Thread: Eigenvectors

I have the matrix



With the eigenvalues



then the eigenvectors





Now I understand that for the eigenvector is since .

However when it comes to the eigenvector for all I have (in reduced form) .
Why is the eigenvector for ?

Is it because the first column is a free variable?

because any vector of the form (a,0) = a(1,0) satisfies x₂ = 0, so this is an eigenvector iff a ≠ 0.

yes this is the same as being a "free variable", any choice of x₁ (freely made) will do (but of course, the fact that an eigenvector must be non-zero means we can't choose x₁ = 0, although (0,0) *is* in the eigenspace E_λ1).

First, do you understand that there is no such thing as "the eigenvector" for given eigenvalue? Given any eigenvector, any (non-zero) multiple of it is also an eigenvector. In fact, the set of all eigenvectors for a given eigenvalue form a subspace. You say that "the eigenvector" for eigenvalue is [1, -1]. In fact, any vector of the form [a, -a], for a any (non-zero) number is an eigenvector.

I had to include "(non-zero)" because Deveno says "but of course, the fact that an eigenvector must be non-zero". Some text (but, admittedly a minority) define an eigenvalue by saying " is an eigenvalue for linear operator A if and only if there exist a non-zero vector " but then define an "eigenvector" as any vector satisfying . That has the advantage, in my opinion, of including the 0 vector as an "eigenvector" so that the set of all eigenvector corresponding to a given eigenvalue is a subspace without having to add the 0 vector separately.