LAT of sboxes, sum of coloms and rows

let we have sbox s: Vn -> Vn.

If we make LAT table for s, fix any row and get a sum by columns, that sum would be $+-2^{n-1}$.

And vice versa, if we fix any column and get a sum by rows, that sum would be $+-2^{n-1}$ too. Why is it so?

Element in "a" row, "b" column of LAT is $#{<a, x>=<b,s(x)>} - 2^{n-1}$. Where <,> is scalar product.

Sum is a sum of integers that are in one matrix column/ one matrix row.

First of all, a row/column of LAT correspond to a component function of the S-box/its inverse (a linear combination of outputs). So, let's derive the value of the sum of all Walsh coefficients of any Boolean function.

I will use this definition of Walsh transform. The results for others can be adapted easily.

$$W_f(a) = \sum_{x\in F_n} (-1)^{\langle a, x\rangle + f(x)},$$ $$\sum_{a \in F_n}W_f(a) = \sum_{a \in F_n}\sum_{x\in F_n} (-1)^{\langle a, x\rangle + f(x)} = \sum_{x\in F_n}\big((-1)^{f(x)}\sum_{a \in F_n}(-1)^{\langle a, x\rangle}\big).$$ The inner sum is equal to zero whenever $x\ne 0$ (linear functions are balanced), and is equal to $2^n$ when $x=0$. We get $$\sum_{a \in F_n}W_f(a) = 2^n\cdot (-1)^{f(0)}.$$

This you can observe e.g. on SageMath's BooleanFunction.walsh_hadamard_transform.