A real number $\lambda$ is an eigenvalue of $A$ if there is a nonzero vector $v$ with $Av=\lambda v$; such a $v$ is an eigenvector for $\lambda$. The nonzero condition is essential.

For distinct eigenvalues ${\lambda }_1,\dots ,{\lambda }_r$, choose corresponding nonzero eigenvectors $v_1,\dots ,v_r$. Prove their linear independence by induction on $r$. The claim for one vector is immediate. Given a relation ${\sum }_{i=1}^r{a}_i{v}_i=0$, apply $A-{\lambda }_{r}I$ to getBy the induction hypothesis, every coefficient here is zero. Distinctness gives $a_i=0$ for $i<r$, and the original relation then gives $a_r=0$ since $v_r\ne 0$. Thus the eigenvectors are linearly independent.

With $n$ distinct real eigenvalues, the $n$ real eigenvectors form a basis of ${\mathbb{R}}^n$. Put them into the columns of $P$. Then $P$ is invertible andThis proves diagonalizability directly.

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 $(t-2)I$.

An elementary row operation is one of three operations: interchange two rows, multiply one row by a nonzero scalar, or add a scalar multiple of one row to a different row. Each operation is reversed by another operation of the same kind.

Performing such an operation on an $m\times n$ matrix $A$ is equivalent to left multiplication by the $m\times m$ elementary matrix obtained by applying that operation to the identity matrix. Each elementary matrix is invertible, with inverse corresponding to the reverse operation. Thus a sequence of operations givesFor square matrices, if $A$ were invertible, then $B=PA$ would have inverse $A^{-1}{P}^{-1}$. Hence if $B$ is singular, $A$ must be singular as well. In fact the two are invertible together, because $A=P^{-1}B$ gives the converse.

For a linear map $T:V\to V$, its kernel of a linear map and image of a linear map areIts nullity of a linear map is $n(T)=\dim \ker T$, and its rank of a linear map is $r(T)=\dim \mathrm{im}T$.

Choose a basis $u_1,\dots ,u_k$ of $\ker T$ and extend it to a basis $u_1,\dots ,u_k,v_1,\dots ,v_m$ of $V$. The vectors $T(v_1),\dots ,T(v_m)$ span $\mathrm{im}T$, because applying $T$ to an arbitrary linear combination of the full basis eliminates the $u_i$ terms. They are also linearly independent: if ${\sum }_j{b}_{j}T(v_j)=0$, then ${\sum }_j{b}_j{v}_j\in \ker T$, so ${\sum }_j{b}_j{v}_j={\sum }_i{a}_i{u}_i$ for some coefficients. Independence of the extended basis forces all $a_i,b_j$ to vanish.

Consequently $r(T)=m$, $n(T)=k$ and $\dim V=k+m$. This proves the rank-nullity theorem:

Choose a basis $u_1,\dots ,u_r$ of $U$, extend it to a basis $u_1,\dots ,u_r,w_1,\dots ,w_s$ of $W$, and extend again to a basis of $V$ by adding $v_1,\dots ,v_t$. DefineEvery member of the full basis lies in $W$ or $W^{\prime }$, so the sum of vector subspaces is $W+W^{\prime }=V$. If $x\in W\cap W^{\prime }$, express it in both displayed bases. Uniqueness of the coordinates in the full basis forces its $w_j$ and $v_j$ coefficients to vanish. Thus $x\in U$. The reverse inclusion is built into the construction, proving $W\cap W^{\prime }=U$. This is a subspace supplement with a prescribed intersection.

For the matrix example, letThe skew-symmetric matrices are exactly $U=\mathrm{span}\{J\}$, and the traceless matrices form $W$. A suitable answer isThe matrix trace of $aI+bJ$ is $2a$, so $W\cap W^{\prime }=\mathrm{span}\{J\}=U$. Every $2\times 2$ real matrix has the decompositionwhose first term lies in $W$ and second in $W^{\prime }$. Therefore $W+W^{\prime }=M_2(\mathbb{R})$, as required.

For a finite set $S=\{s_1,\dots ,s_k\}$ of vectors in a real vector space, its span is the set of all linear combinations:The span of the empty set is $\{0\}$. The set $S$ is linearly independent if ${\sum }_i{a}_i{s}_i=0$ implies $a_i=0$ for every $i$. It is a spanning set for $V$ if $\mathrm{Sp}(S)=V$. A basis is both a linearly independent set and a spanning set.

For basis extension, start with the given linearly independent set. If it does not span $V$, choose $v\notin \mathrm{Sp}(S)$ and adjoin it. The enlarged set remains linearly independent: a relation $av+{\sum }_i{a}_i{s}_i=0$ with $a\ne 0$ would express $v$ as a linear combination of $S$, a contradiction; if $a=0$, independence of $S$ forces all other coefficients to be zero.

Repeat whenever the current span is proper. Each step increases the set's size by one, while the allowed bound says its size cannot exceed $\dim V$. Thus the process terminates with a linearly independent set spanning $V$, hence a basis containing the original set. This also covers an initially empty set and the zero-dimensional vector space.

