Basis of a $n \times n$ matrix which trace is zero

Question

For a matrix $2 \times 2$ it's easy. But for$n \times n$, I don't understand. Someone can help me?

Answer 1

Recall that there are $n^2$ total entries in the matrix. The matrix (which we call $A$) is subject to the condition that if

$$A = [a_{i, j}]_{1 \leq i, j \leq n}$$

then $\sum\limits_{i = 1}^{n} a_{i, i} = 0$. Hence, it doesn't matter what the elements which are not diagonal entries are, so each of those choices is free.

Furthermore, we can select the first few diagonal elements, so long as we adjust the last one to make the sum zero. That is, you can select anything you want for $a_{1, 1}, a_{2, 2}$ and so on, provided that you define

$$a_{n, n} = -(a_{1, 1} + ... + a_{n - 1, n - 1})$$

So you have $n^2$ entries, and all but one of them are free... So what does that mean about the basis?