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Let $M_n(\Bbb C)$ be the linear space of all $n\times n$ complex matrices, then

1). the set $\{MN-NM|M,N\in M_n(\Bbb C)\}$ is a subspace of $M_n(\Bbb C)$;

2). $\{MN-NM|M,N\in M_n(\Bbb C)\}=\{A\in M_n(\Bbb C)|\operatorname{Tr}(A)=0 \}$.

I made some attempt.

Let $U=\{MN-NM|M,N\in M_n(\Bbb C)\}$.

For 1), if $A,B\in U$, and $A=M_1N_1-N_1M_1, B=M_2N_2-N_2M_2$, how to find two matrices $M,N$ such that

$$MN-NM=A+B$$

For 2), if $A$ such that $\operatorname{Tr}(A)=0$, what are the two matrices $M,N$ such that

$$MN-NM=A$$

Thanks a lot!

Clin
  • 2,353

1 Answers1

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The important part of this problem is (2), which to my knowledge does not have a quick solution. You may find reading this previous answer helpful.

Once you've proven (2), you may easily prove (1).

Let $$ \mathfrak{sl}_n(\Bbb C)=\left\{A\in M_n(\Bbb C):\operatorname{tr}(A)=0\right\} $$ Then (1) asks to prove that $\mathfrak{sl}_n(\Bbb C)$ is a subspace of $M_n(\Bbb C)$. To do so, let $\lambda_1,\lambda_2\in\Bbb C$ and let $A_1,A_2\in\mathfrak{sl}_n(\Bbb C)$. Then $$ \operatorname{tr}(\lambda_1 A_1+\lambda_2 A_2)=\lambda_1\operatorname{tr}(A_1)+\lambda_2\operatorname{tr}(A_2)=\lambda_1\cdot 0+\lambda_2\cdot 0=0 $$ so $\lambda_1 A_1+\lambda_2 A_2\in\mathfrak{sl}_n(\Bbb C)$. Hence $\mathfrak{sl}_n(\Bbb C)$ is a subspace of $M_n(\Bbb C)$.

Community
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Brian Fitzpatrick
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  • Thanks! If (1) is right, one can easily prove (2),too: For $E_{ij}(=E_{i1}E_{1j}-E_{1j}E_{i1}),E_{11}-E_{ii}\in U$ linearly independent – Clin May 06 '14 at 21:17