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Let $M(d)_{d\in \mathbb{R}}\subset M_3(\mathbb{R})$, consisting of all the matrices such that the sum of the elements in each row, column and on the two diagonals is $d$. Such an element $M(d)$ is called a magic square with magic number $d$.

Show that $M(0)$ is a vector subspace of $M_3(\mathbb{R})$ and find a basis for $M(0)$. Conclude that the center entry of any element in $M(0)$ is $0$.

Is there another way to doing this problem other than guessing the form of $M(0)$? I thought about writing down $\left(\begin{array}{ccc} \lambda_1 & \lambda_2 & \lambda_3 \\ \lambda_4 & \lambda_5 & \lambda_6 \\ \lambda_7 & \lambda_8 & \lambda_9 \end{array}\right)$ and summing up the rows and columns to zero, but seems unproductive

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The sum of the first and third rows plus the sum of the first and third columns, minus the sum of the two diagonals, gives $$ 2\lambda_2 + 2\lambda_8 + 2\lambda_4 + 2 \lambda_6 -2 \lambda_5 = 0. $$

Similarly, the sum of the middle row and middle column is zero, but it's also $$ 4 \lambda_5 + Q $$ where $Q$ is the sum of the four "mid-edge" entries (2, 4, 6, 8).

Taking twice the second equation minus the first gives $$ 8 \lambda_5 + 2Q - (2Q - 2 \lambda_5) = 0 \\ 10 \lambda_5 = 0. $$

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