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

Question

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

Sol: I thought of this matrix where every row and column sums to $s$ and since it has $10$ variables I think the dim is 10. By separating and taking out the variables I could come up with a $10$ element basis. Through an obvious but lengthy process I could show its linear independence and the fact that it's a spanning set is obvious from the construction. Is this correct? \begin{bmatrix} a & b & c & s-(a+b+c)\\ d & e & f & s-(d+e+f)\\ g & h & i & s-(g+h+i)\\ s-(a+d+g) & s-(b+e+h) & s-(c+f+i) &-2s+(a+b+c+d+e+f+g+h+i) \end{bmatrix}

Answer 1

In general, the dimension of the subspace of $M_n(K)$ with equal row and column sum is $(n-1)^2+1$, see here:

Dimension of vector space of matrices with zero row and column sum.

Actually, if the value is supposed to be zero, the dimension is $(n-1)^2$. This follows for $m=n$ from the duplicate. We have to add plus $1$, if the value is not prescribed.

Answer 2

You were definitely on the right track, but maybe identifying a clear basis of size $10$ for the Vector Space would confidently help you say that the dimension is indeed $10$.

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Notice that in your matrix construction above, the $3\times3$ submatrix given by: $$\begin{bmatrix} a&b&c\\ d&e&f\\ g&h&i\\ \end{bmatrix}$$

has the interesting property that all the $9$ elements are free to take any value because we can always choose the $4th$ element of each row/column such that the sum across the row/column still becomes $S$.

This also makes clear that these $9$ elements have a huge role to play, since the remaining elements are adjusted as per the values that these $9$ elements take.

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If the condition were that $S=0$ (i.e. all the rows/columns have a sum of $0$), then keeping in mind the point made above, basis members for a Vector Space of $4\times4 $ matrices with each row/column sum equal to zero are easily given by: $$\begin{bmatrix} 1&0&0&-1\\ 0&0&0&0\\ 0&0&0&0\\ -1&0&0&1\\ \end{bmatrix}, \begin{bmatrix} 0&1&0&-1\\ 0&0&0&0\\ 0&0&0&0\\ 0&-1&0&1\\ \end{bmatrix}, \begin{bmatrix} 0&0&1&-1\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&-1&1\\ \end{bmatrix}$$

$$\begin{bmatrix} 0&0&0&0\\ 1&0&0&-1\\ 0&0&0&0\\ -1&0&0&1\\ \end{bmatrix}, \begin{bmatrix} 0&0&0&0\\ 0&1&0&-1\\ 0&0&0&0\\ 0&-1&0&1\\ \end{bmatrix}, \begin{bmatrix} 0&0&0&0\\ 0&0&1&-1\\ 0&0&0&0\\ 0&0&-1&1\\ \end{bmatrix}$$

$$\begin{bmatrix} 0&0&0&0\\ 0&0&0&0\\ 1&0&0&-1\\ -1&0&0&1\\ \end{bmatrix}, \begin{bmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&1&0&-1\\ 0&-1&0&1\\ \end{bmatrix}, \begin{bmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&0&1&-1\\ 0&0&-1&1\\ \end{bmatrix}$$

Now, these $9$ matrices are very clearly linearly independent. Also, any $4\times4$ matrix with each row/column sum equal to zero can be given by a Linear Combination of these $9$ matrices.

For example if $A=\begin{bmatrix} a&b&c&\alpha_1\\ d&e&f&\alpha_2\\ g&e&h&\alpha_3\\ \alpha_4&\alpha_5&\alpha_6&\alpha_7\\ \end{bmatrix}$ is one such matrix, then $$A=a\begin{bmatrix} 1&0&0&-1\\ 0&0&0&0\\ 0&0&0&0\\ -1&0&0&1\\ \end{bmatrix}+b\begin{bmatrix} 0&1&0&-1\\ 0&0&0&0\\ 0&0&0&0\\ 0&-1&0&1\\ \end{bmatrix}+\cdots+h\begin{bmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&0&1&-1\\ 0&0&-1&1\\ \end{bmatrix}$$

(I encourage you to actually construct a $4\times4$ matrix and verify that this works.)

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Finally, we arrive at the last and simple bit of our task. What if $S\neq0$? If you have been able to keep up so far, I think this should strike you as a little obvious. The small change here is to include $\begin{bmatrix} 0&0&0&1\\ 0&0&0&1\\ 0&0&0&1\\ 1&1&1&-2\\ \end{bmatrix}$ in our existing basis of size $9$.

For example if we consider $B=\begin{bmatrix}a&b&c&\alpha_1\\ d&e&f&\alpha_2\\ g&e&h&\alpha_3\\ \alpha_4&\alpha_5&\alpha_6&\alpha_7\\ \end{bmatrix}$ to be a matrix with each row/column sum equal to $S$ (not necessarily equal to $0$), then

$$B=a\begin{bmatrix} 1&0&0&-1\\ 0&0&0&0\\ 0&0&0&0\\ -1&0&0&1\\ \end{bmatrix}+\cdots+h\begin{bmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&0&1&-1\\ 0&0&-1&1\\ \end{bmatrix}+S\begin{bmatrix} 0&0&0&1\\ 0&0&0&1\\ 0&0&0&1\\ 1&1&1&-2\\ \end{bmatrix}$$

(I again encourage you to construct a $4\times4$ matrix and check the validity of the result)

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Hence, we can now confidently say that we have obtained a basis of size $10$ and the dimension of the vector space of all $4\times4$ matrices such that the sum of the elements in any row or any column is the same must be $10$.