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Here $L_1=1$ and $L_2=1/2$, so a perturbation varying in both coordinates has $n,m\ge 1$ andWith the given unit coefficients the determinant becomes $D(s)=(1+s{)}^2-\beta s$. The trace remains negative, so instability occurs precisely when $D(s)<0$, orThe right-hand side is strictly increasing for $s>1$, since its derivative is $1-s^{-2}>0$. Its smallest allowed value is attained by the mode $(n,m)=(1,1)$, giving the onset thresholdAt this value the first mixed mode has a zero eigenvalue; it becomes unstable for $\beta >{\beta }_{\mathrm{crit}}$. Thus the critical value is the infimum of the strictly unstable range, rather than a strictly unstable point itself. Its spatial factor is $\cos (\pi x)\cos (2\pi y)$.

For integers $k,l$,Thus reducing coordinates modulo one gives a well-defined function on the torus ${\mathbb{R}}^2/{\mathbb{Z}}^2$. Its determinant is one and its inverse matrix is the integer matrix $\left(\begin{array}{cc}1& -1\\ -1& 2\end{array}\right)$, so it induces a bijection of the torus. The Jacobian determinant one makes this bijection area-preserving.

Its eigenvalues solve ${\lambda }^2-3\lambda +1=0$, givingThere are therefore expanding and contracting eigendirections and no eigenvalue on the unit circle: the function is a hyperbolic toral automorphism. The Lyapunov exponents per iterate areThey sum to zero, as required by area preservation.

The four vertices of the unit square map to $(0,0),(2,1),(3,2),(1,1)$, the vertices of an area-one parallelogram. To reduce it modulo one, cut it by the integer grid. For each piece in the cell $[j,j+1]\times [k,k+1]$, translate by $(-j,-k)$. Up to boundary identifications these translated pieces partition the unit square, because the inverse integer matrix gives exactly one torus preimage to every point. The colored pieces in the original diagram show the same pieces before and after these translations.

For the Fibonacci identity, at $n=1$ the claimed matrix is $\left(\begin{array}{cc}{F}_2& F_1\\ F_1& F_0\end{array}\right)=A$. If the identity holds at $n$, multiplication on the right by $A$ givesEach equality uses $F_{j+1}=F_j+F_{j-1}$, with the specified indexing $F_0=F_1=1$. This proves the formula for every positive integer $n$.

A fixed point satisfies $(A-I)v\in {\mathbb{Z}}^2$. Since $A-I=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ has an inverse integer matrix, this forces $v\in {\mathbb{Z}}^2$, so the origin is the only fixed point modulo one.

For period dividing two,The second congruence gives $y=-3x(\mathrm{mod}1)$; inserting it into the first gives $5x=0(\mathrm{mod}1)$. Thus the complete list isThe first is the fixed origin. Direct application of $A$ exchanges the other four in exactly two pairs:This both verifies the periodic orbits and excludes additional ones.

Finally, if $v$ is fixed by $F^p$, then $(A^p-I)v=m$ for an integer vector $m$. The eigenvalues of $A^p-I$ are ${\lambda }_{+}^p-1$ and ${\lambda }_{-}^p-1$, neither zero. Hence its inverse exists and has rational entries by the adjugate matrix formula for an integer matrix. Therefore $v=(A^p-I{)}^{-1}m$ has rational coordinates. The derivative of the return map is $A^p$, with eigenvalues ${\lambda }_{+}^p>1$ and $0<{\lambda }_{-}^p<1$. Every periodic point is a hyperbolic saddle fixed point of a map and is unstable, although perturbations in its stable eigendirection contract.

Let $S(x,y,z)=(-x,-y,z)$. In the first two components of the vector field, replacing $x,y$ by their negatives changes the sign of the whole component; in the third, the product $xy$ is unchanged. Thus $F(Sv)=SF(v)$, establishing the symmetry, with solutions mapped to solutions by $S$.

