Suppose T is a linear map and dim(Im(T))=k

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In the discussion, it is established that for a linear map T with the property that dim(Im(T))=k, T can have at most k+1 distinct eigenvalues. The reasoning is based on the relationship between the dimension of the image and the kernel of T, where the kernel has a dimension of n-k. Additionally, every vector in the kernel corresponds to an eigenvalue of 0, leading to the conclusion that the maximum number of distinct eigenvalues is limited by the dimensions of the eigenspaces associated with the non-zero eigenvalues.

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Please, help me!
Suppose T is a linear map and dim(Im(T))=k. Prove that T has at most k+1 distinct eigenvalues.
Thank you in advance!
 
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If T maps an n dimensional space into an m dimensional space, then the kernel of T must be of dimension n- k. And, of course, every vector in the kernel of T is an eigenvector with eigenvalue 0. Now, since each eigenvalue has a corresponding "eigen"space of dimension at least 1, how many other eigenvalues can there be?
 

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