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Math Help - Problem 38

  1. #1
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    Problem 38

    Show the center of the general linear group are the proportional diagnol matrices.
    Meaning, Z(\mbox{GL}_n(\mathbb{R})) = \{ k I| k\in \mathbb{R}^{*} \mbox{ and }I \mbox{ identity matrix} \}.

    I try to explain this so it makes it more elementary. The set \mbox{GL}_n(\mathbb{R}) is the set of all n\times n invertible matrices having \mathbb{R} as their entries. This set is called the "general linear group". The "center" of this set are all the matrices which commute with everything else. So you need to show if a matrix commute with all invertible matrices then it must be a proportional diagnol matrix.
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    sorry, but i dont understand a thing
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    I am not sure but is this what you are saying?

    Let A be any n\times n invertible matrix. Now if B is an n\times n matrix such that AB=BA for any choice of A then B is a scalar multiple of the identity matrix. That is B=kI.
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    Quote Originally Posted by putnam120 View Post
    I am not sure but is this what you are saying?

    Let A be any n\times n invertible matrix. Now if B is an n\times n matrix such that AB=BA for any choice of A then B is a scalar multiple of the identity matrix. That is B=kI.
    Exactly. (Just note that A,B are invertible).
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    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.
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    MHF Contributor Bruno J.'s Avatar
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    Let I_{(j)} be the identity matrix with row j multiplied by 2. Then I_{(j)}A is A with row j multiplied by 2, and AI_{(j)} is A with column j multiplied by 2. From

    I_{(j)}A=AI_{(j)}

    we conclude that A has a diagonal entry and zeroes everywhere else on the jth row and the jth column. In this fashion we conclude that A is diagonal.

    It is then easy to show that the diagonal entries are all equal.
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