# Proving a property of the dimension of eigenspaces in a finite dimensional space

• cloverforce
In summary, if A is a linear map on a vector space V with dimension n, and h1,...,hk are pairwise different eigenvalues of A whose geometric multiplicities sum to n, then A does not have any other eigenvalues. This can be proven by contradiction, as the existence of another eigenvalue would result in more than n independent eigenvectors, which is not possible.

## Homework Statement

Prove that if A: V - >V is a linear map, dim V = n, and h1,...,hk (where 1,...,k are subscripts) are pairwise different eigenvalues of A such that their geometric multiplicities sum to n, then A does not have any other eigenvalues.

## Homework Equations

Note sure if this is relevant or not, but if hj and hk are distinct eigenvalues, then the intersection of V(hj) and V(hk) is {0}.

## The Attempt at a Solution

My attempt has been by contradiction. Suppose that there exists an eigenvalue h distinct from the k terms already given. In that case, I want to show that dim(V(h)) = 0, which would mean that its basis is the empty set and thus no such eigenvalue can exist. I'm not sure why this would have to be the case, though.

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One important property you will need is that eigenvectors corresponding to distinct eigenvalues are independent.

If "their geometric multiplicities sum to n" then, by definition of "geometric multiplicity" you already have n independent vectors. If there existed another eigenvalue, distinct from any of the previous ones, then an eigenvector corresponding to it would be independent of the first n, giving you n+1 independent eigenvectors.