# The Unapologetic Mathematician

## Distinct Eigenvalues

Today I’d like to point out a little fact that applies over any field (not just the algebraically-closed ones). Let $T$ be a linear endomorphism on a vector space $V$, and for $1\leq i\leq n$, let $v_i$ be eigenvectors with corresponding eigenvalues $\lambda_i$. Further, assume that $\lambda_i\neq\lambda_i$ for $i\neq j$. I claim that the $v_i$ are linearly independent.

Suppose the collection is linearly dependent. Then for some $k$ we have a linear relation

$\displaystyle v_k=c_1v_1+c_2v_2+...+c_{k-1}v_{k-1}$

We can assume that $k$ is the smallest index so that we get such a relation involving only smaller indices.

Hit both sides of this equation by $T$, and use the eigenvalue properties to find

$\displaystyle\lambda_kv_k=c_1\lambda_1v_1+c_2\lambda_2v_2+...+c_{k-1}\lambda_{k-1}v_{k-1}$

On the other hand, we could just multiply the first equation by $\lambda_k$ to get

$\displaystyle\lambda_kv_k=c_1\lambda_kv_1+c_2\lambda_kv_2+...+c_{k-1}\lambda_kv_{k-1}$

Subtracting, we find the equation

$\displaystyle0=c_1(\lambda_k-\lambda_1)v_1+c_2(\lambda_k-\lambda_2)v_2+...+c_{k-1}(\lambda_k-\lambda_{k-1})v_{k-1}$

But we this would contradict the minimality of $k$ we assumed before. Thus there can be no such linear relation, and eigenvectors corresponding to distinct eigenvalues are linearly independent.

February 4, 2009 Posted by | Algebra, Linear Algebra | 5 Comments