At an equilibrium point, positivity of $a$ makes $y=x$. The remaining equations give $x(2c-a-z)=0$ and $z=x^2/b$. Hencewhere the two nonzero equilibrium points exist when $0<a<2c$. They merge with the origin at $a_p=2c$, in a pitchfork bifurcation.

At the origin the Jacobian matrix and its eigenvalues areAt $a=2c$ these are $0,-c,-b$. To see the local pitchfork bifurcation dynamics more explicitly put $\delta =2c-a$, $y=x+v$, and use $z=x^2/b+\cdots$ on the center manifold. The equations becomeAt $\delta =0$, invariance at cubic order gives $v=-x^3/(bc)+\cdots$, hence the reduced equation isIt has the supercritical pitchfork bifurcation form as $\delta$ increases, with the two branches $x\simeq \pm \sqrt{b\delta }$ attracting near onset while the origin becomes unstable.

For $0<a<2c$, the determinant of the planar block is $a(a-2c)<0$, so the origin has one positive and two negative eigenvalues and is a saddle equilibrium. At $a=c$, the planar eigenvalues are explicitly $\pm c$. For $a>2c$, the planar trace $c-a$ is negative and its determinant positive, so the origin is asymptotically stable. To distinguish its type putFor $2c<a<a_{\ast }$ both planar eigenvalues are real negative, giving a stable node. At $a=a_{\ast }$ they coincide at $-c/\sqrt{3}$; the planar block is not a scalar matrix, so it is a defective stable node. For $a>a_{\ast }$ they are nonreal with negative real part, giving a stable spiral in the planar block together with the stable third direction. This refines the requested node-versus-saddle classification.

The origin cannot have a Hopf bifurcation: a planar imaginary pair would require trace zero, namely $a=c$, but the determinant there is $-c^2<0$, so the pair is real and of opposite signs.

For either nonzero equilibrium point put $s=\pm \sqrt{b(2c-a)}$. The Jacobian matrix isIts characteristic polynomial isA nonzero imaginary root of an equation $i\omega$ requires, by separating real and imaginary parts, ${\omega }^2=B$ and $AB=C$. Equivalentlyso the candidate Hopf bifurcation values areAt such an admissible value the factorization is $p(\lambda )=(\lambda +A)({\lambda }^2+bc)$, giving the pair $\pm i\sqrt{bc}$ and a negative third eigenvalue $-A$, since $A=C/B>0$ on the nonzero branches.

The mathematical model parameter restrictions matter. A real candidate requires $b/c\le 17/8$, and the nonzero equilibrium points also require $0<a_h<2c$. The upper-sign value is admissible for $0<b/c<17/8$; the lower-sign value additionally requires $b/c>1$. For fixed $b,c$, differentiation of the characteristic equation at the imaginary root of an equation gives the eigenvalue crossing speedThus the transverse eigenvalue crossing condition holds except at the coincident-root case $b/c=17/8$, $a_h=3c/4$. At a transverse eigenvalue crossing, a generic nonzero first Lyapunov coefficient gives a Hopf bifurcation. The displayed radical is the spectral threshold; it does not justify an unconditional generic-Hopf claim for all positive mathematical model parameters, nor does it by itself exclude a degenerate Hopf bifurcation point. For example $b>17c/8$ yields no real candidate at all. The cubic Routh-Hurwitz stability criterion gives local Lyapunov stability on the side with $A>0$ and $AB-C>0$, where

For the modified equation, the planar block at the origin is $\left(\begin{array}{cc}-a& a\\ c& c\end{array}\right)$, with determinant $-2ac<0$ for the stipulated positive mathematical model parameters. Thus the origin is always a saddle equilibrium for nonzero $a,c$ in that mathematical model parameter range. The nonzero equilibrium points now satisfy $y=x$, $z=2c$ and $x^2=2bc$, so they areFor $b,c>0$ these branches stay a positive distance from the origin and do not merge with it as $a$ varies. The origin also has no zero eigenvalue for $a,c>0$. Therefore the pitchfork bifurcation is absent in the modified system.

