Linear Algebra Proof 3

Question

Let v ∈ Rⁿ be a unit vector. Define G = I - vv^T .

(a) Show G is symmetric and G² = G.

(b) Prove v is an eigenvector, find the associated eigenvalue.

(c) Prove that if < u; v >= 0 then u is also an eigenvector of G.

(d) Prove that G is diagonalizable.

Answer 1

(a) Recall that for a square matrix M to be symmetric, M = M^τ. We examine G and show that it is equal to its transpose. G^τ = (I - vv^τ)^τ = I^τ - (vv^τ)^τ = I - (v^τ)^τ(v)^τ = I - vv^τ = G, so G is symmetric.

Now, we expand G² to show it equals G. Observe that v^τv = 〈v,v〉= 1 since v is a unit vector.

G² = (I - vv^τ)(I - vv^τ) = I - vv^τ - vv^τ + vv^τvv^τ = I - 2vv^τ + v(1)v^τ = I - 2vv^τ + vv^τ = I - vv^τ = G.

(b) A vector v is an eigenvalue of operator (or square matrix) T if Tv = rv for r a value of the underlying field. We examine Gv. Since v is a unit vector, v^τv is equal to 1, so Gv = Iv - vv^τv = Iv - v(1) = v - 1v = 0. The eigenvalue is 0.

(c) Suppose that u is a vector such that 〈u,v〉= 0. Since the inner product is symmetric, v^τu = 〈v,u〉 = 0, it follows that Gu = Iu - vv^τu = u - v(0) = u. Thus u is an eigenvector with eigenvalue 1.

(d) If we can exhibit an basis of Rⁿ consisting of eigenvectors of G, then we can diagonalize G. We already have shown that v is an eigenvector of G and that any vector u orthogonal to v is also an eigenvector. Thus, if we pick v and a basis of the space orthogonal to the subspace spanned by v, we can exhibit such a basis of Rⁿ. Take an orthogonal decomposition of Rⁿ in terms of v ⊕ v^⊥ where v^⊥ denotes the hyperplane orthogonal to v. Pick a basis u₁,...,u_n-1 of v^⊥. By (c), these are all eigenvectors of G with eigenvalue 1. Hence, G written in the basis {v, u₁,..., u_n-1} will be a diagonal matrix with first entry equal to 0 and subsequent entries equal to 1.