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halbard · #5 March 28th, 2009, 04:18 PM

OK, let's run this past the General and see if it salutes.

Suppose that $B$ commutes with all $A\in\mbox{GL}_n(\mathbb{R})$ . Assume that $B\neq0$ and choose any $u\in\mathbb{R}^n$ with $Bu\neq0$ . Clearly $u\neq0$ .

Claim: The set $\{u, Bu\}$ is linearly dependent.

Assume for contradiction purposes that $u$ and $Bu$ are linearly independent. Then so are $u$ and $u-Bu$ .

The linearly independent sets $\{u, Bu\}\mbox{ and }\{u, u-Bu\}$ can each be extended to a basis.

Therefore there exists an invertible matrix $A$ with $Au=u$ and $A(Bu)=u-Bu$ .

But then $u-Bu=ABu=BAu=Bu$ which shows that $u$ and $Bu$ are linearly dependent after all, contradiction.

Therefore $\{u, Bu\}$ is l.d. and there must exist a non-zero scalar $k$ such that $Bu=ku$ .

For any non-zero $v\in\mathbb{R}^n$ there is an invertible matrix $C$ such that $v=Cu$ .

Therefore $Bv=B(Cu)=BCu=CBu=C(ku)=kCu=kv$ .

Thus $Bv=kv\mbox{ for all non-zero }v\in\mathbb{R}^n$ which means that $B=kI$ .