# DIN A4 aspect ratio seen as an eigenvalue problem

In Math.

Have you ever wondered about the odd dimensions of a DIN A4 sheet? In this blog post, I will show you how to derive the aspect ratio with the help of the eigenvalue decompostion of geometric operations acting on the sheet. Consider $x=(x^0, x^1)^t \in \mathbb R^2$ being the vector of lengths for each dimension i.e. $x^0$ represents the length for the x-axis and $x^1$ the length for the y-axis accordingly. Let us now define two linear operations $F$ and $T$ (more explicitely two automorphism $\mathbb R^2 \rightarrow \mathbb R^2$) acting on the sheet dimensions: $$ F := \begin{pmatrix} \tfrac{1}{2} & 0 \\ 0 & 1 \end{pmatrix} \quad \text{and} \quad T:= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\quad. $$ The matrix $F$ folds the sheet of paper in direction of the x-axis which causes its length to be halved and the transposition map $T$ interchanges the coordinate functions (i.e. turns the sheet by 90 degrees). You may have noticed that a DIN A5 sheet can be obtained by the composition of the two maps $F$ and $T$ (folding and rotating by 90 degrees) and that the aspect ratio will remain constant. As a result, the DIN A5 format is just a scaled variant of DIN A4. An equivalent description can be given by the following equation: $( T \circ F)\, x = \lambda \cdot x$ where $\lambda$ is a (positive) scaling factor. The non-trivial solutions for $x$ can be obtained by the eigenvalue decomposition of the matrix $T\circ F$: $$ \det\bigl(T\cdot F - \lambda \cdot \boldsymbol 1\bigr) = \det \begin{pmatrix} -\lambda & 1 \\ \tfrac{1}{2} & -\lambda \end{pmatrix} = \lambda^2 - \frac{1}{2} \stackrel{!}{=} 0\quad\Rightarrow\quad \lambda_\pm=\pm \frac{1}{\sqrt 2} $$ The associated eigenspaces are spanned by the eigenvectors $u$ and $v$: $$ \mathop{\mathrm{ker}}\bigl(T\cdot F - \lambda_+ \cdot \boldsymbol 1 \bigr) =\mathop{\mathrm{ker}}\begin{pmatrix} -\tfrac{1}{\sqrt 2} & 1 \\ \tfrac{1}{2} & -\tfrac{1}{\sqrt 2} \end{pmatrix}= \mathop{\mathrm{span}}\left\{\begin{pmatrix} \sqrt 2 \\ 1 \end{pmatrix} \right\} =: \mathop{\mathrm{span}}\left\{u \right\} $$ $$ {\mathrm{ker}}\bigl(T\cdot F - \lambda_- \cdot \boldsymbol 1 \bigr) ={\mathrm{ker}}\begin{pmatrix} +\tfrac{1}{\sqrt 2} & 1 \\ \tfrac{1}{2} & +\tfrac{1}{\sqrt 2} \end{pmatrix}= {\mathrm{span}}\left\{\begin{pmatrix} \sqrt 2 \\ -1 \end{pmatrix} \right\} =: {\mathrm{span}}\left\{v \right\} $$ Finally, let us define the aspect ratio $\rho(x) = \tfrac{x^0}{x^1}$ for both eigenvectors $u$ and $v$. $$ \rho(u) = \sqrt{2} \quad\text{and}\quad \rho(v)= -\sqrt 2 $$ Please note, $\rho$ is invariant under scaling operations i.e. $\rho(x) = \rho(\lambda \cdot x)$ for all $\lambda \in \mathbb R^\times$. Furthermore, we discard the solution for the eigenvector $v$ since negative aspect ratios do not apply to our setting. Obviously, the proper aspect ratio for a DIN A4 sheet should be $\sqrt 2 : 1$ (landscape) or $1 : \sqrt 2$ (portrait). The actual dimensions are specified by the quotient $\tfrac{297}{210}$ which is a very close approximation of $\sqrt 2$ since $$ \rho(\text{DIN A4 sheet}) - \rho(u) = \frac{297}{210} - \sqrt 2 \approx 7 \cdot 10^{-5} \quad. $$ We conclude that the dimensions of DIN A4 are governed by the eigenvectors of the fold and transposition map.