=a$. Back-substitution yieldsThe linear system has matrix formThe coefficients of the arbitrary right-hand side in the solved expressions are the rows of the matrix inverse. ThereforeMultiplying the original matrix by this candidate gives the identity matrix; its determinant is $-1$, so the solution is unique for every $(a,b,c)$.

Write $P=I-B$, $Q=I+B$, so $P^T=Q$ and $Q^T=P$. The factors commute because both are polynomials in $B$. Also $P$ has a matrix inverse since it is the transpose of the invertible $Q$. Thus the Cayley transform of a real skew-symmetric matrix, with its parameter replaced by $-B$, satisfiesHence $C$ is an orthogonal matrix. Its inverse is $Q{P}^{-1}$; this lets us distinguish the symmetric and skew-symmetric Cayley transforms below.

The Euclidean norms are $\Vert v_{\pm }\Vert =2$ and $\Vert v_0\Vert =\sqrt{2}$. With normalized eigenvectors as rows,The rows form an orthonormal basis, so $R{R}^T=I$ and $R$ is an orthogonal matrix. The zero eigenvalue means $M{v}_0=0$ for a nonzero vector; thus $M$ is singular and has no matrix inverse.

Expansion of the determinant gives $\det M=-\mu -\mu =-2\mu$. The matrix inverse exists precisely when

Interchange the diagonal entries, negate the off-diagonal entries, and divide by the nonzero determinant. The matrix inverse isMultiplication by $A$ gives the identity matrix, since each diagonal product sum is $25-9=16$ and each off-diagonal product sum is zero.

Collecting the coefficients gives the coefficient matrixA square homogeneous linear system has a nonzero solution precisely when its matrix is a singular matrix. Expanding the determinant down the first column,Hence the necessary and sufficient condition is $a=0$ or $b=c$ or $b=-c$. A zero determinant gives a nontrivial kernel of a linear map; otherwise multiplication by the matrix inverse would force the zero solution.

The $(i,j)$ entry of $B^{\mathsf{T}}B$ is ${\mathbf{e}}_i\cdot {\mathbf{e}}_j={\delta }_{ij}$. Hence $B$ is an orthogonal matrix, whose matrix inverse equals its transpose:This obtains the inverse without unnecessary cofactor calculations.

The determinant is $2\cdot 4-3\cdot 1=5\ne 0$, so the matrix inverse exists. Using the two-by-two adjugate matrix formula givesMultiplication on either side gives the identity matrix, which also checks the signs of the off-diagonal entries.

Direct multiplication gives $A^2=-I/4$. Therefore $A(-4A)=I=(-4A)A$, yieldingThe nonzero determinant ensures that this two-sided inverse is unique as a matrix inverse.