In coordinate order $(n,m)$ the Jacobian matrix isAt the origin it is $\mathrm{diag}(1,q)$, giving a repelling node equilibrium; when $q=1$ the repeated positive eigenvalue is semisimple. At coexistence it isIts trace is $-2(1+q)/3$, determinant $q/3$, and characteristic discriminant $4(q^2-q+1)/9>0$. Hence both eigenvalues are real and negative, proving a stable node equilibrium for every $q>0$.

The other equilibria are $(1,0)$ and $(0,1)$. Their triangular Jacobians have eigenvalues $(-1,q/2)$ and $(1/2,-q)$ respectively. Thus both single-species equilibria are saddles, with their attracting directions on the corresponding coordinate axes and unstable directions allowing invasion by the other species. There are no further nonnegative equilibria because the two non-axis nullclines intersect only at $(2/3,2/3)$.

The coexistence Jacobian matrix calculated in part (a) has negative diagonal entries and positive determinant. For any spatial mode with squared wave number $k^2\ge 0$, the linearized matrix is $J_{\ast }-k^2\mathrm{diag}(1,\eta )$. Its trace is more negative than before, and its determinant isHence both eigenvalues retain negative real parts for every mode and every $\eta >0$. This is the mechanism by which negative self-response excludes two-species Turing instability, excluding diffusion-driven instability using only the reaction Jacobian matrix already computed, together with the diagonal diffusive damping. In particular unequal diffusion is not enough to produce a Turing pattern when both local self-responses are negative in this two-species competitive model.

The equilibria satisfy $y=0$ and $x-x^3=0$, giving $(0,0)$ and $(\pm 1,0)$. At the origin,Their product is $-1$, so the origin is a saddle equilibrium for every real $\Omega$.

At either nonzero equilibrium,The determinant is positive and the trace is $-(\Omega +1)$. Thus both equilibria are attracting for $\Omega >-1$, repelling for $\Omega <-1$, and have purely imaginary eigenvalues $\pm i\sqrt{2}$ at $\Omega =-1$. More precisely they are stable foci for $-1<\Omega <-1+2\sqrt{2}$ and stable nodes for $\Omega >-1+2\sqrt{2}$; they are unstable foci for $-1-2\sqrt{2}<\Omega <-1$ and unstable nodes for $\Omega <-1-2\sqrt{2}$. At the two discriminant-zero endpoints they are repeated-eigenvalue degenerate nodes, with a nondiagonalizable Jacobian. At $\Omega =-1$, linearization alone does not determine nonlinear stability or prove a nonlinear centre.

The equilibrium points areThe third is biologically interior only for $\beta >1$, and coincides with $(1,0)$ at $\beta =1$. The Jacobian matrix isAt $(0,0)$ the eigenvalues are $r,1$, so it is a repelling node. At $(1,0)$ they are $-r,1-\beta$, giving a stable node for $\beta >1$ and a saddle for $\beta <1$.

At the interior equilibrium,so coexistence is a saddle. The formal equilibrium with negative $v$ for $\beta <1$ is not a population state; there its trace is negative and determinant positive.

At $\beta =1$ the boundary equilibrium is nonhyperbolic. To determine its behavior into the positive quadrant, put $x=u-1$. A local center manifold has $x=-v+v^2/r+O(v^3)$, so $\dot{v}=v^2+O(v^3)>0$ for small positive $v$. Thus it is unstable to introduction of species $v$, though attracting along the $v=0$ axis. The boundary/interior branches undergo a transcritical bifurcation at this parameter.

The nonzero homogeneous equilibrium is $\boxed{n_{\ast }=1,a_{\ast }=\alpha /\beta }$. Without transport its Jacobian matrix iswith negative eigenvalues $-1,-\beta$.

