[${AB-BA }$ , where $A_{n\times n}$ and $B_{n\times n}$ are square matrices] is this set a subspace of the vector space of all square matrices of order $n$.
If yes then can you please make me understand how it is closed under addition.
If not then how traceless matrices become commutators.
I have become confused after reading the comments by Alex Wertheim here , Can anyone make me understand?
Let $k$ be an infinite field. Then we have the following fact:
Proposition. Suppose $C\in\operatorname{Mat}_{n}(k)$ is an $n\times n$ matrix with entries in $k$ such that $\operatorname{Tr}(C)=0$. Then $C$ is a commutator, i.e. $C=AB-BA$ for some $A, B\in\operatorname{Mat}_{n}(k)$.
I learnt the following elegant proof from the book Mathematical Bridges by Andreescu, Mortici and Tetiva.
Proof. We first prove the following lemma:
Lemma. If $a_1, a_2, ..., a_n\in k$ are elements such that $a_1+a_2+\cdots + a_n=0$, then there exist distinct elements $b_1, b_2, ..., b_n\in k$ and a permutation $\pi: \{1, 2, ..., n\}\to \{1, 2, ..., n\}$ such that $a_i=b_i-b_{\pi(i)}$ for each $i=1, 2, ..., n$.
Proof of the lemma. We proceed by strong induction on $n$. When $n=1$, this is trivial as $a_1=0$ and so we can take $b_1=1$ and $\pi$ to be the identity permutation. For the inductive step, we have two cases:
Case 1. No proper subset of $\{a_1, ..., a_n\}$ adds up to zero.
Let $b_1$ be arbitrary. Define $b_2=b_1-a_1$. Similarly, define $b_3=b_2-a_2$, and so on. In other words, $b_{i+1} := b_{i}-a_{i}$ for $i=1, 2, ..., n-1$. By construction, $a_{i}=b_{i}-b_{i+1}$ for each $i$, so we can take the permutation $\pi: \{1, 2, ..., n\}\to \{1, 2, ..., n\}$ to be $\pi(i)=i+1$ for $i<n$ and $\pi(n)=1$. Note the assumption $a_1+a_2+\cdots + a_n=0$ was used to conclude that $a_{n}=b_{n}-b_{1}$, so there is no issue when we "loop back" from $n$ to $1$. All we need to show is that $b_1, b_2, ..., b_n$ are distinct. If $b_i=b_j$ for $1\leq i<j\leq n$, then summing up the equalities $a_{i}=b_{i}-b_{i+1}$, $a_{i+1}=b_{i+1}-b_{i+2}$, ..., $a_{j-1}=b_{j-1}-b_{j}$ and taking advantage of the telescoping, we get $a_{i}+a_{i+1}+...+a_{j-1}=b_{i}-b_{j}=0$ which contradicts the assumption that no proper subset of ${a_1, ..., a_n}$ adds up to zero.
Case 2. Some proper subset of $\{a_1, ..., a_n\}$ adds up to zero.
In this case, we have $\{a_1, ..., a_n\}=\{c_1, c_2, ..., c_k\}\cup \{d_1, d_2, ..., d_{n-k}\}$ such that $c_1+c_2+\cdots + c_k=0$ where $1\leq k<n$. Since $a_1+a_2+\cdots + a_n=0$, this forces $d_1+d_2+\cdots + d_{n-k}=0$ as well. By induction hypothesis, we know that there exist permutations $\sigma: \{1, ..., k\}\to \{1, ..., k\}$, $\tau: \{1, ..., n-k\}\to \{1, ..., n-k\}$, distinct elements $\lambda_1, ..., \lambda_k$, and distinct elements $\mu_1, \mu_2, ..., \mu_{n-k}$ such that $c_i=\lambda_i-\lambda_{\sigma(i)}$ and $d_j=\mu_j-\mu_{\tau(j)}$. We can find some $r$ such that $r+d_j \neq c_i$ for every $i$ and $j$. This is where the assumption $k$ is infinite comes in. Anyways, now it is easy to "stitch" together these permutations. Simply define $b_i=\lambda_i$ for each $i=1,2, ..., k$, and define $b_{k+j} = r+d_{j}$ for each $j=1, ..., n-k$. Finally, define $\pi$ to be the permutation $\pi(i)=\sigma(i)$ for each $I=1,..., k$, and $\pi(k+j)=\tau(j)$ for each $j=1, ..., n-k$. The proof of the lemma is complete.
Proof of the Proposition. Let $C$ be a matrix with trace zero. Set $a_i:=c_{ii}$. Since $\operatorname{Tr}(C)=0$, we get $a_1 + a_2 + ... + a_n=0$. By the lemma above, we can write $a_i=b_{i}-b_{\pi(i)}$ for distinct elements $b_1, ..., b_n$ and some permutation $\pi$. It is straightforward to rewrite our matrix $C$ as a difference of two matrices $C=U-L$ where $U$ is an upper triangular matrix with diagonal entries $b_1, ..., b_n$, and $L$ is a lower triangular matrix with diagonal entries $b_{\pi(1)}, ..., b_{\pi(n)}$. Since $U$ and $L$ have distinct eigenvalues (namely, $b_1$, ..., $b_n$), they are both diagonalizable. In fact, they share the same set of eigenvalues, so if we set $D$ to be the diagonal matrix consisting of $b_1, ..., b_n$, then $U=PDP^{-1}$ and $L=QDQ^{-1}$ for some invertible matrices $P$, and $Q$. Write $D=Q^{-1}LQ$ and substitute this into $U=PDP^{-1}$ to get $$U=P(Q^{-1}LQ)P^{-1}=(PQ^{-1})L(PQ^{-1})^{-1}=RLR^{-1}$$ where $R=PQ^{-1}$. Thus, $$ C = U - L = RLR^{-1}-L = RLR^{-1} - (R^{-1}R)L = (RL)R^{-1}-R^{-1}(RL) $$ So we set $A=RL$ and $B=R^{-1}$ to get $C=AB-BA$.
As I mentioned in the comments, the proposition implies that $$ \{AB-BA: A, B \in\operatorname{Mat}_{n}(k)\} = \{C: \operatorname{Tr}(C)=0\} $$ Since the right hand side is clearly a subspace, it follows that the set on the left hand side is actually a subspace! This is kind of surprising! It says that given $AB-BA$ and $CD-DC$, it follows that $(AB-BA) + (CD - DC) = RS-SR$ for some matrices $R, S$.
