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(λ2−λ1)v2=α2(λ2−λ1)v2.

Since λ2−λ1≠0, and since v2≠0 (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 v1≠0).

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*