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

\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

   

Follow

Get every new post delivered to your Inbox.

Join 224 other followers