The Unapologetic Mathematician

Mathematics for the interested outsider

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


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


Subtracting, we find the equation


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