# Physically understanding Dilation Symmetry of Eigenvalue-Eigenvector Identity

In the recent paper by Tao concerning the famous eigenvalue-eigenvector identity, I need some insight regarding the "Dilation Symmetry" from a physical perspective.

It is mentioned that if we assign units to the entries of $$A$$, then the eigenvalues of $$A$$, $$M_j$$ acquire the same units, while $$v_{i,j}$$ remains dimensionless, and the identity (2) is dimensionally consistent.

The actual identity (2) given in the paper is :
(Eigenvector-eigenvalue identity). $$\left|v_{i, j}\right|^{2} \prod_{k=1 ; k \neq i}^{n}\left(\lambda_{i}(A)-\lambda_{k}(A)\right)=\prod_{k=1}^{n-1}\left(\lambda_{i}(A)-\lambda_{k}\left(M_{j}\right)\right) \tag{2}$$ where $$A$$ is an $$n \times n$$ Hermitian matrix, and its $$n$$ real eigenvalues are $$\lambda_{1}(A), \ldots, \lambda_{n}(A)$$, $$M_{j}$$ denote the $$n-1 \times n-1$$ minor formed from $$A$$ by deleting the $$j^{\text {th }}$$ row and column from $$A$$ and $$v_{i, j}$$ denote the $$j^{\text {th }}$$ component of $$v_{i}$$.

And Dilation Symmetry is defined as "If one multiplies the matrix $$A$$ by a real scalar $$c$$, then the eigenvalues of $$A$$ and $$M_j$$ also get multiplied by c, while the coefficients $$v_{i,j}$$ remain unchanged, which does not affect the equation (2)

It would be very helpful of someone can actually provide an physical example of this "Dilation Symmetry". Because personally I couldn't understand what kind of If you could provide me with some physical example of this "Dilation Symmetry", that would be extremely useful.