Prove that for the $\mathbb R^{n\times n}$ vector space doesn't exist a basis from $n^2$ matrices, such that AB=BA for every two matrices in base.

Question

Prove that for the $\mathbb R^{n\times n}$ vector space doesn't exist a basis containing $n^2$ matrices, such that AB=BA for every two matrices $A$, $B$ in the base. $(n>1)$

I think it is true because I did not get any example which makes it false. Thanks in advance.

Well, if that is the case then every pair of matrices would commute. Agreed? — Pedro, Dec 14 '17 at 16:23
There it is shown that not all $n\times n$-matrices commute for $n>1$. Together with Pedro's remark then, it is a duplicate. — Dietrich Burde, Dec 14 '17 at 16:29

score 6 · Accepted Answer · answered Dec 14 '17 at 16:27

6

Take two matrices $A$ and $B$ that do not commute; this is clearly possible for $n>1$.

Suppose there exists a basis $\{E_1,\dots,E_{n^2}\}$ so that $E_iE_j=E_jE_i$ for all $i$ and $j$.

Express $A$ and $B$ as linear combinations of the basis.

Compute $AB$ and $BA$.

Conclude.

answered Dec 14 '17 at 16:27

egreg

238,574

I was really expecting this answer to end with "Profit". – Pedro Dec 14 '17 at 17:07

Prove that for the $\mathbb R^{n\times n}$ vector space doesn't exist a basis from $n^2$ matrices, such that AB=BA for every two matrices in base.

1 Answers1