Zero chemical flux gives $a_x=0$ at the endpoints; zero bacterial flux then gives $n_x=0$. Linear perturbations therefore have Neumann boundary conditions and modes $\cos (k_{m}x)$, $k_m=m\pi$. For each mode, the amplitude matrix isIts trace is always negative. Its determinant isA mode grows exactly when $\Delta (k_m)<0$. In particular, any nonzero mode is destabilized by sufficiently large $\chi$. For continuously variable $k$, a nonempty unstable band exists precisely whenWriting $C=\alpha \chi -D_a-\beta D_n$, its squared-wavenumber endpoints are $[C\pm \sqrt{C^2-4\beta D_n{D}_a}]/(2D_n{D}_a)$. On the finite interval, at least one allowed $k_m^2$ must lie inside this band. The uniform $m=0$ mode stays stable. This is the chemotactic pattern-instability criterion.

There is always the fixed point $(0,0)$. For $P>0$, the first fixed-point equation gives $e^{-P}=1/r$, and the second gives $P=N(1-1/r)$. Thus for $r>1$ there is exactly one positive fixed point,For $r=1$, every point $(N,0)$, $N\ge 0$, is fixed. There are no other nonnegative fixed points.

At the origin the Jacobian matrix has eigenvalues $r,0$. It is asymptotically stable for $0<r<1$ and a saddle for $r>1$. For $r<1$, the bounds $N_{t+1}\le r{N}_t$ and $P_{t+1}\le N_t$ also prove global extinction in the nonnegative quadrant.

At the positive fixed point,Here $\log r<r-1$ and $\log r>1-1/r$ for $r>1$. The discriminant is negative because the positive trace is less than two and determinant exceeds one. The conjugate multipliers have modulus $\sqrt{N_{\ast }}>1$, so the positive fixed point is always unstable. This is the instability of the Nicholson-Bailey equilibrium.

For completeness, on the $r=1$ fixed line,For $0\le N<1$, linear perturbations are bounded but have a neutral component along the fixed line; there is no asymptotic stability of an individual point. For $N>1$ there is transverse instability. At $N=1$, the repeated unit eigenvalue has a nontrivial Jordan block, so powers have an unbounded off-diagonal entry and the linearized system is unstable. This degenerate case is not covered by a distinct-eigenvalue assumption.

If a square matrix $A$ is an invertible matrix, multiplicativity of the determinant givesso $\det A\ne 0$.

Conversely, suppose $\det A\ne 0$. Use Gaussian elimination to reduce $A$ to row echelon form $U$. Each elementary row operation multiplies the determinant by a nonzero factor, so $\det U\ne 0$. If this square matrix had fewer than $n$ pivots, its row echelon form would have a zero row and hence zero determinant, a contradiction. Therefore it has a pivot in every row and column, and further elementary row operations reduce it to $I$.

Each elementary row operation is multiplication on the left by an invertible matrix, its inverse being the reverse operation. Thus $E_k\cdots E_{1}A=I$ implies $A=E_1^{-1}\cdots E_k^{-1}$, a product of invertible matrices. Hence

For a square matrix, the corresponding changes in its determinant areThese follow respectively from the alternating property, multilinearity in the rows, and the fact that a determinant with two equal rows vanishes.

A square matrix is an invertible matrix exactly when its determinant is nonzero. The preceding recurrence relation therefore gives

Let $A_n$ denote the given tridiagonal matrix and put $D_n=\det A_n$, including $D_0=1$ for the empty determinant. The starting values areFor $n\ge 3$, apply the following elementary row operations in the indicated order, always using the updated rows:The first three rows now beginwith the fourth column omitted if $n=3$. When $n\ge 4$, also use $R_4\leftarrow R_4-R_3$. The resulting block upper triangular matrix has a $3\times 3$ upper triangular matrix with diagonal entries all one in its top left corner, and $A_{n-3}$ in its bottom right corner. All elementary row operations preserved the determinant except the single row interchange. ThereforeApplying this recurrence relation to the three starting values proves by mathematical induction thatEquivalently, the sequence is periodic with period six, with values $1,1,0,-1,-1,0$ starting at $n=0$. A compact closed form is $D_n=\sin ((n+1)\pi /3)/\sin (\pi /3)$.

