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Describe all elements $A \in GL_n(\mathbb{R})$ with the property that $AB = BA$ for all $B \in GL_n(\mathbb{R})$.

I would really appreciate any help. I am not sure where to start.

I almost want to say that $AB$ is an identity, but I am not sure. But I do know matrices have no multiplicative inverses.

BlackAdder
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Bob Eccain
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1 Answers1

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A standard way of doing this is: if $A$ is in the center of $GL(n)$ it commutes, in particular, with all the elementary matrices. That means that multiplying the $i$-th row of $A$ by some constant $a$ gives the same result as multiplying the $i$-th column by $a.$ This means that $A$ is diagonal. Further, flipping the $i$-th row and the $j$-th row gives the same result as flipping the $i$th column and the $j$-th column. So, the diagonal entries are equal. This works over any field (actually, I guess integral domain is enough).

Igor Rivin
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  • Wait- the elementary matrices have det = 0, and so are not in GL(n). Right? – voldemort Jan 15 '14 at 05:12
  • Wrong. They generally have determinant $\pm 1.$ – Igor Rivin Jan 15 '14 at 05:13
  • But this might help: http://en.wikipedia.org/wiki/Koch_snowflake – Igor Rivin Jan 15 '14 at 05:14
  • What's your definition of elementary matrix? The definition I know is that an elementary matrix $E_{ij}$ has (i,j)th entry=1, rest =0. – voldemort Jan 15 '14 at 05:15
  • @voldemort they are elementary row operations (as you can deduce without knowing the definition from my answer). So, they are some multiple of the matrix you describe plus the identity OR the "transposition" matrices of the second half of my answer. – Igor Rivin Jan 15 '14 at 05:17
  • I have actually seen the notation you describe (in a French paper), but this seems to be nonstandard. – Igor Rivin Jan 15 '14 at 05:18
  • I just wanted to know your definition :). I+$E_{ij}$ does the trick (for this problem). – voldemort Jan 15 '14 at 05:20