Final answer:
The sum of the dimensions of distinct eigenspaces is less than the dimension of the range of a linear transformation because eigenspaces are independent and only constitute part of the range, which includes images of non-eigenvectors.
Step-by-step explanation:
The question pertains to Linear Algebra, specifically the relationship between eigenvalues and the dimension of the range of a linear transformation T from a finite-dimensional vector space V. According to the question, T has distinct nonzero eigenvalues, which are represented by λ1, ..., λm. Let Eλi be the eigenspace corresponding to the eigenvalue λi. The claim is that the sum of the dimensions of these eigenspaces is less than the dimension of the range of T, denoted as dim R(T). We know that each eigenspace is a subspace of V, and the range of T, R(T), is a subspace of V that contains the images of all vectors in V under T.
To prove the claim, consider that since λi are distinct eigenvalues, their corresponding eigenspaces are independent, meaning that no vector in one Eλi can be expressed as a combination of vectors from the other eigenspaces. This independence implies that the sum of their dimensions gives the dimension of their direct sum. Also, each nonzero eigenvector corresponds to a nonzero vector in R(T) since Tv = λv implies that v is in R(T) when λ ≠ 0. Therefore, the dimension of the direct sum of the eigenspaces can be seen as part of the dimension of R(T).
The actual dimension of R(T) is greater because it also contains the images of vectors that are not necessarily eigenvectors. Hence, the sum of the dimensions of the eigenspaces is a subset of the dimension of R(T), and cannot exceed the dimension of R(T). Therefore, m dim Eλi < dim R(T).