A magic matrix is a square matrix such that the sums of the elements of each row, each column and diagonal equal to a same number, the magic constant.
As reported here, the largest singular value of a magic matrix is its magic constant.
Can someone provide a proof or an intuition for that?
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Consider a magic square matrix $\mathbf{A}\in \mathbb{R}^{n\times n}$. Denote the magic constant by $$ s=\frac{n(1+n^2)}{2}. $$ Since the sum of each row or column equals to $s$, we know that $s$ is a singular value of $\mathbf{A}$, i.e.: $$ \mathbf{A}^\text{T}\mathbf{Av}=s^2\mathbf{v} \text{, where } \mathbf{v}=[1,\dots,1]^\text{T}. $$
We know that the singular values of $\mathbf{A}$ satisfy that $$ \sum_i\sigma_i^2=\text{tr}(\mathbf{A}^\text{T}\mathbf{A})=\sum_i\sum_j a_{ij}^2. $$ The sum of squares of all entries equals to $\sum_{k=1}^{n^2}k^2=\frac{1}{3}n^6+\frac{1}{2}n^4+\frac{1}{6}n^2$. We have $s^2=\frac{1}{4}n^6+\frac{1}{2}n^4+\frac{1}{4}n^2$, so $$ \sum_i\sigma_i^2 < 2s^2. $$ This indicates that all other singular values are smaller than $s$.
Hope this could be helpful.
It sounds like a magic matrix is a special case of a doubly stochastic matrix, (and in turn a special case of a stochastic matrix), scaled by a scalar value: the magic constant.
A stochastic matrix is just a square matrix of nonnegative real numbers with each row summing to $1$. The Perron–Frobenius theorem ensures that every stochastic matrix has an eigenvalue equal to $1$, and that is the eigenvalue with the largest absolute value.