How Do You Determine the Orthogonal Complement in a Linear Algebra Problem?

[Felafel]

Homework Statement

Write a selfadjoint endomorphism $f : E^3 → E^3$ such that $ker(f ) = L((1, 2, 1))$ and $λ_1 = 1, λ_2 = 2$ are eigenvalues of f

The Attempt at a Solution

I know $λ_3=0$ because ́$ker(f ) ≠ {(0, 0, 0)}$ and $(ker(f ))^⊥ = (V0 )^⊥ = V1 ⊕ V2$ due to the definition of selfadjoint.
Then, my book gives the solution:
$(ker(f ))^⊥ = (L((1, 2, 1))^⊥ = {(α, β, −α − 2β) | α, β ∈ R} = L((1, 0, −1), (a, b, c))$
but i don't understand where did it get that (α, β, −α − 2β) from.
Could you please help me? thanks in advance :)

 

[Dick]

Homework Statement

Write a selfadjoint endomorphism $f : E^3 → E^3$ such that $ker(f ) = L((1, 2, 1))$ and $λ_1 = 1, λ_2 = 2$ are eigenvalues of f

The Attempt at a Solution

I know $λ_3=0$ because ́$ker(f ) ≠ {(0, 0, 0)}$ and $(ker(f ))^⊥ = (V0 )^⊥ = V1 ⊕ V2$ due to the definition of selfadjoint.
Then, my book gives the solution:
$(ker(f ))^⊥ = (L((1, 2, 1))^⊥ = {(α, β, −α − 2β) | α, β ∈ R} = L((1, 0, −1), (a, b, c))$
but i don't understand where did it get that (α, β, −α − 2β) from.
Could you please help me? thanks in advance :)

(α, β, −α − 2β) is the subspace orthogonal to (1,2,1). You know that the eigenvectors of a self adjoint operator corresponding to different eigenvectors are orthogonal, yes? So you'll have to pick the other two eigenvectors from that space.