How to prove that eigenvectors from different eigenvalues are linearly independent

How can I prove that if I have n eigenvectors from different eigenvalues, they are all linearly independent?

Answer

I’ll do it with two vectors. I’ll leave it to you do it in general.

Suppose v1 and v2 correspond to distinct eigenvalues λ1 and λ2, respectively.

Take a linear combination that is equal to 0, α1v1+α2v2=0. We need to show that α1=α2=0.

Applying T to both sides, we get
0=T(0)=T(α1v1+α2v2)=α1λ1v1+α2λ2v2.
Now, instead, multiply the original equation by λ1:
0=λ1α1v1+λ1α2v2.
Now take the two equations,
0=α1λ1v1+α2λ2v20=α1λ1v1+α2λ1v2
and taking the difference, we get:
0=0v1+α2(λ2λ1)v2=α2(λ2λ1)v2.

Since λ2λ10, and since v20 (because v2 is an eigenvector), then α2=0. Using this on the original linear combination 0=α1v1+α2v2, we conclude that α1=0 as well (since v10).

So v1 and v2 are linearly independent.

Now try using induction on n for the general case.

Attribution
Source : Link , Question Author : Corey L. , Answer Author : Community

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