At critical damping, the characteristic polynomial has the repeated root $r=-{\omega }_0$. Its independent solutions are $e^{-{\omega }_{0}t}$ and $t{e}^{-{\omega }_{0}t}$, givingThe factor $t$ supplies the second independent solution that would be lost by using just one exponential. There is no oscillatory factor at this boundary between the underdamped and overdamped regimes.

For the damped harmonic oscillator, the characteristic polynomial is $r^2+2\gamma r+{\omega }_0^2$. When $0\le \gamma <{\omega }_0$, its roots are $-\gamma \pm i\Omega$, with $\Omega =\sqrt{{\omega }_0^2-{\gamma }^2}$. ThusHere $A$ and $B$ are arbitrary real constants. The frequency is $\Omega$, and the amplitude envelope decays with time constant $1/\gamma$ when $\gamma >0$; the zero-damping limit recovers simple harmonic motion.

Matrix multiplication is associative, soThe uniqueness of the matrix inverse then givesThe order is reversed; $A^{-1}{B}^{-1}$ does not generally work because matrix multiplication need not be commutative.

If $A$ and $B$ are Hermitian matrices, then $(AB{)}^{†}=B^{†}{A}^{†}=BA$. ThereforeThe required condition is that the matrices commute.

For a Hermitian matrix $H$, $H^{†}=H$. Reversing the product of $n$ identical factors has no effect, soHence every positive integer power is Hermitian. The identical-factor condition matters; a product of different Hermitian matrices need not be Hermitian.

The conjugate transpose also reverses factor order. If $U$ and $V$ are unitary matrices,Thus their product is unitary. This is closure under multiplication in the unitary group.

The transpose of a matrix product reverses factor order:For the requested two-dimensional check, put $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ and $B=\left(\begin{array}{cc}e& f\\ g& h\end{array}\right)$. Both sides areMore generally, the entry formula $(AB{)}_{ji}={\sum }_k{A}_{jk}{B}_{ki}$ is exactly $(B^T{A}^T{)}_{ij}$.

Here $m=2$, so $z=1/y$ obeys $z^{\prime }-2z\tan x=-{\tan }^{2}x$. An integrating factor is ${\cos }^{2}x$, givingThe nonzero solutions are thereforeThe additional solution is $\boxed{y=0}$. Each formula is interpreted on an interval where $\tan x$ is defined and its displayed denominator is nonzero.

For the Bernoulli differential equation, work on a branch where $y\ne 0$ and the required powers are defined. Multiplication by $(1-m)y^{-m}$ makes the chain rule useful:Thus $\alpha =1-m$, $h=(1-m)f$ and $k=(1-m)g$. The case $m=1$ is already linear and must be treated separately, because the proposed substitution would be constant. Solutions lost by dividing by a power of $y$, particularly $y=0$ when it solves the original equation, must also be checked separately.

The second gradient is ${\nabla }_{v}b=[A+B-(\alpha +\beta )I]v$. Combining its vanishing with $(A-\alpha I)v=0$ givesThe zero vector always works. A nonzero common stationary vector exists exactly when $A,B$ have a common eigenvector with respective eigenvalues $\alpha ,\beta$; both stationary values are then zero. Commuting symmetric matrices have a common orthonormal eigenbasis, so commutation is sufficient to find such vectors for the paired eigenvalues in that basis. It does not guarantee an arbitrary prescribed pair.

Commutation is also not necessary for one common eigenvector. For example,are real symmetric matrices sharing $e_1$ with eigenvalues $1,\beta$, but $AB\ne BA$. This separates the question's common-vector condition from the stronger simultaneous diagonalization of the entire space.

Because $A$ is a real symmetric matrix, differentiating the quadratic form givesThus the stationary vectors are precisely $\ker (A-\alpha I)$, and at every one of themThe zero vector is always stationary. Nonzero stationary vectors exist exactly when $\alpha$ is an eigenvalue of $A$, and are exactly its nonzero eigenvectors with that eigenvalue. There is no norm constraint here; introducing one would change the stationary-value problem.

For a diagonalizable matrix, write $M=P\mathrm{diag}({\lambda }_j)P^{-1}$. Relative to an existing fixed point,Convergence to the same fixed point for every initial vector therefore holds exactly whenHere $\rho (M)$ is the spectral radius. Necessity follows by choosing an initial displacement in each eigenspace; sufficiency follows because every displayed scalar power tends to zero. The condition also makes $I-M$ an invertible matrix, so that fixed point exists uniquely for every $c$.

If instead different initial vectors may converge to different constant solutions, an eigenvalue $1$ is allowed: all other eigenvalues must have modulus less than one, and the component of $c$ in the eigenspace for $1$ must vanish. That component of the initial vector then remains unchanged. No other unit-modulus eigenvalue is compatible with convergence for arbitrary initial vectors.

Repeated substitution in the affine iteration givesTherefore the required matrix polynomials areThese formulas do not require an inverse. If $I-M$ is an invertible matrix, multiplying the finite geometric sum by $I-M$ gives $g(M)=(I-M^{k-1})(I-M{)}^{-1}$.

A fixed point obeys $(I-M)X=c$. Thus the necessary and sufficient existence condition isIf $1$ is not an eigenvalue of $M$, then $I-M$ is an invertible matrix and the unique constant solution is $\boxed{X=(I-M{)}^{-1}c}$. If $I-M$ is singular but the equation is consistent, all constant solutions are $X_0+\ker (I-M)$ for any particular solution $X_0$. For an existing fixed point, $x_{k+1}-X=M(x_k-X)$; the successive difference in the question likewise obeys $\delta x_{k+1}=M\delta x_k$.

If $b,c$ are linearly independent, $S$ is the plane through $a$ with normal $b\times c$. Its equation isso it contains the origin exactly when $a\cdot (b\times c)=0$. If $b,c$ are proportional, $S$ is the line through $a$ in direction $b$, containing the origin exactly when $a$ is proportional to $b$. The linear span of two nonzero vectors has dimension one or two, so neither a single point nor all of ${\mathbb{R}}^3$ occurs.

The affine subspace is $S=a+\mathrm{span}(b,c)$. If $b,c$ are linearly independent, their linear span is all of ${\mathbb{R}}^2$, so $S$ is the whole plane. If they are proportional, $S$ is the line through $a$ in direction $b$. This line passes through the origin exactly when $a\in \mathrm{span}(b)$. Because $b$ is nonzero, the set cannot be a single point.

In one dimension the nonzero vector $b$ already spans the real line. Hence $\mathrm{span}(b,c)=\mathbb{R}$, and translating it by $a$ leavesThis is an affine subspace containing the origin for every allowed choice of the three vectors.