For the first matrix, the characteristic polynomial isThus the real eigenvalues are $4,-1,3$. Corresponding eigenvectors are $(3,1,0{)}^{\mathsf{T}}$, $(-2,1,0{)}^{\mathsf{T}}$ and $(0,0,1{)}^{\mathsf{T}}$. They are linearly independent; the determinant of the matrix with these columns is $5$. ThereforeFor example, multiplying the first column by $A$ gives $(12,4,0{)}^{\mathsf{T}}$, four times that column; multiplying the second gives $(2,-1,0{)}^{\mathsf{T}}$, its negative. These also check the chosen eigenvectors directly.

For the second matrix,The quadratic factor has negative discriminant $-3$, so $1$ is the only real eigenvalue. Solving $Av=v$ gives $v=(a,a,a{)}^{\mathsf{T}}$, a one-dimensional eigenspace. It cannot provide a basis of ${\mathbb{R}}^3$, so

Apply the elementary row operation $R_1\leftarrow R_1-R_2$, which preserves the determinant. Expanding along the new first row givesA square matrix is an invertible matrix exactly when its determinant is nonzero. Thereforeare precisely the three values making $A$ a singular matrix. They also have direct explanations: at $c=0$ the second row is zero, at $c=2$ the first two rows coincide, and at $c=7$ the last two columns coincide.

For either an upper triangular matrix or a lower triangular matrix,In the permutation formula for the determinant, every contribution except the diagonal product includes an entry on the zero side of the diagonal. Equivalently, successive cofactor expansions along the first column of an upper triangular matrix prove the formula by mathematical induction.

For $i\ne j$, the elementary row operation $R_i\leftarrow R_i-k{R}_j$ leaves the determinant unchanged:By multilinearity in row $i$, the left side is $\det A$ minus $k$ times a determinant with row $j$ appearing twice. That second determinant is zero by alternation. No row scaling or row interchange is involved.

The elementary row operations preserve the determinant, and the diagonal product of the upper triangular matrix found above isConsequentlyAt $t=1$, all four rows are equal. At $t=-1$, successive rows alternate between one row and its negative. In either case the matrix rank is one, confirming the vanishing determinant.

Apply the elementary row operations from bottom to top: $R_4\leftarrow R_4-R_3$, $R_3\leftarrow R_3-R_2$, and $R_2\leftarrow R_2-R_1$. This order ensures that each subtraction uses the original preceding row. The result is the upper triangular matrixPut ${\delta }_1=a$, ${\delta }_2=b-a$, ${\delta }_3=c-b$, ${\delta }_4=d-c$. Row $i$ is either zero if ${\delta }_i=0$, or has its first nonzero entry in column $i$. Thus, for arbitrary real parameters, a row echelon form is obtained by retaining the nonzero rows of $T$ in their existing order and moving every zero row to the bottom. This deals explicitly with vanishing parameters; an upper triangular matrix with an intermediate zero row is not yet a row echelon form.

The subtractions preserve the determinant, so the original matrix hasA square matrix is invertible exactly when its determinant is nonzero. Consequently,These conditions concern consecutive parameters: no ordering, positivity, or pairwise distinctness of all four is required. More generally the rank of a matrix here is the number of nonzero ${\delta }_i$, because the retained nonzero rows have distinct leading columns.

The pivot columns of the reduced row echelon form are columns 1 and 3. Take those columns of the original matrix, since elementary row operations need not preserve the original column space:For an explicit spanning check, column 2 is $-2a_1$, column 4 is $-a_1+2a_3$, and column 5 is $3a_1-2a_3$. The first two coordinates of the displayed columns form a $2\times 2$ matrix of determinant $-5\ne 0$, proving their linear independence. They are therefore a basis of the column space.