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
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 $$ 8 \lambda_5 + 2Q - (2Q - 2 \lambda_5) = 0 \\ 10 \lambda_5 = 0. $$