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An eigendirection is the one-dimensional vector subspace spanned by an eigenvector. Multiplying the eigenvector by a nonzero scalar leaves this direction unchanged. In a planar linear dynamical system, a negative real eigenvalue gives a direction of exponential approach to the equilibrium point, while a positive one gives exponential departure. For a hyperbolic saddle equilibrium, the nonlinear stable manifold and unstable manifold are tangent to the corresponding eigendirections.

The eigenvalues are real and have opposite signs: $\boxed{{\lambda }_1{\lambda }_2<0}$. One eigendirection contracts and the other expands, giving a saddle.

The eigenvalues are real and have opposite signs: $\boxed{{\lambda }_1{\lambda }_2<0}$. One eigendirection contracts and the other expands, giving a saddle.

At an equilibrium point, $x(1-x-y)=0$ and $2-2x-y=0$. The first equation gives $x=0$ or $y=1-x$. Substituting in the second gives, respectively,These are all the equilibrium points. The Jacobian matrix of the vector field isAt $(0,2)$,The eigenvalue $-1$ has algebraic multiplicity two and eigenspace $\mathrm{span}\{(0,1{)}^T\}$. For the linearized coordinates $u=x$, $v=y-2$,Both components tend to zero. The repeated eigenvalue with one independent eigenvector makes this a stable improper degenerate node, a sink equilibrium. The term $t{e}^{-t}$ is the signature of the nontrivial Jordan block. Since it is a hyperbolic equilibrium point, the linearization stability theorem gives local asymptotic stability of the nonlinear system. Approaching trajectories are tangent to the vertical eigendirection.

At $(1,0)$,One eigenvalue is positive and one negative, so this point is a saddle equilibrium. Its stable manifold is tangent to $y=\sqrt{2}(x-1)$, and its unstable manifold is tangent to $y=-\sqrt{2}(x-1)$.

For the phase portrait, the nullclines areAbove the second line the vertical flow is downward, and below it upward. For $x>0$, horizontal flow is to the right below $y=1-x$ and to the left above it; for $x<0$ the directions reverse. The line $x=0$ is invariant, with $\dot{y}=2-y$, so it flows toward $(0,2)$. Together with the saddle's two invariant directions, these signs determine the sketch.

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.