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The cat map is the area-preserving toral automorphism induced by the integer matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$. Its eigenvalues are $(3\pm \sqrt{5})/2$, giving one expanding and one contracting eigendirection. Cutting its image parallelogram by the integer grid and translating the pieces modulo one gives a concrete geometric construction on the unit square.

For a square integer matrix $A$ with nonzero determinant, the image $A{\mathbb{Z}}^n$ is a subgroup of the integer lattice. Its index of a subgroup isIn particular, $A=\mathrm{diag}(2,1)$ gives all points with even first coordinate. The quotient has two cosets, represented by $(0,0)$ and $(1,0)$, and the fundamental cell has area two. A determinant of two therefore counts how much integer information is missing from the image.

The integrality hypothesis is essential. The rational matrix $R=\mathrm{diag}(2,1/2)$ has determinant one but sends $(0,1)$ to $(0,1/2)$. Its generated Euclidean lattice has the same covolume as ${\mathbb{Z}}^2$ without being the same integer lattice. The unimodular matrix criterion combines an arithmetic condition with a geometric one.

A short proof of the index formula uses Smith normal form. Write $PAQ=\mathrm{diag}(d_1,\dots ,d_n)$, with $P,Q$ unimodular matrices and positive integers $d_i$. They preserve the relevant integer structure, so the quotient has the same size as ${\prod }_i\mathbb{Z}/d_i\mathbb{Z}$, namely ${\prod }_i{d}_i=|\det A|$. No assumption that the matrix is orthogonal or its basis vectors are short is needed.

For a square integer matrix $U$ with nonzero determinant, the inverse matrix has integer entries exactly when $\det U=\pm 1$. The adjugate matrix formula proves one direction:Every entry of $\mathrm{adj}(U)$ is an integer cofactor, so division by $1$ or $-1$ leaves integer entries. Conversely, if both $U$ and $U^{-1}$ are integer matrices, their determinants are integers with product one. The only possibilities are that both are one or both are minus one. This explains why a general integer matrix with nonzero determinant need not have an integer inverse matrix.

An integer matrix is a matrix whose entries are integers. Sums and products of compatible integer matrices remain integer matrices, but a nonsingular square integer matrix can have a nonintegral inverse matrix. The unimodular matrix condition characterizes exactly when that inverse matrix also has integer entries.

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 $B_1,B_2$ be real square invertible matrices whose columns are lattice bases. They generate the same Euclidean lattice precisely when^[ref]

If $B_2=B_{1}U$, multiplication by $U$ merely permutes the integer coordinate vectors, so the two generated sets agree. Conversely, equality of the generated sets means every column of either basis is an integer linear combination of the other. Hence $B_2=B_{1}U$ and $B_1=B_{2}V$ for integer matrices $U,V$. Substitution followed by multiplication by $B_1^{-1}$ gives $UV=I$, so the integer inverse criterion for a unimodular matrix applies.

Equal real linear spans are a weaker condition: any two real square invertible matrices span the same ambient vector space. Even equal covolumes do not force the same Euclidean lattice. The requirement that the change matrix and its inverse matrix have integer entries provides the missing arithmetic information.

A unimodular matrix is a square integer matrix whose determinant is $1$ or $-1$. The decisive property is that its inverse matrix also has integer entries: multiplication by $U$ gives a bijection of the integer lattice ${\mathbb{Z}}^n$ onto itself. Thus an invertible linear map can preserve not only real coordinates but the requirement that every coordinate is an integer. These matrices form the integral general linear group.^[ref]

For an original two-dimensional example, takeIts determinant is one, and it maps $(a,b)$ to $(a+b,b)$. Conversely, the integer point $(x,y)$ comes from $(x-y,y)$. For instance, $(3,2)$ has new coordinates $(1,2)$ relative to the columns $(1,0)$ and $(1,1)$. No integer point disappears, although the basis vectors and their fundamental cell change shape.

This is useful whenever coordinates must remain integral. A unimodular change of lattice basis can give a shorter or more convenient description without changing the objects represented. Smith normal form and Hermite normal form use controlled integer operations to simplify integer matrices; ordinary Gaussian elimination may divide by numbers that destroy integrality. The geometric freedom is substantial: a sheared basis may look completely different while describing exactly the same infinite set of integer points.

A determinant of $-1$ is allowed too. Swapping two basis columns reverses orientation while preserving the integer lattice. The condition is therefore preservation of the complete integer structure, rather than a requirement to preserve orientation.