I've tried to look for the answer to this curiosity over the internet, but I couldn't find any explanation that is clear enough.
So here's the question: consider a linear transformation $L\colon\mathbf{V}\to\mathbf{W}$. This linear transformation is represented by an $m\times n$-matrix $A$ of rank $r\leq\min(m,n)$, so $A^{\top}$ is the representation of $L^{-1}\colon\mathbf{W}\to\mathbf{V}$. The book told us that the singular values of $A$ are the positive square root of the eigenvalues of $A^{\top}A$, i.e., let's say $\delta_{1},\ldots,\delta_{r}$, with $\delta_{1}\geq\cdots\geq\delta_{r}>0$, are the eigenvalues of $A^{\top}A$ such that $\sigma_{1},\ldots,\sigma_{r}$ are the singular values of $A$, where $\sigma_{i}=\sqrt{\delta_{i}}$.
But why is it like that? What is the intuitive explanation as to why we choose that the eigenvalues of $A^{\top}A$ are related to the eigenvalues of $A$?