Centre of the general linear group

Scalar transformation criterion

Let form a basis of a vector space , and let . Then is a scalar transformation iff every for is an eigenvector. #m/thm/linalg Thus, if all are eigenvectors then they all have the same nonzero eigenvalue.

Proof

Let be the eigenvalue corresponding to . Then

and since the decomposition of a vector into basis vectors is unique, it follows for all . The converse is trivial, since every nonzero vector is an eigenvalue of a scalar transformation.

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