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


  1. […] vectors here, they span a subspace of dimension , which must be all of . And we know, by an earlier lemma that a collection of eigenvectors corresponding to distinct eigenvalues must be linearly […]

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  2. […] Eigenspaces are Disjoint We know that eigenspaces of distinct eigenvalues are disjoint. That is, no vector is an eigenvector of two different eigenvalues. This extends to generalized […]

    Pingback by Generalized Eigenspaces are Disjoint « The Unapologetic Mathematician | February 25, 2009 | Reply

  3. […] as a corollary we can find that not only are the eigenvectors corresponding to distinct eigenvalues linearly independent, they are actually orthogonal! Indeed, if and are eigenvectors of with […]

    Pingback by Eigenvalues and Eigenvectors of Normal Transformations « The Unapologetic Mathematician | August 6, 2009 | Reply

  4. Is the dimension of V, n? ie dim(V)=n, ie their contains n elements in a basis for V. since you have n eigenvectors that are linearly independent and any set of n linearly independent vectors is a basis for an n dimensional vectorspace.

    Comment by H | April 18, 2010 | Reply

  5. Yes, H, that’s true. But all this post was intended to prove was their linear independence.

    Oh, and I think you mean “there”.

    Comment by John Armstrong | April 19, 2010 | Reply

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