## 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.