Write $e=E^{\prime }(1)>0$, $q_1=q^{\prime }(1)>0$ and introduce perturbations $(H,U,C,S)e^{\lambda t+ikx}$ to water depth, velocity, concentration and bed height. Take the leading quasi-steady approximation $ϵ=0$ for the fast water and suspension response, retaining the small friction parameter $\delta$. For $k\ne 0$, the linearized equations areHere surface height is $H+S$, and linearization of $1-u^2/h$ gives $H-2U=-3U$. Eliminating $H,C,U$ gives the dispersion relationIn particular, putting $B=F^2-1\ne 0$ and expanding at large $|k|$,Thus the temporal growth rate tends to a finite constant, while the phase velocity isFor supercritical flow $F>1$, the waves propagate upstream. They grow if $\beta e(F^2-1)>3\delta q_1$; at leading order in small $\delta$, this holds for fixed positive $\beta ,e$ away from critical flow. They are antidunes, upstream-migrating bedforms coupled to the free surface. Indeed $H=-U$ and $S\simeq -(F^2-1)U$, so surface height $H+S\simeq -F^{2}U$ is in phase with the bed perturbation in this regime.

For clarity, neglecting $\delta$ completely gives the exact leading-order expressionsThese also identify the missing qualification in the source: the unstable upstream branch requires $F>1$. For $F<1$ this leading model gives decay and downstream propagation. The near-critical case $F^2-1\simeq 0$ needs a different balance; it is not covered by the expansion. Small nonzero $ϵ$ perturbs the slow bedform branch and adds fast water modes, so the stated result concerns the leading slow morphodynamic branch. This is the high-wavenumber antidune dispersion calculation.

For the travelling wave $\xi =s+ct$, continuity gives $(ch+hu{)}^{\prime }=0$. The upstream state fixes the constant as $1+c$, henceSubstituting in momentum yieldssoA constant downstream state $H=h_{+}>1$ requires $H^3=(1+c-cH{)}^2$. The positive square-root choice would give $c<0$, so the upstream-moving wave uses $1+c-cH=-H^{3/2}$. ThusThe downstream flow reverses. Where the numerator is positive in $1<h<H$, a smooth increasing profile requires a positive denominator. Its minimum occurs at $h=1$, givingEquality does not give the prescribed asymptotic upstream state: cancellation of the two simple zeros at $h=1$ leaves a nonzero slope there. For the strict inequality, separation of variables constructs the monotone profile whenever the numerator remains positive between its endpoint zeros.

There is an additional restriction hidden in the printed positivity hint. Put $r=\sqrt{h}$ and $R=\sqrt{H}$. Since $c=R+1/(R-1)$, the potentially vanishing factor of the numerator isThe other factor, $h^{3/2}-(1+c-ch)$, is positive for $h>1$. Therefore the asserted positivity throughout $1<h<H$ holds precisely for $1<R\le 2$, that is $1<H\le 4$. For $H>4$, the zero at $h=[R/(R-1){]}^2$ intervenes before $H$, preventing the requested monotone connection. For example $H=9$, $c=3.5$, $h=4$ gives $h^3-(1+c-ch{)}^2=-26.25$. Thus the smooth upstream tidal-wave admissibility conditions areAt $h_{+}=4$ the downstream zero is double, giving an algebraic approach; below four the endpoint approaches are exponential. The displayed Froude number bound is the requested necessary condition, with the source's missing amplitude qualification made explicit.

Let $B$ be the constant canal width, large compared with water depth. Then $A\simeq Bh$ and $l\simeq B$, so friction is $f{u}^2/h$. Choose a reference uniform water depth $H$, normal-flow speed $U$ satisfying $f{U}^2/H=gS$, and scalesThe continuity equation becomes ${\hat{h}}_{\hat{t}}+(\hat{h}\hat{u}{)}_{\hat{s}}=0$. Dividing momentum by $gS$ and dropping hats givesThe Froude number compares velocity to long gravity-wave speed. At the reference state, $F<1$ is subcritical flow, $F=1$ critical flow, and $F>1$ supercritical flow. In dimensional variables the wave characteristics travel at $u\pm \sqrt{gh}$: one can propagate upstream in subcritical flow, while both move downstream in supercritical flow. Away from the reference state the local Froude number is $F|u|/\sqrt{h}$.

These are the one-dimensional shallow water equations with bed slope and friction. The coordinate $s$ measures distance downstream, $t$ is time, $A(s,t)$ is the wetted cross-sectional area, $u(s,t)$ is section-averaged downstream velocity, and $h(s,t)$ is water water depth above the bed. Thus $Au$ is water volume flux and the first equation expresses mass conservation.

In the momentum equation, $u_t+u{u}_s$ is the material acceleration, $g$ is gravitational acceleration, and $S$ is the positive downward bed slope, so $gS$ drives the flow. The wetted perimeter is $l$ and $f$ is a dimensionless friction coefficient in the stated convention. The frictional deceleration is $fl{u}^2/A$ for downstream flow, while $-g{h}_s$ is the hydrostatic pressure contribution from surface-height variation relative to the sloping bed. The formula $u^2$ assumes the chosen positive-flow direction; a bidirectional friction model uses $u|u|$.