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Let $V$ be the vector space of all $4 \times 4$ matrices such that the sum of the elements in any row or any column is the same. Then the dimension of $V$ is

1. $8$
2. $10$
3. $12$
4. $14$

An easier approach : There are in total 16 variables in 4 X 4 matrix. There are 4 rows ,4 columns satisfying the fixed sum property. So 16 variables and 8 equations, that means 8 variables are independent and 8 are not. So dimension of vector space is 8.
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https://math.stackexchange.com/questions/3218784/let-v-be-the-vector-space-of-all-4x4-matrices-such-that-the-sum-of-the-ele, the same question is discussed here, the answer came out to be 10. The method of proving is applied to finding the dimension of the zero columns and zero-sum, which can be extended to the case of any sum of rows and columns is the same.

$\begin{bmatrix} X_{1} &X_{2} & X_{3} &Y_{1} \\ X_{4}& X_{5} & X_{6} & Y_{2}\\ X_{7} & X_{8}& a& b\\ Y_{3} &Y_{4} &c & d \end{bmatrix}$

Now in this matrix, $X_{1} ... X_{8}$ can be chosen arbitrarily. When you have chosen these 8 independent elements, $Y_{1}.... Y_{4}$ are automatically fixed because row sum and column sum constrains has to be maintained. Now I will prove that $a$,$b$,$c$ and $d$ are also fixed so the rank of the vector space is 8(number of independently chosen elements).

The given four equations i.e.

$a + b = F- X_{7}- X_{8}$

$a+ c= F- X_{3}- X_{6}$

$b + d = F- Y_{1}- Y_{2}$

$c + d = F- Y_{3}- Y_{4}$

have to be satisfied. and these are 4 independent equations in 4 unknowns. So can be solved uniquely which means a,b,c,d are also fixed. Here $F$ is the  fixed row sum or column sum.
Hence $X_{i}$ are the only free elements. so the dimension is 8.
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