Eigenvectors for distinct eigenvalues are linearly independent

Let F be a field, V a vector space over F, and \varphi : V \rightarrow V a linear transformation. Let \{ \lambda_i \}_{i=1}^k be a finite set of eigenvalues for \varphi, and let v_i be a nonzero eigenvector for \lambda_i for each i. (That is, \varphi(v_i) = \lambda_i v_i.) Prove that the v_i are linearly independent. Conclude that any linear transformation on a vector space of finite dimension n has at most n eigenvalues.

We proceed by induction on k.

For the base case k = 1, certainly \{v_1\} is linearly independent since V is torsion-free.

Suppose now that any set of k eigenvectors for distinct eigenvalues is linearly independent, and let \{v_i\}_{i=1}^{k+1} be a set of eigenvectors for the distinct eigenvalues \{\lambda_i\}_{i=1}^{k+1}. Suppose we have a_i \in F such that \sum a_i v_i = 0. Now 0 = \varphi(0) = \varphi(\sum a_iv_i) = \sum a_i\lambda_iv_i. Suppose, without loss of generality, that a_1 \neq 0. Now \lambda_1 \sum a_iv_i = \sum a_i \lambda_i v_i, so that \sum a_i(\lambda_i - \lambda_1)v_i = 0. Now \sum_{i=2}^{k+1} a_i(\lambda_i - \lambda_1)v_i = 0, and by our hypothesis, a_i = 0 for all 2 \leq i \leq k+1. But then we have \alpha_1 v_1 = 0, a contradiction since \alpha_1 \neq 0 and v_1 \neq 0. Thus in the sum \sum a_iv_i = 0, all of the a_i are zero. Hence \{v_i\}_{i=1}^{k+1} is linearly independent.

The dimension of a vector space is the maximal cardinality of a linearly independent subset. In particular, we see that if V has finite dimension n, then \varphi can have at most n eigenvalues.

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